https://wiki.puella-magi.net/w/api.php?action=feedcontributions&user=Anon-kun&feedformat=atomPuella Magi Wiki - User contributions [en]2021-07-27T15:25:20ZUser contributionsMediaWiki 1.32.0https://wiki.puella-magi.net/w/index.php?title=Talk:April_Fool%27s_Day&diff=8451Talk:April Fool's Day2011-03-31T19:27:06Z<p>Anon-kun: /* Other April Fool's Spoofs */ +1</p>
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<div>For those of us who don't know, who is Sonico? -[[User:Universalperson|Universalperson]] 17:16, 31 March 2011 (UTC)<br />
:The character Super Soniko (an idol) from the upcoming Nitroplus game SoniComi [http://supersonico.jp/sonicomi/]. Here is the PV: [http://www.youtube.com/watch?v=MqOFWDGHOqo]. See also [[wikipedia:ja:すーぱーそに子]]. The other characters in the OP are also characters from other games -- [[User:Anon-kun|Anon-kun]] 17:34, 31 March 2011 (UTC)<br />
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== Other April Fool's Spoofs ==<br />
* http://baseson.nexton-net.jp/msha/index.html Puella Magi Hawawa Awawa<br />
* http://ataraxia.blog20.fc2.com/ - Puella Magi Maria Magica</div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=Talk:April_Fool%27s_Day&diff=8449Talk:April Fool's Day2011-03-31T17:58:52Z<p>Anon-kun: </p>
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<div>For those of us who don't know, who is Sonico? -[[User:Universalperson|Universalperson]] 17:16, 31 March 2011 (UTC)<br />
:The character Super Soniko (an idol) from the upcoming Nitroplus game SoniComi [http://supersonico.jp/sonicomi/]. Here is the PV: [http://www.youtube.com/watch?v=MqOFWDGHOqo]. See also [[wikipedia:ja:すーぱーそに子]]. The other characters in the OP are also characters from other games -- [[User:Anon-kun|Anon-kun]] 17:34, 31 March 2011 (UTC)<br />
<br />
== Other April Fool's Spoofs ==<br />
* http://baseson.nexton-net.jp/msha/index.html</div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=April_Fool%27s_Day&diff=8448April Fool's Day2011-03-31T17:42:41Z<p>Anon-kun: Staff and Cast</p>
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<div>__NOTOC__<br />
[[File:Magical girl sonico magica.png|thumb|Magical Girl Sonico Magica]]<br />
On April 1, 2011, Nitroplus ''announced'' a new anime '''Magical Girl Sonico Magica''' (魔法少女そにこ☆マギカ) based on Magical Girl Madoka Magica.<br />
<br />
== Summary ==<br />
大好きな親友がいて、音楽があって、時には笑い、時には泣く、そんなどこにでもある日常。吉祥寺の大学に通う、普通の大学一年生・すーぱーそに子も、そんな日常の中で暮らす一人。ある日、彼女に不思議な出会いが訪れる。この出会いは偶然なのか、必然なのか、彼女はまだ知らない。それは、彼女の運命を変えてしまうような出会い──それは、次なる魔法少女物語の始まり──<br />
<br />
== Staff ==<br />
*原作 ： Sonico Quartet<br />
*監督 ： 徒歩十分<br />
*脚本 ： 虚淵玄？<br />
*キャラクター原案 ： 津路参汰<br />
*キャラクターデザイン/総作画監督 ： 南向春風(M2)<br />
*シリーズディレクター ： 嶋流<br />
*アクションディレクター ： おがみけいち<br />
*3D監督 ： オガタガクオ(ポリゴン番長)<br />
*モーションディレクター ： 森夕晴（ポリゴン番長）<br />
*異空間設定 ： ゆーぽん<br />
*美術監督 ： ぺはら塗装<br />
*美術設定 ： せんこうどっき<br />
*色彩設計 ： 縞うどん/大熊猫介<br />
*ビジュアルエフェクト ： usao.exe<br />
*撮影監督/編集 ： 南向春風(M2)<br />
*音響監督 ： 榎本覚(ブレイブハーツ)<br />
*音楽 ： 平田博信<br />
*音響制作 ： GEORIDE<br />
*アニメーション制作 ： M2<br />
<br />
== Cast ==<br />
*すーぱーそに子 ： すーぱーそに子<br />
*キュゥべえ’s ： すーぱーそに子<br />
<br />
== External links ==<br />
* http://supersonico.jp/<br />
* [http://www.youtube.com/watch?v=23C3tjF14nA OP for Magical Girl Sonico Magica]</div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=Talk:April_Fool%27s_Day&diff=8447Talk:April Fool's Day2011-03-31T17:37:51Z<p>Anon-kun: </p>
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<div>For those of us who don't know, who is Sonico? -[[User:Universalperson|Universalperson]] 17:16, 31 March 2011 (UTC)<br />
:The character Super Soniko (an idol) from the upcoming Nitroplus game SoniComi [http://supersonico.jp/sonicomi/]. Here is the PV: [http://www.youtube.com/watch?v=MqOFWDGHOqo]. See also [[wikipedia:ja:すーぱーそに子]]. The other characters in the OP are also characters from other games -- [[User:Anon-kun|Anon-kun]] 17:34, 31 March 2011 (UTC)</div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=Talk:April_Fool%27s_Day&diff=8446Talk:April Fool's Day2011-03-31T17:36:57Z<p>Anon-kun: </p>
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<div>For those of us who don't know, who is Sonico? -[[User:Universalperson|Universalperson]] 17:16, 31 March 2011 (UTC)<br />
:The character Super Soniko (a gravure idol) from the upcoming Nitroplus game SoniComi [http://supersonico.jp/sonicomi/]. Here is the PV: [http://www.youtube.com/watch?v=MqOFWDGHOqo]. See also [[wikipedia:ja:すーぱーそに子]]. The other characters in the OP are also characters from other games -- [[User:Anon-kun|Anon-kun]] 17:34, 31 March 2011 (UTC)</div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=Talk:April_Fool%27s_Day&diff=8445Talk:April Fool's Day2011-03-31T17:34:00Z<p>Anon-kun: </p>
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<div>For those of us who don't know, who is Sonico? -[[User:Universalperson|Universalperson]] 17:16, 31 March 2011 (UTC)<br />
:The character Super Soniko (a gravure idol) from the upcoming Nitroplus game SoniComi [http://supersonico.jp/sonicomi/]. See also [[wikipedia:ja:すーぱーそに子]]. The other characters in the OP are also characters from other games -- [[User:Anon-kun|Anon-kun]] 17:34, 31 March 2011 (UTC)</div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=April_Fool%27s_Day&diff=8443April Fool's Day2011-03-31T17:13:28Z<p>Anon-kun: Summary</p>
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<div>[[File:Magical girl sonico magica.png|thumb|Magical Girl Sonico Magica]]<br />
On April 1, 2011, Nitroplus ''announced'' a new anime '''Magical Girl Sonico Magica''' (魔法少女そにこ☆マギカ) based on Magical Girl Madoka Magica.<br />
<br />
== Summary ==<br />
大好きな親友がいて、音楽があって、時には笑い、時には泣く、そんなどこにでもある日常。吉祥寺の大学に通う、普通の大学一年生・すーぱーそに子も、そんな日常の中で暮らす一人。ある日、彼女に不思議な出会いが訪れる。この出会いは偶然なのか、必然なのか、彼女はまだ知らない。それは、彼女の運命を変えてしまうような出会い──それは、次なる魔法少女物語の始まり──<br />
<br />
== External links ==<br />
* http://supersonico.jp/<br />
* [http://www.youtube.com/watch?v=23C3tjF14nA OP for Magical Girl Sonico Magica]</div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=April_Fool%27s_Day&diff=8442April Fool's Day2011-03-31T17:02:32Z<p>Anon-kun: image</p>
<hr />
<div>[[File:Magical girl sonico magica.png|thumb|Magical Girl Sonico Magica]]<br />
On April 1, 2011, Nitroplus ''announced'' a new anime '''Magical Girl Sonico Magica''' (魔法少女そにこ☆マギカ) based on Magical Girl Madoka Magica.<br />
<br />
== External links ==<br />
* http://supersonico.jp/<br />
* [http://www.youtube.com/watch?v=23C3tjF14nA OP for Magical Girl Sonico Magica]</div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=File:Magical_girl_sonico_magica.png&diff=8441File:Magical girl sonico magica.png2011-03-31T16:58:22Z<p>Anon-kun: </p>
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<div></div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=April_Fool%27s_Day&diff=8440April Fool's Day2011-03-31T16:43:06Z<p>Anon-kun: April Fool's Day</p>
<hr />
<div>On April 1, 2011, Nitroplus ''announced'' a new anime '''Magical Girl Sonico Magica''' (魔法少女そにこ☆マギカ) based on Magical Girl Madoka Magica.<br />
<br />
== External links ==<br />
* http://supersonico.jp/<br />
* [http://www.youtube.com/watch?v=23C3tjF14nA OP for Magical Girl Sonico Magica]</div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=Madoka_Magica_Episode_10:_I_Won%27t_Depend_on_Anyone_Anymore&diff=8122Madoka Magica Episode 10: I Won't Depend on Anyone Anymore2011-03-25T15:52:35Z<p>Anon-kun: link</p>
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<div>{{Infobox_episode<br />
|name = I Won't Depend on Anyone Anymore<br />
|jname = もう誰にも頼らない<br />
|date = 10 March 2011<br />
|script = [[Gen Urobuchi]]<br />
|storyboard = Shinsaku Sasaki<br />
|director = Yuki Yase<br />
|adirector = Yoshiaki Itou<br />Kazuya Shiotsuki<br />
}}<br />
<br />
== Summary ==<br />
[[File:Ep10 timeline chart.jpg|thumb]]<br />
This episode centers around [[Homura]]. Most of the mystery behind her origin and behavior in previous episodes is explained.<br />
<br />
=== Timeline 1 ===<br />
In the original timeline for Homura, she was a normal human girl suffering from a heart disease. She is shown just released from the hospital and transferring to the [[Mitakihara Middle School]], similar to the [[Episode 1|first episode]]. She differs in appearance to the Homura we know, though - she wears glasses and wears her long hair tied in two braids. Her personality also feels different - she is a lot more shy and clumsy.<br />
<br />
[[Madoka]], [[Sayaka]] and other new classmates try to befriend Homura, which makes her feel uneasy. During a talk with Madoka, she admits finding herself worthless, not unlike Madoka does in the timeline of other episodes. Homura's problems during math and sports classes (probably related to being out of school for a long time) contribute to her depression, which makes her get suicidal thoughts on the way from school.<br />
<br />
Presumably because of these thoughts, she wanders into the barrier of witch [[Izabel]]. She is quickly saved by [[Puella Magi]] Madoka and [[Mami]], with Madoka asking her to keep their jobs a secret from schoolmates. Later, during a visit at Mami's apartment, Madoka reveals that she's been contracted only for a very short time. Homura feels fascinated about her magical friends, similar to how Madoka acts in initial episodes.<br />
<br />
Some time later, Madoka and Mami fight [[Walpurgis Night]] together, with Homura following them as a human. Mami is defeated quickly, then Madoka apparently manages to defeat the witch, but at the expense of her own life. Homura cries over Madoka's dead body, and accepts [[Kyuubey]]'s proposal to become a Puella Magi, her wish being ''"I want to redo my meeting with Kaname-san. Instead of being protected by her, I want to protect her!"'' The contract is formed, with Kyuubey commenting that the wish has "surpassed entropy".<br />
<br />
=== Timeline 2 ===<br />
Homura wakes up in a hospital, on the date she was released in the previous iteration, having a [[Soul Gem]] in her hand. After transferring to school, she instantly tells Madoka about becoming a Puella Magi, hoping for good cooperation.<br />
<br />
Madoka and Mami train Homura the Puella Magi, who turns out to be extremely clumsy, despite having the very powerful ability of stopping time. Meanwhile, Homura reads online guides about making explosives, contributing to improving her skills. While still clumsy, she plays the most important role when the trio - now good friends - fights with the witch [[Patricia]].<br />
<br />
The next scene shown is after the Walpurgis Night - after defeating the witch, Madoka's Soul Gem turns into a [[Grief Seed]] and she becomes a witch.<br />
<br />
=== Timeline 3 ===<br />
The time goes back again, and Homura wakes up at the exact same place and time as previously. Scared and angered by the discovery of the true nature of the contract with Kyuubey, she decides to tell the other girls immediately. Sayaka, who in this timeline became a Puella Magi, too, doesn't believe her and accuses her of collaborating with [[Kyouko]], whom apparently only Homura has met yet.<br />
<br />
Sayaka also tells that she feels afraid of Homura's explosives, which makes Homura go to a meeting of some armed formation and steal some firearms, all on frozen time. Since Sayaka turns into a [[Oktavia von Seckendorff|witch]], presumably for similar reasons as in [[Episode 8]], Homura uses the firearms during the fight with her to protect Madoka, and also defeats her, even though Madoka and Kyouko try to make Sayaka human again in a fashion similar to [[Episode 9]].<br />
<br />
Upon seeing the true nature of Puella Magi herself, Mami breaks down and attempts to kill all the magical girls, including herself, in order to prevent their eventual transformation into witches. She kills Kyouko first by shooting her Soul Gem and attempts to kill Homura by binding her, but before she can do so, Madoka uses her beam arrows to kill her. The remaining two fight on the Walpurgis Night together, and defeat the witch, having their Soul Gems fatally tainted, though. Madoka heals Homura with a Grief Seed she was hiding, and asks Homura to prevent her from becoming a Puella Magi next time she'll go back in time. Homura agrees, and grants Madoka one more wish of hers - killing her to prevent her from becoming a witch.<br />
<br />
=== Timeline 4 ===<br />
Upon waking up in the familiar hospital, Homura changes her presence by magically healing her eyes, wearing the hair down and taming her emotions - just how she looked in all the previous episodes. She warns Madoka at night (with Madoka recognizing her to some extent as she says "Homu...") and goes hunting all the witches.<br />
<br />
On Walpurgis Night, a sequence very similar to the dream from [[Episode 1]] is shown. Homura watches Kyuubey persuading Madoka and screams to prevent her from forming the contract, but Madoka doesn't seem to hear it.<br />
<br />
Off-camera, Madoka forms the contract and defeats the witch with just one shot, but becomes a witch ([[Kriemhild Gretchen]]) herself shortly after. Kyuubey comments that she may destroy the whole Earth in near future. He has collected an enormous amount of energy from her already, though, so he doesn't care about it anymore.<br />
<br />
Homura refuses to fight Madoka the witch, which make Kyuubey realize that she's from a different timeline.<br />
<br />
=== Timeline 5 ===<br />
Homura insists of repeating that particular period of time (between her transferring and the Walpurgis Night), no matter how many times, until she saves Madoka. She kills Kyuubey, but he respawns in a new body, which she chases, but he is saved by Madoka, similar to [[Episode 1]].<br />
<br />
Homura's monologue reveals that she is devoted to saving Madoka no matter how many times she'll have to retry.<br />
<br />
== Runes ==<br />
See [[Runes:Episode 10|Runes of episode 10]]<br />
<br />
==Speculations and Observations==<br />
<br />
===About Walpurgis Night and magic in general===<br />
* According to Kyuubey, Madoka as a Witch has the power to destroy the world but it is of no concern to him as long as they meet their quota.<br />
**It is speculated that because of the nature of Madoka's wish in Timeline 4, that wish gave her the power to not only destroy the Witch during Walpurgis Night with just one shot but that same power will destroy the world once she becomes a Witch herself. For that to happen Madoka will have to make the same wish on Walpurgis Night in Timeline 5.<br />
*Given the nature of Walpurgis Night and the presence of gears and the "falling" figure of the witch, Walpurgis Night could be the witch form of Homura. Although Homura's magic is very limited, her experience in multiple timelines makes the witch a formidable foe.<br />
*Since Walpurgis Night didn't drop a Grief Seed in Timeline 3, it is possible that Walpurgis Night could be a familar of an even more powerful witch. However, this would conflict with QB's statement of Walpurgis Night being the most powerful witch.<br />
*Walpurgis Night could be the witch of a magic girl who defeated the previous most powerful witch. As QB mentioned in Episode 10, the only choice left for a magic girl after defeating the most powerful witch "is to become a (even more powerful) witch herself".<br />
*Walpurgis Night could be Homura's witch form from the future, where she despairs with the fact that her involvement with Madoka only made her suffer some more. This would make sense if Timeline 1 was the best possible ending and the witch's goal is to kill Madoka alone (Timeline 1) or to kill Homura and end her fate.<br />
<br />
===About the timelines and time travel===<br />
* It seems like in every timeline the last ones left to fight Walpurgis Night are Homura and Madoka.<br />
**It is assumed that Mami, Kyouko, and Sayaka always die in every timeline.<br />
***However, some evidence suggests that Kyouko's and Sayaka's absence on Walpurgis Night in the other timelines could be explained by Kyouko never arriving in town (or joining forces) or Sayaka never making her wish.<br />
****Under this assumption we are lead to believe that Mami dies in every timeline.<br />
* It is possible that an imprint or a residue of past timelines are still within Madoka's memory. One example is the dream that Madoka had in Episode 1 and that strange image static in Episode 8.<br />
*It was speculated in [[Episode 9]] that if Kyuubey's people have the technology to travel back in time, they could fix the entropy problem by reversing it. However, in Episode 10, we learn that Kyuubey can only grant the ability to travel back in time. He is only part of the process to create it, so they do not posses the ability itself. The power of time travel rests only within Homura because of her wish.<br />
*Homura's Past Timeline Experiences:<br />
**While powerful, Mami can become mentally and emotionally unstable. If she learns the truth, she's likely to break down from the pressure and kill everyone (she is unreliable as an ally).<br />
**Sayaka refuses to believe or listen to Homura's advice. She will also be prone to making the same mistakes and become a witch (her fate is inevitable no matter what).<br />
**From past experience, Homura believes Kyouko may be the only ally left to help her in stopping Walpurgis Night. However, she did not anticipate her actions that lead to her death since this didnt happen in previous timelines.<br />
**Leaving Madoka alone or keeping her distance from her will not stop her from making her wish. Nor can Homura stop Walpurgis Night on her own.<br />
**The only constant in previous timelines is that Madoka is the only one with a pure soul, she was the only one who called her a friend and trusted her with her life.<br />
*Every time Homura tries to change the past she seems to make it worse:<br />
**Original timeline: Mami and Madoka die like heroines and they are the only known casualties, Earth is saved (and there are no more threats to the planet). The Truth is never revealed.<br />
***In this version Madoka dies without regrets in her heart, she remains pure and innocent in the end thus avoids the fate of becoming a witch.<br />
**Second timeline: Mami dies as a heroine, but Madoka becomes a normal Witch.<br />
**Third timeline: Sayaka becomes a witch and is killed by Homura, Mami dies but not before she loses her sanity and kills Kyouko. Madoka still dies and becomes a witch, but not before she saves Homura from becoming a witch and Homura mercy kills her before that happens.<br />
**Fourth timeline: Madoka becomes a powerful Witch that will destroy the Earth, everyone dies.<br />
**Fifth timeline: Mami dies before learning the truth, Sayaka still becomes a witch and dies with Kyouko.<br />
*Homura's reason to go back in time seems to change in each timeline, even though it's possible that only her methods of trying to save Madoka change and her final goal is still the same:<br />
**Original Timeline: Goes back to help Madoka and stop her from dying.<br />
**Timeline 2: Homura witnesses the truth and sees Madoka die and become a witch. Homura decides to go back and warn the other girls about the truth.<br />
**Timeline 3: Homura seems to resign to her fate, she wants to die alongside with Madoka and become witches together, she does not mind if they bring more destruction. However, Madoka saves her life and asks her to stop her from making the stupid mistake of becoming a Puella Magi/Witch, she also tells her that there are still good things left in the world so she refuses to become a Witch and bring more destruction.<br />
**Timeline 4: Homura realizes that Madoka has the power to destroy the world if left alone. Her mission is to fulfill Madoka's previous request and stop her from becoming a Puella Magi/Witch that will lead to the destruction of the Earth (something that Madoka refused to become part of in Timeline 3).<br />
* Madoka may be getting stronger on each timeline for unknown reasons.<br />
** Timeline 1 Madoka triumphs on Walpurgis night, possibly only after Mami has already weakened the witch. Madoka somehow dies from her wounds from the battle. This seems to be the weakest of all the Madokas.<br />
** Timeline 2 Madoka is victorious on Walpurgis night and does not die from her battle injuries. However, she immediately transforms into a witch afterwards. We do not know if Madoka as a witch is especially strong or not.<br />
***This one can be misleading, we must not forget that in this timeline she has the assistance of Homura and possibly Mami, which is why Madoka didnt die.<br />
** Timeline 3 Madoka wins with only the assistance of Homura. They would be capable of surviving if it were not for the blackening of their Soul Gems. Madoka clings to her humanity longer compared to her last incarnation.<br />
***Why Madoka lasted longer could be a result of her unwillingness to transform as she is now aware of what will happen next. The previous Madoka version was probably scared and confused as to what was happening to her so it is possible she didn't fight back the transformation. Also, she may have more energy left because of Homura's increased power and experience.<br />
** Timeline 4 Madoka is able to stop Walpurgis night in just one hit. Obviously stronger compared to the timeline 1 version. Kyubbey mentions how strong Madoka had become, surpassing his expectations. Her witch form has the power to destroy the whole planet.<br />
***It is stated previously that the strength and power of a Puella Magi is reflected by the nature of their wish, since Madoka witnessed the power of the witch in Walpurgis Night, she could have wished for the power to defeat her thus receiving equal or greater power than the witch.<br />
****Because Madoka has always stated how Puella Magi should protect everyone, it is possible her wish was simply to protect everyone and thus the corresponding witch curse would be to kill everyone.<br />
** Timeline 5 Kyuubey mentions that Madoka could become a god or godlike if she wished, although this may simply be him egging her on to become a magical girl.<br />
* In timeline 5 Homura kills kyuubey once at the beggining and fails to do it again until episode 9<br />
** That's why kyuubey mentions that she already killed him twice and he realizes that Homura can use Time-Control magic<br />
<br />
=== Continuity to previous episodes ===<br />
*Although the assumption is that the prologue from [[Episode 1]] is from Timeline 4, when Homura was fighting Walpurgis Night, that same Witch seems to differ in her shape in Episode 10 during the actual Timeline 4.<br />
**Walpurgis Night from Episode 10's Timeline 4 has slimmer body and longer dress.<br />
**Walpurgis Night of Timeline 1 is also slightly different from other two.<br />
**On top of that, the whole sequence looks different: it is a lot darker in Episode 1, and some backgrounds from the same episode are colored, while it is entirely black and white (and mostly white) in Timeline 4. See [[:File:Walpurgis ep1 versus ep10.jpg|a visual comparison (Episode 10 on the left, Episode 1 on the right)]].<br />
*Likewise, while Timeline 5 looks very similar to a scene from Episode 1 (contributing to an interpretation that it is the timeline of all previous episodes), there are small diferences:<br />
**There is a short conversation between Homura and Madoka in Episode 1, but in Episode 10, Madoka only says "Homura-chan". In episode 1, when Madoka encounters Homura in the closed section of the mall, Sayaka surprises Homura with a fire extinguisher; in episode 10, Sayaka doesn't make a surprise attack and Homura simply runs away when she appears.<br />
**Additionally, in Episode 1 Homura is seen with one of her legs propped on a bag and is never seen taking a step back.<br />
*Homura comments in [[Episode 4]] that she has witnessed "countless" magical girls die. In Episode 10, we see her witness (or kill herself) at most 9 deaths (the exact number depends on whether to count becoming a witch as dying, or whether to count killing the witch form). Even though some events may have happened off-camera, it is still too low number to consider it "too many to count". This suggests that there were indeed more loops than seen in the episode.<br />
<br />
=== Other observations and speculations ===<br />
* The calendar in Homura's room shows that she's getting discharged from the hospital on the 16th and returns to school on the 25th. From the format of the calendar, the 25th is apparently a Friday. The only month in 2011 that has 31 days and the 25th falling on a Friday is March. This means that the anime is set during the same time the final episode will be airing.<br />
*Homura's only way to get guns and weapons is by stealing them, it is assumed that she must have stolen them from a group of criminals.<br />
*Although it is shown that Homura can time travel, the limitations are not clear. She is always shown returning to the hospital only after something happens to Madoka, whether it be her dying or becoming a witch. This does not indicate whether or not Homura can continue after Madoka dies or if she can start over in the middle of one timeline.<br />
**In addition, it is unclear if Homura retains all of her physical improvements between time traveling. We see obvious contrasts between episode 1 and 10 where Homura is certainly stronger in timeline 5. However, this could be due to Homura retraining her weak body every time she starts again in the hospital and not be a result of accumulated strength.<br />
***As she was able to use magic to heal her eyes, it is quite possible she did the same with her body.<br />
*In Timeline 3, the Grief Seed Madoka uses to heal Homura maybe Sayaka's Grief Seed. She might have obtained it after defeating Oktavia at the station. Madoka might have kept the Grief Seed until the end to remember Sayaka.<br />
** The Grief Seed from Episode 10 looks different from that of Episode 8. However, this can be explained if the Grief Seed from Episode 8 is an immature form. Both Sayaka and Madoka's Grief Seed does not show dark parts as other Witches' Grief Seeds (except Charlotte's Seed which was partially dark since Charlotte had revived recently).<br />
*Fans have complained that they think Madoka is a weak character and doesn't have a lot of character development, regardless of their criticism, arguably one of the reasons can be Homura. Homura's mission to stop Madoka from becoming a Mahou Shoujo has also halted some of her character development. In each separate timeline we see a different side of Madoka: she can be strong, resilient, and self-sacrificing when she is a Mahou Shoujo. The character trait that remains constant in every timeline compared to the current one is her kindness and trust, in which she's willing to do what she can to help those in need.<br />
