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<div>You know, I don't like red links. So I created this user page. While you're at it, why not go to [[Mathematics of Madoka Magica]]?</div>LanceLongFeihttps://wiki.puella-magi.net/w/index.php?title=User:LanceLongFei&diff=106032User:LanceLongFei2021-02-01T01:07:05Z<p>LanceLongFei: The CREATION of my ULTIMATE user page!</p>
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<div>You know, I don't link red links. So I created this user page. While you're at it, why not go to [[Mathematics of Madoka Magica]]?</div>LanceLongFeihttps://wiki.puella-magi.net/w/index.php?title=Mathematics_of_Madoka_Magica&diff=106031Mathematics of Madoka Magica2021-02-01T01:04:20Z<p>LanceLongFei: /* Question 4 */ Replaced non-TeX with TeX and minor cleanup</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]]. Even though it is 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 that a has a remainder of 6 and b has a remainder of 1 when divided by 14, what is the remainder of x when divided by 14, given x is an integer solution to <math>x^2-2ax+b=0</math>?<br />
<br />
'''Solution:'''<br />
<br />
This problem can be solved simply with [[wikipedia:Modular arithmetic|modular arithmetic]]:<br />
<br />
*<math>a \equiv 6 \pmod{14}</math>;<br />
*<math>b \equiv 1 \pmod{14}</math>.<br />
<br />
*<math>\displaystyle{x^2 - 2ax + b = 0}</math> implies<br />
<br />
*<math>x^2 - 12x + 1 \equiv 0 \pmod{14}</math><br />
<br />
*<math>x^2 + 2x + 1 \equiv 0 \pmod{14} </math> (because <math>-12 \equiv 2 \pmod{14}</math>)<br />
<br />
*<math>(x + 1)^2 \equiv 0 \pmod{14} </math><br />
<br />
*<math> x + 1 \equiv 0 \pmod{14} </math> (because 14 is square-free)<br />
<br />
*<math> x \equiv 13 \pmod{14}</math> (because <math> -1 \equiv 13 \pmod{14}</math>).<br />
<br />
'''Second Solution:'''<br />
<br />
Homura used in Episode 1 a basic approach with usual integer [[wikipedia:arithmetic|arithmetic]].<br />
<br />
Let <math> \displaystyle{x = 14q + r, a = 14s + 6, b = 14t + 1} </math>.<br />
<br />
Substitute into <math> \displaystyle{x^2 - 2ax + b = 0} </math> to get after some calculations<br />
<br />
<math> \displaystyle{x^2 - 2ax + b = 14c + (r+1)^2 = 0} \mbox{ with } \displaystyle{c = 14q^2 + 2q - 28qs - 2rs - 12q + t - r}</math><br />
<br />
14 divides 14c and 0, hence it also divides <math>(r+1)^2</math>. This implies that <math>r+1</math> 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://archive.foolz.us/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 <math>(1+n)^p - n^p - 1 \,</math> 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 />
<math> a^p \equiv a \pmod{p} \,</math><br />
<br />
Thus:<br />
<br />
<math>(1+n)^p - n^p - 1 \pmod{p} \,</math> is equivalent to<br />
<br />
<math>(1+n) - n - 1 \pmod{p} \,</math> is equivalent to<br />
<br />
<math>0 \pmod{p} \,</math><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 <math>a, b</math> not equal to <math>0</math> and nonnegative integer <math>p</math> it holds that:<br />
<br />
<math>\displaystyle (a + b)^p = \sum_{k=0}^{p} \binom{p}{k} a^{p-k} b^k</math><br />
<br />
where the binomial coefficient <math>\binom{p}{k}</math> is the integer <math>\frac{p(p - 1) \cdots (p - k + 1)}{k(k-1) \cdots 1} </math> for <math>0 \leq k \leq p</math>.<br />
<br />
Therefore,<br />
<br />
<math>\displaystyle (1 + n)^p = \sum_{k=0}^{p} \binom{p}{k} n^k</math><br />
<br />
and<br />
<br />
<math>\displaystyle (1 + n)^p - n^p - 1 = \sum_{k = 1}^{p - 1} \binom{p}{k} n^k</math><br />
<br />
since <math>\binom{p}{0} = \binom{p}{p} = 1</math>.<br />
<br />
<br />
It holds that <math>\binom{p}{1} = p </math>.<br />
Since <math>p</math> is a prime number, the factor <math>p</math> in <math>\binom{p}{k}</math> is not divisible by <math>k, k-1,\ldots, 2</math> for <math>2 \leq k \leq p-1</math>.<br />
Therefore, <math>p</math> divides <math>\binom{p}{k}</math> for <math>1 \leq k \leq p-1</math>.<br />
<br />
Since each summand on the right side is divisible by <math>p</math>, the whole sum, i.e. the left side is also divisible by <math>p</math>.<br />
<br />
===Question 3===<br />
[[file:Episode One Math Q3.jpg|thumb|Episode one Math question 3]]<br />
<br />
Find the integer solutions <math>(a,b)</math> with <math>(a^3 + a^2 - 1) - (a - 1)b = 0 \,</math>.<br />
<br />
'''Solution:'''<br />
[[file:Episode One Math S3.jpg|thumb|Episode one Math solution to question 3]]<br />
<br />
<math><br />
\begin{align}<br />
(&a^3 + a^2 - 1) - (a - 1)b \\<br />
& = (a^3 - 1) + (a^2 - 1) + 1 - (a - 1)b \\<br />
& = (a - 1)(a^2 + a + 1) + (a - 1)(a + 1) + 1 - (a - 1)b \\<br />
& = (a - 1)(a^2 + 2a - b + 2) + 1 = 0<br />
\end{align}<br />
</math><br />
<br />
Therefore,<br> <br />
<math>(a - 1)(a^2 + 2a - b + 2) = -1 \,</math><br />
<br />
<math>a, b</math> are integers, therefore the factors <math>a - 1</math> and <math>a^2 + 2a - b + 2 \,</math> are integers too. Since the product is <math>-1</math>, one of the factors must be equal to <math>-1</math>, the other to <math>1</math>.<br />
<br />
If <math>a - 1 = -1</math> then <math>a = 0</math> and from <math>a^2 + 2a - b + 2 = 1 \,</math> we obtain <math>b = 1</math>.<br />
<br />
If <math>a - 1 = 1</math> then <math>a = 2</math> and <math>a^2 + 2a - b + 2 = -1 \,</math> implies that <math>b = 11</math>.<br />
<br />
There are two integer solutions: <math>(a,b) = (0,1) \,</math> or <math>(a,b) = (2,11) \,</math>.<br />
<br />
'''Second Solution'''<br />
<br />
The question can also be solved by polynomial division.<br />
<br />
<math>(a^3 + a^2 - 1) - (a - 1)b = 0 \,</math><br />
<br />
<math> b = \frac{a^3 + a^2 - 1}{a - 1} = a^2 + 2a + 2 + \frac{1}{a - 1} \,</math><br />
<br />
As <math>b \in \mathbb{Z}</math>, <math> a^2 + 2a + 2 + \frac{1}{a - 1} \in \mathbb{Z}</math>, so <math> \frac{1}{a - 1} \in \mathbb{Z}</math><br />
<br />
As <math> 1 </math> has only two factors: <math>1</math> and <math>-1</math>, <br />
<br />
<math> a-1 = 1 \,</math> or <math> a-1 = -1 \,</math><br />
<br />
<math> a = 2 \,</math> or <math> a = 0 \,</math><br />