*Homura's shield can bend space and time; it is also useful for storing her arsenal of weapons. It is also a symbol of her will to protect Madoka.<br />
**Inside Homura's shield there are mechanisms like gears, springs, and a escapement wheel; these are parts typically found in mechanical clocks.<br />
***Everytime Homura uses her power her shield produces a small drop of red material. It's composition and function is unknown but it may have something to do with Homura's ability to surpass entropy to stop or travel back in time.<br />
*It is possible that the reason Kyuubey made Homura a Puella Magi in timeline 1 was to meet his energy quota. The Madoka version of that timeline died before she could become a witch.<br />
* This time, the only change spotted in the opening animation was the addition of Homura and Kyouko to the group shot. The opening animation has also been used as the ending sequence. (this episode has no proper opening)<br />
* The building from which Homura steals weapons in Timeline 3 has a banner reading 「射太興業」. 興業 means "industrial enterprise", and 射太 can be read as "Shafuto", the Japanese pronounciation of "[[wikipedia:Shaft (company)|Shaft]]".<br />
*The new version of Episode 10 that was broadcast at Nico Nico shows a tree in Timeline 4. Some speculate that this tree could somehow be related to the [http://en.wikipedia.org/wiki/Yggdrasil Yggdrasil Tree] (world tree) of Norse Mythology.<br />
<br />
== Images ==<br />
<gallery position="center" captionalign="left"><br />
File:Madoka-grief-seed.png|Madoka's grief seed, the cutest grief seed yet.<br />
File:Oktavia-ep10.JPG|Oktavia as seen in Episode 10.<br />
File:Oktavia Hitomi like minions.jpg|Oktavia's minions look like Hitomi<br />
File:Madoka-witch-first.jpg|Madoka's first witch transformation.<br />
File:Kaname witch form power.jpg|This is either Madoka's witch form or her display of power.<br />
File:Ep10-calendar.jpg|Homura's calendar and Madoka revealing the timeframe in which she contracted.<br />
File:Happy Homura.jpg|This is a happier and innocent Homura from Timeline 1. By Timeline 5 her demeanor will change.<br />
File:Homurabomb.jpg|How to make bombs.<br />
File:Mami crying mental breakdown.jpg|Mami suffers from a mental breakdown once she learns the truth. She decides suicide is the only option.<br />
File:Ep10-ending-group.jpg|Group shot with Kyouko and Homura.<br />
File:Bakemonogatari Madoka.jpg|Comparison between Bakemonogatari and Madoka<br />
File:Sailor uniform witch ep 10.jpg|This Witch is PATRICIA<br />
File:Homura Madoka the Witch Walpurgis Night.jpg|Walpurgis Night from Timeline 1<br />
File:Worlf after Walpurgis Night.png|The world after Walpurgis Night in Timeline 4<br />
File:Madoka10 art references.jpg| Paintings referenced in this episode.<br />
File:Grief Seed from episode 10.PNG|The Grief Seed maybe Sayaka's<br />
File:Walpurgis Comparison.PNG|Walpurgis Night looks different in each fight.<br />
File:Madoka-witch-forms.jpg|Comparison of Madoka's witch form in episode 10 and the image at the beginning of episode 1<br />
File:Different timelines.png|Contrast of the first day of school between timeline 1 and timeline 5<br />
File:Mami apt name.png |Mami's apartment name plate was changed in Episode 10<br />
File:Homura shield clock.png|The inside of Homura's shield resembles the mechanism of a mechanical clock<br />
File:Walpurgis ep1 versus ep10.jpg|The sequence in Episode 10 (left) is ''not'' the dream from [[Episode 1]] (right).<br />
File:Homura_madoka_eyes.jpg|Comparison of Madoka's eyes from the opening secuence and homura's eyes in the first time travel.<br />
File:Ep10 Tree.png|Tree added in the Niconico Channel broadcast version.<br />
File:Ep10 Kriemhild Name.png|SHAFT updated Episode 10 for a rerun stream. This updated version shows Madoka's witch name as [[Kriemhild Gretchen]]. Gretchen happened to be Faust's lover.<br />
File:Episode 10 Nico Changes.jpg|2ch comparison chart between MBS broadcast and Nico channel broadcast.<br />
</gallery><br />
<br />
==External links== <br />
* [http://www.madoka-magica.com/story/index.html ''Episode 10'' summary at official website] (Japanese)<br />
<!--*{{summary_link|10}} Remove line above and comment when official site is updated--><br />
<br />
{{Episode navigation<br />
|title = I Won't Depend on Anyone Anymore<br />
|previous_episode = 9<br />
|previous_title = I'll Never Allow That<br />
|next_episode = 11<br />
|next_title = To The Only Path Remaining<br />
}}<br />
<br />
[[Category:Episodes|Episode A]]<!--Next to 9--></div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=Madoka_Magica_Episode_9:_I%27ll_Never_Allow_That&diff=8121Madoka Magica Episode 9: I'll Never Allow That2011-03-25T15:48:28Z<p>Anon-kun: link to homepage</p>
<hr />
<div>{{Infobox_episode<br />
|name = I'll Never Allow That<br />
|jname = そんなの、あたしが許さない<br />
|date = 3 March 2011<br />
|script = [[Gen Urobuchi]]<br />
|storyboard = Noriko Shichishima<br />
|director = Masahiro Mukai<br />
|adirector = Miyuki Katayama<br />
}}<br />
<br />
== Summary ==<br />
[[Sayaka]] becomes a [[Oktavia von Seckendorff|witch]] and [[Kyouko]] enters her maze, holding Sayaka's now soulless body. [[Homura]] saves Kyouko and while they are walking off, they meet [[Madoka]], worried upon seeing Sayaka's body. Homura explains everything to Kyouko and Madoka.<br />
<br />
When Madoka is crying alone in her room, [[Kyuubey]] enters and explains everything in detail to her. He reveals that he is a member of an extraterrestial, emotionless society, worried about the total energy in the universe lowering due to entropy<ref>What Kyuubey is describing is called the [[wikipedia:Heat death of the universe|Heat Death of the Universe]]: To elaborate, this term describes the point at which the universe has reached a state of maximum entropy -- when all available energy has moved to places of less energy. Once this has happened, no more work can be extracted from the universe. Since heat ceases to flow, no more work can be acquired from heat transfer. This same kind of equilibrium state will also happen with all other forms of energy. Since no more work can be extracted from the universe at that point, it is effectively dead.</ref>. Puella Magi and witches are a way to counter this, creating energy from wavering emotions of young girls. As Kyuubey and his peers are emotionless, they are unable to produce such energy themselves. Because of this, they also don't understand the concept of human morals. Madoka is saddened and angry because of all those revelations she has just heard.<br />
<br />
Kyouko is maintaining natural heat of Sayaka's body using her magic. She asks Kyuubey whether it is possible to recover Sayaka's Soul Gem. He dodges the question, saying that nobody ever tried it before and due to the very existence of Puella Magi is contradictory, such achievement wouldn't surprise him.<br />
<br />
The following day, Hitomi attempts to speak to Madoka but Madoka avoids her. Kyouko makes Madoka skip school, with a quest to rescue Sayaka instead. She explains her plan to ignite human emotions in the witch whom Sayaka became, by having her best human friend talk to her. Madoka agrees, and they go to witch-Sayaka's maze. Madoka tries talking with Kyouko protecting her, but with no effect. Kyouko commits suicide, also killing the witch, while Homura (who went out from school and entered the maze to protect Madoka) takes fainted Madoka out.<br />
<br />
The episode ends with Homura and Kyuubey talking in Homura's house, with Kyuubey revealing that rescuing Sayaka was in fact impossible, but he didn't stop Kyouko from doing so to make Madoka become a Puella Magi, given that Homura alone will not be able to win against Walpurgis. Homura says that she won't allow it.<br />
<br />
== Runes ==<br />
<br />
See [[Runes:Episode 9|Runes of episode 9]]<br />
<br />
==Speculations and Observations==<br />
[[File:Chairs s.jpg|thumb|190px|A comparison between Madoka's bedroom and Bokurano.]]<br />
*Per the image at the right, the chairs seen in [[Madoka's House|Madoka's bedroom]] may be a homage to [[wikipedia:Bokurano:_Ours|Bokurano]], a manga and anime series that [[wikipedia:Deconstruction|deconstructed]] a children's mecha anime called [[wikipedia:Zettai Muteki Raijin-Oh|Zettai Muteki Raijin-Oh]]. In Bokurano, children are called by a strange man and an evil-looking mascot character to pilot a giant robot and defend their world from invaders from an alternate time-line. If they lose, the loser's universe is destroyed. They are also told by the mascot only after the first battle that the robot's fuel source is the life of the person who pilots -- in other words, each of them will only be able to pilot one time. Accordingly, the first pilot slumps down and dies right after the end of his battle. The chairs in Bokurano are reflections of the personalities of the selected pilots, drawn from their memories and manifested in the robot's cockpit. After a pilot dies, their vacant seat remains.<br />
<br />
*During the fight with Kyouko, Sayaka's lower body turns into one of a mermaid. Many speculate that Sayaka's tragedy is similar to that of [[wikipedia:The_Little_Mermaid|the Little Mermaid's story]] because of the similarities. You can read a summary of the story [http://pastebin.com/gCGG8Ljf here].<br />
<br />
*The mermaid tail may also be referencing the story of [[wikipedia:Melusine|Melusine]]. Goethe wrote a version of Melusine titled Die Neue Melusine or [http://books.google.com/books?id=J8Pb0oeVAaIC&pg=PA347&dq=the+new+melusina&hl=en&ei=u09zTc3mA4K4twfAyv35Dg&sa=X&oi=book_result&ct=result&resnum=1&ved=0CCgQ6AEwAA#v=onepage&q=the%20new%20melusina&f=false The New Melusina].<br />
<br />
*Sayaka's witch name is [[Oktavia von Seckendorff]]. In real life, Karl Siegmund von Seckendorff was a german poet who wrote a novel called Das Rad des Schicksals = The Wheel of Fate (which explains the wheels, there was also mention of fate in her barrier). He also put 'The King in Thule' (a poem by Goethe, part of Faust1) to music.<br />
<br />
*The wheels used in Oktavia's attacks may also symbolize the Wheel of Fortune from tarot, whose popular interpretations include "possibilities, opportunities, new developments, sudden changes", which fit Sayaka's life from when she encountered Kyubey to when she became a Witch.<br />
<br />
*Alternatively, Sayaka was at a train station when she transformed. Her attacks may be inspired by train wheels.<br />
<br />
*Kyouko's use of [[wikipedia:Umaibō|Umaibou]] as a sign of friendship with Madoka in this episode can be seen as baton passing, as her main role ended in this episode. It's also possible that she choose Umaibou due to its extremely low price (9 yen) being something that she can afford, since Sayaka didn't accept stolen goods back in [[episode 7]].<br />
<br />
*In Episode 7 the Witch Elsa Maria was praying with her back turned (as if ignoring Sayaka) facing a monstrance. In Episode 9 Kyouko almost looks like she is praying while facing [[Oktavia von Seckendorff]], as if the prayer is for Sayaka's soul. See the gallery below for a visual comparison.<br />
<br />
*Some suggest that Kyubei's solution for entropy is just flawed and downright impossible. Defenders like to point out that Kyubei is from an advanced alien civilization, so maybe their understanding of the universe is far superior than that of humans. [[:File:Infinite power.png|This picture]] summarizes the paradox in a simplified way.<br />
**[http://www.darkmirage.com/2011/03/06/the-physics-of-puella-magi/ This blog explains] why Kyubei's solution regarding entropy is just downright irrational<ref>At this point some fans suspect that Kyubei is actually lying just as his explanation that he can not understand how to trick humans sounds spurious (or the time he tricked Kyouko that Sayaka could be saved so she would die in the process), but it is to be seen in future Episodes where the truth lies. There is speculation that Urobuchi got the entropy idea from another show ([http://en.wikipedia.org/wiki/Star_Driver:_Kagayaki_no_Takuto Star Driver]) as a away to mislead his fans. It is believed that Urobuchi introduced the changes [http://forums.animesuki.com/showthread.php?t=102397&page=24 at the last minute] for Episode 9.</ref>.<br />
<br />
*The [[:File:Sayaka Sheet.PNG|musical score seen in the beginning of the episode]] is the beginning of the background music playing during the scene in [[Oktavia]]'s barrier.<br />
<br />
*A hidden hairpin on Kyoko merges with her Soul Gem kept on her chest before she unleashes her suicide attack.<br />
<br />
*This episode reverted back to no ending sequence like in episodes 1 and 2.<br />
<br />
== Images ==<br />
<gallery position="center" captionalign="left"><br />
File:Ep 9syrenka.png|Some information regarding images and symbols in Episode 9<br />
File:Love me do (beatles).jpg|The image reads: ＬＯＶＥ ＭＥ ＤＯ！！ ＼(*´3｀*)／ "Love me Do" could be a reference to a Beatles song or a h-manga by Kensoh Ogawa (aka Fukudahda). <br />
File:Sayaka Sheet.PNG|Transcript of the musical sheets seen in the episode; link to MIDI on file page<br />
File:Sayaka witch form.jpg|Sayaka's witch form. The armor could be a reference to the von Seckendorff family's knightly heritage.<br />
File:Close up Sayaka coats of arm and the Seckendorff family coats of arm.png|comparison between Sayaka's coat of arms and the Seckendorff family coats of arm<br />
File:Kamijou silhouette background.png|a silhouette of what looks like Kamijou with his back turned holding a violin<br />
File:Revolutionary Girl Utena and Madoka.png|The image on the left is from Revolutionary Girl Utena, the image on the right is from Episode 9 with Sayaka and Kyouko<br />
File:Sayaka wheel attack fortune tarot.png|Sayaka's witch name is [[Oktavia von Seckendorff]], a possible reference to the poet Karl von Seckendorff (see above).<br />
File:Human emotion energy end entropy.png|According to Kyubey, to solve the entropy problem human emotions must be used as a form of energy. In exchange for their sacrifices, girls are given one wish.<br />
File:Ep9 Kyouko SG.jpg|Kyouko's Soul Gem is dimmed prior to fighting with Oktavia presumably due to using her power to maintain Sayaka's body.<br />
File:Praying kyouko sayaka.png|Parallel between [[Elsa Maria]] (episode 7) and [[Kyouko]] (episode 9)<br />
File:Infinite power.png|Kyubey's flawed solution to the entropy, explained with a no break. See further explanations [[Episode_9#Speculations_and_Observations|above]].<br />
</gallery><br />
<br />
==Annotations & References==<br />
<references/><br />
<br />
==See also==<br />
*[[Madoka Magica and Science]]<br />
<br />
==External links==<br />
*{{summary_link|09}}<br />
<br />
{{Episode navigation<br />
|title = I'll Never Allow That<br />
|previous_episode = 8<br />
|previous_title = I'm Such a Fool<br />
|next_episode = 10<br />
|next_title = I Won't Depend on Anyone Anymore<br />
}}<br />
<br />
[[Category:Episodes]]</div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=Mathieu&diff=8120Mathieu2011-03-25T15:44:22Z<p>Anon-kun: cat</p>
<hr />
<div>#REDIRECT [[Patricia#Mathieu]]<br />
<br />
[[Category:Minions]]</div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=Michaela&diff=8119Michaela2011-03-25T15:44:05Z<p>Anon-kun: cat</p>
<hr />
<div>#REDIRECT [[Izabel#Michaela]]<br />
<br />
[[Category:Minions]]</div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=Klarissa&diff=8118Klarissa2011-03-25T15:43:08Z<p>Anon-kun: cat</p>
<hr />
<div>#REDIRECT [[Oktavia von Seckendorff#Klarissa]]<br />
<br />
[[Category:Minions]]</div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=Gotz&diff=8117Gotz2011-03-25T15:40:35Z<p>Anon-kun: cat</p>
<hr />
<div>#REDIRECT [[Roberta#Gotz]]<br />
<br />
[[Category:Minions]]</div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=Madoka_Magica_Products&diff=7618Madoka Magica Products2011-03-20T11:37:04Z<p>Anon-kun: manga 3 delayed too</p>
<hr />
<div>{{Note|For Madoka-related free content, please refer to [[Media]].}}<br />
<br />
{| class="wikitable sortable collapsible" style="width: 100%;"<br />
! Product type<br />
! Title<br />
! Release date<br />
! Price<br />
|-<br />
| CD<br />
| [http://www.amazon.co.jp/dp/B004DGD400/ ClariS: Connect Limited Edition]<br />
| 2011-02-02<br />
| ¥ 1,575<br />
|-<br />
| CD<br />
| [http://www.amazon.co.jp/dp/B004DGD4FU/ ClariS: Connect Anime Edition]<br />
| 2011-02-02<br />
| ¥ 1,300<br />
|-<br />
| CD<br />
| [http://www.amazon.co.jp/dp/B004DGD4AA/ ClariS: Connect Regular Edition]<br />
| 2011-02-02<br />
| ¥ 1,223<br />
|-<br />
| Manga<br />
| [http://www.amazon.co.jp/dp/4832279904/ Puella Magi Madoka☆Magica Vol.1]<br />
| 2011-02-12<br />
| ¥ 690<br />
|-<br />
| CD<br />
| [http://www.amazon.co.jp/dp/B004DGD45A/ Kalafina: Magia Limited Edition]<br />
| 2011-02-16<br />
| ¥ 1,575<br />
|-<br />
| CD<br />
| [http://www.amazon.co.jp/dp/B004DGD4QE/ Kalafina: Magia Anime Edition]<br />
| 2011-02-16<br />
| ¥ 1,300<br />
|-<br />
| CD<br />
| [http://www.amazon.co.jp/dp/B004DGD4NM/ Kalafina: Magia Regular Edition]<br />
| 2011-02-16<br />
| ¥ 1,223<br />
|-<br />
| Manga<br />
| [http://www.amazon.co.jp/dp/483224003X/ Puella Magi Madoka☆Magica Vol.2]<br />
| 2011-03-12<br />
| ¥ 690<br />
|-<br />
| Blu-ray<br />
| [http://www.amazon.co.jp/dp/B004INGZAE/ Puella Magi Madoka☆Magica Vol.1 Limited Edition]<br />
| TBD<br />
| ¥ 7,350<br />
|-<br />
| DVD<br />
| [http://www.amazon.co.jp/dp/B004INGZCC/ Puella Magi Madoka☆Magica Vol.1 Limited Edition]<br />
| TBD<br />
| ¥ 6,300<br />
|-<br />
| DVD<br />
| [http://www.amazon.co.jp/dp/B004INGZLS/ Puella Magi Madoka☆Magica Vol.1 Regular Edition]<br />
| TBD<br />
| ¥ 5,250<br />
|-<br />
| Manga<br />
| [http://www.amazon.co.jp/dp/4832240145/ Puella Magi Madoka☆Magica Vol.3 (End)]<br />
| TBD<br />
| ¥ 690<br />
|-<br />
| Blu-ray<br />
| [http://www.amazon.co.jp/dp/B004L7A7WO/ Puella Magi Madoka☆Magica Vol.2 Limited Edition]<br />
| 2011-04-27<br />
| ¥ 7,350<br />
|-<br />
| DVD<br />
| [http://www.amazon.co.jp/dp/B004L7A7Q0/ Puella Magi Madoka☆Magica Vol.2 Limited Edition]<br />
| 2011-04-27<br />
| ¥ 6,300<br />
|-<br />
| DVD<br />
| [http://www.amazon.co.jp/dp/B004L7A7UG/ Puella Magi Madoka☆Magica Vol.2 Regular Edition]<br />
| 2011-04-27<br />
| ¥ 5,250<br />
|-<br />
| Blu-ray<br />
| [http://www.amazon.co.jp/dp/B004MMFTMQ/ Puella Magi Madoka☆Magica Vol.3 Limited Edition]<br />
| 2011-05-25<br />
| ¥ 7,350<br />
|-<br />
| DVD<br />
| [http://www.amazon.co.jp/dp/B004MMFTN0/ Puella Magi Madoka☆Magica Vol.3 Limited Edition]<br />
| 2011-05-25<br />
| ¥ 6,300<br />
|-<br />
| DVD<br />
| [http://www.amazon.co.jp/dp/B004MMFTNA/ Puella Magi Madoka☆Magica Vol.3 Regular Edition]<br />
| 2011-05-25<br />
| ¥ 5,250<br />
|-<br />
| Blu-ray<br />
| [http://www.amazon.co.jp/dp/B004P4RTKQ/ Puella Magi Madoka☆Magica Vol.4 Limited Edition]<br />
| 2011-06-22<br />
| ¥ 7,350<br />
|-<br />
| DVD<br />
| [http://www.amazon.co.jp/dp/B004P4RTLK/ Puella Magi Madoka☆Magica Vol.4 Limited Edition]<br />
| 2011-06-22<br />
| ¥ 6,300<br />
|-<br />
| DVD<br />
| [http://www.amazon.co.jp/dp/B004P4RTL0/ Puella Magi Madoka☆Magica Vol.4 Regular Edition]<br />
| 2011-06-22<br />
| ¥ 5,250<br />
|-<br />
| Blu-ray<br />
| [http://www.amazon.co.jp/dp/B004RCJJFE/ Puella Magi Madoka☆Magica Vol.5 Limited Edition]<br />
| 2011-07-27<br />
| ¥ 7,350<br />
|-<br />
| DVD<br />
| [http://www.amazon.co.jp/dp/B004RCJJ54/ Puella Magi Madoka☆Magica Vol.5 Limited Edition]<br />
| 2011-07-27<br />
| ¥ 6,300<br />
|-<br />
| DVD<br />
| [http://www.amazon.co.jp/dp/B004RCJJWC/ Puella Magi Madoka☆Magica Vol.5 Regular Edition]<br />
| 2011-07-27<br />
| ¥ 5,250<br />
|-<br />
|}<br />
<br />
==Blu-Ray Discs==<br />
{| class="wikitable" style="width: 100%; vertical-align: top;"<br />
! style="width: 300px;" | Cover<br />
! Description<br />
|-<br />
|[[File:Disc 1.jpg|300px]]<br />
| style="vertical-align:top; padding: 10px;" |<br />
;[http://www.amazon.co.jp/dp/B004INGZAE Puella Magi Madoka☆Magica Vol.1 Blu-ray]<br />
:'''Release date''': TBD<br />
:'''Price''': ¥ 7,350 (~90 USD)<br />
:'''Running time''': 60 minutes<br />
:'''Package contents''':<br />
:*Blu-ray disc (episode 1, 2)<br />
:*Special CD<br />
:*#Urobuchi Gen-Supervised Drama CD<br />
:*#Kaname Madoka Character Song (ED theme for episode 1, 2)<br />
:*Tri-fold clear case<br />
:*Package illustration by Kishida Takahiro<br />
:*Special booklet with Aoki Ume's 4koma Manga<br />
:'''Extra contents''':<br />
:#Episode 1 audio commentary (Yūki Aoi, Saitō Chiwa, Kitamura Eri)<br />
:#Episode 2 audio commentary (Yūki Aoi, Saitō Chiwa, Aoki Ume)<br />
:#Unbroadcasted original ending theme version<br />
|-<br />
|[[File:Disc 2.jpg|300px]]<br />
| style="vertical-align:top; padding: 10px;"|<br />
;[http://www.amazon.co.jp/dp/B004L7A7WO/ Puella Magi Madoka☆Magica Vol.2 Blu-ray]<br />
:'''Release date''': April 27, 2011<br />
:'''Price''': ¥ 7,350 (~90 USD)<br />
:'''Running time''': 48 minutes<br />
:'''Package contents''':<br />
:*Blu-ray disc (episode 3, 4)<br />
:*Original soundtrack CD Volume 1<br />
:*Tri-fold clear case<br />
:*Package illustration by Kishida Takahiro<br />
:*Special booklet with Aoki Ume's 4koma Manga<br />
:'''Extra contents''':<br />
:#Audio commentary<br />
|-<br />
| style="text-align: center;"| NO IMAGE<br />
| style="vertical-align:top; padding: 10px;"|<br />
;[http://www.amazon.co.jp/dp/B004MMFTMQ/ Puella Magi Madoka☆Magica Vol.3 Blu-ray]<br />
:'''Release date''': May 25, 2011<br />
:'''Price''': ¥ 7,350 (~90 USD)<br />
:'''Running time''': 48 minutes<br />
:'''Package contents''':<br />
:*Blu-ray disc (episode 5, 6)<br />
:*Special CD<br />
:*Tri-fold clear case<br />
:*Package illustration by Kishida Takahiro<br />
:*Special booklet with Aoki Ume's 4koma Manga<br />
:'''Extra contents''':<br />
:#Audio commentary<br />
|-<br />
| style="text-align: center;"| NO IMAGE<br />
| style="vertical-align:top; padding: 10px;"|<br />
;[http://www.amazon.co.jp/dp/B004P4RTKQ/ Puella Magi Madoka☆Magica Vol.4 Blu-ray]<br />
:'''Release date''': June 22, 2011<br />
:'''Price''': ¥ 7,350 (~90 USD)<br />
:'''Running time''': 48 minutes<br />
:'''Package contents''':<br />
:*Blu-ray disc (episode 7, 8)<br />
:*Special CD<br />
:*Tri-fold clear case<br />
:*Package illustration by Kishida Takahiro<br />
:*Special booklet with Aoki Ume's 4koma Manga<br />
:'''Extra contents''':<br />
:#Audio commentary<br />
|-<br />
| style="text-align: center;"| NO IMAGE<br />
| style="vertical-align:top; padding: 10px;"|<br />
;[http://www.amazon.co.jp/dp/B004RCJJFE/ Puella Magi Madoka☆Magica Vol.5 Blu-ray]<br />
:'''Release date''': July 27, 2011<br />
:'''Price''': ¥ 7,350 (~90 USD)<br />
:'''Running time''': 48 minutes<br />
:'''Package contents''':<br />
:*Blu-ray disc (episode 9, 10)<br />
:*Special CD<br />
:*Tri-fold clear case<br />
:*Package illustration by Kishida Takahiro<br />
:*Special booklet with Aoki Ume's 4koma Manga<br />
:'''Extra contents''':<br />
:#Audio commentary<br />
|}<br />
<br />
==Singles==<br />
===Connect===<br />
:''See also [[Connect]] for lyrics, credits and observations.''<br />
This is the opening song for the series.<br />
{| class="wikitable" style="width: 100%; vertical-align: top;"<br />
! style="width: 300px;" | Cover<br />
! Description<br />
|-<br />
|[[File:Connect Anime Cover.jpg|300px]]<br />
| style="vertical-align:top; padding: 10px;" |<br />
;[http://www.amazon.co.jp/dp/B004DGD4FU ClariS: Connect (Anime Edition)]<br />
:'''Release date''': February 2, 2011<br />
:'''Price''': ¥ 1,300 (~15 USD)<br />
:'''Tracklist''':<br />
:#コネクト (Connect)<br />
:#Dreamin’<br />
:#キミとふたり (Kimi to futari)<br />
:#コネクト (Connect) -TV MIX-<br />
:'''Package contents''':<br />
:*Audio CD<br />
:*Anime character design package illustration<br />
|-<br />
|[[File:Connect OP Cover.jpg|300px]]<br />
| style="vertical-align:top; padding: 10px;" |<br />
;[http://www.amazon.co.jp/dp/B004DGD400 ClariS: Connect (Limited Edition)]<br />
:'''Release date''': February 2, 2011<br />
:'''Price''': ¥ 1,575 (~19 USD)<br />
:'''Tracklist''':<br />
:#コネクト (Connect)<br />
:#Dreamin’<br />
:#キミとふたり (Kimi to futari)<br />
:#コネクト (Connect) -Instrumental-<br />
:'''Package contents''':<br />
:*Audio CD<br />
:*Music video DVD<br />
:*Anime character design illustration sticker<br />
|}<br />
===Magia===<br />
:''See also [[Magia]] for lyrics and credits, and [[Speculah:Ending Analysis]] for analysis of the lyrics and ending animation.''<br />
This is the ending song of the series.<br />
{| class="wikitable" style="width: 100%; vertical-align: top;"<br />
! style="width: 300px;" | Cover<br />
! Description<br />
|-<br />
|[[File:Magia Anime Cover.jpg|300px]]<br />
| style="vertical-align:top; padding: 10px;" |<br />
;[http://www.amazon.co.jp/dp/B004DGD4QE Kalafina: Magia ED (Anime Edition)]<br />
:'''Release date''': February 16, 2011<br />
:'''Price''': ¥ 1,300 (~15 USD)<br />
:'''Tracklist''':<br />
:#Magia<br />
:#Magia (magic mix)<br />
:#Magia (TV Version)<br />
:#Magia (instrumental)<br />
:'''Package contents''':<br />
:*Audio CD<br />
:*Digipak case<br />
:*Anime character design package illustration<br />
|-<br />
|[[File:Magia ED Cover.jpg|300px]]<br />
| style="vertical-align:top; padding: 10px;" |<br />
;[http://www.amazon.co.jp/dp/B004DGD45A Kalafina: Magia ED (Limited Edition)]<br />
:'''Release date''': February 16, 2011<br />
:'''Price''': ¥ 1,575 (~19 USD)<br />
:'''Tracklist''':<br />
:#Magia<br />
:#snow falling<br />
:#Magia (instrumental)<br />
:'''Package contents''':<br />
:*Audio CD<br />
:*Music video DVD<br />
|}<br />
<br />
==Manga==<br />
{| class="wikitable" style="width: 100%; vertical-align: top;"<br />
! style="width: 300px;" | Cover<br />
! Description<br />
|-<br />
| [[File:Manga Vol.1 Cover.jpg|300px]]<br />
| style="vertical-align:top; padding: 10px;" |<br />
;[http://www.amazon.co.jp/dp/4832279904 Puella Magi Madoka☆Magica Vol.1]<br />
:'''Release date''': February 12, 2011<br />
:'''Price''': ¥ 690 (~8 USD)<br />
:'''Artist''': Hanokage<br />
:'''Publisher''': Houbunsha<br />
|-<br />
| [[File:Manga Vol.2 Cover.jpg|300px]]<br />
| style="vertical-align:top; padding: 10px;" |<br />
;[http://www.amazon.co.jp/dp/483224003X Puella Magi Madoka☆Magica Vol.2]<br />
:'''Release date''': March 12, 2011<br />
:'''Price''': ¥ 690 (~8 USD)<br />
:'''Artist''': Hanokage<br />
:'''Publisher''': Houbunsha<br />
|-<br />
| style="text-align: center;"| NO IMAGE<br />
| style="vertical-align:top; padding: 10px;" |<br />
;[http://www.amazon.co.jp/dp/4832240145 Puella Magi Madoka☆Magica Vol.3 (End)]<br />
:'''Release date''': April 12, 2011<br />
:'''Price''': ¥ 690 (~8 USD)<br />
:'''Artist''': Hanokage<br />
:'''Publisher''': Houbunsha<br />
|}<br />
<br />
== External links ==<br />
* [http://www.madoka-magica.com/bddvd/ Official BD, DVD, and CD page] (Japanese)<br />
* [http://www.dokidokivisual.com/madokamagica/ Official manga page] (Japanese)</div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=Mathematics_of_Madoka_Magica&diff=7406Mathematics of Madoka Magica2011-03-19T00:12:14Z<p>Anon-kun: she probably encountered each witch a fair number of times</p>
<hr />
<div>[[File:Homura using statistics.jpg|thumb|[[Homura]] said that she uses applied statistics in witch hunts]]<br />
The cast of Mahou Shoujo Madoka Magica are students of [[Mitakihara Middle School]]. Despite merely a middle school, coursework in Mitakihara can be quite challenging. The following are Math problems that have appeared thus far in the show, and their solutions:<br />
<br />
==Episode 1==<br />
[[file:Episode One Algebra.jpg|thumb|Episode one Math question 1]]<br />
<br />
===Question 1===<br />
Any integer divided by 14 will have a remainder between 0 and 13. Given a is has remainder of 6 and b has remainder of 1 when divided by 14, what is the remainder of x when divided by 14, given x is an integer solution to x^2-2ax+b=0?<br />
<br />
'''Solution:'''<br />
<br />
This problem can be solved simply with [[wikipedia:Modular arithmetic|modular arithmetic]]:<br />
<br />
a = 6 mod 14;<br />
b = 1 mod 14;<br />
x = r mod 14.<br />
<br />
x^2 - 2ax + b = 0 is equivalent to<br />
<br />
x^2 - 12x + 1 = 0 mod 14<br />
<br />
x(x - 12) = -1 mod 14<br />
<br />
-1 has no proper divisors modulo 14, thus factors x and x-12 must correspond to either 1 or -1 mod 14.<br />
<br />
If x = 1 mod 14, then x-12 = 3 mod 14: false. If x = -1 mod 14, then x - 12 = 1 mod 14: true.<br />
<br />
Therefore, x = -1 mod 14 = 13 mod 14, i.e., the remainder r of x is 13.<br />
<br />
<br />
'''Second Solution:'''<br />
<br />
Homura used in Episode 1 a basic approach with usual integer [[wikipedia:arithmetic|arithmetic]].<br />
<br />
Let x = 14q + r, a = 14s + 6, b = 14t + 1.<br />
<br />
Substitute into x^2 - 2ax + b = 0 to get after some calculations<br />
<br />
x^2 - 2ax + b = 14c + (r+1)^2 = 0 with c = 14q^2 + 2q - 28qs - 2rs - 12q + t - r<br />
<br />
14 divides 14c and 0, hence it also divides (r+1)^2. This implies that r+1 is divisible by 14.<br />
<br />
The question asks for a remainder r between 0 and 13, so we obtain r = 13.<br />
<br />
See also: http://green-oval.net/cgi-board.pl/a/thread/44824340<br />
<br />
===Question 2===<br />
[[file:Episode One Math Q2.jpg|thumb|Episode one Math question 2]]<br />
Assuming that p is a prime number and n is an arbitrary natural number, prove that (1+n)^p - n^p - 1 is divisible by p.<br />
<br />
'''Solution:'''<br />
<br />
By [[wikipedia:Fermat's Little Theorem|Fermat's Little Theorem]], for any prime p and integer a,<br />
<br />
a^p = a (mod p)<br />
<br />
Thus:<br />
<br />
(1+n)^p - n^p - 1 (mod p) is equivalent to<br />
<br />
(1+n) - n - 1 (mod p) is equivalent to<br />
<br />
0 (mod p)<br />
<br />
So the overall expression is divisible by p.<br />
<br />
<br />
'''Second solution:'''<br />
<br />
The problem can be solved with the [[wikipedia:binomial theorem|binomial theorem]]:<br />
<br />
For a, b not equal to 0 and nonnegative integer p it holds that:<br />
<br />
<!--(a + b)^p = sum (for k from 0 to p) of (p choose k) a^(p-k) b^k-->[[file:Q2 binomial theorem.png]]<br />
<br />
where the binomial coefficient (p choose k) is the integer <!-- p(p-1)...(p-k+1)/(k(k-1)...1) --> [[file:Q2 binomial coefficient.png]] for 0 ≤ k ≤ p.<br />
<br />
Therefore,<br />
<br />
<!--(1 + n)^p = sum (for k from 0 to p) of (p choose k) n^k-->[[file:Q2 equation 1.png]]<br />
<br />
and<br />
<br />
<!-- (1+n)^p - n^p - 1 = sum (for k from 1 to p-1) of (p choose k) n^k -->[[file:Q2 equation 2.png]]<br />
<br />
since (p choose 0) = (p choose p) = 1.<br />
<br />
It holds that (p choose 1) = p.<br />
Since p is a prime number, the factor p in (p choose k) is not divisible by k, k-1,...,2 for 2 ≤ k ≤ p-1.<br />
Therefore, p divides (p choose k) for 1 ≤ k ≤ p-1.<br />
<br />
Since each summand on the right side is divisible by p, the whole sum, i.e. the left side is also divisible by p.<br />
<br />
<br />
===Question 3===<br />
[[file:Episode One Math Q3.jpg|thumb|Episode one Math question 3]]<br />
<br />
<br />
Find the integer solutions (a,b) with (a^3 + a^2 - 1) - (a - 1)b = 0<br />
<br />
'''Solution:'''<br />
[[file:Episode One Math S3.jpg|thumb|Episode one Math solution to question 3]]<br />
<br />
(a^3 + a^2 - 1) - (a - 1)b<br><br />
= (a^3 - 1) + (a^2 - 1) + 1 - (a - 1)b<br><br />
= (a - 1)(a^2 + a + 1) + (a - 1)(a + 1) + 1 - (a - 1)b<br><br />
= (a - 1)(a^2 + 2a - b + 2) + 1 = 0<br />
<br />
Therefore,<br> <br />
(a - 1)(a^2 + 2a - b + 2) = -1<br />
<br />
a, b are integers, therefore the factors a - 1 and a^2 + 2a - b + 2 are integers too. Since the product is -1, one of the factors must be equal to -1, the other to 1.<br />
<br />
If a - 1 = -1 then a = 0 and from a^2 + 2a - b + 2 = 1 we obtain b = 1.<br />
<br />
If a - 1 = 1 then a = 2 and a^2 + 2a - b + 2 = -1 implies that b = 11.<br />
<br />
There are two integer solutions: (a,b) = (0,1) and (a,b) = (2,11).<br />
<br />
<br />
===Question 4===<br />
[[file:Episode One Math Q4.jpg|thumb|Episode one Math question 4]]<br />
Given [[file:Math episode one question four.png]],<br />
<br />
find the sum of F(1)+F(2)+F(3)+...+F(60).<br />
<br />
'''Solution:'''<br />
[[file:Episode One Math S4.jpg|thumb|Episode one Math, partial solution to question 4. Note that there are errors.]]<br />
<br />
Simplying the fraction:<br />
<br />
Reference that for any variables a & b, (a-b) (a+b) = (a-b)^2. Let (2x + 1)^.5 = a and (2x - 1)^.5 = b. Multiply F(x) by 1 or (a-b)/(a-b) or ((2x + 1)^.5 - (2x - 1)^.5) / ((2x + 1)^.5 - (2x - 1)^.5). The denominator becomes:<br />
<br />
((2x + 1)^.5 + (2x - 1)^.5) * ((2x + 1)^.5 - (2x - 1)^.5) <br />
<br />
= (2x + 1) - (2x - 1)<br />
<br />
= 2<br />
<br />
The numerator becomes:<br />
<br />
(4x + (4x^2 - 1)^.5 ) ((2x + 1)^.5 - (2x - 1)^.5) <br />
<br />
=(4x + ((2x)^2 - 1)^.5 ) (a - b) <br />
<br />
=(4x + (2x - 1)^.5 (2x + 1)^.5 ) (a -b) <br />
<br />
=(4x + a b ) (a - b) <br />
<br />
=4x (a - b) + (a^2) b - a (b^2)<br />
<br />
=4x (a - b) + (2x + 1) b - a (2x - 1)<br />
<br />
=4x (a - b) - 2x (a - b) + a + b<br />
<br />
=2x ( a - b) + a + b<br />
<br />
=(2x + 1) a - (2x - 1) b<br />
<br />
=a^3 - b^3<br />
<br />
<br />
Therefore:<br />
<br />
F(x) = (a^3 - b^3 ) / 2<br />
<br />
Note that when x= 1, 2, 3, 4, ..., 60;<br />
<br />
a^3 = 3^1.5, 5^1.5, 7^1.5, 9^1.5, ..., 119^1.5, 121^1.5;<br />
<br />
b^3 = 1^1.5, 3^1.5, 5^1.5, 7^1.5, 9^1.5, ..., 119^1.5;<br />
<br />
When summating over F(x) for x between 1 and 60, observe that the majority of the terms in a^3 and b^3 cancel out, leaving:<br />
<br />
(121^1.5 - 1^1.5)/2 =((11*11)^1.5 - 1)/2 = 665<br />
<br />
Thus, sum of F(x) for x between 1 and 60 is 665. The solution can be generalized as equal to (a(x_max)^3 - b(x_min)^3)/2.<br />
<br />
==Episode 8==<br />
Homura triangulated the likely location of [[Walpurgis Night]] to be the clock tower using [[wikipedia:Statistics|Statistics]], a branch of Mathematics that determines the probably of a future event by analyzing collective data of similar events.<br />
<br />
==Episode 9==<br />
[[File:Episode 9 Math.png|thumb|Fibonacci sequence question that appeared in Episode 9]]<br />
<br />
This advanced [[wikipedia:Algebra|Algebra]] problem previously appeared in the first '''[[wikipedia:Tokyo University|Tokyo University]]''' [http://www.j3e.info/ojyuken/math/php.php?name=tokyo&v1=0&v2=1992&v3=1&v4=3&y=1992&n=0_3 1992 Entrance Exam]:<br />
<br />
===Problem===<br />
A number sequence {F(n)} that can be defined as F(1) = 1, F(2) = 1, F(n+2) = F(n) + F(n+1) (where n is any natural number) is called the [[wikipedia:Fibonacci number|Fibonacci sequence]] and its general solution is given by, <br />
<br />
[[file:fibonacci.png]]<br />
<br />
[[file:golden ratio.png]]<br />
<br />
Answer the following questions by using this fact if needed:<br />
<br />
<br />
Define a sequence of natural numbers X(n) (where n is any natural number), in which each digit is either 0 or 1, set by the following rules:<br />
<br />
(i) X(1) = 1<br />
<br />
(ii) We define X(n+1) as a natural number, which can be gotten by replacing the digits of X(n) with 1 if the digit is 0, and with 10 if the digit is 1.<br />
<br />
For example, X(1) = 1, X(2) = 10, X(3) = 101, X(4) = 10110, X(5) = 10110101, ... <br />
<br />
'' '''Anecdote''': X(n) can be considered a clever analogy to the show, where [[Episode 10]] should replaces [[Episode 1]] and [[Episode 1]] replaces Episode 0 in the viewer's next viewing.''<br />
<br />
(1) Find A(n), defined as the number of digits of X(n)?<br />
<br />
(2) Find B(n),defined as the numbers of times '01' appears in X(n)?<br />
For example, B(1) = 0, B(2) = 0, B(3) = 1, B(4) = 1, B(5) = 3,...<br />
<br />
<br />
===Solution===<br />
====Part 1====<br />
<br />
Let A(n) equal to the number of digits in X(n), which consists solely of 1s and 0s. Let's suppose x(n) is the number of 0s in X(n) at n iteration (poor choice of variable by the student). Let's suppose y(n) is the number of 1s in X(n) at n iteration, then x(n) + y(n) = A(n). Since<br />
*Every time a 0 appears, it is replaced with 1 at the next iteration, contributing to a single 1 in X(n+1).<br />
*Every time a 1 appears, it is replaced with 10 at t he next iteration, contributing to a single 1 and a single 0 in X(n+1).<br />
it follows that the number of 0s in the next iteration is equal to the number of 1s previously: <br />
<br />
x(n+1)= y(n) ;<br />
<br />
and the number of 1s in the next iteration is equal to the number of 0s AND the numbers of 1s previously:<br />
<br />
y(n+1)=x(n)+y(n) .<br />
<br />
<br />
Next, prove that x(n) is a fibonacci sequence, since we know that:<br />
<br />
*y(n+1) = x(n) + y(n) ;<br />
*x(n+2) = y(n+1) ;<br />
*x(n+1) = y(n) ;<br />
<br />
by substition we can show,<br />
<br />
x(n+2) = y(n+1) = x(n) + y(n) = x(n) + x(n+1)<br />
<br />
x(n+2) = x(n+1) + x(n)<br />
<br />
x(n) fits the definition of a fibonacci sequence. Since x(n) is a fibonacci sequence, it follows that y(n) is also a fibonacci sequence (since x(n+1) = y(n)). Therefore:<br />
<br />
x(n+2)=x(n+1)+x(n)<br />
<br />
y(n+2)=y(n+1)+y(n)<br />
<br />
x(n+2)+y(n+2)=x(n+1)+y(n+1)+x(n)+y(n)<br />
<br />
Recall: A(n)=x(n)+y(n), therefore<br />
<br />
A(n+2)=A(n+1)+A(n) .<br />
<br />
Since, X(1) = 1, X(2) = 10, X(3) = 101, X(4) = 10110, X(5) = 10110101, ...<br />
<br />
It follows that A(1) = 1, A(2) = 2, A(3) = 3, A(4) = 5, A(5) = 8,...<br />
<br />
Recall fibonacci sequence F(n)= {1, 1, 2, 3, 5, 8, ...}. Therefore A(n) = F(n+1), or<br />
<br />
[[file:An.png]]<br />
<br />
====Part 2====<br />
<br />
Let B(n) be the number of times '01' appears in X(n). Any two digits in X(n) may be 00, 01, 10, 11, and the corresponding digits in the next iteration X(n+1) will be<br />
<br />
*00 -> 11<br />
*01 -> 110<br />
*10 -> 101<br />
*11 -> 1010<br />
<br />
Thus, any two digit sequence in X(n) that begins with 1* will contribute to a 01 sequence in the next iteration. In other words, B(n+1) equals y(n), except when unit digit of X(n) is 1, or:<br />
<br />
B(n+1) = y(n) - odd(X(n))<br />
<br />
where<br />
*odd(n) = 1 if n is odd (unit digit is 1)<br />
*odd(n) = 0 if n is even (unit digit is 0)<br />
<br />
y(n) is a fibonacci sequence with starting values: y(1) = 1, y(2) = 1, y(3) = 2<br />
thus y(n) = F(n)<br />
<br />
B(n+1) = F(n) - odd(X(n)) <br />
<br />
B(n) = F(n-1) - odd(X(n-1)), or<br />
<br />
[[file:Bn.png]]<br />
<br />
==Episode 10==<br />
Same as [[Mathematics_of_Madoka_Magica#Episode_1|Episode 1]].<br />
[[Category:Articles]]<br />
[[Category:Fanmade analysis]]<br />
[[Category:Science]]</div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=Talk:Characters&diff=7171Talk:Characters2011-03-17T18:39:04Z<p>Anon-kun: /* Refactoring */</p>
<hr />
<div>==Walpurgis Night==<br />
Walpurgis Night has been confirmed to be a witch on the official website. Hence, I added a link to the speculah article here. When more info on the witch is gathered, I think we should make a separate page for the witch herself. --[[User:Homerun-chan|Homerun-chan]] 18:18, 9 March 2011 (UTC)<br />
<br />
== Refactoring ==<br />
<br />
I like that new design. We should apply it to the witches section too. The problem is that using the cards as images renders pretty meh (see below -- the biggest problem is that we lack a lot of cards). Any idea? --[[User:Homerun-chan|Homerun-chan]] 16:42, 17 March 2011 (UTC)<br />
<br />
{|class="wikitable" style="width:100%;text-align:center;"<br />
!colspan="6"|Witches and minions<br />
|-<br />
|style="width:16.5%;"|[[File:Card Gertrud.png|120px|link=Gertrud]]<br />
|style="width:16.5%;"|[[File:Card Suleika.png|120px|link=Suleika]]<br />
|style="width:16.5%;"|[[File:Card Charlotte.png|120px|link=Charlotte]]<br />
|style="width:16.5%;"|[[File:Card Elly.png|120px|link=Elly]]<br />
|style="width:16.5%;"|[[File:Card Albertine.png|120px|link=Albertine]]<br />
|style="width:16.5%;"|[[File:Card Gisela.png|120px|link=Gisela]]<br />
|-<br />
|'''[[Gertrud]]'''<br />
|'''[[Suleika]]'''<br />
|'''[[Charlotte]]'''<br />
|'''[[Elly|Elly (Kristen)]]'''<br />
|'''[[Albertine]]'''<br />
|'''[[Gisela]]'''<br />
|-<br />
|[[Anthony]] -- [[Adelbert]]<br />
|[[Ulla]]<br />
|[[Pyotr]]<br />
|[[Elly#Daniyyel_.26_Jennifer|Danyyiel -- Jennifer]]<br />
|[[Anja]]<br />
|[[Elsa_Maria#Sebastians|Sebastian's]]<br />
|-<br />
|style="width:16.5%;"|[[File:Card Uhrmann.png|120px|link=Uhrmann]]<br />
|style="width:16.5%;"|[[File:Card Oktavia.png|120px|link=Oktavia von Seckendorff]]<br />
|style="width:16.5%;"|NO IMAGE<br />
|style="width:16.5%;"|NO IMAGE<br />
|style="width:16.5%;"|NO IMAGE<br />
|style="width:16.5%;"|NO IMAGE<br />
|-<br />
|'''[[Uhrmann]]'''<br />
|'''[[Oktavia von Seckendorff]]'''<br />
|'''[[Isadel]]'''<br />
|'''[[Patricia]]'''<br />
|'''[[Roberta]]'''<br />
|'''[[Kriemhild Gretchen]]'''<br />
|-<br />
|[[Holger]]<br />
|<br />
|<br />
|<br />
|<br />
|<br />
|-<br />
|style="width:16.5%; height: 200px;"|NO IMAGE<br />
|<br />
|<br />
|<br />
|<br />
|<br />
|-<br />
|'''[[Speculah:Walpurgis Night|Walpurgis Night]]'''<br />
|<br />
|<br />
|<br />
|<br />
|<br />
|}<br />
<br />
I like it, but the lines between cards and links should be removed. We could use normal screenshots for the missing pictures or a ? pic like Suleika if the character did not appear in the series. On a related note, the parts about the witch in the [[Sayaka]] article should be moved to the [[Oktavia]] article. -- [[User:Anon-kun|Anon-kun]] 18:39, 17 March 2011 (UTC)</div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=Mathematics_of_Madoka_Magica&diff=6654Mathematics of Madoka Magica2011-03-14T00:27:30Z<p>Anon-kun: linkfix</p>
<hr />
<div>The cast of Mahou Shoujo Madoka Magica are students of [[Mitakihara Middle School]]. Despite merely a middle school, coursework in Mitakihara can be quite challenging. The following are Math problems that have appeared thus far in the show, and their solutions:<br />
<br />
==Episode 1==<br />
[[file:Episode One Algebra.jpg|thumb|Episode one Math question 1]]<br />
<br />
===Question 1===<br />
Any integer divided by 14 will have a remainder between 0 and 13. Given a is has remainder of 6 and b has remainder of 1 when divided by 14, what is the remainder of x when divided by 14, given x is an integer solution to x^2-2ax+b=0?<br />
<br />
'''Solution:'''<br />
<br />
This problem can be solved simply with [[wikipedia:Modular arithmetic|modular arithmetic]]:<br />
<br />
a = 6 mod 14;<br />
b = 1 mod 14;<br />
x = r mod 14.<br />
<br />
x^2 - 2ax + b = 0 is equivalent to<br />
<br />
x^2 - 12x + 1 = 0 mod 14<br />
<br />
x(x - 12) = -1 mod 14<br />
<br />
-1 has no proper divisors modulo 14, thus factors x and x-12 must correspond to either 1 or -1 mod 14.<br />
<br />
If x = 1 mod 14, then x-12 = 3 mod 14: false. If x = -1 mod 14, then x - 12 = 1 mod 14: true.<br />
<br />
Therefore, x = -1 mod 14 = 13 mod 14, i.e., the remainder r of x is 13.<br />
<br />
<br />
'''Second Solution:'''<br />
<br />
Homura used in Episode 1 a basic approach with usual integer [[wikipedia:arithmetic|arithmetic]].<br />
<br />
Let x = 14q + r, a = 14s + 6, b = 14t + 1.<br />
<br />
Substitute into x^2 - 2ax + b = 0 to get after some calculations<br />
<br />
x^2 - 2ax + b = 14c + (r+1)^2 = 0 with c = 14q^2 + 2q - 28qs - 2rs - 12q + t - r<br />
<br />
14 divides 14c and 0, hence it also divides (r+1)^2. This implies that r+1 is divisible by 14.<br />
<br />
The question asks for a remainder r between 0 and 13, so we obtain r = 13.<br />
<br />
See also: http://green-oval.net/cgi-board.pl/a/thread/44824340<br />
<br />
===Question 2===<br />
[[file:Episode One Math Q2.jpg|thumb|Episode one Math question 2]]<br />
Assuming that p is a prime number and n is an arbitrary natural number, prove that (1+n)^p - n^p - 1 is divisible by p.<br />
<br />
'''Solution:'''<br />
<br />
By [[wikipedia:Fermat's Little Theorem|Fermat's Little Theorem]], for any prime p and integer a,<br />
<br />
a^p = a (mod p)<br />
<br />
Thus:<br />
<br />
(1+n)^p - n^p - 1 (mod p) is equivalent to<br />
<br />
(1+n) - n - 1 (mod p) is equivalent to<br />
<br />
0 (mod p)<br />
<br />
So the overall expression is divisible by p.<br />
<br />
<br />
'''Second solution:'''<br />
<br />
The problem can be solved with the [[wikipedia:binomial theorem|binomial theorem]]:<br />
<br />
For a, b not equal to 0 and nonnegative integer p it holds that:<br />
<br />
<!--(a + b)^p = sum (for k from 0 to p) of (p choose k) a^(p-k) b^k-->[[file:Q2 binomial theorem.png]]<br />
<br />
where the binomial coefficient (p choose k) is the integer <!-- p(p-1)...(p-k+1)/(k(k-1)...1) --> [[file:Q2 binomial coefficient.png]] for 0 ≤ k ≤ p.<br />
<br />
Therefore,<br />
<br />
<!--(1 + n)^p = sum (for k from 0 to p) of (p choose k) n^k-->[[file:Q2 equation 1.png]]<br />
<br />
and<br />
<br />
<!-- (1+n)^p - n^p - 1 = sum (for k from 1 to p-1) of (p choose k) n^k -->[[file:Q2 equation 2.png]]<br />
<br />
since (p choose 0) = (p choose p) = 1.<br />
<br />
It holds that (p choose 1) = p.<br />
Since p is a prime number, the factor p in (p choose k) is not divisible by k, k-1,...,2 for 2 ≤ k ≤ p-1.<br />
Therefore, p divides (p choose k) for 1 ≤ k ≤ p-1.<br />
<br />
Since each summand on the right side is divisible by p, the whole sum, i.e. the left side is also divisible by p.<br />
<br />
<br />
===Question 3===<br />
[[file:Episode One Math Q3.jpg|thumb|Episode one Math question 3]]<br />
<br />
<br />
Find the integer solutions (a,b) with (a^3 + a^2 - 1) - (a - 1)b = 0<br />
<br />
'''Solution:'''<br />
[[file:Episode One Math S3.jpg|thumb|Episode one Math solution to question 3]]<br />
<br />
(a^3 + a^2 - 1) - (a - 1)b<br><br />
= (a^3 - 1) + (a^2 - 1) + 1 - (a - 1)b<br><br />
= (a - 1)(a^2 + a + 1) + (a - 1)(a + 1) + 1 - (a - 1)b<br><br />
= (a - 1)(a^2 + 2a - b + 2) + 1 = 0<br />
<br />
Therefore,<br> <br />
(a - 1)(a^2 + 2a - b + 2) = -1<br />
<br />
a, b are integers, therefore the factors a - 1 and a^2 + 2a - b + 2 are integers too. Since the product is -1, one of the factors must be equal to -1, the other to 1.<br />
<br />
If a - 1 = -1 then a = 0 and from a^2 + 2a - b + 2 = 1 we obtain b = 1.<br />
<br />
If a - 1 = 1 then a = 2 and a^2 + 2a - b + 2 = -1 implies that b = 11.<br />
<br />
There are two integer solutions: (a,b) = (0,1) and (a,b) = (2,11).<br />
<br />
<br />
===Question 4===<br />
[[file:Episode One Math Q4.jpg|thumb|Episode one Math question 4]]<br />
Given [[file:Math episode one question four.png]],<br />
<br />
find the sum of F(1)+F(2)+F(3)+...+F(60).<br />
<br />
'''Solution:'''<br />
[[file:Episode One Math S4.jpg|thumb|Episode one Math, partial solution to question 4. Note that there are errors.]]<br />
<br />
Simplying the fraction:<br />
<br />
Reference that for any variables a & b, (a-b) (a+b) = (a-b)^2. Let (2x + 1)^.5 = a and (2x - 1)^.5 = b. Multiply F(x) by 1 or (a-b)/(a-b) or ((2x + 1)^.5 - (2x - 1)^.5) / ((2x + 1)^.5 - (2x - 1)^.5). The denominator becomes:<br />
<br />
((2x + 1)^.5 + (2x - 1)^.5) * ((2x + 1)^.5 - (2x - 1)^.5) <br />
<br />
= (2x + 1) - (2x - 1)<br />
<br />
= 2<br />
<br />
The numerator becomes:<br />
<br />
(4x + (4x^2 - 1)^.5 ) ((2x + 1)^.5 - (2x - 1)^.5) <br />
<br />
=(4x + ((2x)^2 - 1)^.5 ) (a - b) <br />
<br />
=(4x + (2x - 1)^.5 (2x + 1)^.5 ) (a -b) <br />
<br />
=(4x + a b ) (a - b) <br />
<br />
=4x (a - b) + (a^2) b - a (b^2)<br />
<br />
=4x (a - b) + (2x + 1) b - a (2x - 1)<br />
<br />
=4x (a - b) - 2x (a - b) + a + b<br />
<br />
=2x ( a - b) + a + b<br />
<br />
=(2x + 1) a - (2x - 1) b<br />
<br />
=a^3 - b^3<br />
<br />
<br />
Therefore:<br />
<br />
F(x) = (a^3 - b^3 ) / 2<br />
<br />
Note that when x= 1, 2, 3, 4, ..., 60;<br />
<br />
a^3 = 3^1.5, 5^1.5, 7^1.5, 9^1.5, ..., 119^1.5, 121^1.5;<br />
<br />
b^3 = 1^1.5, 3^1.5, 5^1.5, 7^1.5, 9^1.5, ..., 119^1.5;<br />
<br />
When summating over F(x) for x between 1 and 60, observe that the majority of the terms in a^3 and b^3 cancel out, leaving:<br />
<br />
(121^1.5 - 1^1.5)/2 =((11*11)^1.5 - 1)/2 = 665<br />
<br />
Thus, sum of F(x) for x between 1 and 60 is 665. The solution can be generalized as equal to (a(x_max)^3 - b(x_min)^3)/2.<br />
<br />
==Episode 8==<br />
Homura triangulated the likely location of [[Walpurgis Night]] to be the clock tower using [[wikipedia:Statistics|Statistics]], a branch of Mathematics that determines the probably of a future event by analyzing collective data of similar events.<br />
<br />
==Episode 9==<br />
[[File:Episode 9 Math.png|thumb|Fibonacci sequence question that appeared in Episode 9]]<br />
<br />
This advanced [[wikipedia:Algebra|Algebra]] problem previously appeared in the first '''[[wikipedia:Tokyo University|Tokyo University]]''' [http://www.j3e.info/ojyuken/math/php.php?name=tokyo&v1=0&v2=1992&v3=1&v4=3&y=1992&n=0_3 1992 Entrance Exam]:<br />
<br />
===Problem===<br />
A number sequence {F(n)} that can be defined as F(1) = 1, F(2) = 1, F(n+2) = F(n) + F(n+1) (where n is any natural number) is called the [[wikipedia:Fibonacci number|Fibonacci sequence]] and its general solution is given by, <br />
<br />
[[file:fibonacci.png]]<br />
<br />
[[file:golden ratio.png]]<br />
<br />
Answer the following questions by using this fact if needed:<br />
<br />
<br />
Define a sequence of natural numbers X(n) (where n is any natural number), in which each digit is either 0 or 1, set by the following rules:<br />
<br />
(i) X(1) = 1<br />
<br />
(ii) We define X(n+1) as a natural number, which can be gotten by replacing the digits of X(n) with 1 if the digit is 0, and with 10 if the digit is 1.<br />
<br />
For example, X(1) = 1, X(2) = 10, X(3) = 101, X(4) = 10110, X(5) = 10110101, ... <br />
<br />
'' '''Anecdote''': X(n) can be considered a clever analogy to the show, where [[Episode 10]] should replaces [[Episode 1]] and [[Episode 1]] replaces Episode 0 in the viewer's next viewing.''<br />
<br />
(1) Find A(n), defined as the number of digits of X(n)?<br />
<br />
(2) Find B(n),defined as the numbers of times '01' appears in X(n)?<br />
For example, B(1) = 0, B(2) = 0, B(3) = 1, B(4) = 1, B(5) = 3,...<br />
<br />
<br />
===Solution===<br />
====Part 1====<br />
<br />
Let A(n) equal to the number of digits in X(n), which consists solely of 1s and 0s. Let's suppose x(n) is the number of 0s in X(n) at n iteration (poor choice of variable by the student). Let's suppose y(n) is the number of 1s in X(n) at n iteration, then x(n) + y(n) = A(n). Since<br />
*Every time a 0 appears, it is replaced with 1 at the next iteration, contributing to a single 1 in X(n+1).<br />
*Every time a 1 appears, it is replaced with 10 at t he next iteration, contributing to a single 1 and a single 0 in X(n+1).<br />
it follows that the number of 0s in the next iteration is equal to the number of 1s previously: <br />
<br />
x(n+1)= y(n) ;<br />
<br />
and the number of 1s in the next iteration is equal to the number of 0s AND the numbers of 1s previously:<br />
<br />
y(n+1)=x(n)+y(n) .<br />
<br />
<br />
Next, prove that x(n) is a fibonacci sequence, since we know that:<br />
<br />
*y(n+1) = x(n) + y(n) ;<br />
*x(n+2) = y(n+1) ;<br />
*x(n+1) = y(n) ;<br />
<br />
by substition we can show,<br />
<br />
x(n+2) = y(n+1) = x(n) + y(n) = x(n) + x(n+1)<br />
<br />
x(n+2) = x(n+1) + x(n)<br />
<br />
x(n) fits the definition of a fibonacci sequence. Since x(n) is a fibonacci sequence, it follows that y(n) is also a fibonacci sequence (since x(n+1) = y(n)). Therefore:<br />
<br />
x(n+2)=x(n+1)+x(n)<br />
<br />
y(n+2)=y(n+1)+y(n)<br />
<br />
x(n+2)+y(n+2)=x(n+1)+y(n+1)+x(n)+y(n)<br />
<br />
Recall: A(n)=x(n)+y(n), therefore<br />
<br />
A(n+2)=A(n+1)+A(n) .<br />
<br />
Since, X(1) = 1, X(2) = 10, X(3) = 101, X(4) = 10110, X(5) = 10110101, ...<br />
<br />
It follows that A(1) = 1, A(2) = 2, A(3) = 3, A(4) = 5, A(5) = 8,...<br />
<br />
Recall fibonacci sequence F(n)= {1, 1, 2, 3, 5, 8, ...}. Therefore A(n) = F(n+1), or<br />
<br />
[[file:An.png]]<br />
<br />
====Part 2====<br />
<br />
Let B(n) be the number of times '01' appears in X(n). Any two digits in X(n) may be 00, 01, 10, 11, and the corresponding digits in the next iteration X(n+1) will be<br />
<br />
*00 -> 11<br />
*01 -> 110<br />
*10 -> 101<br />
*11 -> 1010<br />
<br />
Thus, any two digit sequence in X(n) that begins with 1* will contribute to a 01 sequence in the next iteration. In other words, B(n+1) equals y(n), except when unit digit of X(n) is 1, or:<br />
<br />
B(n+1) = y(n) - odd(X(n))<br />
<br />
where<br />
*odd(n) = 1 if n is odd (unit digit is 1)<br />
*odd(n) = 0 if n is even (unit digit is 0)<br />
<br />