<br />
Substituting the above values into the equation to find the corresponding <math> b </math>, we have <br />
<br />
<math> (a, b) = (0, 1) \,</math> or <math> (2, 11) </math>.<br />
<br />
===Question 4===<br />
[[file:Episode One Math Q4.jpg|thumb|Episode one Math question 4]]<br />
Given <math>\displaystyle F(x) = \frac{4x + \sqrt{4x^2 - 1}}{\sqrt{2x + 1} + \sqrt{2x-1}}</math>,<br />
<br />
find the sum of <math> \displaystyle{F(1)+F(2)+F(3)+...+F(60)}</math>.<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 />
Notice that for any variables <math>a</math> and <math>b</math>:<br />
<br />
<math> \displaystyle{(a-b)(a+b) = a^2-b^2} </math>. <br />
<br />
Let <math> \displaystyle{a = \sqrt{2x + 1}} </math> and <math> \displaystyle{b = \sqrt{2x - 1}} </math>. <br />
<br />
Multiply <math>\displaystyle{F(x)}</math> by 1 or <math>\displaystyle{\frac{a-b}{a-b}}</math> which is equal to <math> \displaystyle{\frac{\sqrt{2x + 1} - \sqrt{2x - 1}}{\sqrt{2x + 1} - \sqrt{2x - 1}}} </math>. The denominator becomes:<br />
<br />
<math><br />
\begin{align}<br />
&(\sqrt{2x + 1} + \sqrt{2x - 1})(\sqrt{2x + 1} - \sqrt{2x - 1})\\<br />
&= (2x + 1) - (2x - 1)\\<br />
&= 2<br />
\end{align}<br />
</math><br />
<br />
The numerator becomes:<br />
<br />
<math><br />
\begin{align}<br />
&(4x + \sqrt{4x^2 - 1} ) (\sqrt{2x + 1} - \sqrt{2x - 1})\\ <br />
<br />
&=(4x + \sqrt{(2x)^2 - 1)}) (a - b)\\<br />
<br />
&=(4x + \sqrt{2x - 1} \sqrt{2x + 1} ) (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 />
\end{align}<br />
<br />
</math><br />
<br />
Therefore:<br />
<br />
<math> \displaystyle {F(x) = \frac{a^3 - b^3 }{2}} </math><br />
<br />
Note that when <math> \displaystyle{ x = 1, 2, 3, 4, \ldots, 60} </math>;<br />
<br />
<math> \displaystyle{ a^3 = 3^\frac{3}{2}, 5^\frac{3}{2}, 7^\frac{3}{2}, 9^\frac{3}{2}, \ldots, 119^\frac{3}{2}, 121^\frac{3}{2}} </math>;<br />
<br />
<math> \displaystyle{ b^3 = 1^\frac{3}{2}, 3^\frac{3}{2}, 5^\frac{3}{2}, 7^\frac{3}{2}, 9^\frac{3}{2}, \ldots, 119^\frac{3}{2} } </math>;<br />
<br />
Taking the sum over <math> \displaystyle{F(x)} </math> for <math> \displaystyle{x} </math> between <math>1</math> and <math>60</math>, observe that the majority of the terms in <math> \displaystyle{a^3} </math> and <math> \displaystyle{b^3} </math> cancel out, leaving:<br />
<br />
<math><br />
\begin{align}<br />
\sum_{x=1}^{60}F(x) &= \frac{121^\frac{3}{2} - 1^\frac{3}{2}}{2}\\<br />
&=\frac{(11^2)^\frac{3}{2} - 1}{2}\\<br />
&= 665<br />
\end{align}<br />
</math><br />
<br />
Thus, sum of <math> \displaystyle{F(x)} </math> for <math> \displaystyle{x} </math> between <math>1</math> and <math>60</math> is <math>665</math>. The solution can be generalized as equal to <math> \displaystyle{\frac{a(x_{\max}^3) - b(x_{\min}^3)}{2}} </math>. Where <math> \displaystyle{a} </math> and <math> \displaystyle{b} </math> are interpreted as functions of <math> \displaystyle{x} </math>.<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 [[wikipedia:probability|probability]] of a future event by analysis of data from past events of similar nature.<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]]''' Entrance Exam:<br />
<br />
===Problem===<br />
A number sequence <math> \displaystyle{ \{F(n)\} }</math> that can be defined as <math> \displaystyle{F(1) = 1, F(2) = 1}</math>, <math> \displaystyle{F(n+2) = F(n) + F(n+1)} </math> (where <math> \displaystyle{n \in \mathbb{N}} </math>) 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 <math> \displaystyle{ \{X(n)\} } </math> (where n is any natural number), in which each digit is either 0 or 1, set by the following rules:<br />
<br />
(i) <math> \displaystyle{X(1) = 1} </math><br />
<br />
(ii) We define <math> \displaystyle{X(n+1)} </math> as a natural number, which can be obtained by replacing the digits of <math> \displaystyle{X(n)} </math> with 1 if the digit is 0, and with 10 if the digit is 1.<br />
<br />
For example, <math> \displaystyle{X(1) = 1, X(2) = 10, X(3) = 101, X(4) = 10110, X(5) = 10110101, \ldots} </math> <br />
<br />
'' '''Anecdote''': <math> \displaystyle{X(n)} </math> 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 <math> \displaystyle{A(n)} </math>, defined as the number of digits of <math> \displaystyle{X(n)} </math>.<br />
<br />
(2) Find <math> \displaystyle{B(n)} </math>, defined as the numbers of times '01' appears in <math> \displaystyle{X(n)} </math>?<br />
For example, <math> \displaystyle{B(1) = 0, B(2) = 0, B(3) = 1, B(4) = 1, B(5) = 3, \ldots} </math><br />
<br />
<br />
===Solution===<br />
====Part 1====<br />
<br />
Let <math> \displaystyle{A(n)} </math> equal to the number of digits in X(n), which consists solely of 1s and 0s. Let's suppose <math> \displaystyle{x(n)} </math> is the number of 0s in <math> \displaystyle{X(n)} </math> at the n-th iteration (poor choice of variable by the student). Let's suppose <math> \displaystyle{y(n)} </math> is the number of 1s in <math> \displaystyle{X(n)} </math> in the n-th iteration, then <math> \displaystyle{x(n) + y(n) = A(n)} </math>. Since<br />
*Every time a 0 appears, it is replaced with 1 at the next iteration, contributing to a single 1 in <math> \displaystyle{X(n+1)} </math>.<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 <math>X(n+1)</math>.<br />
it follows that the number of 0s in the next iteration is equal to the number of 1s previously: <br />
<br />
<math> \displaystyle{x(n+1)= y(n)}</math> ;<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 />
<math>y(n+1)= \displaystyle{x(n)+y(n)} </math> .<br />
<br />
<br />
Next, prove that x(n) is a Fibonacci sequence, since we know that:<br />
<br />
*<math> \displaystyle{y(n+1) = x(n) + y(n)} </math> ;<br />
*<math> \displaystyle{x(n+2) = y(n+1)} </math> ;<br />
*<math> \displaystyle{x(n+1) = y(n)} </math> ;<br />
<br />
by substition we can show,<br />
<br />
<math><br />
\begin{align}<br />
&x(n+2)\\ <br />
&= y(n+1)\\<br />
&= x(n) + y(n)\\<br />
&= x(n) + x(n+1)<br />
\end{align}<br />
</math><br />
<br />