y(n) is a fibonacci sequence with starting values: y(1) = 1, y(2) = 1, y(3) = 2<br />
thus y(n) = F(n)<br />
<br />
B(n+1) = F(n) - odd(X(n)) <br />
<br />
B(n) = F(n-1) - odd(X(n-1)), or<br />
<br />
[[file:Bn.png]]<br />
<br />
==Episode 10==<br />
Same as [[Mathematics_of_Madoka_Magica#Episode_1|Episode 1]].<br />
[[Category:Articles]]<br />
[[Category:Fanmade analysis]]<br />
[[Category:Science]]</div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=Manga&diff=6479Manga2011-03-12T15:19:35Z<p>Anon-kun: </p>
<hr />
<div>There are three official manga adaptations that will be published during or after the anime broadcast:<br />
<br />
;''[[Madoka Magica (manga)|Mahou Shoujo Madoka Magica]]'' (魔法少女まどか☆マギカ)<br />
:A direct adaptation of the anime. It will be published in three volumes on February 12, March 12 and April 12, 2011.<br />
;''[[Kazumi Magica|Mahou Shoujo Kazumi Magica: The Innocent Malice]]'' (魔法少女かずみ☆マギカ〜The innocent malice〜)<br />
:The first spin-off manga featuring different characters. It started serialization in the ''Manga Time Kirara Forward'' magazine on January 24, 2011.<br />
;''[[Oriko Magica|Mahou Shoujo Oriko Magica]]'' (魔法少女おりこ☆マギカ)<br />
:The the second spin-off manga will be released after the anime broadcast, the first volume on April 12, 2011. This might be a prequel of the original series featuring Mami and Kyouko before their meeting with Madoka and the other characters.<br />
<br />
== External links ==<br />
* [http://www.dokidokivisual.com/madokamagica/ Official website for all manga series] (japanese)<br />
* [http://www.dokidokivisual.com/madokamagica/madoka/ Official website for Mahou Shoujo Madoka Magica] (japanese)<br />
* [http://www.dokidokivisual.com/madokamagica/kazumi/ Official website for Mahou Shoujo Kazumi Magica] (japanese)<br />
<br />
[[Category:Manga| ]]</div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=Madoka_Magica_Products&diff=6420Madoka Magica Products2011-03-12T00:31:57Z<p>Anon-kun: DVD/BD 5 on homepage, content seems to be the same as the other volumes</p>
<hr />
<div>{{Note|For Madoka-related free content, please refer to [[Media]].}}<br />
<br />
{| class="wikitable sortable collapsible" style="width: 100%;"<br />
! Product type<br />
! Title<br />
! Release date<br />
! Price<br />
|-<br />
| CD<br />
| [http://www.amazon.co.jp/dp/B004DGD400/ ClariS: Connect Limited Edition]<br />
| 2011-02-02<br />
| ¥ 1,575<br />
|-<br />
| CD<br />
| [http://www.amazon.co.jp/dp/B004DGD4FU/ ClariS: Connect Anime Edition]<br />
| 2011-02-02<br />
| ¥ 1,300<br />
|-<br />
| CD<br />
| [http://www.amazon.co.jp/dp/B004DGD4AA/ ClariS: Connect Regular Edition]<br />
| 2011-02-02<br />
| ¥ 1,223<br />
|-<br />
| Manga<br />
| [http://www.amazon.co.jp/dp/4832279904/ Puella Magi Madoka☆Magica Vol.1]<br />
| 2011-02-12<br />
| ¥ 690<br />
|-<br />
| CD<br />
| [http://www.amazon.co.jp/dp/B004DGD45A/ Kalafina: Magia Limited Edition]<br />
| 2011-02-16<br />
| ¥ 1,575<br />
|-<br />
| CD<br />
| [http://www.amazon.co.jp/dp/B004DGD4QE/ Kalafina: Magia Anime Edition]<br />
| 2011-02-16<br />
| ¥ 1,300<br />
|-<br />
| CD<br />
| [http://www.amazon.co.jp/dp/B004DGD4NM/ Kalafina: Magia Regular Edition]<br />
| 2011-02-16<br />
| ¥ 1,223<br />
|-<br />
| Manga<br />
| [http://www.amazon.co.jp/dp/483224003X/ Puella Magi Madoka☆Magica Vol.2]<br />
| 2011-03-12<br />
| ¥ 690<br />
|-<br />
| Blu-ray<br />
| [http://www.amazon.co.jp/dp/B004INGZAE/ Puella Magi Madoka☆Magica Vol.1 Limited Edition]<br />
| 2011-03-30<br />
| ¥ 7,350<br />
|-<br />
| DVD<br />
| [http://www.amazon.co.jp/dp/B004INGZCC/ Puella Magi Madoka☆Magica Vol.1 Limited Edition]<br />
| 2011-03-30<br />
| ¥ 6,300<br />
|-<br />
| DVD<br />
| [http://www.amazon.co.jp/dp/B004INGZLS/ Puella Magi Madoka☆Magica Vol.1 Regular Edition]<br />
| 2011-03-30<br />
| ¥ 5,250<br />
|-<br />
| Manga<br />
| [http://www.amazon.co.jp/dp/4832240145/ Puella Magi Madoka☆Magica Vol.3 (End)]<br />
| 2011-04-12<br />
| ¥ 690<br />
|-<br />
| Blu-ray<br />
| [http://www.amazon.co.jp/dp/B004L7A7WO/ Puella Magi Madoka☆Magica Vol.2 Limited Edition]<br />
| 2011-04-27<br />
| ¥ 7,350<br />
|-<br />
| DVD<br />
| [http://www.amazon.co.jp/dp/B004L7A7Q0/ Puella Magi Madoka☆Magica Vol.2 Limited Edition]<br />
| 2011-04-27<br />
| ¥ 6,300<br />
|-<br />
| DVD<br />
| [http://www.amazon.co.jp/dp/B004L7A7UG/ Puella Magi Madoka☆Magica Vol.2 Regular Edition]<br />
| 2011-04-27<br />
| ¥ 5,250<br />
|-<br />
| Blu-ray<br />
| [http://www.amazon.co.jp/dp/B004MMFTMQ/ Puella Magi Madoka☆Magica Vol.3 Limited Edition]<br />
| 2011-05-25<br />
| ¥ 7,350<br />
|-<br />
| DVD<br />
| [http://www.amazon.co.jp/dp/B004MMFTN0/ Puella Magi Madoka☆Magica Vol.3 Limited Edition]<br />
| 2011-05-25<br />
| ¥ 6,300<br />
|-<br />
| DVD<br />
| [http://www.amazon.co.jp/dp/B004MMFTNA/ Puella Magi Madoka☆Magica Vol.3 Regular Edition]<br />
| 2011-05-25<br />
| ¥ 5,250<br />
|-<br />
| Blu-ray<br />
| [http://www.amazon.co.jp/dp/B004P4RTKQ/ Puella Magi Madoka☆Magica Vol.4 Limited Edition]<br />
| 2011-06-22<br />
| ¥ 7,350<br />
|-<br />
| DVD<br />
| [http://www.amazon.co.jp/dp/B004P4RTLK/ Puella Magi Madoka☆Magica Vol.4 Limited Edition]<br />
| 2011-06-22<br />
| ¥ 6,300<br />
|-<br />
| DVD<br />
| [http://www.amazon.co.jp/dp/B004P4RTL0/ Puella Magi Madoka☆Magica Vol.4 Regular Edition]<br />
| 2011-06-22<br />
| ¥ 5,250<br />
|-<br />
| Blu-ray<br />
| [http://www.amazon.co.jp/dp/B004RCJJFE/ Puella Magi Madoka☆Magica Vol.5 Limited Edition]<br />
| 2011-07-27<br />
| ¥ 7,350<br />
|-<br />
| DVD<br />
| [http://www.amazon.co.jp/dp/B004RCJJ54/ Puella Magi Madoka☆Magica Vol.5 Limited Edition]<br />
| 2011-07-27<br />
| ¥ 6,300<br />
|-<br />
| DVD<br />
| [http://www.amazon.co.jp/dp/B004RCJJWC/ Puella Magi Madoka☆Magica Vol.5 Regular Edition]<br />
| 2011-07-27<br />
| ¥ 5,250<br />
|-<br />
|}<br />
<br />
==Blu-Ray Discs==<br />
{| class="wikitable" style="width: 100%; vertical-align: top;"<br />
! style="width: 300px;" | Cover<br />
! Description<br />
|-<br />
|[[File:Disc 1.jpg|300px]]<br />
| style="vertical-align:top; padding: 10px;" |<br />
;[http://www.amazon.co.jp/dp/B004INGZAE Puella Magi Madoka☆Magica Vol.1 Blu-ray]<br />
:'''Release date''': March 30, 2011<br />
:'''Price''': ¥ 7,350 (~90 USD)<br />
:'''Running time''': 60 minutes<br />
:'''Package contents''':<br />
:*Blu-ray disc (episode 1, 2)<br />
:*Special CD<br />
:*#Urobuchi Gen-Supervised Drama CD<br />
:*#Kaname Madoka Character Song (ED theme for episode 1, 2)<br />
:*Tri-fold clear case<br />
:*Package illustration by Kishida Takahiro<br />
:*Special booklet with Aoki Ume's 4koma Manga<br />
:'''Extra contents''':<br />
:#Episode 1 audio commentary (Yūki Aoi, Saitō Chiwa, Kitamura Eri)<br />
:#Episode 2 audio commentary (Yūki Aoi, Saitō Chiwa, Aoki Ume)<br />
:#Unbroadcasted original ending theme version<br />
|-<br />
|[[File:Disc 2.jpg|300px]]<br />
| style="vertical-align:top; padding: 10px;"|<br />
;[http://www.amazon.co.jp/dp/B004L7A7WO/ Puella Magi Madoka☆Magica Vol.2 Blu-ray]<br />
:'''Release date''': April 27, 2011<br />
:'''Price''': ¥ 7,350 (~90 USD)<br />
:'''Running time''': 48 minutes<br />
:'''Package contents''':<br />
:*Blu-ray disc (episode 3, 4)<br />
:*Original soundtrack CD Volume 1<br />
:*Tri-fold clear case<br />
:*Package illustration by Kishida Takahiro<br />
:*Special booklet with Aoki Ume's 4koma Manga<br />
:'''Extra contents''':<br />
:#Audio commentary<br />
|-<br />
| style="text-align: center;"| NO IMAGE<br />
| style="vertical-align:top; padding: 10px;"|<br />
;[http://www.amazon.co.jp/dp/B004MMFTMQ/ Puella Magi Madoka☆Magica Vol.3 Blu-ray]<br />
:'''Release date''': May 25, 2011<br />
:'''Price''': ¥ 7,350 (~90 USD)<br />
:'''Running time''': 48 minutes<br />
:'''Package contents''':<br />
:*Blu-ray disc (episode 5, 6)<br />
:*Special CD<br />
:*Tri-fold clear case<br />
:*Package illustration by Kishida Takahiro<br />
:*Special booklet with Aoki Ume's 4koma Manga<br />
:'''Extra contents''':<br />
:#Audio commentary<br />
|-<br />
| style="text-align: center;"| NO IMAGE<br />
| style="vertical-align:top; padding: 10px;"|<br />
;[http://www.amazon.co.jp/dp/B004P4RTKQ/ Puella Magi Madoka☆Magica Vol.4 Blu-ray]<br />
:'''Release date''': June 22, 2011<br />
:'''Price''': ¥ 7,350 (~90 USD)<br />
:'''Running time''': 48 minutes<br />
:'''Package contents''':<br />
:*Blu-ray disc (episode 7, 8)<br />
:*Special CD<br />
:*Tri-fold clear case<br />
:*Package illustration by Kishida Takahiro<br />
:*Special booklet with Aoki Ume's 4koma Manga<br />
:'''Extra contents''':<br />
:#Audio commentary<br />
|-<br />
| style="text-align: center;"| NO IMAGE<br />
| style="vertical-align:top; padding: 10px;"|<br />
;[http://www.amazon.co.jp/dp/B004RCJJFE/ Puella Magi Madoka☆Magica Vol.5 Blu-ray]<br />
:'''Release date''': July 27, 2011<br />
:'''Price''': ¥ 7,350 (~90 USD)<br />
:'''Running time''': 48 minutes<br />
:'''Package contents''':<br />
:*Blu-ray disc (episode 9, 10)<br />
:*Special CD<br />
:*Tri-fold clear case<br />
:*Package illustration by Kishida Takahiro<br />
:*Special booklet with Aoki Ume's 4koma Manga<br />
:'''Extra contents''':<br />
:#Audio commentary<br />
|}<br />
<br />
==Singles==<br />
{| class="wikitable" style="width: 100%; vertical-align: top;"<br />
! style="width: 300px;" | Cover<br />
! Description<br />
|-<br />
|[[File:Connect Anime Cover.jpg|300px]]<br />
| style="vertical-align:top; padding: 10px;" |<br />
;[http://www.amazon.co.jp/dp/B004DGD4FU ClariS: Connect (Anime Edition)]<br />
:'''Release date''': February 2, 2011<br />
:'''Price''': ¥ 1,300 (~15 USD)<br />
:'''Tracklist''':<br />
:#コネクト (Connect)<br />
:#Dreamin’<br />
:#キミとふたり (Kimi to futari)<br />
:#コネクト (Connect) -TV MIX-<br />
:'''Package contents''':<br />
:*Audio CD<br />
:*Anime character design package illustration<br />
|-<br />
|[[File:Connect OP Cover.jpg|300px]]<br />
| style="vertical-align:top; padding: 10px;" |<br />
;[http://www.amazon.co.jp/dp/B004DGD400 ClariS: Connect (Limited Edition)]<br />
:'''Release date''': February 2, 2011<br />
:'''Price''': ¥ 1,575 (~19 USD)<br />
:'''Tracklist''':<br />
:#コネクト (Connect)<br />
:#Dreamin’<br />
:#キミとふたり (Kimi to futari)<br />
:#コネクト (Connect) -Instrumental-<br />
:'''Package contents''':<br />
:*Audio CD<br />
:*Music video DVD<br />
:*Anime character design illustration sticker<br />
|-<br />
! Cover<br />
! Description<br />
|-<br />
|[[File:Magia Anime Cover.jpg|300px]]<br />
| style="vertical-align:top; padding: 10px;" |<br />
;[http://www.amazon.co.jp/dp/B004DGD4QE Kalafina: Magia ED (Anime Edition)]<br />
:'''Release date''': February 16, 2011<br />
:'''Price''': ¥ 1,300 (~15 USD)<br />
:'''Tracklist''':<br />
:#Magia<br />
:#Magia (magic mix)<br />
:#Magia (TV Version)<br />
:#Magia (instrumental)<br />
:'''Package contents''':<br />
:*Audio CD<br />
:*Digipak case<br />
:*Anime character design package illustration<br />
|-<br />
|[[File:Magia ED Cover.jpg|300px]]<br />
| style="vertical-align:top; padding: 10px;" |<br />
;[http://www.amazon.co.jp/dp/B004DGD45A Kalafina: Magia ED (Limited Edition)]<br />
:'''Release date''': February 16, 2011<br />
:'''Price''': ¥ 1,575 (~19 USD)<br />
:'''Tracklist''':<br />
:#Magia<br />
:#snow falling<br />
:#Magia (instrumental)<br />
:'''Package contents''':<br />
:*Audio CD<br />
:*Music video DVD<br />
|}<br />
<br />
==Manga==<br />
{| class="wikitable" style="width: 100%; vertical-align: top;"<br />
! style="width: 300px;" | Cover<br />
! Description<br />
|-<br />
| [[File:Manga Vol.1 Cover.jpg|300px]]<br />
| style="vertical-align:top; padding: 10px;" |<br />
;[http://www.amazon.co.jp/dp/4832279904 Puella Magi Madoka☆Magica Vol.1]<br />
:'''Release date''': February 12, 2011<br />
:'''Price''': ¥ 690 (~8 USD)<br />
:'''Artist''': Hanokage<br />
:'''Publisher''': Houbunsha<br />
|-<br />
| [[File:Manga Vol.2 Cover.jpg|300px]]<br />
| style="vertical-align:top; padding: 10px;" |<br />
;[http://www.amazon.co.jp/dp/483224003X Puella Magi Madoka☆Magica Vol.2]<br />
:'''Release date''': March 12, 2011<br />
:'''Price''': ¥ 690 (~8 USD)<br />
:'''Artist''': Hanokage<br />
:'''Publisher''': Houbunsha<br />
|-<br />
| style="text-align: center;"| NO IMAGE<br />
| style="vertical-align:top; padding: 10px;" |<br />
;[http://www.amazon.co.jp/dp/4832240145 Puella Magi Madoka☆Magica Vol.3 (End)]<br />
:'''Release date''': April 12, 2011<br />
:'''Price''': ¥ 690 (~8 USD)<br />
:'''Artist''': Hanokage<br />
:'''Publisher''': Houbunsha<br />
|}<br />
<br />
== External links ==<br />
* [http://www.madoka-magica.com/bddvd/ Official BD, DVD, and CD page] (Japanese)<br />
* [http://www.dokidokivisual.com/madokamagica/ Official manga page] (Japanese)</div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=Runes:Episode_10&diff=6419Runes:Episode 102011-03-12T00:26:15Z<p>Anon-kun: +1</p>
<hr />
<div>Runes listing and translation for [[Episode 10]].<br />
<br />
{|class="wikitable" style="line-height: 1.6em; width: 100%;"<br />
! style="text-align: center; background: #efefef; width: 35%;"|Screencap<br />
! style="text-align: center; background: #efefef; width: 30%;"|Transcription<br />
! style="text-align: center; background: #efefef; width: 30%;"|Explanation/Translation<br />
! style="text-align: left; width: 5%; background: #efefef;"|Time<br />
|-<br />
|[[File:Isadel.jpg|300px]]<br />
|[Archaic]<br />{{Runes|ISADEL}}<br />Isadel<br />
|The name of the witch who attacks Homura at the beginning of the episode.<br />
|3:33<br />
|-<br />
|[[File:ep10-09-00.jpg|300px]]<br />
|[Archaic]<br />{{Runes|HO}}<br />Ho<br />
|The first two runes of Homura's name on her ring.<br />
|9:00<br />
|-<br />
|[[File:Patricia.jpg|300px]]<br />
|[Archaic]<br />{{Runes|PATRICIA}}<br />Patricia<br />
|The name of the flying laundry witch.<br />
|10:15<br />
|-<br />
|[[File:Oktavia-ep10.JPG|300px]]<br />
|[Musical]<br />{{Runes|OKTAVIA<br />VON}}<br />Oktavia<br />Von<br>(...)<br />
|Oktavia von Seckendorff's second appearance<br />
|13:10<br />
|-<br />
|[[File:Roberta.png|300px]]<br />
|[Archaic]<br />{{Runes|ROBERTA}}<br />Roberta<br />
|The name of the witch Homura fights alone later in the episode.<br />
|18:29<br />
|}</div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=File:Ep10-09-00.jpg&diff=6418File:Ep10-09-00.jpg2011-03-12T00:24:32Z<p>Anon-kun: Detail of Homura's ring</p>
<hr />
<div>Detail of Homura's ring</div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=Mathematics_of_Madoka_Magica&diff=6382Mathematics of Madoka Magica2011-03-11T09:51:06Z<p>Anon-kun: ≤</p>
<hr />
<div>The cast of Mahou Shoujo Madoka Magica are students of [[Mitakihara Middle School]]. Despite merely a middle school, coursework in Mitakihara can be quite challenging. The following are Math problems that have appeared thus far in the show, and their solutions:<br />
<br />
==Episode 1==<br />
[[file:Episode One Algebra.jpg|thumb|Episode one Math question 1]]<br />
<br />
===Question 1===<br />
Any integer divided by 14 will have a remainder between 0 and 13. Given a is has remainder of 6 and b has remainder of 1 when divided by 14, what is the remainder of x when divided by 14, given x is an integer solution to x^2-2ax+b=0?<br />
<br />
'''Solution:'''<br />
<br />
This problem can be solved simply with modular arithmetrics: <br />
a = 6 mod 14;<br />
b = 1 mod 14;<br />
x = r mod 14.<br />
<br />
x^2 - 2ax + b = 0 is equivalent to<br />
<br />
x^2 - 12x + 1 = 0 mod 14<br />
<br />
x(x - 12) = -1 mod 14<br />
<br />
-1 has no proper divisors modulo 14, thus factors x and x-12 must correspond to either 1 or -1 mod 14.<br />
<br />
If x = 1 mod 14, then x-12 = 3 mod 14: false. If x = -1 mod 14, then x - 12 = 1 mod 14: true.<br />
<br />
Therefore, x = -1 mod 14 = 13 mod 14, i.e., the remainder r of x is 13.<br />
<br />
<br />
'''Second Solution:'''<br />
<br />
Homura used in Episode 1 a basic approach with usual integer arithmetic.<br />
<br />
Let x = 14q + r, a = 14s + 6, b = 14t + 1.<br />
<br />
Substitute into x^2 - 2ax + b = 0 to get after some calculations<br />
<br />
x^2 - 2ax + b = 14c + (r+1)^2 = 0 with c = 14q^2 + 2q - 28qs - 2rs - 12q + t - r<br />
<br />
14 divides 14c and 0, hence it also divides (r+1)^2. This implies that r+1 is divisible by 14.<br />
<br />
The question asks for a remainder r between 0 and 13, so we obtain r = 13.<br />
<br />
See also: http://green-oval.net/cgi-board.pl/a/thread/44824340<br />
<br />
<br />
===Question 2===<br />
[[file:Episode One Math Q2.jpg|thumb|Episode one Math question 2]]<br />
Assuming that p is a prime number and n is an arbitrary natural number, prove that (1+n)^p - n^p - 1 is divisible by p.<br />
<br />
'''Solution:'''<br />
<br />
By Fermat's Little theorem, for any prime p and integer a,<br />
<br />
a^p = a (mod p)<br />
<br />
Thus:<br />
<br />
(1+n)^p - n^p - 1 (mod p) is equivalent to<br />
<br />
(1+n) - n - 1 (mod p) is equivalent to<br />
<br />
0 (mod p)<br />
<br />
So the overall expression is divisible by p.<br />
<br />
<br />
'''Second solution:'''<br />
<br />
The problem can be solved with the binomial theorem:<br />
<br />
For a, b not equal to 0 and nonnegative integer p it holds that:<br />
<br />
<!--(a + b)^p = sum (for k from 0 to p) of (p choose k) a^(p-k) b^k-->[[file:Q2 binomial theorem.png]]<br />
<br />
where the binomial coefficient (p choose k) is the integer <!-- p(p-1)...(p-k+1)/(k(k-1)...1) --> [[file:Q2 binomial coefficient.png]] for 0 ≤ k ≤ p.<br />
<br />
Therefore,<br />
<br />
<!--(1 + n)^p = sum (for k from 0 to p) of (p choose k) n^k-->[[file:Q2 equation 1.png]]<br />
<br />
and<br />
<br />
<!-- (1+n)^p - n^p - 1 = sum (for k from 1 to p-1) of (p choose k) n^k -->[[file:Q2 equation 2.png]]<br />
<br />
since (p choose 0) = (p choose p) = 1.<br />
<br />
It holds that (p choose 1) = p.<br />
Since p is a prime number, the factor p in (p choose k) is not divisible by k, k-1,...,2 for 2 ≤ k ≤ p-1.<br />
Therefore, p divides (p choose k) for 1 ≤ k ≤ p-1.<br />
<br />
Since each summand on the right side is divisible by p, the whole sum, i.e. the left side is also divisible by p.<br />
<br />
<br />
===Question 3===<br />
[[file:Episode One Math Q3.jpg|thumb|Episode one Math question 3]]<br />
(Guessing from the solution shown)<br />
<br />
Find the integer solutions (a,b) with (a^3 + a^2 - 1) - (a - 1)b = 0<br />
<br />
'''Solution:'''<br />
[[file:Episode One Math S3.jpg|thumb|Episode one Math solution to question 3]]<br />
<br />
(a^3 + a^2 - 1) - (a - 1)b<br><br />
= (a^3 - 1) + (a^2 - 1) + 1 - (a - 1)b<br><br />
= (a - 1)(a^2 + a + 1) + (a - 1)(a + 1) + 1 - (a - 1)b<br><br />
= (a - 1)(a^2 + 2a - b + 2) + 1 = 0<br />
<br />
Therefore,<br> <br />
(a - 1)(a^2 + 2a - b + 2) = -1<br />
<br />
a, b are integers, therefore the factors a - 1 and a^2 + 2a - b + 2 are integers too. Since the product is -1, one of the factors must be equal to -1, the other to 1.<br />
<br />
If a - 1 = -1 then a = 0 and from a^2 + 2a - b + 2 = 1 we obtain b = 1.<br />
<br />
If a - 1 = 1 then a = 2 and a^2 + 2a - b + 2 = -1 implies that b = 11.<br />
<br />
There are two integer solutions: (a,b) = (0,1) and (a,b) = (2,11).<br />
<br />
<br />
===Question 4===<br />
[[file:Episode One Math Q4.jpg|thumb|Episode one Math question 4]]<br />
Given [[file:Math episode one question four.png]],<br />
<br />
find the sum of F(1)+F(2)+F(3)+...+F(60).<br />
<br />
'''Solution:'''<br />
[[file:Episode One Math S4.jpg|thumb|Episode one Math, partial solution to question 4. Note that there are errors.]]<br />
<br />
Simplying the fraction:<br />
<br />
Reference that for any variables a & b, (a-b) (a+b) = (a-b)^2. Let (2x + 1)^.5 = a and (2x - 1)^.5 = b. Multiply F(x) by 1 or (a-b)/(a-b) or ((2x + 1)^.5 - (2x - 1)^.5) / ((2x + 1)^.5 - (2x - 1)^.5). The denominator becomes:<br />
<br />
((2x + 1)^.5 + (2x - 1)^.5) * ((2x + 1)^.5 - (2x - 1)^.5) <br />
<br />
= (2x + 1) - (2x - 1)<br />
<br />
= 2<br />
<br />
The numerator becomes:<br />
<br />
(4x + (4x^2 - 1)^.5 ) ((2x + 1)^.5 - (2x - 1)^.5) <br />
<br />
=(4x + ((2x)^2 - 1)^.5 ) (a - b) <br />
<br />
=(4x + (2x - 1)^.5 (2x + 1)^.5 ) (a -b) <br />
<br />
=(4x + a b ) (a - b) <br />
<br />
=4x (a - b) + (a^2) b - a (b^2)<br />
<br />
=4x (a - b) + (2x + 1) b - a (2x - 1)<br />
<br />
=4x (a - b) - 2x (a - b) + a + b<br />
<br />
=2x ( a - b) + a + b<br />
<br />
=(2x + 1) a - (2x - 1) b<br />
<br />
=a^3 - b^3<br />
<br />
<br />
Therefore:<br />
<br />
F(x) = (a^3 - b^3 ) / 2<br />
<br />
Note that when x= 1, 2, 3, 4, ..., 60;<br />
<br />
a^3 = 3^1.5, 5^1.5, 7^1.5, 9^1.5, ..., 119^1.5, 121^1.5;<br />
<br />
b^3 = 1^1.5, 3^1.5, 5^1.5, 7^1.5, 9^1.5, ..., 119^1.5;<br />
<br />
When summating over F(x) for x between 1 and 60, observe that the majority of the terms in a^3 and b^3 cancel out, leaving:<br />
<br />
(121^1.5 - 1^1.5)/2 =((11*11)^1.5 - 1)/2 = 665<br />
<br />
Thus, sum of F(x) for x between 1 and 60 is 665. The solution can be generalized as equal to (a(x_max)^3 - b(x_min)^3)/2.<br />
<br />
==Episode 9==<br />
[[File:Episode 9 Math.png|thumb|Fibonacci sequence question that appeared in Episode 9]]<br />
<br />
This advanced Algebra problem previously appeared in the first [http://www.j3e.info/ojyuken/math/php.php?name=tokyo&v1=0&v2=1992&v3=1&v4=3&y=1992&n=0_3 1992 Tokyo University Entrance Exam]:<br />
<br />
===Problem===<br />
A number sequence {F(n)} that can be defined as F(1) = 1, F(2) = 1, F(n+2) = F(n) + F(n+1) (where n is any natural number) is called the Fibonacci sequence and its general solution is given by, <br />
<br />
[[file:fibonacci.png]]<br />
<br />
[[file:golden ratio.png]]<br />
<br />
Answer the following questions by using this fact if needed:<br />
<br />
<br />
Define a sequence of natural numbers X(n) (where n is any natural number), in which each digit is either 0 or 1, set by the following rules:<br />
<br />
(i) X(1) = 1<br />
<br />
(ii) We define X(n+1) as a natural number, which can be gotten by replacing the digits of X(n) with 1 if the digit is 0, and with 10 if the digit is 1.<br />
<br />
For example, X(1) = 1, X(2) = 10, X(3) = 101, X(4) = 10110, X(5) = 10110101, ... <br />
<br />
<br />
(1) Find A(n), defined as the number of digits of X(n)?<br />
<br />
(2) Find B(n),defined as the numbers of times '01' appears in X(n)?<br />
For example, B(1) = 0, B(2) = 0, B(3) = 1, B(4) = 1, B(5) = 3,...<br />
<br />
<br />
'' '''Side note''': X(n) can be considered a clever analogy to the show: where [[Episode 10]] replaces [[Episode 1]], and [[Episode 1]] replaces Episode 0 in the next iteration.''<br />
<br />
===Solution===<br />
====Part 1====<br />
<br />
Let A(n) equal to the number of digits in X(n), which consists solely of 1s and 0s. Let's suppose x(n) is the number of 0s in X(n) at n iteration (poor choice of variable by the student). Let's suppose y(n) is the number of 1s in X(n) at n iteration.<br />
<br />
<br />
1) Every time a 0 appears, it is replaced with 1 at the next iteration.<br />
<br />
2) Every time a 1 appears, it is replaced with 10 at t he next iteration.<br />
<br />
Therefore, the number of 0s in the next iteration is equal to the number of 1s previously:<br />
<br />
x(n+1)= y(n)<br />
<br />
and the number of 1s in the next iteration is equal to the number of 0s AND the numbers of 1s previously:<br />
<br />
y(n+1)=x(n)+y(n)<br />
<br />
<br />
Prove: x(n) is a fibonacci sequence:<br />
<br />
i) y(n+1) = x(n) + y(n)<br />
<br />
ii) x(n+2) = y(n+1)<br />
<br />
iii) x(n+1) = y(n)<br />
<br />
<br />
complete the substition shows:<br />
<br />
x(n+2) = y(n+1) = x(n) + y(n) = x(n) + x(n+1)<br />
<br />
x(n+2) = x(n+1) + x(n) ie definition of fibonacci sequence<br />
<br />
Since x(n) is a fibonacci sequence, y(n) is also a fibonacci sequence since x(n+1) = y(n). Therefore:<br />
<br />
x(n+2)=x(n+1)+x(n)<br />
<br />
y(n+2)=y(n+1)+y(n)<br />
<br />
x(n+2)+y(n+2)=x(n+1)+y(n+1)+x(n)+y(n)<br />
<br />
Recall: A(n)=x(n)+y(n), therefore<br />
<br />
A(n+2)=A(n+1)+A(n)<br />
<br />
X(1) = 1, X(2) = 10, X(3) = 101, X(4) = 10110, X(5) = 10110101, ...<br />
<br />
therefore: A(1) = 1, A(2) = 2, A(3) = 3, A(4) = 5, A(5) = 8,...<br />
<br />
Recall fibonacci sequence F(n): 1, 1, 2, 3, 5, 8.<br />
Therefore A(n) = F(n+1), or<br />
<br />
[[file:An.png]]<br />
<br />
<br />
====Part 2====<br />
<br />
Let B(n) be the number of times '01' appears in X(n).<br />
<br />
Any two digits in X(n) may be 00, 01, 10, 11<br />
<br />
the corresponding digits in the next iteration X(n+1) will be<br />
<br />
00 -> 11<br />
<br />
01 -> 110<br />
<br />
10 -> 101<br />
<br />
11 -> 1010<br />
<br />
Thus, any two digit sequence in X(n) that begins with 1* will contribute to a 01 sequence in the next iteration. In other words, B(n+1) equals y(n), except when unit digit of X(n) is 1, or:<br />
<br />
B(n+1) = y(n) - odd(X(n))<br />
<br />
y(n) is a fibonacci sequence with starting values: y(1) = 1, y(2) = 1, y(3) = 2<br />
thus y(n) = F(n)<br />
<br />
B(n+1) = F(n) - odd(X(n)) <br />
<br />
B(n) = F(n-1) - odd(X(n-1)), or<br />
<br />
[[file:Bn.png]]<br />
<br />
==Episode 10==<br />
Same as [[Mathematics_of_Madoka_Magica#Episode_1|Episode 1]].<br />
[[Category:Articles]]<br />
[[Category:Fanmade analysis]]<br />
[[Category:Science]]</div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=Mathematics_of_Madoka_Magica&diff=6381Mathematics of Madoka Magica2011-03-11T09:49:55Z<p>Anon-kun: replaced formulas with images</p>
<hr />
<div>The cast of Mahou Shoujo Madoka Magica are students of [[Mitakihara Middle School]]. Despite merely a middle school, coursework in Mitakihara can be quite challenging. The following are Math problems that have appeared thus far in the show, and their solutions:<br />
<br />
==Episode 1==<br />
[[file:Episode One Algebra.jpg|thumb|Episode one Math question 1]]<br />
<br />
===Question 1===<br />
Any integer divided by 14 will have a remainder between 0 and 13. Given a is has remainder of 6 and b has remainder of 1 when divided by 14, what is the remainder of x when divided by 14, given x is an integer solution to x^2-2ax+b=0?<br />
<br />
'''Solution:'''<br />
<br />
This problem can be solved simply with modular arithmetrics: <br />
a = 6 mod 14;<br />
b = 1 mod 14;<br />
x = r mod 14.<br />
<br />
x^2 - 2ax + b = 0 is equivalent to<br />
<br />
x^2 - 12x + 1 = 0 mod 14<br />
<br />
x(x - 12) = -1 mod 14<br />
<br />
-1 has no proper divisors modulo 14, thus factors x and x-12 must correspond to either 1 or -1 mod 14.<br />
<br />
If x = 1 mod 14, then x-12 = 3 mod 14: false. If x = -1 mod 14, then x - 12 = 1 mod 14: true.<br />
<br />
Therefore, x = -1 mod 14 = 13 mod 14, i.e., the remainder r of x is 13.<br />
<br />
<br />
'''Second Solution:'''<br />
<br />
Homura used in Episode 1 a basic approach with usual integer arithmetic.<br />
<br />
Let x = 14q + r, a = 14s + 6, b = 14t + 1.<br />
<br />
Substitute into x^2 - 2ax + b = 0 to get after some calculations<br />
<br />
x^2 - 2ax + b = 14c + (r+1)^2 = 0 with c = 14q^2 + 2q - 28qs - 2rs - 12q + t - r<br />
<br />
14 divides 14c and 0, hence it also divides (r+1)^2. This implies that r+1 is divisible by 14.<br />
<br />
The question asks for a remainder r between 0 and 13, so we obtain r = 13.<br />
<br />
See also: http://green-oval.net/cgi-board.pl/a/thread/44824340<br />
<br />
<br />
===Question 2===<br />
[[file:Episode One Math Q2.jpg|thumb|Episode one Math question 2]]<br />
Assuming that p is a prime number and n is an arbitrary natural number, prove that (1+n)^p - n^p - 1 is divisible by p.<br />
<br />
'''Solution:'''<br />
<br />
By Fermat's Little theorem, for any prime p and integer a,<br />
<br />
a^p = a (mod p)<br />
<br />
Thus:<br />
<br />
(1+n)^p - n^p - 1 (mod p) is equivalent to<br />
<br />
(1+n) - n - 1 (mod p) is equivalent to<br />
<br />
0 (mod p)<br />
<br />
So the overall expression is divisible by p.<br />
<br />
<br />
'''Second solution:'''<br />
<br />
The problem can be solved with the binomial theorem:<br />
<br />
For a, b not equal to 0 and nonnegative integer p it holds that:<br />
<br />
<!--(a + b)^p = sum (for k from 0 to p) of (p choose k) a^(p-k) b^k-->[[file:Q2 binomial theorem.png]]<br />
<br />
where the binomial coefficient (p choose k) is the integer <!-- p(p-1)...(p-k+1)/(k(k-1)...1) --> [[file:Q2 binomial coefficient.png]] for 0 ≤ k ≤ p.<br />
<br />
Therefore,<br />
<br />
<!--(1 + n)^p = sum (for k from 0 to p) of (p choose k) n^k-->[[file:Q2 equation 1.png]]<br />
<br />
and<br />
<br />
<!-- (1+n)^p - n^p - 1 = sum (for k from 1 to p-1) of (p choose k) n^k -->[[file:Q2 equation 2.png]]<br />
<br />
since (p choose 0) = (p choose p) = 1.<br />
<br />
It holds that (p choose 1) = p.<br />
Since p is a prime number, the factor p in (p choose k) is not divisible by k, k-1,...,2 for 2 ≤ k ≤ p-1.<br />
Therefore, p divides (p choose k) for 1 <= k <= p-1.<br />
<br />
Since each summand on the right side is divisible by p, the whole sum, i.e. the left side is also divisible by p.<br />
<br />
<br />
===Question 3===<br />
[[file:Episode One Math Q3.jpg|thumb|Episode one Math question 3]]<br />
(Guessing from the solution shown)<br />
<br />
Find the integer solutions (a,b) with (a^3 + a^2 - 1) - (a - 1)b = 0<br />
<br />
'''Solution:'''<br />
[[file:Episode One Math S3.jpg|thumb|Episode one Math solution to question 3]]<br />
<br />
(a^3 + a^2 - 1) - (a - 1)b<br><br />
= (a^3 - 1) + (a^2 - 1) + 1 - (a - 1)b<br><br />
= (a - 1)(a^2 + a + 1) + (a - 1)(a + 1) + 1 - (a - 1)b<br><br />
= (a - 1)(a^2 + 2a - b + 2) + 1 = 0<br />
<br />
Therefore,<br> <br />
(a - 1)(a^2 + 2a - b + 2) = -1<br />
<br />
a, b are integers, therefore the factors a - 1 and a^2 + 2a - b + 2 are integers too. Since the product is -1, one of the factors must be equal to -1, the other to 1.<br />
<br />
If a - 1 = -1 then a = 0 and from a^2 + 2a - b + 2 = 1 we obtain b = 1.<br />
<br />
If a - 1 = 1 then a = 2 and a^2 + 2a - b + 2 = -1 implies that b = 11.<br />
<br />
There are two integer solutions: (a,b) = (0,1) and (a,b) = (2,11).<br />
<br />
<br />
===Question 4===<br />
[[file:Episode One Math Q4.jpg|thumb|Episode one Math question 4]]<br />
Given [[file:Math episode one question four.png]],<br />
<br />
find the sum of F(1)+F(2)+F(3)+...+F(60).<br />
<br />
'''Solution:'''<br />
[[file:Episode One Math S4.jpg|thumb|Episode one Math, partial solution to question 4. Note that there are errors.]]<br />
<br />
Simplying the fraction:<br />
<br />
Reference that for any variables a & b, (a-b) (a+b) = (a-b)^2. Let (2x + 1)^.5 = a and (2x - 1)^.5 = b. Multiply F(x) by 1 or (a-b)/(a-b) or ((2x + 1)^.5 - (2x - 1)^.5) / ((2x + 1)^.5 - (2x - 1)^.5). The denominator becomes:<br />
<br />
((2x + 1)^.5 + (2x - 1)^.5) * ((2x + 1)^.5 - (2x - 1)^.5) <br />
<br />
= (2x + 1) - (2x - 1)<br />
<br />
= 2<br />
<br />
The numerator becomes:<br />
<br />
(4x + (4x^2 - 1)^.5 ) ((2x + 1)^.5 - (2x - 1)^.5) <br />
<br />
=(4x + ((2x)^2 - 1)^.5 ) (a - b) <br />
<br />
=(4x + (2x - 1)^.5 (2x + 1)^.5 ) (a -b) <br />
<br />
=(4x + a b ) (a - b) <br />
<br />
=4x (a - b) + (a^2) b - a (b^2)<br />
<br />
=4x (a - b) + (2x + 1) b - a (2x - 1)<br />
<br />
=4x (a - b) - 2x (a - b) + a + b<br />
<br />
=2x ( a - b) + a + b<br />
<br />
=(2x + 1) a - (2x - 1) b<br />
<br />
=a^3 - b^3<br />
<br />
<br />
Therefore:<br />
<br />
F(x) = (a^3 - b^3 ) / 2<br />
<br />
Note that when x= 1, 2, 3, 4, ..., 60;<br />
<br />
a^3 = 3^1.5, 5^1.5, 7^1.5, 9^1.5, ..., 119^1.5, 121^1.5;<br />
<br />
b^3 = 1^1.5, 3^1.5, 5^1.5, 7^1.5, 9^1.5, ..., 119^1.5;<br />
<br />
When summating over F(x) for x between 1 and 60, observe that the majority of the terms in a^3 and b^3 cancel out, leaving:<br />
<br />
(121^1.5 - 1^1.5)/2 =((11*11)^1.5 - 1)/2 = 665<br />
<br />
Thus, sum of F(x) for x between 1 and 60 is 665. The solution can be generalized as equal to (a(x_max)^3 - b(x_min)^3)/2.<br />
<br />
==Episode 9==<br />
[[File:Episode 9 Math.png|thumb|Fibonacci sequence question that appeared in Episode 9]]<br />
<br />
This advanced Algebra problem previously appeared in the first [http://www.j3e.info/ojyuken/math/php.php?name=tokyo&v1=0&v2=1992&v3=1&v4=3&y=1992&n=0_3 1992 Tokyo University Entrance Exam]:<br />
<br />
===Problem===<br />
A number sequence {F(n)} that can be defined as F(1) = 1, F(2) = 1, F(n+2) = F(n) + F(n+1) (where n is any natural number) is called the Fibonacci sequence and its general solution is given by, <br />
<br />
[[file:fibonacci.png]]<br />
<br />
[[file:golden ratio.png]]<br />
<br />
Answer the following questions by using this fact if needed:<br />
<br />
<br />
Define a sequence of natural numbers X(n) (where n is any natural number), in which each digit is either 0 or 1, set by the following rules:<br />
<br />
(i) X(1) = 1<br />
<br />
(ii) We define X(n+1) as a natural number, which can be gotten by replacing the digits of X(n) with 1 if the digit is 0, and with 10 if the digit is 1.<br />
<br />
For example, X(1) = 1, X(2) = 10, X(3) = 101, X(4) = 10110, X(5) = 10110101, ... <br />
<br />
<br />
(1) Find A(n), defined as the number of digits of X(n)?<br />
<br />
(2) Find B(n),defined as the numbers of times '01' appears in X(n)?<br />
For example, B(1) = 0, B(2) = 0, B(3) = 1, B(4) = 1, B(5) = 3,...<br />
<br />
<br />
'' '''Side note''': X(n) can be considered a clever analogy to the show: where [[Episode 10]] replaces [[Episode 1]], and [[Episode 1]] replaces Episode 0 in the next iteration.''<br />
<br />
===Solution===<br />
====Part 1====<br />
<br />
Let A(n) equal to the number of digits in X(n), which consists solely of 1s and 0s. Let's suppose x(n) is the number of 0s in X(n) at n iteration (poor choice of variable by the student). Let's suppose y(n) is the number of 1s in X(n) at n iteration.<br />
<br />
<br />
1) Every time a 0 appears, it is replaced with 1 at the next iteration.<br />
<br />
2) Every time a 1 appears, it is replaced with 10 at t he next iteration.<br />
<br />
Therefore, the number of 0s in the next iteration is equal to the number of 1s previously:<br />
<br />
x(n+1)= y(n)<br />
<br />
and the number of 1s in the next iteration is equal to the number of 0s AND the numbers of 1s previously:<br />
<br />
y(n+1)=x(n)+y(n)<br />
<br />
<br />
Prove: x(n) is a fibonacci sequence:<br />
<br />
i) y(n+1) = x(n) + y(n)<br />
<br />
ii) x(n+2) = y(n+1)<br />
<br />
iii) x(n+1) = y(n)<br />
<br />
<br />
complete the substition shows:<br />
<br />
x(n+2) = y(n+1) = x(n) + y(n) = x(n) + x(n+1)<br />
<br />
x(n+2) = x(n+1) + x(n) ie definition of fibonacci sequence<br />
<br />
Since x(n) is a fibonacci sequence, y(n) is also a fibonacci sequence since x(n+1) = y(n). Therefore:<br />
<br />
x(n+2)=x(n+1)+x(n)<br />
<br />
y(n+2)=y(n+1)+y(n)<br />
<br />
x(n+2)+y(n+2)=x(n+1)+y(n+1)+x(n)+y(n)<br />
<br />
Recall: A(n)=x(n)+y(n), therefore<br />
<br />
A(n+2)=A(n+1)+A(n)<br />
<br />
X(1) = 1, X(2) = 10, X(3) = 101, X(4) = 10110, X(5) = 10110101, ...<br />
<br />
therefore: A(1) = 1, A(2) = 2, A(3) = 3, A(4) = 5, A(5) = 8,...<br />
<br />
Recall fibonacci sequence F(n): 1, 1, 2, 3, 5, 8.<br />
Therefore A(n) = F(n+1), or<br />
<br />
[[file:An.png]]<br />
<br />
<br />
====Part 2====<br />
<br />
Let B(n) be the number of times '01' appears in X(n).<br />
<br />
Any two digits in X(n) may be 00, 01, 10, 11<br />
<br />
the corresponding digits in the next iteration X(n+1) will be<br />
<br />
00 -> 11<br />
<br />
01 -> 110<br />
<br />
10 -> 101<br />
<br />
11 -> 1010<br />
<br />
Thus, any two digit sequence in X(n) that begins with 1* will contribute to a 01 sequence in the next iteration. In other words, B(n+1) equals y(n), except when unit digit of X(n) is 1, or:<br />
<br />
B(n+1) = y(n) - odd(X(n))<br />
<br />
y(n) is a fibonacci sequence with starting values: y(1) = 1, y(2) = 1, y(3) = 2<br />
thus y(n) = F(n)<br />
<br />
B(n+1) = F(n) - odd(X(n)) <br />
<br />
B(n) = F(n-1) - odd(X(n-1)), or<br />
<br />
[[file:Bn.png]]<br />
<br />
==Episode 10==<br />
Same as [[Mathematics_of_Madoka_Magica#Episode_1|Episode 1]].<br />
[[Category:Articles]]<br />
[[Category:Fanmade analysis]]<br />
[[Category:Science]]</div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=File:Q2_equation_2.png&diff=6380File:Q2 equation 2.png2011-03-11T09:37:34Z<p>Anon-kun: </p>
<hr />
<div></div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=File:Q2_equation_1.png&diff=6379File:Q2 equation 1.png2011-03-11T09:37:14Z<p>Anon-kun: </p>
<hr />
<div></div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=File:Q2_binomial_coefficient.png&diff=6378File:Q2 binomial coefficient.png2011-03-11T09:36:42Z<p>Anon-kun: </p>
<hr />
<div></div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=File:Q2_binomial_theorem.png&diff=6376File:Q2 binomial theorem.png2011-03-11T09:36:15Z<p>Anon-kun: </p>
<hr />
<div></div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=Homura_Akemi&diff=6270Homura Akemi2011-03-11T02:08:11Z<p>Anon-kun: weapons</p>
<hr />
<div>{{spoilers}}<br />
{{Infobox_character<br />
|name=Homura Akemi<br />
|image=[[File:Homura Chara Sheet.jpg|300px]]<br />
|jname=暁美 ほむら (Akemi Homura) <br />
|seiyuu=[http://anidb.net/perl-bin/animedb.pl?show=creator&creatorid=305 Chiwa Saitou]<br />
|age=13-14 (estimated)<br />
|weapon=Shield (storage and timejump device), various mundane weapons like grenades, pipe bombs, pistols, machine guns<br />
}}<br />
Homura is first seen by Madoka in a dream of her fighting a monster in a desolate landscape of skyscrapers. The day after the dream, Homura has transferred to Madoka's school where she gives her ominous warnings.<br />
<br />
==Official Info==<br />
A smart, graceful, athletically-talented girl. Cool if not chilly, her personality even can even be seen in her posture. Her weapon appears to be a bow, the same as Madoka?<br />
<br />
She seems to be a magical girl, but there are a lot of mysteries surrounding the way she acts. Her fighting style is also unidentified.<br />
* Ms. Aoki’s Design Check: A blunt girl. I tried to let this shine through in her design. Her distinguishing point is her silky long hair.<br />
* Magical girl outfit: Homura was designed with her personality and combat style in mind. I found her fascinating right away. (Aoki)<br />
* The sharp lines of her upper back and jacket evoke her personality. Overall rather monochromatic. Her black tights have a diamond pattern.<br />
* School uniform: She wears black tights in uniform too, leaving her overall feel just like in her magical girl outfit. She gives off the air of a silent beauty.<br />
* Regardless of her reasons, Homura'a priority is preventing Madoka from becoming a Mahou Shoujo. Since that is not understandable by other Mahou Shoujos or Madoka, she decided to act alone.<br />
* Homura is a magical girl from another timeline, as revealed by Kyubey. She knows that Kyubey is behind a sinister plot, and it mostly involves Madoka turning into a magical girl.<br />
* Homura's special ability is revealed to be some kind of time-travel, which explains why she knows so much to the point of knowing exactly when and where certain things will happen. This also may be why she can move at blistering speed.<br />
* Her [[Homura_Residence|house]] has a wall display of documents relating to witches.<br />
* In episode 9, it was revealed that Homura can stop time. However, anyone she touches while the ability is under effect will not be stopped.<br />
<br />
==Observations==<br />
*She wants to stop [[Madoka Kaname|Madoka]] from making a contract with [[Kyuubey]]<br />
*She has much knowledge about [[Mitakihara Middle School]].<br />
*Is annoyed when Madoka uses her last name or honorifics instead of the more intimate "Homura".<br />
*She was openly pursuing [[Kyuubey]] until Kyuubey made contact with Madoka.<br />
*She appears obsessed with Madoka's happiness and/or survival. <br />
<br />
==Speculations==<br />
[[File:Homura_Move2.jpg|right|thumb|Homura is able to move extremely fast in comparison with water drops.]]<br />
[[File:Homuragrenade.jpg|right|thumb|Homura's grenade in Episode 8 looks like an M61 grenade, used during the Vietnam War.]]<br />
[[File:Homugun.jpg|right|thumb|Homura's pistol in Episode 8 looks like an M9.]]<br />
*In the first two episodes, Homura appeared to be the antagonist. It was later explained that she only wanted to prevent Madoka becoming a magical girl. She saved Madoka and Sayaka multiple times and may become an ally and/or mentor of Madoka in the future.<br />
*If the theory about witches and magical girls is true (see [[Speculah:Witches as Fallen Magical Girls Theory|this article]]), then it is fairly possible that Homura is on the brink of becoming a witch, which could be why Mami wanted to share her Grief Seed with her in [[Episode 2]].<br />
*She might also wish to prevent Madoka becoming a magical girl because of the risk that she could turn into an overwhelmingly powerful witch.<br />
*She may be a former magical girl who has had her contract revoked.<br />
*In [[Episode 5]], Homura saves Sayaka from Kyouko by suddenly manifesting as Kyouko's attack is about to land, somehow moving Sayaka out of the path of the attack. The ability seems to involve massive acceleration rather than teleportation, as she is able to act in a [[:File:Homura_Move1.jpg|split second]]. As there are hints that a Puella Magi's powers may relate to her wish (e.g. Sayaka's regeneration), this may be a clue as to Homura's true motives. As of [[Episode 8]] it has been confirmed that Homura's ability has something to do with time.<br />
*Per the image in the gallery below, the pipe layout on the corridor wall is completely different after Homura uses her 'teleportation' ability. Assuming this is not a production error, it has been suggested that the ability actually involves magically reconstituting the entire corridor while moving the people in it around. For more information, please refer to the entry on this phenomenon in the [[Speculah:Magical_Powers#Spatial_Reconstruction_.28unconfirmed.29|Magical Powers speculah]].<br />
*Homura may be Madoka's pet cat who was transformed into a human through magic. She's the same black cat that appears in the anime's opening and the Homura cover of Megami magazine. Homura often shows cat-like mannerisms and poor understanding of/disinterest in human social interaction that doesn't involve Madoka. Black cats are traditional familiars for witches, and Homura frequently plays the role of Madoka's familiar.<br />
*Homura may be Madoka's mother at a younger age. She seemed familiar to Madoka, seemed to know her plight in [[Episode 3]] and possibly travelled through time to warn her not to become a magical girl, a job she probably still currently holds.<br />
*Homura may also be Madoka from the future, since Sayaka says that Homura talks like a person that lost everything ([[Episode 8]]), maybe she lost her family after becoming a Mahō Shōjo and came back in time to stop herself, and she looks a lot like the Madoka with larger hair that apears in the opening.<br />
*Homura may be a homunculus created by Madoka.<ref>See [http://gall.dcinside.com/anigallers/259050 DCinside] and [http://www.angelhalowiki.com/r1/wiki.php/%EB%A7%88%EB%B2%95%EC%86%8C%EB%85%80%20%EB%A7%88%EB%8F%84%EC%B9%B4%E2%98%86%EB%A7%88%EA%B8%B0%EC%B9%B4/%EB%96%A1%EB%B0%A5%EB%AA%A8%EC%9D%8C#s-7.4 Angelhalo Wiki](Documents are in Korean)</ref><br />
**The last line of [[Magia]] reads 命を作るのは願い(What creates life is a wish).<br />
**'Akemi(アケミ)' and 'alchemy(アルケミー)' reads similar in Japanese.<br />
**'Homu' in 'Homura' is also the first four letters in the word 'homunculus'.<br />
**Dr. Faust in Goethe's Faust created a homunculus, which has superhuman perception to see through Faust's dream. Mephistopheles could be subtituted by Kyuubey, and Dr. Faust by Madoka. See [[Speculah:Madoka Magica and Faust]] on Shaft's usage of Faust.<br />
*Homura might be in possession of the soul gems of magical girls from an alternate timeline, such as the alternate Madoka's. See also [[Speculah:Homura Identity Swap Theory]].<br />
*One of the speculations as for why Homura does not tell Madoka upfront of the consequences of becoming a magical girl, is because Madoka might just become one in order to save Homura or Mami. This could also be the reason why she's trying to keep a distance between her and Madoka.<br />
<br />
==Trivia==<br />
*Homura is frequently referred to as "Homerun-chan" amongst Western fans, due to the similarity of the two names in Japanese, however she is frequently referred as "Homuhomu" (ほむほむ) within Japanese community.<br />
*The first character for Akemi (暁) means "daybreak", while the second one (美) means beauty.<br />
*When her full name is written vertically in hiragana (あけみほむら) and with left half covered in certain angle, it could be read as カナメまどか which is Kaname written in katakana and Madoka written in hiragana. <br />
*Her last name can also be used as a first name.<br />
*Her first name is written in hiragana, which has no special meaning. However, when written in kanji, it can mean:<br />
**inflammation, blaze (炎 or 焔)<br />
**heart on fire (心火)<br />
**mass of flame (炎叢)<br />
**village protector (保村)<br /><br />
*Some folks in 2ch nicknamed [[Homura Residence|Homura's Residence]] "Homuhōmu," (ほむホーム) since Homu is a nickname for Homura, and Hōmu is "home" rendered in the Japanese syllabary. It should be noted that "homu" is more often used for dog houses and similar structures, with the English word "home" expressed using "uchi" (うち, which refers to one's own house or household), "ie" (いえ, which can refer to a physical house or a family lineage), or the honorific "otaku" (お宅, which refers to someone else's house or household).<br />
<br />
==Gallery==<br />
<gallery captionalign="left" position="center"><br />
File:Homura Akemi Anime Design.jpg<br />
File:Homura Akemi Original Design.jpg|Original character design by Aoki Ume<br />
File:Why_Homerun-chan.jpg|How Homura became Homerun-chan<br />
File:Homerun-chan2.png|2ch's reaction<br />
File:Ep5 runes mura.png|Homura's ring and fingernail<br />
File:1296786066859 2.jpg|Spatial Reconstitution Theory (空間再構成)<br />
File:Homura-magazine.jpg|The cover of Megami magazine featuring Homura with a black cat.<br />
File:Hunter-x-hunter-1641920 2.jpg|Comparison of unnamed Nen ability in Hunter x Hunter and Homura's 'teleportation.' <br />
</gallery><br />
<br />
==References==<br />
{{reflist}}<br />
<br />
==External links==<br />
*{{pixiv_tag|暁美ほむら}}<br />
*{{booru_tag|danbooru|akemi_homura}}<br />
<br />
[[Category:Characters]]</div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=Roberta&diff=6269Roberta2011-03-11T01:54:36Z<p>Anon-kun: placeholder</p>
<hr />
<div>__NOTOC__<br />
'''Roberta''' ({{Runes|ROBERTA}}) is one of the [[Witch|witches]] that appeared in [[Episode 10]].<br />
<br />
==Gallery==<br />
<gallery captionalign="left" position="center"><br />
File:Roberta.png|Roberta<br />
</gallery><br />
<br />
[[Category:Witches]]</div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=Patricia&diff=6268Patricia2011-03-11T01:52:22Z<p>Anon-kun: placeholder</p>
<hr />
<div>__NOTOC__<br />
'''Patricia''' ({{Runes|PATRICIA}}) is one of the [[Witch|witches]] that appeared in [[Episode 10]].<br />
<br />
==Gallery==<br />
<gallery captionalign="left" position="center"><br />
File:Patricia.jpg|Patricia<br />
</gallery><br />
<br />
[[Category:Witches]]</div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=Izabel&diff=6267Izabel2011-03-11T01:50:39Z<p>Anon-kun: placeholder</p>
<hr />
<div>__NOTOC__<br />
'''Isadel''' ({{Runes|ISADEL}}) is one of the [[Witch|witches]] that appeared in [[Episode 10]].<br />
<br />
==Gallery==<br />
<gallery captionalign="left" position="center"><br />
File:Isadel.jpg|Isadel<br />
</gallery><br />
<br />
[[Category:Witches]]</div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=Characters&diff=6266Characters2011-03-11T01:45:56Z<p>Anon-kun: witches from episode 10</p>
<hr />
<div>[[File:Characters.png|440px|right]]<br />
*[[Madoka Kaname]]<br />
*[[Homura Akemi]]<br />
*[[Sayaka Miki]]<br />
*[[Mami Tomoe]]<br />
*[[Kyouko Sakura]]<br />
*[[Kyuubey]]<br />
<br />
== Witches ==<br />
<!-- Only a test at the moment. See [[Talk:Witch from the dream in episode one]] for the details<br />
*[[Witch from the dream in episode one]] --><br />
*[[Gertrud]]<br />
**[[Gertrud#Anthony|Anthony]]<br />
**[[Gertrud#Adelbert|Adelbert]]<br />
*[[Suleika]]<br />
**[[Suleika#Ulla|Ulla]]<br />
*[[Charlotte]]<br />
**[[Charlotte#Pyotr|Pyotr]]<br />
*[[Elly]] (Kirsten)<br />
** [[Elly#Daniyyel & Jennifer|Daniyyel and Jennifer]]<br />
*[[Albertine]]<br />
** [[Albertine#Anja|Anja]]<br />
*[[Elsa Maria]]<br />
**[[Elsa Maria#Sebastians|Sebastians]]<br />
*[[Gisela]]<br />
**[[Gisela#Dora|Dora]]<br />
*[[Uhrmann]]<br />
**[[Uhrmann#Bartels|Bartels]]<br />
*[[Oktavia von Seckendorff]]<br />
**[[Oktavia von Seckendorff#Holger|Holger]]<br />
*[[Isadel]]<br />
*[[Patricia]]<br />
*[[Roberta]]<br />
*[[Speculah:Walpurgis Night|Walpurgis Night]]<br />
<br />
==Support Characters==<br />
*Madoka's Family<br />
**[[Junko Kaname]]<br />
**[[Tomohisa Kaname]]<br />
**[[Tatsuya Kaname]]<br />
*[[Hitomi Shizuki]]<br />
*[[Kazuko Saotome]]<br />
*[[Kyousuke Kamijou]]<br />
*[[Inu Daruma Brothers]]<br />
<br />
[[Category:Characters| ]]</div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=Mathematics_of_Madoka_Magica&diff=6216Mathematics of Madoka Magica2011-03-11T00:15:36Z<p>Anon-kun: :</p>
<hr />
<div>The cast of Mahou Shoujo Madoka Magica are students of [[Mitakihara Middle School]]. Despite merely a middle school, coursework in Mitakihara can be quite challenging. The following are Math problems that have appeared thus far in the show, and their solutions:<br />
<br />
==Episode 1==<br />
[[file:Episode One Algebra.jpg|thumb|Episode one Math question 1]]<br />
<br />
===Question 1===<br />
Any integer divided by 14 will have a remainder between 0 and 13. Given a is has remainder of 6 and b has remainder of 1 when divided by 14, what is the remainder of x when divided by 14, given x is an integer solution to x^2-2ax+b=0?<br />
<br />
'''Solution:'''<br />
<br />
This problem can be solved simply with modular arithmetrics: <br />
a = 6 mod 14;<br />
b = 1 mod 14;<br />
x = r mod 14.<br />
<br />
x^2 - 2ax + b = 0 is equivalent to<br />
<br />
x^2 - 12x + 1 = 0 mod 14<br />
<br />
x(x - 12) = -1 mod 14<br />
<br />
-1 has no proper divisors modulo 14, thus factors x and x-12 must correspond to either 1 or -1 mod 14.<br />
<br />
If x = 1 mod 14, then x-12 = 3 mod 14: false. If x = -1 mod 14, then x - 12 = 1 mod 14: true.<br />
<br />
Therefore, x = -1 mod 14 = 13 mod 14, i.e., the remainder r of x is 13.<br />
<br />
<br />
'''Second Solution:'''<br />
<br />
Homura used in Episode 1 a basic approach with usual integer arithmetic.<br />
<br />
Let x = 14q + r, a = 14s + 6, b = 14t + 1.<br />
<br />
Substitute into x^2 - 2ax + b = 0 to get after some calculations<br />
<br />
x^2 - 2ax + b = 14c + (r+1)^2 = 0 with c = 14q^2 + 2q - 28qs - 2rs - 12q + t - r<br />
<br />
14 divides 14c and 0, hence it also divides (r+1)^2. This implies that r+1 is divisible by 14.<br />
<br />
The question asks for a remainder r between 0 and 13, so we obtain r = 13.<br />
<br />
See also: http://green-oval.net/cgi-board.pl/a/thread/44824340<br />
<br />
<br />
===Question 2===<br />
[[file:Episode One Math Q2.jpg|thumb|Episode one Math question 2]]<br />
Assuming that p is a prime number and n is an arbitrary natural number, prove that (1+n)^p - n^p - 1 is divisible by p.<br />
<br />
'''Solution:'''<br />
<br />
By Fermat's Little theorem, for any prime p and integer a,<br />
<br />
a^p = a (mod p)<br />
<br />
Thus:<br />
<br />
(1+n)^p - n^p - 1 (mod p) is equivalent to<br />
<br />
(1+n) - n - 1 (mod p) is equivalent to<br />
<br />
0 (mod p)<br />
<br />
So the overall expression is divisible by p.<br />
<br />
<br />
'''Second solution:'''<br />
<br />
The problem can be solved with the binomial theorem:<br />
<br />
For a, b not equal to 0 and nonnegative integer p it holds that:<br />
<br />
(a + b)^p = sum (for k from 0 to p) of (p choose k) a^(p-k) b^k where the binomial coefficient (p choose k) is the integer p(p-1)...(p-k+1)/(k(k-1)...1) for 0 <= k <= p<br />
<br />
Therefore,<br />
<br />
(1 + n)^p = sum (for k from 0 to p) of (p choose k) n^k<br />
<br />
and<br />
<br />
(1+n)^p - n^p - 1 = sum (for k from 1 to p-1) of (p choose k) n^k<br />
<br />
since (p choose 0) = (p choose p) = 1.<br />
<br />
It holds that (p choose 1) = p.<br />
Since p is a prime number, the factor p in (p choose k) is not divisible by k, k-1,...,2 for 2 <= k <= p-1.<br />
Therefore, p divides (p choose k) for 1 <= k <= p-1.<br />
<br />
Since each summand on the right side is divisible by p, the whole sum, i.e. the left side is also divisible by p.<br />
<br />
<br />
===Question 3===<br />
[[file:Episode One Math Q3.jpg|thumb|Episode one Math question 3]]<br />
(Guessing from the solution shown)<br />
<br />
Find the integer solutions (a,b) with (a^3 + a^2 - 1) - (a - 1)b = 0<br />
<br />
'''Solution:'''<br />