Hence, <math> \displaystyle{x(n+2) = x(n+1) + x(n)}</math>.<br />
<br />
Thus <math> \displaystyle{x(n)} </math> fits the definition of a Fibonacci sequence. Since <math> \displaystyle{x(n)} </math> is a Fibonacci sequence, it follows that <math> \displaystyle{y(n)} </math> is also a Fibonacci sequence (given that <math> \displaystyle{x(n+1) = y(n)}</math>). Therefore, since it has already been shown that:<br />
<br />
*<math> \displaystyle{x(n+2)=x(n+1)+x(n)} </math><br />
<br />
*<math> \displaystyle{y(n+2)=y(n+1)+y(n)} </math><br />
<br />
It follows that <math> \displaystyle {x(n+2)+y(n+2)=x(n+1)+y(n+1)+x(n)+y(n)} </math>.<br />
<br />
Recall that <math> \displaystyle{A(n)=x(n)+y(n)} </math>, therefore<br />
<br />
:<math> \displaystyle{A(n+2)=A(n+1)+A(n)} </math> .<br />
<br />
Since, <math> \displaystyle{X(1) = 1, X(2) = 10, X(3) = 101, X(4) = 10110, X(5) = 10110101, \ldots} </math><br />
<br />
It follows that <math> \displaystyle{A(1) = 1, A(2) = 2, A(3) = 3, A(4) = 5, A(5) = 8, \ldots} </math><br />
<br />
Recall by definition of the Fibonacci sequence: <math> \displaystyle{\{F(n)\} = \{1, 1, 2, 3, 5, 8, \ldots\}} </math>. Therefore <math> \displaystyle{A(n) = F(n+1)} </math>, 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 />
It can be seen that odd(n) = odd(X(n)) for all n. To prove this inductively: observe that odd(X(1)) = odd(1). Now let odd(X(n)) = odd(n), then if n is even, X(n) is even, thus X(n) ends with 0, which gets mapped to 1, making X(n+1) odd and so odd(X(n+1)) = odd(n+1). And if n is odd, then X(n) is odd, thus X(n) ends with 1, which gets mapped to 10, making X(n+1) even, and so odd(X(n+1)) = odd(n+1).<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(n) <br />
<br />
B(n) = F(n-1) - odd(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]]<br />
<br />
==Episode 11==<br />
<br />
[[File:Homura Walpurgis Ballistics Calculations.jpg|thumb|250px]]<br />
Pieces of papers floats by Homura as she faces [[Walpurgis Night]]. On it are calculations done by hand by [[Homura]]. Most likely, these are [[wikipedia:ballistic]] of the big guns. Two pairs of Xs and Ys on each of the four pieces of paper, making it 8 shots in all.<br />
<br />
== Movie 3: Rebellion ==<br />
[[File:Rebellion Math 1.jpg|thumb]]<br />
<br />
=== Problem A ===<br />
<br />
The first problem posed, (a), was to calculate <math>\int\frac{\arcsin^3x}{\sqrt{1-x^2}}\,dx</math><br />
<br />
We can perform the substitution <math>u=\arcsin x \longrightarrow \frac{du}{dx}=\frac{1}{\sqrt{1-x^2}}</math> and so rewrite the integral as <math>\int u^3\,du</math><br />
<br />
This can be easily solved via basic integration formulas: <math>\int u^3\,du=\frac{1}{4}u^4+C</math><br />
<br />
Undoing the substitution we get the solution: <math>\frac{1}{4}\arcsin^4x + C</math><br />
<br />
=== Problem B ===<br />
<br />
The second problem, (b), was to calculate <math>\int x\ln(x^2+y)\,dx</math><br />
In order to solve it we'll assume <math>\displaystyle y</math> to be a parameter independent from the value of <math>\displaystyle x</math> and <math>\displaystyle\ln x</math> to denote the natural logarithm of <math>\displaystyle x</math>.<br />
<br />
Performing the substitution <math>u=x^2+y\longrightarrow\frac{du}{dx}=2x</math> we can rewrite the integral as <math>\frac{1}{2}\int\ln u\,du</math><br />
<br />
This can be solved integrating by parts: <math>\frac{1}{2}\int \ln u\,du=\frac{1}{2}(u\ln u - \int1\,du)=\frac{1}{2}u(\ln u - 1)</math><br />
<br />
And the final answer in obtained undoing the substitution: <math>\frac{1}{2}(x^2+y)\ln((x^2+y)-1)+C</math><br />
<br />
=== Problem C ===<br />
<br />
One of the problems posed, (c), was to calculate <math>\int\frac{x^3+2x^2+10x}{x^2-x+1}\, dx</math>.<br />
<br />
The answer is to first simplify the function:<br />
<math>\frac{x^3+2x^2+10x}{x^2-x+1} = x + 3 + \frac{12x - 3}{x^2-x+1} = x + 3 + 6\frac{2x - 1}{x^2-x+1} + \frac{3}{x^2-x+1}</math>.<br />
Note that <math>x^2-x+1</math> has only simple complex zeros.<br />
<br />
It holds <math>\int x + 3 \, dx = \frac{1}{2}x^2 + 3x + C</math>. <br />
<br />
The general formula <math>\int\frac{f'(x)}{f(x)}\, dx = \ln|f(x)| dx + C</math> can be used to calculate <math>\int \frac{2x - 1}{x^2-x+1} \, dx = \ln(x^2-x+1) + C</math>.<br />
<br />
<br />
For the last term, <math>\int\frac{dx}{a^2 + x^2} = \frac{1}{a}\arctan \frac{x}{a} + C</math> can be used (with some additional calculations) to obtain the solution<br />
<br />
<math>\int\frac{x^3+2x^2+10x}{x^2-x+1}\, dx = \frac{1}{2}x^2 + 3x + 6\ln(x^2-x+1) + 2\sqrt{3} \arctan \frac{2x-1}{\sqrt{3}} + C</math></div>LanceLongFeihttps://wiki.puella-magi.net/w/index.php?title=Mathematics_of_Madoka_Magica&diff=106030Mathematics of Madoka Magica2021-02-01T00:59:20Z<p>LanceLongFei: /* Question 3 */ Cleaned up. \Z was causing problems with the converter so it has been replaced with \mathbb{Z} which seems to work fine.</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]]. Even though it is 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 that a has a remainder of 6 and b has a remainder of 1 when divided by 14, what is the remainder of x when divided by 14, given x is an integer solution to <math>x^2-2ax+b=0</math>?<br />
<br />
'''Solution:'''<br />
<br />
This problem can be solved simply with [[wikipedia:Modular arithmetic|modular arithmetic]]:<br />
<br />
*<math>a \equiv 6 \pmod{14}</math>;<br />
*<math>b \equiv 1 \pmod{14}</math>.<br />
<br />
*<math>\displaystyle{x^2 - 2ax + b = 0}</math> implies<br />
<br />
*<math>x^2 - 12x + 1 \equiv 0 \pmod{14}</math><br />
<br />
*<math>x^2 + 2x + 1 \equiv 0 \pmod{14} </math> (because <math>-12 \equiv 2 \pmod{14}</math>)<br />
<br />
*<math>(x + 1)^2 \equiv 0 \pmod{14} </math><br />
<br />
*<math> x + 1 \equiv 0 \pmod{14} </math> (because 14 is square-free)<br />
<br />
*<math> x \equiv 13 \pmod{14}</math> (because <math> -1 \equiv 13 \pmod{14}</math>).<br />
<br />
'''Second Solution:'''<br />
<br />
Homura used in Episode 1 a basic approach with usual integer [[wikipedia:arithmetic|arithmetic]].<br />
<br />
Let <math> \displaystyle{x = 14q + r, a = 14s + 6, b = 14t + 1} </math>.<br />
<br />
Substitute into <math> \displaystyle{x^2 - 2ax + b = 0} </math> to get after some calculations<br />
<br />
<math> \displaystyle{x^2 - 2ax + b = 14c + (r+1)^2 = 0} \mbox{ with } \displaystyle{c = 14q^2 + 2q - 28qs - 2rs - 12q + t - r}</math><br />
<br />
14 divides 14c and 0, hence it also divides <math>(r+1)^2</math>. This implies that <math>r+1</math> 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://archive.foolz.us/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 <math>(1+n)^p - n^p - 1 \,</math> 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 />
<math> a^p \equiv a \pmod{p} \,</math><br />
<br />
Thus:<br />
<br />
<math>(1+n)^p - n^p - 1 \pmod{p} \,</math> is equivalent to<br />
<br />
<math>(1+n) - n - 1 \pmod{p} \,</math> is equivalent to<br />
<br />
<math>0 \pmod{p} \,</math><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 <math>a, b</math> not equal to <math>0</math> and nonnegative integer <math>p</math> it holds that:<br />
<br />
<math>\displaystyle (a + b)^p = \sum_{k=0}^{p} \binom{p}{k} a^{p-k} b^k</math><br />
<br />
where the binomial coefficient <math>\binom{p}{k}</math> is the integer <math>\frac{p(p - 1) \cdots (p - k + 1)}{k(k-1) \cdots 1} </math> for <math>0 \leq k \leq p</math>.<br />
<br />
Therefore,<br />
<br />
<math>\displaystyle (1 + n)^p = \sum_{k=0}^{p} \binom{p}{k} n^k</math><br />
<br />
and<br />
<br />
<math>\displaystyle (1 + n)^p - n^p - 1 = \sum_{k = 1}^{p - 1} \binom{p}{k} n^k</math><br />
<br />
since <math>\binom{p}{0} = \binom{p}{p} = 1</math>.<br />
<br />
<br />
It holds that <math>\binom{p}{1} = p </math>.<br />
Since <math>p</math> is a prime number, the factor <math>p</math> in <math>\binom{p}{k}</math> is not divisible by <math>k, k-1,\ldots, 2</math> for <math>2 \leq k \leq p-1</math>.<br />
Therefore, <math>p</math> divides <math>\binom{p}{k}</math> for <math>1 \leq k \leq p-1</math>.<br />
<br />
Since each summand on the right side is divisible by <math>p</math>, the whole sum, i.e. the left side is also divisible by <math>p</math>.<br />
<br />
===Question 3===<br />
[[file:Episode One Math Q3.jpg|thumb|Episode one Math question 3]]<br />
<br />
Find the integer solutions <math>(a,b)</math> with <math>(a^3 + a^2 - 1) - (a - 1)b = 0 \,</math>.<br />
<br />
'''Solution:'''<br />
[[file:Episode One Math S3.jpg|thumb|Episode one Math solution to question 3]]<br />
<br />
<math><br />
\begin{align}<br />
(&a^3 + a^2 - 1) - (a - 1)b \\<br />
& = (a^3 - 1) + (a^2 - 1) + 1 - (a - 1)b \\<br />
& = (a - 1)(a^2 + a + 1) + (a - 1)(a + 1) + 1 - (a - 1)b \\<br />
& = (a - 1)(a^2 + 2a - b + 2) + 1 = 0<br />
\end{align}<br />
</math><br />
<br />
Therefore,<br> <br />
<math>(a - 1)(a^2 + 2a - b + 2) = -1 \,</math><br />
<br />
<math>a, b</math> are integers, therefore the factors <math>a - 1</math> and <math>a^2 + 2a - b + 2 \,</math> are integers too. Since the product is <math>-1</math>, one of the factors must be equal to <math>-1</math>, the other to <math>1</math>.<br />
<br />
If <math>a - 1 = -1</math> then <math>a = 0</math> and from <math>a^2 + 2a - b + 2 = 1 \,</math> we obtain <math>b = 1</math>.<br />
<br />
If <math>a - 1 = 1</math> then <math>a = 2</math> and <math>a^2 + 2a - b + 2 = -1 \,</math> implies that <math>b = 11</math>.<br />
<br />
There are two integer solutions: <math>(a,b) = (0,1) \,</math> or <math>(a,b) = (2,11) \,</math>.<br />
<br />
'''Second Solution'''<br />
<br />
The question can also be solved by polynomial division.<br />
<br />
<math>(a^3 + a^2 - 1) - (a - 1)b = 0 \,</math><br />
<br />
<math> b = \frac{a^3 + a^2 - 1}{a - 1} = a^2 + 2a + 2 + \frac{1}{a - 1} \,</math><br />
<br />
As <math>b \in \mathbb{Z}</math>, <math> a^2 + 2a + 2 + \frac{1}{a - 1} \in \mathbb{Z}</math>, so <math> \frac{1}{a - 1} \in \mathbb{Z}</math><br />
<br />
As <math> 1 </math> has only two factors: <math>1</math> and <math>-1</math>, <br />
<br />
<math> a-1 = 1 \,</math> or <math> a-1 = -1 \,</math><br />
<br />
<math> a = 2 \,</math> or <math> a = 0 \,</math><br />
<br />
Substituting the above values into the equation to find the corresponding <math> b </math>, we have <br />
<br />
<math> (a, b) = (0, 1) \,</math> or <math> (2, 11) </math>.<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 <math> \displaystyle{F(1)+F(2)+F(3)+...+F(60)}</math>.<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 />
Notice that for any variables a & b:<br />
<br />
<math> \displaystyle{(a-b)(a+b) = a^2-b^2} </math>. <br />
<br />
Let <math> \displaystyle{a = \sqrt{2x + 1}} </math> and <math> \displaystyle{b = \sqrt{2x - 1}} </math>. <br />
<br />
Multiply <math>\displaystyle{F(x)}</math> by 1 or <math>\displaystyle{\frac{a-b}{a-b}}</math> which is equal to <math> \displaystyle{\frac{\sqrt{2x + 1} - \sqrt{2x - 1}}{\sqrt{2x + 1} - \sqrt{2x - 1}}} </math>. The denominator becomes:<br />
<br />
<math><br />
\begin{align}<br />
&(\sqrt{2x + 1} + \sqrt{2x - 1})(\sqrt{2x + 1} - \sqrt{2x - 1})\\<br />
&= (2x + 1) - (2x - 1)\\<br />
&= 2<br />
\end{align}<br />
</math><br />
<br />
The numerator becomes:<br />
<br />
<math><br />
\begin{align}<br />
&(4x + \sqrt{4x^2 - 1} ) (\sqrt{2x + 1} - \sqrt{2x - 1})\\ <br />
<br />
&=(4x + \sqrt{(2x)^2 - 1)}) (a - b)\\<br />
<br />
&=(4x + \sqrt{2x - 1} \sqrt{2x + 1} ) (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 />
\end{align}<br />
<br />
</math><br />
<br />
Therefore:<br />
<br />
<math> \displaystyle {F(x) = \frac{a^3 - b^3 }{2}} </math><br />
<br />
Note that when <math> \displaystyle{ x = 1, 2, 3, 4, \ldots, 60} </math>;<br />
<br />
<math> \displaystyle{ a^3 = 3^\frac{3}{2}, 5^\frac{3}{2}, 7^\frac{3}{2}, 9^\frac{3}{2}, \ldots, 119^\frac{3}{2}, 121^\frac{3}{2}} </math>;<br />
<br />
<math> \displaystyle{ b^3 = 1^\frac{3}{2}, 3^\frac{3}{2}, 5^\frac{3}{2}, 7^\frac{3}{2}, 9^\frac{3}{2}, \ldots, 119^\frac{3}{2} } </math>;<br />
<br />