[[file:Episode One Math S3.jpg|thumb|Episode one Math solution to question 3]]<br />
<br />
(a^3 + a^2 - 1) - (a - 1)b<br><br />
= (a^3 - 1) + (a^2 - 1) + 1 - (a - 1)b<br><br />
= (a - 1)(a^2 + a + 1) + (a - 1)(a + 1) + 1 - (a - 1)b<br><br />
= (a - 1)(a^2 + 2a - b + 2) + 1 = 0<br />
<br />
Therefore,<br> <br />
(a - 1)(a^2 + 2a - b + 2) = -1<br />
<br />
a, b are integers, therefore the factors a - 1 and a^2 + 2a - b + 2 are integers too. Since the product is -1, one of the factors must be equal to -1, the other to 1.<br />
<br />
If a - 1 = -1 then a = 0 and from a^2 + 2a - b + 2 = 1 we obtain b = 1.<br />
<br />
If a - 1 = 1 then a = 2 and a^2 + 2a - b + 2 = -1 implies that b = 11.<br />
<br />
There are two integer solutions: (a,b) = (0,1) and (a,b) = (2,11).<br />
<br />
<br />
===Question 4===<br />
[[file:Episode One Math Q4.jpg|thumb|Episode one Math question 4]]<br />
Given [[file:Math episode one question four.png]],<br />
<br />
find the sum of F(1)+F(2)+F(3)+...+F(60).<br />
<br />
'''Solution:'''<br />
[[file:Episode One Math S4.jpg|thumb|Episode one Math, partial solution to question 4. Note that there are errors.]]<br />
<br />
Simplying the fraction:<br />
<br />
Reference that for any variables a & b, (a-b) (a+b) = (a-b)^2. Let (2x + 1)^.5 = a and (2x - 1)^.5 = b. Multiply F(x) by 1 or (a-b)/(a-b) or ((2x + 1)^.5 - (2x - 1)^.5) / ((2x + 1)^.5 - (2x - 1)^.5). The denominator becomes:<br />
<br />
((2x + 1)^.5 + (2x - 1)^.5) * ((2x + 1)^.5 - (2x - 1)^.5) <br />
<br />
= (2x + 1) - (2x - 1)<br />
<br />
= 2<br />
<br />
The numerator becomes:<br />
<br />
(4x + (4x^2 - 1)^.5 ) ((2x + 1)^.5 - (2x - 1)^.5) <br />
<br />
=(4x + ((2x)^2 - 1)^.5 ) (a - b) <br />
<br />
=(4x + (2x - 1)^.5 (2x + 1)^.5 ) (a -b) <br />
<br />
=(4x + a b ) (a - b) <br />
<br />
=4x (a - b) + (a^2) b - a (b^2)<br />
<br />
=4x (a - b) + (2x + 1) b - a (2x - 1)<br />
<br />
=4x (a - b) - 2x (a - b) + a + b<br />
<br />
=2x ( a - b) + a + b<br />
<br />
=(2x + 1) a - (2x - 1) b<br />
<br />
=a^3 - b^3<br />
<br />
<br />
Therefore:<br />
<br />
F(x) = (a^3 - b^3 ) / 2<br />
<br />
Note that when x= 1, 2, 3, 4, ..., 60;<br />
<br />
a^3 = 3^1.5, 5^1.5, 7^1.5, 9^1.5, ..., 119^1.5, 121^1.5;<br />
<br />
b^3 = 1^1.5, 3^1.5, 5^1.5, 7^1.5, 9^1.5, ..., 119^1.5;<br />
<br />
When summating over F(x) for x between 1 and 60, observe that the majority of the terms in a^3 and b^3 cancel out, leaving:<br />
<br />
(121^1.5 - 1^1.5)/2 =((11*11)^1.5 - 1)/2 = 665<br />
<br />
Thus, sum of F(x) for x between 1 and 60 is 665. The solution can be generalized as equal to (a(x_max)^3 - b(x_min)^3)/2.<br />
<br />
==Episode 9==<br />
[[File:Episode 9 Math.png|thumb|Fibonacci sequence question that appeared in Episode 9]]<br />
<br />
This advanced Algebra problem previously appeared in the first [http://www.j3e.info/ojyuken/math/php.php?name=tokyo&v1=0&v2=1992&v3=1&v4=3&y=1992&n=0_3 1992 Tokyo University Entrance Exam]:<br />
<br />
===Problem===<br />
A number sequence {F(n)} that can be defined as F(1) = 1, F(2) = 1, F(n+2) = F(n) + F(n+1) (where n is any natural number) is called the Fibonacci sequence and its general solution is given by, <br />
<br />
[[file:fibonacci.png]]<br />
<br />
[[file:golden ratio.png]]<br />
<br />
Answer the following questions by using this fact if needed:<br />
<br />
<br />
Define a sequence of natural numbers X(n) (where n is any natural number), in which each digit is either 0 or 1, set by the following rules:<br />
<br />
(i) X(1) = 1<br />
<br />
(ii) We define X(n+1) as a natural number, which can be gotten by replacing the digits of X(n) with 1 if the digit is 0, and with 10 if the digit is 1.<br />
<br />
For example, X(1) = 1, X(2) = 10, X(3) = 101, X(4) = 10110, X(5) = 10110101, ... <br />
<br />
<br />
(1) Find A(n), defined as the number of digits of X(n)?<br />
<br />
(2) Find B(n),defined as the numbers of times '01' appears in X(n)?<br />
For example, B(1) = 0, B(2) = 0, B(3) = 1, B(4) = 1, B(5) = 3,...<br />
<br />
<br />
===Solution===<br />
====Part 1====<br />
<br />
Let A(n) equal to the number of digits in X(n), which consists solely of 1s and 0s. Let's suppose x(n) is the number of 0s in X(n) at n iteration (poor choice of variable by the student). Let's suppose y(n) is the number of 1s in X(n) at n iteration.<br />
<br />
<br />
1) Every time a 0 appears, it is replaced with 1 at the next iteration.<br />
<br />
2) Every time a 1 appears, it is replaced with 10 at t he next iteration.<br />
<br />
Therefore, the number of 0s in the next iteration is equal to the number of 1s previously:<br />
<br />
x(n+1)= y(n)<br />
<br />
and the number of 1s in the next iteration is equal to the number of 0s AND the numbers of 1s previously:<br />
<br />
y(n+1)=x(n)+y(n)<br />
<br />
<br />
Prove: x(n) is a fibonacci sequence:<br />
<br />
i) y(n+1) = x(n) + y(n)<br />
<br />
ii) x(n+2) = y(n+1)<br />
<br />
iii) x(n+1) = y(n)<br />
<br />
<br />
complete the substition shows:<br />
<br />
x(n+2) = y(n+1) = x(n) + y(n) = x(n) + x(n+1)<br />
<br />
x(n+2) = x(n+1) + x(n) ie definition of fibonacci sequence<br />
<br />
Since x(n) is a fibonacci sequence, y(n) is also a fibonacci sequence since x(n+1) = y(n). Therefore:<br />
<br />
x(n+2)=x(n+1)+x(n)<br />
<br />
y(n+2)=y(n+1)+y(n)<br />
<br />
x(n+2)+y(n+2)=x(n+1)+y(n+1)+x(n)+y(n)<br />
<br />
Recall: A(n)=x(n)+y(n), therefore<br />
<br />
A(n+2)=A(n+1)+A(n)<br />
<br />
X(1) = 1, X(2) = 10, X(3) = 101, X(4) = 10110, X(5) = 10110101, ...<br />
<br />
therefore: A(1) = 1, A(2) = 2, A(3) = 3, A(4) = 5, A(5) = 8,...<br />
<br />
Recall fibonacci sequence F(n): 1, 1, 2, 3, 5, 8.<br />
Therefore A(n) = F(n+1), or<br />
<br />
[[file:An.png]]<br />
<br />
<br />
====Part 2====<br />
<br />
Let B(n) be the number of times '01' appears in X(n).<br />
<br />
Any two digits in X(n) may be 00, 01, 10, 11<br />
<br />
the corresponding digits in the next iteration X(n+1) will be<br />
<br />
00 -> 11<br />
<br />
01 -> 110<br />
<br />
10 -> 101<br />
<br />
11 -> 1010<br />
<br />
Thus, any two digit sequence in X(n) that begins with 1* will contribute to a 01 sequence in the next iteration. In other words, B(n+1) equals y(n), except when unit digit of X(n) is 1, or:<br />
<br />
B(n+1) = y(n) - odd(X(n))<br />
<br />
y(n) is a fibonacci sequence with starting values: y(1) = 1, y(2) = 1, y(3) = 2<br />
thus y(n) = F(n)<br />
<br />
B(n+1) = F(n) - odd(X(n)) <br />
<br />
B(n) = F(n-1) - odd(X(n-1)), or<br />
<br />
[[file:Bn.png]]<br />
<br />
==Episode 10==<br />
Same as Episode One.<br />
[[Category:Articles]]<br />
[[Category:Fanmade analysis]]<br />
[[Category:Science]]</div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=Mathematics_of_Madoka_Magica&diff=6215Mathematics of Madoka Magica2011-03-11T00:13:24Z<p>Anon-kun: second solution</p>
<hr />
<div>The cast of Mahou Shoujo Madoka Magica are students of [[Mitakihara Middle School]]. Despite merely a middle school, coursework in Mitakihara can be quite challenging. The following are Math problems that have appeared thus far in the show, and their solutions:<br />
<br />
==Episode 1==<br />
[[file:Episode One Algebra.jpg|thumb|Episode one Math question 1]]<br />
<br />
===Question 1===<br />
Any integer divided by 14 will have a remainder between 0 and 13. Given a is has remainder of 6 and b has remainder of 1 when divided by 14, what is the remainder of x when divided by 14, given x is an integer solution to x^2-2ax+b=0?<br />
<br />
'''Solution:'''<br />
<br />
This problem can be solved simply with modular arithmetrics: <br />
a = 6 mod 14;<br />
b = 1 mod 14;<br />
x = r mod 14.<br />
<br />
x^2 - 2ax + b = 0 is equivalent to<br />
<br />
x^2 - 12x + 1 = 0 mod 14<br />
<br />
x(x - 12) = -1 mod 14<br />
<br />
-1 has no proper divisors modulo 14, thus factors x and x-12 must correspond to either 1 or -1 mod 14.<br />
<br />
If x = 1 mod 14, then x-12 = 3 mod 14: false. If x = -1 mod 14, then x - 12 = 1 mod 14: true.<br />
<br />
Therefore, x = -1 mod 14 = 13 mod 14, i.e., the remainder r of x is 13.<br />
<br />
<br />
'''Second Solution'''<br />
Homura used in Episode 1 a basic approach with usual integer arithmetic.<br />
<br />
Let x = 14q + r, a = 14s + 6, b = 14t + 1.<br />
<br />
Substitute into x^2 - 2ax + b = 0 to get after some calculations<br />
<br />
x^2 - 2ax + b = 14c + (r+1)^2 = 0 with c = 14q^2 + 2q - 28qs - 2rs - 12q + t - r<br />
<br />
14 divides 14c and 0, hence it also divides (r+1)^2. This implies that r+1 is divisible by 14.<br />
<br />
The question asks for a remainder r between 0 and 13, so we obtain r = 13.<br />
<br />
See also: http://green-oval.net/cgi-board.pl/a/thread/44824340<br />
<br />
<br />
===Question 2===<br />
[[file:Episode One Math Q2.jpg|thumb|Episode one Math question 2]]<br />
Assuming that p is a prime number and n is an arbitrary natural number, prove that (1+n)^p - n^p - 1 is divisible by p.<br />
<br />
'''Solution:'''<br />
<br />
By Fermat's Little theorem, for any prime p and integer a,<br />
<br />
a^p = a (mod p)<br />
<br />
Thus:<br />
<br />
(1+n)^p - n^p - 1 (mod p) is equivalent to<br />
<br />
(1+n) - n - 1 (mod p) is equivalent to<br />
<br />
0 (mod p)<br />
<br />
So the overall expression is divisible by p.<br />
<br />
<br />
'''Second solution'''<br />
<br />
The problem can be solved with the binomial theorem:<br />
<br />
For a, b not equal to 0 and nonnegative integer p it holds that:<br />
<br />
(a + b)^p = sum (for k from 0 to p) of (p choose k) a^(p-k) b^k where the binomial coefficient (p choose k) is the integer p(p-1)...(p-k+1)/(k(k-1)...1) for 0 <= k <= p<br />
<br />
Therefore,<br />
<br />
(1 + n)^p = sum (for k from 0 to p) of (p choose k) n^k<br />
<br />
and<br />
<br />
(1+n)^p - n^p - 1 = sum (for k from 1 to p-1) of (p choose k) n^k<br />
<br />
since (p choose 0) = (p choose p) = 1.<br />
<br />
It holds that (p choose 1) = p.<br />
Since p is a prime number, the factor p in (p choose k) is not divisible by k, k-1,...,2 for 2 <= k <= p-1.<br />
Therefore, p divides (p choose k) for 1 <= k <= p-1.<br />
<br />
Since each summand on the right side is divisible by p, the whole sum, i.e. the left side is also divisible by p.<br />
<br />
<br />
===Question 3===<br />
[[file:Episode One Math Q3.jpg|thumb|Episode one Math question 3]]<br />
(Guessing from the solution shown)<br />
<br />
Find the integer solutions (a,b) with (a^3 + a^2 - 1) - (a - 1)b = 0<br />
<br />
'''Solution'''<br />
[[file:Episode One Math S3.jpg|thumb|Episode one Math solution to question 3]]<br />
<br />
(a^3 + a^2 - 1) - (a - 1)b<br><br />
= (a^3 - 1) + (a^2 - 1) + 1 - (a - 1)b<br><br />
= (a - 1)(a^2 + a + 1) + (a - 1)(a + 1) + 1 - (a - 1)b<br><br />
= (a - 1)(a^2 + 2a - b + 2) + 1 = 0<br />
<br />
Therefore,<br> <br />
(a - 1)(a^2 + 2a - b + 2) = -1<br />
<br />
a, b are integers, therefore the factors a - 1 and a^2 + 2a - b + 2 are integers too. Since the product is -1, one of the factors must be equal to -1, the other to 1.<br />
<br />
If a - 1 = -1 then a = 0 and from a^2 + 2a - b + 2 = 1 we obtain b = 1.<br />
<br />
If a - 1 = 1 then a = 2 and a^2 + 2a - b + 2 = -1 implies that b = 11.<br />
<br />
There are two integer solutions: (a,b) = (0,1) and (a,b) = (2,11).<br />
<br />
<br />
===Question 4===<br />
[[file:Episode One Math Q4.jpg|thumb|Episode one Math question 4]]<br />
Given [[file:Math episode one question four.png]],<br />
<br />
find the sum of F(1)+F(2)+F(3)+...+F(60).<br />
<br />
'''Solution:'''<br />
[[file:Episode One Math S4.jpg|thumb|Episode one Math, partial solution to question 4. Note that there are errors.]]<br />
<br />
Simplying the fraction:<br />
<br />
Reference that for any variables a & b, (a-b) (a+b) = (a-b)^2. Let (2x + 1)^.5 = a and (2x - 1)^.5 = b. Multiply F(x) by 1 or (a-b)/(a-b) or ((2x + 1)^.5 - (2x - 1)^.5) / ((2x + 1)^.5 - (2x - 1)^.5). The denominator becomes:<br />
<br />
((2x + 1)^.5 + (2x - 1)^.5) * ((2x + 1)^.5 - (2x - 1)^.5) <br />
<br />
= (2x + 1) - (2x - 1)<br />
<br />
= 2<br />
<br />
The numerator becomes:<br />
<br />
(4x + (4x^2 - 1)^.5 ) ((2x + 1)^.5 - (2x - 1)^.5) <br />
<br />
=(4x + ((2x)^2 - 1)^.5 ) (a - b) <br />
<br />
=(4x + (2x - 1)^.5 (2x + 1)^.5 ) (a -b) <br />
<br />
=(4x + a b ) (a - b) <br />
<br />
=4x (a - b) + (a^2) b - a (b^2)<br />
<br />
=4x (a - b) + (2x + 1) b - a (2x - 1)<br />
<br />
=4x (a - b) - 2x (a - b) + a + b<br />
<br />
=2x ( a - b) + a + b<br />
<br />
=(2x + 1) a - (2x - 1) b<br />
<br />
=a^3 - b^3<br />
<br />
<br />
Therefore:<br />
<br />
F(x) = (a^3 - b^3 ) / 2<br />
<br />
Note that when x= 1, 2, 3, 4, ..., 60;<br />
<br />
a^3 = 3^1.5, 5^1.5, 7^1.5, 9^1.5, ..., 119^1.5, 121^1.5;<br />
<br />
b^3 = 1^1.5, 3^1.5, 5^1.5, 7^1.5, 9^1.5, ..., 119^1.5;<br />
<br />
When summating over F(x) for x between 1 and 60, observe that the majority of the terms in a^3 and b^3 cancel out, leaving:<br />
<br />
(121^1.5 - 1^1.5)/2 =((11*11)^1.5 - 1)/2 = 665<br />
<br />
Thus, sum of F(x) for x between 1 and 60 is 665. The solution can be generalized as equal to (a(x_max)^3 - b(x_min)^3)/2.<br />
<br />
==Episode 9==<br />
[[File:Episode 9 Math.png|thumb|Fibonacci sequence question that appeared in Episode 9]]<br />
<br />
This advanced Algebra problem previously appeared in the first [http://www.j3e.info/ojyuken/math/php.php?name=tokyo&v1=0&v2=1992&v3=1&v4=3&y=1992&n=0_3 1992 Tokyo University Entrance Exam]:<br />
<br />
===Problem===<br />
A number sequence {F(n)} that can be defined as F(1) = 1, F(2) = 1, F(n+2) = F(n) + F(n+1) (where n is any natural number) is called the Fibonacci sequence and its general solution is given by, <br />
<br />
[[file:fibonacci.png]]<br />
<br />
[[file:golden ratio.png]]<br />
<br />
Answer the following questions by using this fact if needed:<br />
<br />
<br />
Define a sequence of natural numbers X(n) (where n is any natural number), in which each digit is either 0 or 1, set by the following rules:<br />
<br />
(i) X(1) = 1<br />
<br />
(ii) We define X(n+1) as a natural number, which can be gotten by replacing the digits of X(n) with 1 if the digit is 0, and with 10 if the digit is 1.<br />
<br />
For example, X(1) = 1, X(2) = 10, X(3) = 101, X(4) = 10110, X(5) = 10110101, ... <br />
<br />
<br />
(1) Find A(n), defined as the number of digits of X(n)?<br />
<br />
(2) Find B(n),defined as the numbers of times '01' appears in X(n)?<br />
For example, B(1) = 0, B(2) = 0, B(3) = 1, B(4) = 1, B(5) = 3,...<br />
<br />
<br />
===Solution===<br />
====Part 1====<br />
<br />
Let A(n) equal to the number of digits in X(n), which consists solely of 1s and 0s. Let's suppose x(n) is the number of 0s in X(n) at n iteration (poor choice of variable by the student). Let's suppose y(n) is the number of 1s in X(n) at n iteration.<br />
<br />
<br />
1) Every time a 0 appears, it is replaced with 1 at the next iteration.<br />
<br />
2) Every time a 1 appears, it is replaced with 10 at t he next iteration.<br />
<br />
Therefore, the number of 0s in the next iteration is equal to the number of 1s previously:<br />
<br />
x(n+1)= y(n)<br />
<br />
and the number of 1s in the next iteration is equal to the number of 0s AND the numbers of 1s previously:<br />
<br />
y(n+1)=x(n)+y(n)<br />
<br />
<br />
Prove: x(n) is a fibonacci sequence:<br />
<br />
i) y(n+1) = x(n) + y(n)<br />
<br />
ii) x(n+2) = y(n+1)<br />
<br />
iii) x(n+1) = y(n)<br />
<br />
<br />
complete the substition shows:<br />
<br />
x(n+2) = y(n+1) = x(n) + y(n) = x(n) + x(n+1)<br />
<br />
x(n+2) = x(n+1) + x(n) ie definition of fibonacci sequence<br />
<br />
Since x(n) is a fibonacci sequence, y(n) is also a fibonacci sequence since x(n+1) = y(n). Therefore:<br />
<br />
x(n+2)=x(n+1)+x(n)<br />
<br />
y(n+2)=y(n+1)+y(n)<br />
<br />
x(n+2)+y(n+2)=x(n+1)+y(n+1)+x(n)+y(n)<br />
<br />
Recall: A(n)=x(n)+y(n), therefore<br />
<br />
A(n+2)=A(n+1)+A(n)<br />
<br />
X(1) = 1, X(2) = 10, X(3) = 101, X(4) = 10110, X(5) = 10110101, ...<br />
<br />
therefore: A(1) = 1, A(2) = 2, A(3) = 3, A(4) = 5, A(5) = 8,...<br />
<br />
Recall fibonacci sequence F(n): 1, 1, 2, 3, 5, 8.<br />
Therefore A(n) = F(n+1), or<br />
<br />
[[file:An.png]]<br />
<br />
<br />
====Part 2====<br />
<br />
Let B(n) be the number of times '01' appears in X(n).<br />
<br />
Any two digits in X(n) may be 00, 01, 10, 11<br />
<br />
the corresponding digits in the next iteration X(n+1) will be<br />
<br />
00 -> 11<br />
<br />
01 -> 110<br />
<br />
10 -> 101<br />
<br />
11 -> 1010<br />
<br />
Thus, any two digit sequence in X(n) that begins with 1* will contribute to a 01 sequence in the next iteration. In other words, B(n+1) equals y(n), except when unit digit of X(n) is 1, or:<br />
<br />
B(n+1) = y(n) - odd(X(n))<br />
<br />
y(n) is a fibonacci sequence with starting values: y(1) = 1, y(2) = 1, y(3) = 2<br />
thus y(n) = F(n)<br />
<br />
B(n+1) = F(n) - odd(X(n)) <br />
<br />
B(n) = F(n-1) - odd(X(n-1)), or<br />
<br />
[[file:Bn.png]]<br />
<br />
==Episode 10==<br />
Same as Episode One.<br />
[[Category:Articles]]<br />
[[Category:Fanmade analysis]]<br />
[[Category:Science]]</div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=Mathematics_of_Madoka_Magica&diff=6211Mathematics of Madoka Magica2011-03-10T23:36:42Z<p>Anon-kun: image</p>
<hr />
<div>The cast of Mahou Shoujo Madoka Magica are students of [[Mitakihara Middle School]]. Despite merely a middle school, coursework in Mitakihara can be quite challenging. The following are Math problems that have appeared thus far in the show, and their solutions:<br />
<br />
==Episode 1==<br />
[[file:Episode One Algebra.jpg|thumb|Episode one Math question 1]]<br />
<br />
===Question 1===<br />
Any integer divided by 14 will have a remainder between 0 and 13. Given a is has remainder of 6 and b has remainder of 1 when divided by 14, what is the remainder of x when divided by 14, given x is an integer solution to x^2-2ax+b=0?<br />
<br />
'''Solution:'''<br />
<br />
This problem can be solved simply with modular arithmetrics: <br />
a = 6 mod 14;<br />
b = 1 mod 14;<br />
x = r mod 14.<br />
<br />
x^2 - 2ax + b = 0 is equivalent to<br />
<br />
x^2 - 12x + 1 = 0 mod 14<br />
<br />
x(x - 12) = -1 mod 14<br />
<br />
-1 has no proper divisors modulo 14, thus factors x and x-12 must correspond to either 1 or -1 mod 14.<br />
<br />
If x = 1 mod 14, then x-12 = 3 mod 14: false. If x = -1 mod 14, then x - 12 = 1 mod 14: true.<br />
<br />
Therefore, x = -1 mod 14 = 13 mod 14, i.e., the remainder r of x is 13.<br />
<br />
<br />
'''Second Solution'''<br />
Homura used in Episode 1 a basic approach with usual integer arithmetic.<br />
<br />
Let x = 14q + r, a = 14s + 6, b = 14t + 1.<br />
<br />
Substitute into x^2 - 2ax + b = 0 to get after some calculations<br />
<br />
x^2 - 2ax + b = 14c + (r+1)^2 = 0 with c = 14q^2 + 2q - 28qs - 2rs - 12q + t - r<br />
<br />
14 divides 14c and 0, hence it also divides (r+1)^2. This implies that r+1 is divisible by 14.<br />
<br />
The question asks for a remainder r between 0 and 13, so we obtain r = 13.<br />
<br />
See also: http://green-oval.net/cgi-board.pl/a/thread/44824340<br />
<br />
<br />
===Question 2===<br />
[[file:Episode One Math Q2.jpg|thumb|Episode one Math question 2]]<br />
Assuming that p is a prime number and n is an arbitrary natural number, prove that (1+n)^p - n^p - 1 is divisible by p.<br />
<br />
'''Solution'''<br />
<br />
<br />
<br />
<br />
<br />
<br />
===Question 3===<br />
[[file:Episode One Math Q3.jpg|thumb|Episode one Math question 3]]<br />
(Guessing from the solution shown)<br />
<br />
Find the integer solutions (a,b) with (a^3 + a^2 - 1) - (a - 1)b = 0<br />
<br />
'''Solution'''<br />
[[file:Episode One Math S3.jpg|thumb|Episode one Math solution to question 3]]<br />
<br />
(a^3 + a^2 - 1) - (a - 1)b<br><br />
= (a^3 - 1) + (a^2 - 1) + 1 - (a - 1)b<br><br />
= (a - 1)(a^2 + a + 1) + (a - 1)(a + 1) + 1 - (a - 1)b<br><br />
= (a - 1)(a^2 + 2a - b + 2) + 1 = 0<br />
<br />
Therefore,<br> <br />
(a - 1)(a^2 + 2a - b + 2) = -1<br />
<br />
a, b are integers, therefore the factors a - 1 and a^2 + 2a - b + 2 are integers too. Since the product is -1, one of the factors must be equal to -1, the other to 1.<br />
<br />
If a - 1 = -1 then a = 0 and from a^2 + 2a - b + 2 = 1 we obtain b = 1.<br />
<br />
If a - 1 = 1 then a = 2 and a^2 + 2a - b + 2 = -1 implies that b = 11.<br />
<br />
There are two integer solutions: (a,b) = (0,1) and (a,b) = (2,11).<br />
<br />
<br />
===Question 4===<br />
[[file:Episode One Math Q4.jpg|thumb|Episode one Math question 4]]<br />
Given [[file:Math episode one question four.png]],<br />
<br />
find the sum of F(1)+F(2)+F(3)+...+F(60).<br />
<br />
'''Solution:'''<br />
[[file:Episode One Math S4.jpg|thumb|Episode one Math, partial solution to question 4. Note that there are errors.]]<br />
<br />
Simplying the fraction:<br />
<br />
Reference that for any variables a & b, (a-b) (a+b) = (a-b)^2. Let (2x + 1)^.5 = a and (2x - 1)^.5 = b. Multiply F(x) by 1 or (a-b)/(a-b) or ((2x + 1)^.5 - (2x - 1)^.5) / ((2x + 1)^.5 - (2x - 1)^.5). The denominator becomes:<br />
<br />
((2x + 1)^.5 + (2x - 1)^.5) * ((2x + 1)^.5 - (2x - 1)^.5) <br />
<br />
= (2x + 1) - (2x - 1)<br />
<br />
= 2<br />
<br />
The numerator becomes:<br />
<br />
(4x + (4x^2 - 1)^.5 ) ((2x + 1)^.5 - (2x - 1)^.5) <br />
<br />
=(4x + ((2x)^2 - 1)^.5 ) (a - b) <br />
<br />
=(4x + (2x - 1)^.5 (2x + 1)^.5 ) (a -b) <br />
<br />
=(4x + a b ) (a - b) <br />
<br />
=4x (a - b) + (a^2) b - a (b^2)<br />
<br />
=4x (a - b) + (2x + 1) b - a (2x - 1)<br />
<br />
=4x (a - b) - 2x (a - b) + a + b<br />
<br />
=2x ( a - b) + a + b<br />
<br />
=(2x + 1) a - (2x - 1) b<br />
<br />
=a^3 - b^3<br />
<br />
<br />
Therefore:<br />
<br />
F(x) = (a^3 - b^3 ) / 2<br />
<br />
Note that when x= 1, 2, 3, 4, ..., 60;<br />
<br />
a^3 = 3^1.5, 5^1.5, 7^1.5, 9^1.5, ..., 119^1.5, 121^1.5;<br />
<br />
b^3 = 1^1.5, 3^1.5, 5^1.5, 7^1.5, 9^1.5, ..., 119^1.5;<br />
<br />
When summating over F(x) for x between 1 and 60, observe that the majority of the terms in a^3 and b^3 cancel out, leaving:<br />
<br />
(121^1.5 - 1^1.5)/2 =((11*11)^1.5 - 1)/2 = 665<br />
<br />
Thus, sum of F(x) for x between 1 and 60 is 665. The solution can be generalized as equal to (a(x_max)^3 - b(x_min)^3)/2.<br />
<br />
==Episode 9==<br />
[[File:Episode 9 Math.png|thumb|Fibonacci sequence question that appeared in Episode 9]]<br />
<br />
This advanced Algebra problem previously appeared in the first [http://www.j3e.info/ojyuken/math/php.php?name=tokyo&v1=0&v2=1992&v3=1&v4=3&y=1992&n=0_3 1992 Tokyo University Entrance Exam]:<br />
<br />
===Problem===<br />
A number sequence {F(n)} that can be defined as F(1) = 1, F(2) = 1, F(n+2) = F(n) + F(n+1) (where n is any natural number) is called the Fibonacci sequence and its general solution is given by, <br />
<br />
[[file:fibonacci.png]]<br />
<br />
[[file:golden ratio.png]]<br />
<br />
Answer the following questions by using this fact if needed:<br />
<br />
<br />
Define a sequence of natural numbers X(n) (where n is any natural number), in which each digit is either 0 or 1, set by the following rules:<br />
<br />
(i) X(1) = 1<br />
<br />
(ii) We define X(n+1) as a natural number, which can be gotten by replacing the digits of X(n) with 1 if the digit is 0, and with 10 if the digit is 1.<br />
<br />
For example, X(1) = 1, X(2) = 10, X(3) = 101, X(4) = 10110, X(5) = 10110101, ... <br />
<br />
<br />
(1) Find A(n), defined as the number of digits of X(n)?<br />
<br />
(2) Find B(n),defined as the numbers of times '01' appears in X(n)?<br />
For example, B(1) = 0, B(2) = 0, B(3) = 1, B(4) = 1, B(5) = 3,...<br />
<br />
<br />
===Solution===<br />
====Part 1====<br />
<br />
Let A(n) equal to the number of digits in X(n), which consists solely of 1s and 0s. Let's suppose x(n) is the number of 0s in X(n) at n iteration (poor choice of variable by the student). Let's suppose y(n) is the number of 1s in X(n) at n iteration.<br />
<br />
<br />
1) Every time a 0 appears, it is replaced with 1 at the next iteration.<br />
<br />
2) Every time a 1 appears, it is replaced with 10 at t he next iteration.<br />
<br />
Therefore, the number of 0s in the next iteration is equal to the number of 1s previously:<br />
<br />
x(n+1)= y(n)<br />
<br />
and the number of 1s in the next iteration is equal to the number of 0s AND the numbers of 1s previously:<br />
<br />
y(n+1)=x(n)+y(n)<br />
<br />
<br />
Prove: x(n) is a fibonacci sequence:<br />
<br />
i) y(n+1) = x(n) + y(n)<br />
<br />
ii) x(n+2) = y(n+1)<br />
<br />
iii) x(n+1) = y(n)<br />
<br />
<br />
complete the substition shows:<br />
<br />
x(n+2) = y(n+1) = x(n) + y(n) = x(n) + x(n+1)<br />
<br />
x(n+2) = x(n+1) + x(n) ie definition of fibonacci sequence<br />
<br />
Since x(n) is a fibonacci sequence, y(n) is also a fibonacci sequence since x(n+1) = y(n). Therefore:<br />
<br />
x(n+2)=x(n+1)+x(n)<br />
<br />
y(n+2)=y(n+1)+y(n)<br />
<br />
x(n+2)+y(n+2)=x(n+1)+y(n+1)+x(n)+y(n)<br />
<br />
Recall: A(n)=x(n)+y(n), therefore<br />
<br />
A(n+2)=A(n+1)+A(n)<br />
<br />
X(1) = 1, X(2) = 10, X(3) = 101, X(4) = 10110, X(5) = 10110101, ...<br />
<br />
therefore: A(1) = 1, A(2) = 2, A(3) = 3, A(4) = 5, A(5) = 8,...<br />
<br />
Recall fibonacci sequence F(n): 1, 1, 2, 3, 5, 8.<br />
Therefore A(n) = F(n+1), or<br />
<br />
[[file:An.png]]<br />
<br />
<br />
====Part 2====<br />
<br />
Let B(n) be the number of times '01' appears in X(n).<br />
<br />
Any two digits in X(n) may be 00, 01, 10, 11<br />
<br />
the corresponding digits in the next iteration X(n+1) will be<br />
<br />
00 -> 11<br />
<br />
01 -> 110<br />
<br />
10 -> 101<br />
<br />
11 -> 1010<br />
<br />
Thus, any two digit sequence in X(n) that begins with 1* will contribute to a 01 sequence in the next iteration. In other words, B(n+1) equals y(n), except when unit digit of X(n) is 1, or:<br />
<br />
B(n+1) = y(n) - odd(X(n))<br />
<br />
y(n) is a fibonacci sequence with starting values: y(1) = 1, y(2) = 1, y(3) = 2<br />
thus y(n) = F(n)<br />
<br />
B(n+1) = F(n) - odd(X(n)) <br />
<br />
B(n) = F(n-1) - odd(X(n-1)), or<br />
<br />
[[file:Bn.png]]<br />
<br />
==Episode 10==<br />
Same as Episode One.<br />
[[Category:Articles]]<br />
[[Category:Fanmade analysis]]<br />
[[Category:Science]]</div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=File:Episode_One_Math_S4.jpg&diff=6204File:Episode One Math S4.jpg2011-03-10T23:28:42Z<p>Anon-kun: Partial solution to Question 4, shown out-of-order at the beginning of the sequence.