Taking the sum over <math> \displaystyle{F(x)} </math> for <math> \displaystyle{x} </math> between 1 and 60, observe that the majority of the terms in <math> \displaystyle{a^3} </math> and <math> \displaystyle{b^3} </math> cancel out, leaving:<br />
<br />
<math><br />
\begin{align}<br />
\sum_{x=1}^{60}F(x) &= \frac{121^\frac{3}{2} - 1^\frac{3}{2}}{2}\\<br />
&=\frac{(11^2)^\frac{3}{2} - 1}{2}\\<br />
&= 665<br />
\end{align}<br />
</math><br />
<br />
Thus, sum of <math> \displaystyle{F(x)} </math> for <math> \displaystyle{x} </math> between 1 and 60 is 665. The solution can be generalized as equal to <math> \displaystyle{\frac{a(x_{max}^3) - b(x_{min}^3)}{2}} </math>. Where <math> \displaystyle{a} </math> and <math> \displaystyle{b} </math> are interpreted as functions of <math> \displaystyle{x} </math>.<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 [[wikipedia:probability|probability]] of a future event by analysis of data from past events of similar nature.<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]]''' Entrance Exam:<br />
<br />
===Problem===<br />
A number sequence <math> \displaystyle{ \{F(n)\} }</math> that can be defined as <math> \displaystyle{F(1) = 1, F(2) = 1}</math>, <math> \displaystyle{F(n+2) = F(n) + F(n+1)} </math> (where <math> \displaystyle{n \in \mathbb{N}} </math>) 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 <math> \displaystyle{ \{X(n)\} } </math> (where n is any natural number), in which each digit is either 0 or 1, set by the following rules:<br />
<br />
(i) <math> \displaystyle{X(1) = 1} </math><br />
<br />
(ii) We define <math> \displaystyle{X(n+1)} </math> as a natural number, which can be obtained by replacing the digits of <math> \displaystyle{X(n)} </math> with 1 if the digit is 0, and with 10 if the digit is 1.<br />
<br />
For example, <math> \displaystyle{X(1) = 1, X(2) = 10, X(3) = 101, X(4) = 10110, X(5) = 10110101, \ldots} </math> <br />
<br />
'' '''Anecdote''': <math> \displaystyle{X(n)} </math> 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 <math> \displaystyle{A(n)} </math>, defined as the number of digits of <math> \displaystyle{X(n)} </math>.<br />
<br />
(2) Find <math> \displaystyle{B(n)} </math>, defined as the numbers of times '01' appears in <math> \displaystyle{X(n)} </math>?<br />
For example, <math> \displaystyle{B(1) = 0, B(2) = 0, B(3) = 1, B(4) = 1, B(5) = 3, \ldots} </math><br />
<br />
<br />
===Solution===<br />
====Part 1====<br />
<br />
Let <math> \displaystyle{A(n)} </math> equal to the number of digits in X(n), which consists solely of 1s and 0s. Let's suppose <math> \displaystyle{x(n)} </math> is the number of 0s in <math> \displaystyle{X(n)} </math> at the n-th iteration (poor choice of variable by the student). Let's suppose <math> \displaystyle{y(n)} </math> is the number of 1s in <math> \displaystyle{X(n)} </math> in the n-th iteration, then <math> \displaystyle{x(n) + y(n) = A(n)} </math>. Since<br />
*Every time a 0 appears, it is replaced with 1 at the next iteration, contributing to a single 1 in <math> \displaystyle{X(n+1)} </math>.<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 <math>X(n+1)</math>.<br />
it follows that the number of 0s in the next iteration is equal to the number of 1s previously: <br />
<br />
<math> \displaystyle{x(n+1)= y(n)}</math> ;<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 />
<math>y(n+1)= \displaystyle{x(n)+y(n)} </math> .<br />
<br />
<br />
Next, prove that x(n) is a Fibonacci sequence, since we know that:<br />
<br />
*<math> \displaystyle{y(n+1) = x(n) + y(n)} </math> ;<br />
*<math> \displaystyle{x(n+2) = y(n+1)} </math> ;<br />
*<math> \displaystyle{x(n+1) = y(n)} </math> ;<br />
<br />
by substition we can show,<br />
<br />
<math><br />
\begin{align}<br />
&x(n+2)\\ <br />
&= y(n+1)\\<br />
&= x(n) + y(n)\\<br />
&= x(n) + x(n+1)<br />
\end{align}<br />
</math><br />
<br />
Hence, <math> \displaystyle{x(n+2) = x(n+1) + x(n)}</math>.<br />
<br />
Thus <math> \displaystyle{x(n)} </math> fits the definition of a Fibonacci sequence. Since <math> \displaystyle{x(n)} </math> is a Fibonacci sequence, it follows that <math> \displaystyle{y(n)} </math> is also a Fibonacci sequence (given that <math> \displaystyle{x(n+1) = y(n)}</math>). Therefore, since it has already been shown that:<br />
<br />
*<math> \displaystyle{x(n+2)=x(n+1)+x(n)} </math><br />
<br />
*<math> \displaystyle{y(n+2)=y(n+1)+y(n)} </math><br />
<br />
It follows that <math> \displaystyle {x(n+2)+y(n+2)=x(n+1)+y(n+1)+x(n)+y(n)} </math>.<br />
<br />
Recall that <math> \displaystyle{A(n)=x(n)+y(n)} </math>, therefore<br />
<br />
:<math> \displaystyle{A(n+2)=A(n+1)+A(n)} </math> .<br />
<br />
Since, <math> \displaystyle{X(1) = 1, X(2) = 10, X(3) = 101, X(4) = 10110, X(5) = 10110101, \ldots} </math><br />
<br />
It follows that <math> \displaystyle{A(1) = 1, A(2) = 2, A(3) = 3, A(4) = 5, A(5) = 8, \ldots} </math><br />
<br />
Recall by definition of the Fibonacci sequence: <math> \displaystyle{\{F(n)\} = \{1, 1, 2, 3, 5, 8, \ldots\}} </math>. Therefore <math> \displaystyle{A(n) = F(n+1)} </math>, 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 />
It can be seen that odd(n) = odd(X(n)) for all n. To prove this inductively: observe that odd(X(1)) = odd(1). Now let odd(X(n)) = odd(n), then if n is even, X(n) is even, thus X(n) ends with 0, which gets mapped to 1, making X(n+1) odd and so odd(X(n+1)) = odd(n+1). And if n is odd, then X(n) is odd, thus X(n) ends with 1, which gets mapped to 10, making X(n+1) even, and so odd(X(n+1)) = odd(n+1).<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(n) <br />
<br />
B(n) = F(n-1) - odd(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]]<br />
<br />
==Episode 11==<br />
<br />
[[File:Homura Walpurgis Ballistics Calculations.jpg|thumb|250px]]<br />
Pieces of papers floats by Homura as she faces [[Walpurgis Night]]. On it are calculations done by hand by [[Homura]]. Most likely, these are [[wikipedia:ballistic]] of the big guns. Two pairs of Xs and Ys on each of the four pieces of paper, making it 8 shots in all.<br />
<br />
== Movie 3: Rebellion ==<br />