Note that the exponents are placed wrong and that a + sign is missing.</p>
<hr />
<div>Partial solution to Question 4, shown out-of-order at the beginning of the sequence.<br />
<br />
Note that the exponents are placed wrong and that a + sign is missing.</div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=Mathematics_of_Madoka_Magica&diff=6200Mathematics of Madoka Magica2011-03-10T23:12:36Z<p>Anon-kun: fixes</p>
<hr />
<div>The cast of Mahou Shoujo Madoka Magica are students of [[Mitakihara Middle School]]. Despite merely a middle school, coursework in Mitakihara can be quite challenging. The following are Math problems that have appeared thus far in the show, and their solutions:<br />
<br />
==Episode One==<br />
[[file:Episode One Algebra.jpg|thumb|Episode one Math question 1]]<br />
<br />
===Question 1===<br />
Any integer divided by 14 will have a remainder between 0 and 13. Given a is has remainder of 6 and b has remainder of 1 when divided by 14, what is the remainder of x when divided by 14, given x is an integer solution to x^2-2ax+b=0?<br />
<br />
'''Solution:'''<br />
<br />
This problem can be solved simply with modular arithmetrics: <br />
a = 6 mod 14;<br />
b = 1 mod 14;<br />
x = r mod 14.<br />
<br />
x^2 - 2ax + b = 0 is equivalent to<br />
<br />
x^2 - 12x + 1 = 0 mod 14<br />
<br />
x(x - 12) = -1 mod 14<br />
<br />
-1 has no proper divisors modulo 14, thus factors x and x-12 must correspond to either 1 or -1 mod 14.<br />
<br />
If x = 1 mod 14, then x-12 = 3 mod 14: false. If x = -1 mod 14, then x - 12 = 1 mod 14: true.<br />
<br />
Therefore, x = -1 mod 14 = 13 mod 14, i.e., the remainder r of x is 13.<br />
<br />
<br />
'''Second Solution'''<br />
Homura used in Episode 1 a basic approach with usual integer arithmetic.<br />
<br />
Let x = 14q + r, a = 14s + 6, b = 14t + 1.<br />
<br />
Substitute into x^2 - 2ax + b = 0 to get after some calculations<br />
<br />
x^2 - 2ax + b = 14c + (r+1)^2 = 0 with c = 14q^2 + 2q - 28qs - 2rs - 12q + t - r<br />
<br />
14 divides 14c and 0, hence it also divides (r+1)^2. This implies that r+1 is divisible by 14.<br />
<br />
The question asks for a remainder r between 0 and 13, so we obtain r = 13.<br />
<br />
See also: http://green-oval.net/cgi-board.pl/a/thread/44824340<br />
<br />
<br />
===Question 2===<br />
[[file:Episode One Math Q2.jpg|thumb|Episode one Math question 2]]<br />
(Illegible/need translation)<br />
<br />
<br />
===Question 3===<br />
[[file:Episode One Math Q3.jpg|thumb|Episode one Math question 3]]<br />
(Guessing from the solution shown)<br />
<br />
Find the integer solutions (a,b) with (a^3 + a^2 - 1) - (a - 1)b = 0<br />
<br />
'''Solution'''<br />
[[file:Episode One Math S3.jpg|thumb|Episode one Math solution to question 3]]<br />
<br />
(a^3 + a^2 - 1) - (a - 1)b<br><br />
= (a^3 - 1) + (a^2 - 1) + 1 - (a - 1)b<br><br />
= (a - 1)(a^2 + a + 1) + (a - 1)(a + 1) + 1 - (a - 1)b<br><br />
= (a - 1)(a^2 + 2a - b + 2) + 1 = 0<br />
<br />
Therefore,<br> <br />
(a - 1)(a^2 + 2a - b + 2) = -1<br />
<br />
a, b are integers, therefore the factors a - 1 and a^2 + 2a - b + 2 are integers too. Since the product is -1, one of the factors must be equal to -1, the other to 1.<br />
<br />
If a - 1 = -1 then a = 0 and from a^2 + 2a - b + 2 = 1 we obtain b = 1.<br />
<br />
If a - 1 = 1 then a = 2 and a^2 + 2a - b + 2 = -1 implies that b = 11.<br />
<br />
There are two integer solutions: (a,b) = (0,1) and (a,b) = (2,11).<br />
<br />
<br />
===Question 4===<br />
[[file:Episode One Math Q4.jpg|thumb|Episode one Math question 4]]<br />
Given [[file:Math episode one question four.png]],<br />
<br />
find the sum of F(1)+F(2)+F(3)+...+F(60).<br />
<br />
'''Solution:'''<br />
<br />
Simplying the fraction:<br />
<br />
Reference that for any variables a & b, (a-b) (a+b) = (a-b)^2. Let (2x + 1)^.5 = a and (2x - 1)^.5 = b. Multiply F(x) by 1 or (a-b)/(a-b) or ((2x + 1)^.5 - (2x - 1)^.5) / ((2x + 1)^.5 - (2x - 1)^.5). The denominator becomes:<br />
<br />
((2x + 1)^.5 + (2x - 1)^.5) * ((2x + 1)^.5 - (2x - 1)^.5) <br />
<br />
= (2x + 1) - (2x - 1)<br />
<br />
= 2<br />
<br />
The numerator becomes:<br />
<br />
(4x + (4x^2 - 1)^.5 ) ((2x + 1)^.5 - (2x - 1)^.5) <br />
<br />
=(4x + ((2x)^2 - 1)^.5 ) (a - b) <br />
<br />
=(4x + (2x - 1)^.5 (2x + 1)^.5 ) (a -b) <br />
<br />
=(4x + a b ) (a - b) <br />
<br />
=4x (a - b) + (a^2) b - a (b^2)<br />
<br />
=4x (a - b) + (2x + 1) b - a (2x - 1)<br />
<br />
=4x (a - b) - 2x (a - b) + a + b<br />
<br />
=2x ( a - b) + a + b<br />
<br />
=(2x + 1) a - (2x - 1) b<br />
<br />
=a^3 - b^3<br />
<br />
<br />
Therefore:<br />
<br />
F(x) = (a^3 - b^3 ) / 2<br />
<br />
Note that when x= 1, 2, 3, 4, ..., 60;<br />
<br />
a^3 = 3^1.5, 5^1.5, 7^1.5, 9^1.5, ..., 119^1.5, 121^1.5;<br />
<br />
b^3 = 1^1.5, 3^1.5, 5^1.5, 7^1.5, 9^1.5, ..., 119^1.5;<br />
<br />
When summating over F(x) for x between 1 and 60, observe that the majority of the terms in a^3 and b^3 cancel out, leaving:<br />
<br />
(121^1.5 - 1^1.5)/2 =((11*11)^1.5 - 1)/2 = 665<br />
<br />
Thus, sum of F(x) for x between 1 and 60 is 665. The solution can be generalized as equal to (a(x_max)^3 - b(x_min)^3)/2.<br />
<br />
==Episode Nine==<br />
[[File:Episode 9 Math.png|thumb|Fibonacci sequence question that appeared in Episode 9]]<br />
<br />
This advanced Algebra problem previously appeared in the first [http://www.j3e.info/ojyuken/math/php.php?name=tokyo&v1=0&v2=1992&v3=1&v4=3&y=1992&n=0_3 1992 Tokyo University Entrance Exam]:<br />
<br />
===Problem===<br />
A number sequence {F(n)} that can be defined as F(1) = 1, F(2) = 1, F(n+2) = F(n) + F(n+1) (where n is any natural number) is called the Fibonacci sequence and its general solution is given by, <br />
<br />
[[file:fibonacci.png]]<br />
<br />
[[file:golden ratio.png]]<br />
<br />
Answer the following questions by using this fact if needed:<br />
<br />
<br />
Define a sequence of natural numbers X(n) (where n is any natural number), in which each digit is either 0 or 1, set by the following rules:<br />
<br />
(i) X(1) = 1<br />
<br />
(ii) We define X(n+1) as a natural number, which can be gotten by replacing the digits of X(n) with 1 if the digit is 0, and with 10 if the digit is 1.<br />
<br />
For example, X(1) = 1, X(2) = 10, X(3) = 101, X(4) = 10110, X(5) = 10110101, ... <br />
<br />
<br />
(1) Find A(n), defined as the number of digits of X(n)?<br />
<br />
(2) Find B(n),defined as the numbers of times '01' appears in X(n)?<br />
For example, B(1) = 0, B(2) = 0, B(3) = 1, B(4) = 1, B(5) = 3,...<br />
<br />
<br />
===Solution===<br />
====Part 1====<br />
<br />
Let A(n) equal to the number of digits in X(n), which consists solely of 1s and 0s. Let's suppose x(n) is the number of 0s in X(n) at n iteration (poor choice of variable by the student). Let's suppose y(n) is the number of 1s in X(n) at n iteration.<br />
<br />
<br />
1) Every time a 0 appears, it is replaced with 1 at the next iteration.<br />
<br />
2) Every time a 1 appears, it is replaced with 10 at t he next iteration.<br />
<br />
Therefore, the number of 0s in the next iteration is equal to the number of 1s previously:<br />
<br />
x(n+1)= y(n)<br />
<br />
and the number of 1s in the next iteration is equal to the number of 0s AND the numbers of 1s previously:<br />
<br />
y(n+1)=x(n)+y(n)<br />
<br />
<br />
Prove: x(n) is a fibonacci sequence:<br />
<br />
i) y(n+1) = x(n) + y(n)<br />
<br />
ii) x(n+2) = y(n+1)<br />
<br />
iii) x(n+1) = y(n)<br />
<br />
<br />
complete the substition shows:<br />
<br />
x(n+2) = y(n+1) = x(n) + y(n) = x(n) + x(n+1)<br />
<br />
x(n+2) = x(n+1) + x(n) ie definition of fibonacci sequence<br />
<br />
Since x(n) is a fibonacci sequence, y(n) is also a fibonacci sequence since x(n+1) = y(n). Therefore:<br />
<br />
x(n+2)=x(n+1)+x(n)<br />
<br />
y(n+2)=y(n+1)+y(n)<br />
<br />
x(n+2)+y(n+2)=x(n+1)+y(n+1)+x(n)+y(n)<br />
<br />
Recall: A(n)=x(n)+y(n), therefore<br />
<br />
A(n+2)=A(n+1)+A(n)<br />
<br />
X(1) = 1, X(2) = 10, X(3) = 101, X(4) = 10110, X(5) = 10110101, ...<br />
<br />
therefore: A(1) = 1, A(2) = 2, A(3) = 3, A(4) = 5, A(5) = 8,...<br />
<br />
Recall fibonacci sequence F(n): 1, 1, 2, 3, 5, 8.<br />
Therefore A(n) = F(n+1), or<br />
<br />
[[file:An.png]]<br />
<br />
<br />
====Part 2====<br />
<br />
Let B(n) be the number of times '01' appears in X(n).<br />
<br />
Any two digits in X(n) may be 00, 01, 10, 11<br />
<br />
the corresponding digits in the next iteration X(n+1) will be<br />
<br />
00 -> 11<br />
<br />
01 -> 110<br />
<br />
10 -> 101<br />
<br />
11 -> 1010<br />
<br />
Thus, any two digit sequence in X(n) that begins with 1* will contribute to a 01 sequence in the next iteration. In other words, B(n+1) equals y(n), except when unit digit of X(n) is 1, or:<br />
<br />
B(n+1) = y(n) - odd(X(n))<br />
<br />
y(n) is a fibonacci sequence with starting values: y(1) = 1, y(2) = 1, y(3) = 2<br />
thus y(n) = F(n)<br />
<br />
B(n+1) = F(n) - odd(X(n)) <br />
<br />
B(n) = F(n-1) - odd(X(n-1)), or<br />
<br />
[[file:Bn.png]]<br />
<br />
[[Category:Articles]]<br />
[[Category:Fanmade analysis]]<br />
[[Category:Science]]</div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=Mathematics_of_Madoka_Magica&diff=6198Mathematics of Madoka Magica2011-03-10T22:52:49Z<p>Anon-kun: solution to one problem</p>
<hr />
<div>The cast of Mahou Shoujo Madoka Magica are students of [[Mitakihara Middle School]]. Despite merely a middle school, coursework in Mitakihara can be quite challenging. The following are Math problems that have appeared thus far in the show, and their solutions:<br />
<br />
==Episode One==<br />
[[file:Episode One Algebra.jpg|thumb|Episode one Math question 1]]<br />
<br />
===Question 1===<br />
Any integer divided by 14 will have a remainder between 0 and 13. Given a is has remainder of 6 and b has remainder of 1 when divided by 14, what is the remainder of x when divided by 14, given x is an integer solution to x^2-2ax+b=0?<br />
<br />
'''Solution:'''<br />
<br />
This problem can be solved simply with modular arithmetrics: <br />
a = 6 mod 14;<br />
b = 1 mod 14;<br />
x = r mod 14.<br />
<br />
x^2 - 2ax + b = 0 is equivalent to<br />
<br />
x^2 - 12x + 1 = 0 mod 14<br />
<br />
x(x - 12) = -1 mod 14<br />
<br />
-1 has no proper divisors modulo 14, thus factors x and x-12 must correspond to either 1 or -1 mod 14.<br />
<br />
If x = 1 mod 14, then x-12 = 3 mod 14: false. If x = -1 mod 14, then x - 12 = 1 mod 14: true.<br />
<br />
Therefore, x = -1 mod 14 = 13 mod 14, i.e., the remainder r of x is 13.<br />
<br />
<br />
'''Second Solution'''<br />
Homura used in Episode 1 a basic approach with usual integer arithmetic.<br />
<br />
Let x = 14q + r, a = 14s + 6, b = 14t + 1.<br />
<br />
Substitute into x^2 - 2ax + b = 0 to get after some calculations<br />
<br />
x^2 - 2ax + b = 14c + (r+1)^2 = 0 with c = 14q^2 + 2q - 28qs - 2rs - 12q + t - r<br />
<br />
14 divides 14c and 0, hence it also divides (r+1)^2. This implies that r+1 is divisible by 14.<br />
<br />
The question asks for a remainder r between 0 and 13, so we obtain r = 13.<br />
<br />
See also: http://green-oval.net/cgi-board.pl/a/thread/44824340<br />
<br />
<br />
===Question 2===<br />
[[file:Episode One Math Q2.jpg|thumb|Episode one Math question 2]]<br />
(Illegible/need translation)<br />
<br />
<br />
===Question 3===<br />
[[file:Episode One Math Q3.jpg|thumb|Episode one Math question 3]]<br />
(Guessing from the solution shown)<br />
<br />
Find the integer solutions (a,b) with (a^3 + a^2 - 1) - (a - 1)b = 0<br />
<br />
'''Solution'''<br />
[[file:Episode One Math S3.jpg|thumb|Episode one Math solution to question 3]]<br />
<br />
(a^3 + a^2 - 1) - (a - 1)b<br><br />
= (a^3 - 1) + (a^2 - 1) + 1 - (a - 1)b<br><br />
= (a - 1)(a^2 + a + 1) + (a - 1)(a + 1) + 1 - (a - 1)b<br><br />
= (a - 1)(a^2 + 2a - b + 2) + 1 = 0<br />
<br />
Therefore,<br> <br />
(a - 1)(a^2 + 2a - b + 2) = -1<br />
<br />
a, b are integers, therefore the factors a - 1 and a^2 + 2a - b + 2 are integers too. Since the product is -1, one of the factors must be equal to -1, the other to 1.<br />
<br />
If a - 1 = -1 then a = 0 and from a^2 + 2a - b + 2 = 1 we obtain b = 1.<br />
<br />
If a - 1 = 1 then a = 2 and a^2 + 2a - b + 2 = -1 implies that b = 11.<br />
<br />
There are two integer solutions: (a,b) = (0,1) and (a,b) = (2,11).<br />
<br />
<br />
===Question 4===<br />
[[file:Episode One Math Q4.jpg|thumb|Episode one Math question 4]]<br />
Given [[file:Math episode one question four.png]],<br />
<br />
find the sum of F(1)+F(2)+F(3)+...+F(60).<br />
<br />
'''Solution:'''<br />
<br />
Simplying the fraction:<br />
<br />
Reference that for any variables a & b, (a-b) (a+b) = (a-b)^2. Let (2x + 1)^.5 = a and (2x - 1)^.5 = b. Multiply F(x) by 1 or (a-b)/(a-b) or ((2x + 1)^.5 - (2x - 1)^.5) / ((2x + 1)^.5 - (2x - 1)^.5). The denominator becomes:<br />
<br />
((2x + 1)^.5 + (2x - 1)^.5) * ((2x + 1)^.5 - (2x - 1)^.5) <br />
<br />
= (2x + 1) - (2x - 1)<br />
<br />
= 2<br />
<br />
The numeriator becomes:<br />
<br />
(4x + (4x^2 - 1)^.5 ) ((2x + 1)^.5 - (2x - 1)^.5) <br />
<br />
=(4x + ((2x)^2 - 1)^.5 ) (a - b) <br />
<br />
=(4x + (2x - 1)^.5 (2x + 1)^.5 ) (a -b) <br />
<br />
=(4x + a b ) (a - b) <br />
<br />
=4x (a - b) + (a^2) b - a (b^2)<br />
<br />
=4x (a - b) + (2x + 1) b - a (2x - 1)<br />
<br />
=4x (a - b) - 2x (a - b) + a + b<br />
<br />
=2x ( a - b) + a + b<br />
<br />
=(2x + 1) a - (2x - 1) b<br />
<br />
=a^3 - b^3<br />
<br />
<br />
Therefore:<br />
<br />
F(x) = (a^3 - b^3 ) / 2<br />
<br />
Note that when x= 1, 2, 3, 4, ..., 60;<br />
<br />
a^3 = 3^1.5, 5^1.5, 7^1.5, 9^1.5, ..., 119^1.5, 121^1.5;<br />
<br />
b^3 = 1^1.5, 3^1.5, 5^1.5, 7^1.5, 9^1.5, ..., 119^1.5;<br />
<br />
When summation over F(x) for x between 1 and 60, observe that majority of the terms in a^3 and b^3 cancel out, leaving:<br />
<br />
(121^1.5 - 1^1.5)/2 =((11*11)^1.5 - 1)/2 = 665<br />
<br />
Thus, sum of F(x) for x between 1 and 60 is 665. The solution can be generalized as equal to a(x_max)^3 - b(x_min)^3.<br />
<br />
==Episode Nine==<br />
[[File:Episode 9 Math.png|thumb|Fibonacci sequence question that appeared in Episode 9]]<br />
<br />
This advanced Algebra problem previously appeared in the first [http://www.j3e.info/ojyuken/math/php.php?name=tokyo&v1=0&v2=1992&v3=1&v4=3&y=1992&n=0_3 1992 Tokyo University Entrance Exam]:<br />
<br />
===Problem===<br />
A number sequence {F(n)} that can be defined as F(1) = 1, F(2) = 1, F(n+2) = F(n) + F(n+1) (where n is any natural number) is called the Fibonacci sequence and its general solution is given by, <br />
<br />
[[file:fibonacci.png]]<br />
<br />
[[file:golden ratio.png]]<br />
<br />
Answer the following questions by using this fact if needed:<br />
<br />
<br />
Define a sequence of natural numbers X(n) (where n is any natural number), in which each digit is either 0 or 1, set by the following rules:<br />
<br />
(i) X(1) = 1<br />
<br />
(ii) We define X(n+1) as a natural number, which can be gotten by replacing the digits of X(n) with 1 if the digit is 0, and with 10 if the digit is 1.<br />
<br />
For example, X(1) = 1, X(2) = 10, X(3) = 101, X(4) = 10110, X(5) = 10110101, ... <br />
<br />
<br />
(1) Find A(n), defined as the number of digits of X(n)?<br />
<br />
(2) Find B(n),defined as the numbers of times '01' appears in X(n)?<br />
For example, B(1) = 0, B(2) = 0, B(3) = 1, B(4) = 1, B(5) = 3,...<br />
<br />
<br />
===Solution===<br />
====Part 1====<br />
<br />
Let A(n) equal to the number of digits in X(n), which consists solely of 1s and 0s. Let's suppose x(n) is the number of 0s in X(n) at n iteration (poor choice of variable by the student). Let's suppose y(n) is the number of 1s in X(n) at n iteration.<br />
<br />
<br />
1) Every time a 0 appears, it is replaced with 1 at the next iteration.<br />
<br />
2) Every time a 1 appears, it is replaced with 10 at t he next iteration.<br />
<br />
Therefore, the number of 0s in the next iteration is equal to the number of 1s previously:<br />
<br />
x(n+1)= y(n)<br />
<br />
and the number of 1s in the next iteration is equal to the number of 0s AND the numbers of 1s previously:<br />
<br />
y(n+1)=x(n)+y(n)<br />
<br />
<br />
Prove: x(n) is a fibonacci sequence:<br />
<br />
i) y(n+1) = x(n) + y(n)<br />
<br />
ii) x(n+2) = y(n+1)<br />
<br />
iii) x(n+1) = y(n)<br />
<br />
<br />
complete the substition shows:<br />
<br />
x(n+2) = y(n+1) = x(n) + y(n) = x(n) + x(n+1)<br />
<br />
x(n+2) = x(n+1) + x(n) ie definition of fibonacci sequence<br />
<br />
Since x(n) is a fibonacci sequence, y(n) is also a fibonacci sequence since x(n+1) = y(n). Therefore:<br />
<br />
x(n+2)=x(n+1)+x(n)<br />
<br />
y(n+2)=y(n+1)+y(n)<br />
<br />
x(n+2)+y(n+2)=x(n+1)+y(n+1)+x(n)+y(n)<br />
<br />
Recall: A(n)=x(n)+y(n), therefore<br />
<br />
A(n+2)=A(n+1)+A(n)<br />
<br />
X(1) = 1, X(2) = 10, X(3) = 101, X(4) = 10110, X(5) = 10110101, ...<br />
<br />
therefore: A(1) = 1, A(2) = 2, A(3) = 3, A(4) = 5, A(5) = 8,...<br />
<br />
Recall fibonacci sequence F(n): 1, 1, 2, 3, 5, 8.<br />
Therefore A(n) = F(n+1), or<br />
<br />
[[file:An.png]]<br />
<br />
<br />
====Part 2====<br />
<br />
Let B(n) be the number of times '01' appears in X(n).<br />
<br />
Any two digits in X(n) may be 00, 01, 10, 11<br />
<br />
the corresponding digits in the next iteration X(n+1) will be<br />
<br />
00 -> 11<br />
<br />
01 -> 110<br />
<br />
10 -> 101<br />
<br />
11 -> 1010<br />
<br />
Thus, any two digit sequence in X(n) that begins with 1* will contribute to a 01 sequence in the next iteration. In other words, B(n+1) equals y(n), except when unit digit of X(n) is 1, or:<br />
<br />
B(n+1) = y(n) - odd(X(n))<br />
<br />
y(n) is a fibonacci sequence with starting values: y(1) = 1, y(2) = 1, y(3) = 2<br />
thus y(n) = F(n)<br />
<br />
B(n+1) = F(n) - odd(X(n)) <br />
<br />
B(n) = F(n-1) - odd(X(n-1)), or<br />
<br />
[[file:Bn.png]]<br />
<br />
[[Category:Articles]]<br />
[[Category:Fanmade analysis]]<br />
[[Category:Science]]</div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=File:Episode_One_Math_S3.jpg&diff=6196File:Episode One Math S3.jpg2011-03-10T22:50:45Z<p>Anon-kun: Solution to one of the problems.