[[File:Rebellion Math 1.jpg|thumb]]<br />
<br />
=== Problem A ===<br />
<br />
The first problem posed, (a), was to calculate <math>\int\frac{\arcsin^3x}{\sqrt{1-x^2}}\,dx</math><br />
<br />
We can perform the substitution <math>u=\arcsin x \longrightarrow \frac{du}{dx}=\frac{1}{\sqrt{1-x^2}}</math> and so rewrite the integral as <math>\int u^3\,du</math><br />
<br />
This can be easily solved via basic integration formulas: <math>\int u^3\,du=\frac{1}{4}u^4+C</math><br />
<br />
Undoing the substitution we get the solution: <math>\frac{1}{4}\arcsin^4x + C</math><br />
<br />
=== Problem B ===<br />
<br />
The second problem, (b), was to calculate <math>\int x\ln(x^2+y)\,dx</math><br />
In order to solve it we'll assume <math>\displaystyle y</math> to be a parameter independent from the value of <math>\displaystyle x</math> and <math>\displaystyle\ln x</math> to denote the natural logarithm of <math>\displaystyle x</math>.<br />
<br />
Performing the substitution <math>u=x^2+y\longrightarrow\frac{du}{dx}=2x</math> we can rewrite the integral as <math>\frac{1}{2}\int\ln u\,du</math><br />
<br />
This can be solved integrating by parts: <math>\frac{1}{2}\int \ln u\,du=\frac{1}{2}(u\ln u - \int1\,du)=\frac{1}{2}u(\ln u - 1)</math><br />
<br />
And the final answer in obtained undoing the substitution: <math>\frac{1}{2}(x^2+y)\ln((x^2+y)-1)+C</math><br />
<br />
=== Problem C ===<br />
<br />
One of the problems posed, (c), was to calculate <math>\int\frac{x^3+2x^2+10x}{x^2-x+1}\, dx</math>.<br />
<br />
The answer is to first simplify the function:<br />
<math>\frac{x^3+2x^2+10x}{x^2-x+1} = x + 3 + \frac{12x - 3}{x^2-x+1} = x + 3 + 6\frac{2x - 1}{x^2-x+1} + \frac{3}{x^2-x+1}</math>.<br />
Note that <math>x^2-x+1</math> has only simple complex zeros.<br />
<br />
It holds <math>\int x + 3 \, dx = \frac{1}{2}x^2 + 3x + C</math>. <br />
<br />
The general formula <math>\int\frac{f'(x)}{f(x)}\, dx = \ln|f(x)| dx + C</math> can be used to calculate <math>\int \frac{2x - 1}{x^2-x+1} \, dx = \ln(x^2-x+1) + C</math>.<br />
<br />
<br />
For the last term, <math>\int\frac{dx}{a^2 + x^2} = \frac{1}{a}\arctan \frac{x}{a} + C</math> can be used (with some additional calculations) to obtain the solution<br />
<br />
<math>\int\frac{x^3+2x^2+10x}{x^2-x+1}\, dx = \frac{1}{2}x^2 + 3x + 6\ln(x^2-x+1) + 2\sqrt{3} \arctan \frac{2x-1}{\sqrt{3}} + C</math></div>LanceLongFeihttps://wiki.puella-magi.net/w/index.php?title=Mathematics_of_Madoka_Magica&diff=106029Mathematics of Madoka Magica2021-02-01T00:45:06Z<p>LanceLongFei: /* Question 2 */ Replaced everything that wasn't TeX before with TeX in solution 2.</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]]. Even though it is 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 that a has a remainder of 6 and b has a remainder of 1 when divided by 14, what is the remainder of x when divided by 14, given x is an integer solution to <math>x^2-2ax+b=0</math>?<br />
<br />
'''Solution:'''<br />
<br />
This problem can be solved simply with [[wikipedia:Modular arithmetic|modular arithmetic]]:<br />
<br />
*<math>a \equiv 6 \pmod{14}</math>;<br />
*<math>b \equiv 1 \pmod{14}</math>.<br />
<br />
*<math>\displaystyle{x^2 - 2ax + b = 0}</math> implies<br />
<br />
*<math>x^2 - 12x + 1 \equiv 0 \pmod{14}</math><br />
<br />
*<math>x^2 + 2x + 1 \equiv 0 \pmod{14} </math> (because <math>-12 \equiv 2 \pmod{14}</math>)<br />
<br />
*<math>(x + 1)^2 \equiv 0 \pmod{14} </math><br />
<br />
*<math> x + 1 \equiv 0 \pmod{14} </math> (because 14 is square-free)<br />
<br />
*<math> x \equiv 13 \pmod{14}</math> (because <math> -1 \equiv 13 \pmod{14}</math>).<br />
<br />
'''Second Solution:'''<br />
<br />
Homura used in Episode 1 a basic approach with usual integer [[wikipedia:arithmetic|arithmetic]].<br />
<br />
Let <math> \displaystyle{x = 14q + r, a = 14s + 6, b = 14t + 1} </math>.<br />
<br />
Substitute into <math> \displaystyle{x^2 - 2ax + b = 0} </math> to get after some calculations<br />
<br />
<math> \displaystyle{x^2 - 2ax + b = 14c + (r+1)^2 = 0} \mbox{ with } \displaystyle{c = 14q^2 + 2q - 28qs - 2rs - 12q + t - r}</math><br />
<br />
14 divides 14c and 0, hence it also divides <math>(r+1)^2</math>. This implies that <math>r+1</math> 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://archive.foolz.us/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 <math>(1+n)^p - n^p - 1 \,</math> 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 />
<math> a^p \equiv a \pmod{p} \,</math><br />
<br />
Thus:<br />
<br />
<math>(1+n)^p - n^p - 1 \pmod{p} \,</math> is equivalent to<br />
<br />
<math>(1+n) - n - 1 \pmod{p} \,</math> is equivalent to<br />
<br />
<math>0 \pmod{p} \,</math><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 <math>a, b</math> not equal to <math>0</math> and nonnegative integer <math>p</math> it holds that:<br />
<br />
<math>\displaystyle (a + b)^p = \sum_{k=0}^{p} \binom{p}{k} a^{p-k} b^k</math><br />
<br />
where the binomial coefficient <math>\binom{p}{k}</math> is the integer <math>\frac{p(p - 1) \cdots (p - k + 1)}{k(k-1) \cdots 1} </math> for <math>0 \leq k \leq p</math>.<br />
<br />
Therefore,<br />
<br />
<math>\displaystyle (1 + n)^p = \sum_{k=0}^{p} \binom{p}{k} n^k</math><br />
<br />
and<br />
<br />
<math>\displaystyle (1 + n)^p - n^p - 1 = \sum_{k = 1}^{p - 1} \binom{p}{k} n^k</math><br />
<br />
since <math>\binom{p}{0} = \binom{p}{p} = 1</math>.<br />
<br />
<br />
It holds that <math>\binom{p}{1} = p </math>.<br />
Since <math>p</math> is a prime number, the factor <math>p</math> in <math>\binom{p}{k}</math> is not divisible by <math>k, k-1,\ldots, 2</math> for <math>2 \leq k \leq p-1</math>.<br />
Therefore, <math>p</math> divides <math>\binom{p}{k}</math> for <math>1 \leq k \leq p-1</math>.<br />
<br />
Since each summand on the right side is divisible by <math>p</math>, the whole sum, i.e. the left side is also divisible by <math>p</math>.<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 <math>(a^3 + a^2 - 1) - (a - 1)b = 0 \,</math><br />
<br />
'''Solution:'''<br />
[[file:Episode One Math S3.jpg|thumb|Episode one Math solution to question 3]]<br />
<br />
<math><br />
\begin{align}<br />
(&a^3 + a^2 - 1) - (a - 1)b \\<br />
& = (a^3 - 1) + (a^2 - 1) + 1 - (a - 1)b \\<br />
& = (a - 1)(a^2 + a + 1) + (a - 1)(a + 1) + 1 - (a - 1)b \\<br />
& = (a - 1)(a^2 + 2a - b + 2) + 1 = 0<br />
\end{align}<br />
</math><br />
<br />
Therefore,<br> <br />
<math>(a - 1)(a^2 + 2a - b + 2) = -1 \,</math><br />
<br />
a, b are integers, therefore the factors a - 1 and <math>a^2 + 2a - b + 2 \,</math> are integers too. Since the product is -1, one of the factors must be equal to -1, the other to 1.<br />
<br />
If <math>a - 1 = -1</math> then a = 0 and from <math>a^2 + 2a - b + 2 = 1 \,</math> we obtain b = 1.<br />
<br />
If <math>a - 1 = 1</math> then a = 2 and <math>a^2 + 2a - b + 2 = -1 \,</math> implies that b = 11.<br />
<br />
There are two integer solutions: <math>(a,b) = (0,1) \,</math> or <math>(a,b) = (2,11) \,</math>.<br />
<br />
'''Second Solution'''<br />
<br />
The question can also be solved by polynomial division.<br />
<br />
<math>(a^3 + a^2 - 1) - (a - 1)b = 0 \,</math><br />
<br />
<math> b = \frac{a^3 + a^2 - 1}{a - 1} = a^2 + 2a + 2 + \frac{1}{a - 1} \,</math><br />
<br />
As <math>b \in \Z</math> is an integer, <math> a^2 + 2a + 2 + \frac{1}{a - 1} \in \Z</math>, so <math> \frac{1}{a - 1} \in \Z</math><br />
<br />
As <math> 1 </math> has only two factors: <math>1</math> and <math>-1</math>, <br />
<br />
<math> a-1 = 1 \,</math> or <math> a-1 = -1 \,</math><br />
<br />
<math> a = 2 \,</math> or <math> a = 0 \,</math><br />
<br />
Substituting the above values into the equation to find the corresponding <math> b </math>, we have <br />
<br />
<math> (a, b) = (0, 1) \,</math> or <math> (2, 11) \,</math><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 <math> \displaystyle{F(1)+F(2)+F(3)+...+F(60)}</math>.<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 />
Notice that for any variables a & b:<br />
<br />
<math> \displaystyle{(a-b)(a+b) = a^2-b^2} </math>. <br />
<br />
Let <math> \displaystyle{a = \sqrt{2x + 1}} </math> and <math> \displaystyle{b = \sqrt{2x - 1}} </math>. <br />
<br />
Multiply <math>\displaystyle{F(x)}</math> by 1 or <math>\displaystyle{\frac{a-b}{a-b}}</math> which is equal to <math> \displaystyle{\frac{\sqrt{2x + 1} - \sqrt{2x - 1}}{\sqrt{2x + 1} - \sqrt{2x - 1}}} </math>. The denominator becomes:<br />
<br />
<math><br />
\begin{align}<br />
&(\sqrt{2x + 1} + \sqrt{2x - 1})(\sqrt{2x + 1} - \sqrt{2x - 1})\\<br />
&= (2x + 1) - (2x - 1)\\<br />
&= 2<br />
\end{align}<br />
</math><br />
<br />
The numerator becomes:<br />
<br />
<math><br />
\begin{align}<br />
&(4x + \sqrt{4x^2 - 1} ) (\sqrt{2x + 1} - \sqrt{2x - 1})\\ <br />
<br />
&=(4x + \sqrt{(2x)^2 - 1)}) (a - b)\\<br />
<br />
&=(4x + \sqrt{2x - 1} \sqrt{2x + 1} ) (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 />
\end{align}<br />
<br />
</math><br />
<br />
Therefore:<br />
<br />
<math> \displaystyle {F(x) = \frac{a^3 - b^3 }{2}} </math><br />
<br />
Note that when <math> \displaystyle{ x = 1, 2, 3, 4, \ldots, 60} </math>;<br />
<br />
<math> \displaystyle{ a^3 = 3^\frac{3}{2}, 5^\frac{3}{2}, 7^\frac{3}{2}, 9^\frac{3}{2}, \ldots, 119^\frac{3}{2}, 121^\frac{3}{2}} </math>;<br />
<br />
<math> \displaystyle{ b^3 = 1^\frac{3}{2}, 3^\frac{3}{2}, 5^\frac{3}{2}, 7^\frac{3}{2}, 9^\frac{3}{2}, \ldots, 119^\frac{3}{2} } </math>;<br />
<br />
Taking the sum over <math> \displaystyle{F(x)} </math> for <math> \displaystyle{x} </math> between 1 and 60, observe that the majority of the terms in <math> \displaystyle{a^3} </math> and <math> \displaystyle{b^3} </math> cancel out, leaving:<br />
<br />
<math><br />
\begin{align}<br />
\sum_{x=1}^{60}F(x) &= \frac{121^\frac{3}{2} - 1^\frac{3}{2}}{2}\\<br />
&=\frac{(11^2)^\frac{3}{2} - 1}{2}\\<br />
&= 665<br />
\end{align}<br />
</math><br />
<br />
Thus, sum of <math> \displaystyle{F(x)} </math> for <math> \displaystyle{x} </math> between 1 and 60 is 665. The solution can be generalized as equal to <math> \displaystyle{\frac{a(x_{max}^3) - b(x_{min}^3)}{2}} </math>. Where <math> \displaystyle{a} </math> and <math> \displaystyle{b} </math> are interpreted as functions of <math> \displaystyle{x} </math>.<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 [[wikipedia:probability|probability]] of a future event by analysis of data from past events of similar nature.<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]]''' Entrance Exam:<br />
<br />
===Problem===<br />
A number sequence <math> \displaystyle{ \{F(n)\} }</math> that can be defined as <math> \displaystyle{F(1) = 1, F(2) = 1}</math>, <math> \displaystyle{F(n+2) = F(n) + F(n+1)} </math> (where <math> \displaystyle{n \in \mathbb{N}} </math>) 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 <math> \displaystyle{ \{X(n)\} } </math> (where n is any natural number), in which each digit is either 0 or 1, set by the following rules:<br />
<br />
(i) <math> \displaystyle{X(1) = 1} </math><br />
<br />
(ii) We define <math> \displaystyle{X(n+1)} </math> as a natural number, which can be obtained by replacing the digits of <math> \displaystyle{X(n)} </math> with 1 if the digit is 0, and with 10 if the digit is 1.<br />
<br />
For example, <math> \displaystyle{X(1) = 1, X(2) = 10, X(3) = 101, X(4) = 10110, X(5) = 10110101, \ldots} </math> <br />
<br />
'' '''Anecdote''': <math> \displaystyle{X(n)} </math> 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 <math> \displaystyle{A(n)} </math>, defined as the number of digits of <math> \displaystyle{X(n)} </math>.<br />
<br />
(2) Find <math> \displaystyle{B(n)} </math>, defined as the numbers of times '01' appears in <math> \displaystyle{X(n)} </math>?<br />
For example, <math> \displaystyle{B(1) = 0, B(2) = 0, B(3) = 1, B(4) = 1, B(5) = 3, \ldots} </math><br />
<br />
<br />
===Solution===<br />
====Part 1====<br />
<br />