Note that the exponent b is wrong, it must be a factor instead.</p>
<hr />
<div>Solution to one of the problems.<br />
<br />
Note that the exponent b is wrong, it must be a factor instead.</div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=Main_Page&diff=6119Main Page2011-03-10T16:32:16Z<p>Anon-kun: episode 11</p>
<hr />
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==External Links==<br />
===Official websites===<br />
*[http://www.madoka-magica.com/ ''Puella Magi Madoka Magica'' official website] (Japanese)<br />
*[http://twitter.com/madoka_magica ''Puella Magi Madoka Magica'' official Twitter] (Japanese)<br />
===Other resources===<br />
*[http://en.wikipedia.org/wiki/Puella_Magi_Madoka_Magica ''Puella Magi Madoka Magica'' at Wikipedia]<br />
*[http://tvtropes.org/pmwiki/pmwiki.php/Main/PuellaMagiMadokaMagica ''Puella Magi Madoka Magica'' at TVTropes.org]<br />
*[http://www22.atwiki.jp/madoka-magica/ ''Puella Magi Madoka Magica'' at @wiki] (Japanese)</div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=Madoka_Magica_Products&diff=6113Madoka Magica Products2011-03-10T13:28:35Z<p>Anon-kun: not on homepage, but in amazon</p>
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<br />
{| class="wikitable sortable collapsible" style="width: 100%;"<br />
! Product type<br />
! Title<br />
! Release date<br />
! Price<br />
|-<br />
| CD<br />
| [http://www.amazon.co.jp/dp/B004DGD400/ ClariS: Connect Limited Edition]<br />
| 2011-02-02<br />
| ¥ 1,575<br />
|-<br />
| CD<br />
| [http://www.amazon.co.jp/dp/B004DGD4FU/ ClariS: Connect Anime Edition]<br />
| 2011-02-02<br />
| ¥ 1,300<br />
|-<br />
| CD<br />
| [http://www.amazon.co.jp/dp/B004DGD4AA/ ClariS: Connect Regular Edition]<br />
| 2011-02-02<br />
| ¥ 1,223<br />
|-<br />
| Manga<br />
| [http://www.amazon.co.jp/dp/4832279904/ Puella Magi Madoka☆Magica Vol.1]<br />
| 2011-02-12<br />
| ¥ 690<br />
|-<br />
| CD<br />
| [http://www.amazon.co.jp/dp/B004DGD45A/ Kalafina: Magia Limited Edition]<br />
| 2011-02-16<br />
| ¥ 1,575<br />
|-<br />
| CD<br />
| [http://www.amazon.co.jp/dp/B004DGD4QE/ Kalafina: Magia Anime Edition]<br />
| 2011-02-16<br />
| ¥ 1,300<br />
|-<br />
| CD<br />
| [http://www.amazon.co.jp/dp/B004DGD4NM/ Kalafina: Magia Regular Edition]<br />
| 2011-02-16<br />
| ¥ 1,223<br />
|-<br />
| Manga<br />
| [http://www.amazon.co.jp/dp/483224003X/ Puella Magi Madoka☆Magica Vol.2]<br />
| 2011-03-12<br />
| ¥ 690<br />
|-<br />
| Blu-ray<br />
| [http://www.amazon.co.jp/dp/B004INGZAE/ Puella Magi Madoka☆Magica Vol.1 Limited Edition]<br />
| 2011-03-30<br />
| ¥ 7,350<br />
|-<br />
| DVD<br />
| [http://www.amazon.co.jp/dp/B004INGZCC/ Puella Magi Madoka☆Magica Vol.1 Limited Edition]<br />
| 2011-03-30<br />
| ¥ 6,300<br />
|-<br />
| DVD<br />
| [http://www.amazon.co.jp/dp/B004INGZLS/ Puella Magi Madoka☆Magica Vol.1 Regular Edition]<br />
| 2011-03-30<br />
| ¥ 5,250<br />
|-<br />
| Manga<br />
| [http://www.amazon.co.jp/dp/4832240145/ Puella Magi Madoka☆Magica Vol.3 (End)]<br />
| 2011-04-12<br />
| ¥ 690<br />
|-<br />
| Blu-ray<br />
| [http://www.amazon.co.jp/dp/B004L7A7WO/ Puella Magi Madoka☆Magica Vol.2 Limited Edition]<br />
| 2011-04-27<br />
| ¥ 7,350<br />
|-<br />
| DVD<br />
| [http://www.amazon.co.jp/dp/B004L7A7Q0/ Puella Magi Madoka☆Magica Vol.2 Limited Edition]<br />
| 2011-04-27<br />
| ¥ 6,300<br />
|-<br />
| DVD<br />
| [http://www.amazon.co.jp/dp/B004L7A7UG/ Puella Magi Madoka☆Magica Vol.2 Regular Edition]<br />
| 2011-04-27<br />
| ¥ 5,250<br />
|-<br />
| Blu-ray<br />
| [http://www.amazon.co.jp/dp/B004MMFTMQ/ Puella Magi Madoka☆Magica Vol.3 Limited Edition]<br />
| 2011-05-25<br />
| ¥ 7,350<br />
|-<br />
| DVD<br />
| [http://www.amazon.co.jp/dp/B004MMFTN0/ Puella Magi Madoka☆Magica Vol.3 Limited Edition]<br />
| 2011-05-25<br />
| ¥ 6,300<br />
|-<br />
| DVD<br />
| [http://www.amazon.co.jp/dp/B004MMFTNA/ Puella Magi Madoka☆Magica Vol.3 Regular Edition]<br />
| 2011-05-25<br />
| ¥ 5,250<br />
|-<br />
| Blu-ray<br />
| [http://www.amazon.co.jp/dp/B004P4RTKQ/ Puella Magi Madoka☆Magica Vol.4 Limited Edition]<br />
| 2011-06-22<br />
| ¥ 7,350<br />
|-<br />
| DVD<br />
| [http://www.amazon.co.jp/dp/B004P4RTLK/ Puella Magi Madoka☆Magica Vol.4 Limited Edition]<br />
| 2011-06-22<br />
| ¥ 6,300<br />
|-<br />
| DVD<br />
| [http://www.amazon.co.jp/dp/B004P4RTL0/ Puella Magi Madoka☆Magica Vol.4 Regular Edition]<br />
| 2011-06-22<br />
| ¥ 5,250<br />
|-<br />
| Blu-ray<br />
| [http://www.amazon.co.jp/dp/B004RCJJFE/ Puella Magi Madoka☆Magica Vol.5 Limited Edition]<br />
| 2011-07-27<br />
| ¥ 7,350<br />
|-<br />
| DVD<br />
| [http://www.amazon.co.jp/dp/B004RCJJ54/ Puella Magi Madoka☆Magica Vol.5 Limited Edition]<br />
| 2011-07-27<br />
| ¥ 6,300<br />
|-<br />
| DVD<br />
| [http://www.amazon.co.jp/dp/B004RCJJWC/ Puella Magi Madoka☆Magica Vol.5 Regular Edition]<br />
| 2011-07-27<br />
| ¥ 5,250<br />
|-<br />
|}<br />
<br />
==Blu-Ray Discs==<br />
{| class="wikitable" style="width: 100%; vertical-align: top;"<br />
! style="width: 300px;" | Cover<br />
! Description<br />
|-<br />
|[[File:Disc 1.jpg|300px]]<br />
| style="vertical-align:top; padding: 10px;" |<br />
;[http://www.amazon.co.jp/dp/B004INGZAE Puella Magi Madoka☆Magica Vol.1 Blu-ray]<br />
:'''Release date''': March 30, 2011<br />
:'''Price''': ¥ 7,350 (~90 USD)<br />
:'''Running time''': 60 minutes<br />
:'''Package contents''':<br />
:*Blu-ray disc (episode 1, 2)<br />
:*Special CD<br />
:*#Urobuchi Gen-Supervised Drama CD<br />
:*#Kaname Madoka Character Song (ED theme for episode 1, 2)<br />
:*Tri-fold clear case<br />
:*Package illustration by Kishida Takahiro<br />
:*Special booklet with Aoki Ume's 4koma Manga<br />
:'''Extra contents''':<br />
:#Episode 1 audio commentary (Yūki Aoi, Saitō Chiwa, Kitamura Eri)<br />
:#Episode 2 audio commentary (Yūki Aoi, Saitō Chiwa, Aoki Ume)<br />
:#Unbroadcasted original ending theme version<br />
|-<br />
|[[File:Disc 2.jpg|300px]]<br />
| style="vertical-align:top; padding: 10px;"|<br />
;[http://www.amazon.co.jp/dp/B004L7A7WO/ Puella Magi Madoka☆Magica Vol.2 Blu-ray]<br />
:'''Release date''': April 27, 2011<br />
:'''Price''': ¥ 7,350 (~90 USD)<br />
:'''Running time''': 48 minutes<br />
:'''Package contents''':<br />
:*Blu-ray disc (episode 3, 4)<br />
:*Original soundtrack CD Volume 1<br />
:*Tri-fold clear case<br />
:*Package illustration by Kishida Takahiro<br />
:*Special booklet with Aoki Ume's 4koma Manga<br />
:'''Extra contents''':<br />
:#Audio commentary<br />
|-<br />
| style="text-align: center;"| NO IMAGE<br />
| style="vertical-align:top; padding: 10px;"|<br />
;[http://www.amazon.co.jp/dp/B004MMFTMQ/ Puella Magi Madoka☆Magica Vol.3 Blu-ray]<br />
:'''Release date''': May 25, 2011<br />
:'''Price''': ¥ 7,350 (~90 USD)<br />
:'''Running time''': 48 minutes<br />
:'''Package contents''':<br />
:*Blu-ray disc (episode 5, 6)<br />
:*Special CD<br />
:*Tri-fold clear case<br />
:*Package illustration by Kishida Takahiro<br />
:*Special booklet with Aoki Ume's 4koma Manga<br />
:'''Extra contents''':<br />
:#Audio commentary<br />
|-<br />
| style="text-align: center;"| NO IMAGE<br />
| style="vertical-align:top; padding: 10px;"|<br />
;[http://www.amazon.co.jp/dp/B004P4RTKQ/ Puella Magi Madoka☆Magica Vol.4 Blu-ray]<br />
:'''Release date''': June 22, 2011<br />
:'''Price''': ¥ 7,350 (~90 USD)<br />
:'''Running time''': 48 minutes<br />
:'''Package contents''':<br />
:*Blu-ray disc (episode 7, 8)<br />
:*Special CD<br />
:*Tri-fold clear case<br />
:*Package illustration by Kishida Takahiro<br />
:*Special booklet with Aoki Ume's 4koma Manga<br />
:'''Extra contents''':<br />
:#Audio commentary<br />
|}<br />
<br />
==Singles==<br />
{| class="wikitable" style="width: 100%; vertical-align: top;"<br />
! style="width: 300px;" | Cover<br />
! Description<br />
|-<br />
|[[File:Connect Anime Cover.jpg|300px]]<br />
| style="vertical-align:top; padding: 10px;" |<br />
;[http://www.amazon.co.jp/dp/B004DGD4FU ClariS: Connect (Anime Edition)]<br />
:'''Release date''': February 2, 2011<br />
:'''Price''': ¥ 1,300 (~15 USD)<br />
:'''Tracklist''':<br />
:#コネクト (Connect)<br />
:#Dreamin’<br />
:#キミとふたり (Kimi to futari)<br />
:#コネクト (Connect) -TV MIX-<br />
:'''Package contents''':<br />
:*Audio CD<br />
:*Anime character design package illustration<br />
|-<br />
|[[File:Connect OP Cover.jpg|300px]]<br />
| style="vertical-align:top; padding: 10px;" |<br />
;[http://www.amazon.co.jp/dp/B004DGD400 ClariS: Connect (Limited Edition)]<br />
:'''Release date''': February 2, 2011<br />
:'''Price''': ¥ 1,575 (~19 USD)<br />
:'''Tracklist''':<br />
:#コネクト (Connect)<br />
:#Dreamin’<br />
:#キミとふたり (Kimi to futari)<br />
:#コネクト (Connect) -Instrumental-<br />
:'''Package contents''':<br />
:*Audio CD<br />
:*Music video DVD<br />
:*Anime character design illustration sticker<br />
|-<br />
! Cover<br />
! Description<br />
|-<br />
|[[File:Magia Anime Cover.jpg|300px]]<br />
| style="vertical-align:top; padding: 10px;" |<br />
;[http://www.amazon.co.jp/dp/B004DGD4QE Kalafina: Magia ED (Anime Edition)]<br />
:'''Release date''': February 16, 2011<br />
:'''Price''': ¥ 1,300 (~15 USD)<br />
:'''Tracklist''':<br />
:#Magia<br />
:#Magia (magic mix)<br />
:#Magia (TV Version)<br />
:#Magia (instrumental)<br />
:'''Package contents''':<br />
:*Audio CD<br />
:*Digipak case<br />
:*Anime character design package illustration<br />
|-<br />
|[[File:Magia ED Cover.jpg|300px]]<br />
| style="vertical-align:top; padding: 10px;" |<br />
;[http://www.amazon.co.jp/dp/B004DGD45A Kalafina: Magia ED (Limited Edition)]<br />
:'''Release date''': February 16, 2011<br />
:'''Price''': ¥ 1,575 (~19 USD)<br />
:'''Tracklist''':<br />
:#Magia<br />
:#snow falling<br />
:#Magia (instrumental)<br />
:'''Package contents''':<br />
:*Audio CD<br />
:*Music video DVD<br />
|}<br />
<br />
==Manga==<br />
{| class="wikitable" style="width: 100%; vertical-align: top;"<br />
! style="width: 300px;" | Cover<br />
! Description<br />
|-<br />
| [[File:Manga Vol.1 Cover.jpg|300px]]<br />
| style="vertical-align:top; padding: 10px;" |<br />
;[http://www.amazon.co.jp/dp/4832279904 Puella Magi Madoka☆Magica Vol.1]<br />
:'''Release date''': February 12, 2011<br />
:'''Price''': ¥ 690 (~8 USD)<br />
:'''Artist''': Hanokage<br />
:'''Publisher''': Houbunsha<br />
|-<br />
| [[File:Manga Vol.2 Cover.jpg|300px]]<br />
| style="vertical-align:top; padding: 10px;" |<br />
;[http://www.amazon.co.jp/dp/483224003X Puella Magi Madoka☆Magica Vol.2]<br />
:'''Release date''': March 12, 2011<br />
:'''Price''': ¥ 690 (~8 USD)<br />
:'''Artist''': Hanokage<br />
:'''Publisher''': Houbunsha<br />
|-<br />
| style="text-align: center;"| NO IMAGE<br />
| style="vertical-align:top; padding: 10px;" |<br />
;[http://www.amazon.co.jp/dp/4832240145 Puella Magi Madoka☆Magica Vol.3 (End)]<br />
:'''Release date''': April 12, 2011<br />
:'''Price''': ¥ 690 (~8 USD)<br />
:'''Artist''': Hanokage<br />
:'''Publisher''': Houbunsha<br />
|}<br />
<br />
== External links ==<br />
* [http://www.madoka-magica.com/bddvd/ Official BD, DVD, and CD page] (Japanese)<br />
* [http://www.dokidokivisual.com/madokamagica/ Official manga page] (Japanese)</div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=Madoka_Magica_Episode_10:_I_Won%27t_Depend_on_Anyone_Anymore&diff=6112Madoka Magica Episode 10: I Won't Depend on Anyone Anymore2011-03-10T13:14:15Z<p>Anon-kun: break</p>
<hr />
<div>{{Infobox_episode<br />
|name = I Won't Trust Anyone Anymore<br />
|jname = もう誰にも頼らない<br />
|date = 10 March 2011<br />
|script = [[Gen Urobuchi]]<br />
|storyboard = Shinsaku Sasaki<br />
|director = Yuki Yase<br />
|adirector = Yoshiaki Itou<br />Kazuya Shiotsuki<br />
}}<br />
<br />
{{-}}<br />
<br />
{{Episode navigation<br />
|title = I Won't Trust Anyone Anymore<br />
|previous_episode = 9<br />
|previous_title = I'll Never Allow That<br />
|next_episode = 11<br />
|next_title = <br />
}}<br />
<br />
[[Category:Episodes]]</div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=Glossary&diff=6023Glossary2011-03-09T18:47:07Z<p>Anon-kun: </p>
<hr />
<div>__NOTOC__<br />
<!-- When editing, please note that the description fields in the glossary templates are only for translations of official site definitions. Anything else goes *OUTSIDE* the template.--><br />
===Barrier===<br />
{{Glossary<br />
|term = {{Nihongo|Barrier|結界|Kekkai}}<br />
|description = An otherworld used by [[Witch]]es to conceal themselves. Should a normal human wander in by accident, it is impossible for them to escape.<br />
|image = File:Witch ep2 entering.jpg<br />
|image_desc = Madoka and friends entering [[Gertrud|Gertrud's]] barrier.<br />
}}<br />
<!-- When editing, please note that the description fields in the glossary templates are only for translations of official site definitions. Anything else goes *OUTSIDE* the template.--><br />
<br />
===Familiar===<br />
{{Glossary<br />
|term = {{Nihongo|Familiar|使い魔|Tsukaima}}<br />
|description = Lesser monsters divided out of the essence of a [[Witch]]. Though normally assigned the task of maintaining and defending their parent-witch's [[barrier]], it seems that they can eventually gain independence and stray to form their own barriers. Further growing, they take on the form of their parent.<br />
|image = File:Ulla_screen.png<br />
|image_desc = [[Ulla]], [[Suleika]]'s familiar in his own [[barrier]].<br />
}}<br />
<!-- When editing, please note that the description fields in the glossary templates are only for translations of official site definitions. Anything else goes *OUTSIDE* the template.--><br />
<br />
===Grief Seed===<br />
{{Glossary|<br />
|term = {{Nihongo|Grief Seed|グリーフシード}}<br />
|description = A [[Witch]]'s egg, which occasionally appears after a Witch is defeated. Puella Magi require these items to restore their depleted magical energies.<br />
|image = File:Grief seed.png<br />
|image_desc = [[Gertrud]]'s Grief Seed.<br />
}}<br />
<!-- When editing, please note that the description fields in the glossary templates are only for translations of official site definitions. Anything else goes *OUTSIDE* the template.--><br />
The odd center of weight in Grief Seeds causes them to balance on their bottom pins. It may indicate some sort of gyroscopic self-stabilization.<br />
<br />
<span id="Magical Girl"></span><!-- [[Magical girl]] redirects here --><br />
===Puella Magi===<br />
{{Glossary<br />
|term = {{Nihongo|Puella Magi|魔法少女|Mahō shōjo|lit. Magical Girl}}<br />
|description = Girls who have formed a contract with [[Kyuubey]], and in exchange gained the power of magic. Existences entrusted with the task of combating [[Witch]]es.<br />
|image = <br />
|image_desc = <br />
}}<br />
<!-- When editing, please note that the description fields in the glossary templates are only for translations of official site definitions. Anything else goes *OUTSIDE* the template.--><br />
<br />
===Soul Gem===<br />
{{Glossary<br />
|term = {{Nihongo|Soul Gem|ソウルジェム}}<br />
|description = A gem birthed from a contract with [[Kyuubey]]. Proof that one is a Puella Magi; the source of her magical power.<br><br />
|image = File:Soul gem.png<br />
|image_desc = [[Mami Tomoe|Mami]]'s Soul gem<br />
}}<br />
<!-- When editing, please note that the description fields in the glossary templates are only for translations of official site definitions. Anything else goes *OUTSIDE* the template.--><br />
[[Episode 6]] reveals {{Spoiler|that Soul Gems contain the magical girl's soul, while the body is merely a husk that is controlled by the Gem.}}<br />
<br />
===Walpurgis Night===<br />
{{Glossary<br />
|term = {{Nihongo|Walpurgis Night|ワルプルギスの夜}}<br />
|description = A gargantuan witch impossible for a single magical girl to handle on her own.<br><br />
|image = <br />
|image_desc =<br />
}}<br />
<!-- When editing, please note that the description fields in the glossary templates are only for translations of official site definitions. Anything else goes *OUTSIDE* the template.--><br />
<br />
===Witch===<br />
{{Glossary<br />
|term = {{Nihongo|Witch|魔女|Majo}}<br />
|description = The negative impulse that is called anxiety or suspicion, or excessive wrath or hatred = existences that spread the seeds of disaster across the world. Normally, they conceal themselves behind [[#Barrier|barriers]].<br><br />
|image = <br />
|image_desc =<br />
}}<br />
<!-- When editing, please note that the description fields in the glossary templates are only for translations of official site definitions. Anything else goes *OUTSIDE* the template.--><br />
[[Episode 8]] reveals {{Spoiler|that Witches are in fact the "grown" forms of Puella Magi, as 魔女, Majo, are evolved forms of 魔法少女, Mahou Shoujo.}}<br />
<br />
===Witch's Kiss===<br />
{{Glossary<br />
|term = {{Nihongo|Witch's Kiss|魔女の口づけ|Majo no kuchizuke}}<br />
|description = A mark that appears on a human targeted by a [[Witch]].<br />
|image = File:Witch's kiss.png<br />
|image_desc = Witch's kiss on a victim's body<br />
}}<br />
<!-- When editing, please note that the description fields in the glossary templates are only for translations of official site definitions. Anything else goes *OUTSIDE* the template.--><br />
<br />
==External links==<br />
*[http://www.madoka-magica.com/special/keyword/junko.html Glossary page on the official website]</div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=Oktavia_von_Seckendorff&diff=6019Oktavia von Seckendorff2011-03-09T18:43:37Z<p>Anon-kun: musical runes needed. familiar appeared in transformation sequence in Ep 8</p>
<hr />
<div>__NOTOC__<br />
'''Oktavia von Seckendorff''' ({{Runes|OKTAVIA VON SECKENDORFF}}) is a [[witch]] that was [[Sayaka Miki]]. She briefly appears at the end of [[Episode 8]] and battles with [[Kyouko Sakura]] in [[Episode 9]]. Her Labyrinth is filled with calls for attention -- probably to [[Kyousuke Kamijou]]. Initially, the outer areas of her Labyrinth are filled with train tracks, perhaps a reflection of the train station where she became a witch. Later on, the entrance area becomes a long corridor filled with concert posters. The center of the Labyrinth appears to be a distorted concert hall, where the seats fill the domed ceiling. Her weapons appear to be manifested wheels, and the concert hall is populated with a spectral string orchestra. <br />
<br />
{{Card<br />
|name = Oktavia<br />
|runes = <!-- OKTAVIA --><br />
|image = File:Card_Oktavia.png<br />
|japanese_desc = 人魚の魔女。その性質は恋慕。 在りし日の感動を夢見ながらコンサートホールごと移動する魔女。 回る運命は思い出だけを乗せてもう未来へは転がらない。 もう何も届かない。もう何も知ることなどない。 今はただ手下達の演奏を邪魔する存在を許さない。<br />
|english_desc = The witch of Mermaid. Her nature is love attachment. She moves around together with concert hall while dreaming her good days. Her rolling fate is only her memories and can no longer move to the future. Nothing is within her reach. She can no longer knows anything. She can't forgive anyone who interrupt her familiar's music.<br />
|type = Witch of Mermaid<br />
|nature = Love attachment<br />
|episodes = [[Episode 8]], [[Episode 9]]<br />
}}<br />
<br />
== Minions ==<br />
{{Card<br />
|name = Holger<br />
|runes = <!-- HOLGER --><br />
|image = File:Card_Holger.png<br />
|japanese_desc = 人魚の魔女の手下。その役割は演奏。 魔女のために音楽を奏で続ける虚ろな楽団。 その音を長く聞き続けた者は魂を抜き取られてしまう。 この楽団は魔女のためだけに存在し、魔女には楽団が全て。<br />
|english_desc = A minion of the witch of mermaid. A hollow orchestra that play music for the witch. Anyone who listen to their music too long will have their soul taken away. The orchestra is everything for the witch, and the witch is everything for the orchestra.<br />
|type = Minion of the Witch of Mermaid<br />
|task = Concert<br />
|episodes = [[Episode 8]], [[Episode 9]]<br />
}}<br />
<br />
==Observations and Facts==<br />
*Sayaka's witch name is Oktavia von Seckendorff. Karl Siegmund von Seckendorff was a german poet who wrote a book called "The Wheel of Fate" (Das Rad des Schicksals), and could explain Sayaka's wheel attack.<br />
*Oktavia is the german form of the latin name ''Octavia'' which means "the eigth" (child, month or musical interval) or "from the family Octavia". It is the direct root of [[Wikipedia:Octave|octave]] in English (an octave being the eight interval of a scale). The name was most likely chosen because of the link between Sayaka and music.<br />
<br />
== Speculations ==<br />
*The runic font doesn't seem to exist in the "real world". Ergo, the strange posters that bear runic text in [[Episode 8]] may have been a "telltale sign" of the beginning of Sayaka's Labyrinth manifestation.<br />
*The pink ribbon is a sign of attraction, since Madoka's mother mentioned in ep 1 that wearing a pink ribbon would attract attention.<br />
**It may though just be a transformed ribbon of Sayaka's school uniform.<br />
*Similarities with the myth of Orion:<br />
**"Orion was the son of the sea-god Poseidon" - the reason for Sayaka's (Octavia's) half-fish form.<br />
**"Odysseus sees him hunting in the underworld with a bronze club" - first time we've seen Sayaka in a barrier (underworld), she carried a golden (bronze) bat.<br />
**"he hunted with the goddess Artemis and her mother Leto" - she hunted with Mami and Madoka.<br />
**"threatened to kill every beast on Earth" - kill every witch on Earth.<br />
**"Mother Earth objected and sent a giant scorpion to kill Orion" - Kyouko appears, Kyouko's spear could refer to the scorpions spear (tail).<br />
**"The creature succeeded, and after his death, the goddesses asked Zeus to place Orion among the constellations. Zeus consented and, as a memorial to the hero's death, added the Scorpion to the heavens as well" - After they died Sayaka's grief seed and Kyouko's Soul Gem turned into raw energy used to fuel the universe (as explained by Kyuubey in episode 9 ) they literally among the stars now.<br />
*The witch's mermaid tail may be a reference to both Hans Christian Andersen's [[wikipedia:The_Little_Mermaid|The Little Mermaid]] and the story of [[wikipedia:Melusine|Melusine]]. Goethe wrote a version of Melusine titled Die Neue Melusine or [http://books.google.com/books?id=J8Pb0oeVAaIC&pg=PA347&dq=the+new+melusina&hl=en&ei=u09zTc3mA4K4twfAyv35Dg&sa=X&oi=book_result&ct=result&resnum=1&ved=0CCgQ6AEwAA#v=onepage&q=the%20new%20melusina&f=false The New Melusina].<br />
<br />
*The wheels used in Oktavia's attack may also symbolize the Wheel of Fortune from tarot, whose popular interpretations include "possibilities, opportunities, new developments, sudden changes", which fit Sayaka's life from when she encountered Kyubey to when she became a Witch.<br />
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*The structure of the final room may be a reflection of Miki Sayaka's subconscious. <br />
*The shade which resembles Kyousuke is hidden under the crowded concert hall, suggesting that Sayaka does not want to admit her ambitions of owning Kyousuke's love.<br />
== Gallery ==<br />
<gallery><br />
File:WitchSayaka-screenshots.jpg<br />
File:Sayaka_witch.PNG<br />
File:Sayaka_witch4.PNG<br />
File:Ep9_poster_runes.jpg<br />
<!-- File:Ep9_poster_back.jpg --><br />
File:SayakaWitch wants attention.jpg|The subtle messages show her eagerness for attention.<br />
File:Kyousuke labyrinth p.jpg|A violinist that resembles Kyousuke.<br />
File:Melusine information.jpg|Information regarding the Melusine<br />
</gallery><br />
<br />
[[Category:Witches]]</div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=Holger&diff=5999Holger2011-03-09T18:32:29Z<p>Anon-kun: Redirected page to Oktavia von Seckendorff#Holger</p>
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<div>#REDIRECT [[Oktavia von Seckendorff#Holger]]<br />
<br />
[[Category:Minions]]</div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=Characters&diff=5998Characters2011-03-09T18:31:15Z<p>Anon-kun: </p>
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<div>[[File:Characters.png|440px|right]]<br />
*[[Madoka Kaname]]<br />
*[[Homura Akemi]]<br />
*[[Sayaka Miki]]<br />
*[[Mami Tomoe]]<br />
*[[Kyouko Sakura]]<br />
*[[Kyuubey]]<br />
<br />
== Witches ==<br />
*[[Witch from the dream in episode one]]<br />
*[[Gertrud]]<br />
**[[Gertrud#Anthony|Anthony]]<br />
**[[Gertrud#Adelbert|Adelbert]]<br />
*[[Suleika]]<br />
**[[Suleika#Ulla|Ulla]]<br />
*[[Charlotte]]<br />
**[[Charlotte#Pyotr|Pyotr]]<br />
*[[Elly]] (Kirsten)<br />
** [[Elly#Daniyyel & Jennifer|Daniyyel and Jennifer]]<br />
*[[Albertine]]<br />
** [[Albertine#Anja|Anja]]<br />
*[[Elsa Maria]]<br />
**[[Elsa Maria#Sebastians|Sebastians]]<br />
*[[Gisela]]<br />
**[[Gisela#Dora|Dora]]<br />
*[[Uhrmann]]<br />
**[[Uhrmann#Bartels|Bartels]]<br />
*[[Oktavia von Seckendorff]]<br />
**[[Oktavia von Seckendorff#Holger|Holger]]<br />
*[[Speculah:Walpurgis Night|Walpurgis Night]]<br />
<br />
==Support Characters==<br />
*Madoka's Family<br />
**[[Junko Kaname]]<br />
**[[Tomohisa Kaname]]<br />
**[[Tatsuya Kaname]]<br />
*[[Hitomi Shizuki]]<br />
*[[Kazuko Saotome]]<br />
*[[Kyousuke Kamijou]]<br />
*[[Inu Daruma Brothers]]<br />
<br />
[[Category:Characters| ]]</div>Anon-kunhttps://wiki.puella-magi.net/w/index.php?title=File:Card_Holger.png&diff=5997File:Card Holger.png2011-03-09T18:30:20Z<p>Anon-kun: </p>
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<div></div>Anon-kun