Let <math> \displaystyle{A(n)} </math> equal to the number of digits in X(n), which consists solely of 1s and 0s. Let's suppose <math> \displaystyle{x(n)} </math> is the number of 0s in <math> \displaystyle{X(n)} </math> at the n-th iteration (poor choice of variable by the student). Let's suppose <math> \displaystyle{y(n)} </math> is the number of 1s in <math> \displaystyle{X(n)} </math> in the n-th iteration, then <math> \displaystyle{x(n) + y(n) = A(n)} </math>. Since<br />
*Every time a 0 appears, it is replaced with 1 at the next iteration, contributing to a single 1 in <math> \displaystyle{X(n+1)} </math>.<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 <math>X(n+1)</math>.<br />
it follows that the number of 0s in the next iteration is equal to the number of 1s previously: <br />
<br />
<math> \displaystyle{x(n+1)= y(n)}</math> ;<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 />
<math>y(n+1)= \displaystyle{x(n)+y(n)} </math> .<br />
<br />
<br />
Next, prove that x(n) is a Fibonacci sequence, since we know that:<br />
<br />
*<math> \displaystyle{y(n+1) = x(n) + y(n)} </math> ;<br />
*<math> \displaystyle{x(n+2) = y(n+1)} </math> ;<br />
*<math> \displaystyle{x(n+1) = y(n)} </math> ;<br />
<br />
by substition we can show,<br />
<br />
<math><br />
\begin{align}<br />
&x(n+2)\\ <br />
&= y(n+1)\\<br />
&= x(n) + y(n)\\<br />
&= x(n) + x(n+1)<br />
\end{align}<br />
</math><br />
<br />
Hence, <math> \displaystyle{x(n+2) = x(n+1) + x(n)}</math>.<br />
<br />
Thus <math> \displaystyle{x(n)} </math> fits the definition of a Fibonacci sequence. Since <math> \displaystyle{x(n)} </math> is a Fibonacci sequence, it follows that <math> \displaystyle{y(n)} </math> is also a Fibonacci sequence (given that <math> \displaystyle{x(n+1) = y(n)}</math>). Therefore, since it has already been shown that:<br />
<br />
*<math> \displaystyle{x(n+2)=x(n+1)+x(n)} </math><br />
<br />
*<math> \displaystyle{y(n+2)=y(n+1)+y(n)} </math><br />
<br />
It follows that <math> \displaystyle {x(n+2)+y(n+2)=x(n+1)+y(n+1)+x(n)+y(n)} </math>.<br />
<br />
Recall that <math> \displaystyle{A(n)=x(n)+y(n)} </math>, therefore<br />
<br />
:<math> \displaystyle{A(n+2)=A(n+1)+A(n)} </math> .<br />
<br />
Since, <math> \displaystyle{X(1) = 1, X(2) = 10, X(3) = 101, X(4) = 10110, X(5) = 10110101, \ldots} </math><br />
<br />
It follows that <math> \displaystyle{A(1) = 1, A(2) = 2, A(3) = 3, A(4) = 5, A(5) = 8, \ldots} </math><br />
<br />
Recall by definition of the Fibonacci sequence: <math> \displaystyle{\{F(n)\} = \{1, 1, 2, 3, 5, 8, \ldots\}} </math>. Therefore <math> \displaystyle{A(n) = F(n+1)} </math>, 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 />
It can be seen that odd(n) = odd(X(n)) for all n. To prove this inductively: observe that odd(X(1)) = odd(1). Now let odd(X(n)) = odd(n), then if n is even, X(n) is even, thus X(n) ends with 0, which gets mapped to 1, making X(n+1) odd and so odd(X(n+1)) = odd(n+1). And if n is odd, then X(n) is odd, thus X(n) ends with 1, which gets mapped to 10, making X(n+1) even, and so odd(X(n+1)) = odd(n+1).<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(n) <br />
<br />
B(n) = F(n-1) - odd(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]]<br />
<br />
==Episode 11==<br />
<br />
[[File:Homura Walpurgis Ballistics Calculations.jpg|thumb|250px]]<br />
Pieces of papers floats by Homura as she faces [[Walpurgis Night]]. On it are calculations done by hand by [[Homura]]. Most likely, these are [[wikipedia:ballistic]] of the big guns. Two pairs of Xs and Ys on each of the four pieces of paper, making it 8 shots in all.<br />
<br />
== Movie 3: Rebellion ==<br />
[[File:Rebellion Math 1.jpg|thumb]]<br />
<br />
=== Problem A ===<br />
<br />
The first problem posed, (a), was to calculate <math>\int\frac{\arcsin^3x}{\sqrt{1-x^2}}\,dx</math><br />
<br />
We can perform the substitution <math>u=\arcsin x \longrightarrow \frac{du}{dx}=\frac{1}{\sqrt{1-x^2}}</math> and so rewrite the integral as <math>\int u^3\,du</math><br />
<br />
This can be easily solved via basic integration formulas: <math>\int u^3\,du=\frac{1}{4}u^4+C</math><br />
<br />
Undoing the substitution we get the solution: <math>\frac{1}{4}\arcsin^4x + C</math><br />
<br />
=== Problem B ===<br />
<br />
The second problem, (b), was to calculate <math>\int x\ln(x^2+y)\,dx</math><br />
In order to solve it we'll assume <math>\displaystyle y</math> to be a parameter independent from the value of <math>\displaystyle x</math> and <math>\displaystyle\ln x</math> to denote the natural logarithm of <math>\displaystyle x</math>.<br />
<br />
Performing the substitution <math>u=x^2+y\longrightarrow\frac{du}{dx}=2x</math> we can rewrite the integral as <math>\frac{1}{2}\int\ln u\,du</math><br />
<br />
This can be solved integrating by parts: <math>\frac{1}{2}\int \ln u\,du=\frac{1}{2}(u\ln u - \int1\,du)=\frac{1}{2}u(\ln u - 1)</math><br />
<br />
And the final answer in obtained undoing the substitution: <math>\frac{1}{2}(x^2+y)\ln((x^2+y)-1)+C</math><br />
<br />
=== Problem C ===<br />
<br />
One of the problems posed, (c), was to calculate <math>\int\frac{x^3+2x^2+10x}{x^2-x+1}\, dx</math>.<br />
<br />
The answer is to first simplify the function:<br />
<math>\frac{x^3+2x^2+10x}{x^2-x+1} = x + 3 + \frac{12x - 3}{x^2-x+1} = x + 3 + 6\frac{2x - 1}{x^2-x+1} + \frac{3}{x^2-x+1}</math>.<br />
Note that <math>x^2-x+1</math> has only simple complex zeros.<br />
<br />
It holds <math>\int x + 3 \, dx = \frac{1}{2}x^2 + 3x + C</math>. <br />
<br />
The general formula <math>\int\frac{f'(x)}{f(x)}\, dx = \ln|f(x)| dx + C</math> can be used to calculate <math>\int \frac{2x - 1}{x^2-x+1} \, dx = \ln(x^2-x+1) + C</math>.<br />
<br />
<br />
For the last term, <math>\int\frac{dx}{a^2 + x^2} = \frac{1}{a}\arctan \frac{x}{a} + C</math> can be used (with some additional calculations) to obtain the solution<br />
<br />
<math>\int\frac{x^3+2x^2+10x}{x^2-x+1}\, dx = \frac{1}{2}x^2 + 3x + 6\ln(x^2-x+1) + 2\sqrt{3} \arctan \frac{2x-1}{\sqrt{3}} + C</math></div>LanceLongFeihttps://wiki.puella-magi.net/w/index.php?title=Mathematics_of_Madoka_Magica&diff=106028Mathematics of Madoka Magica2021-02-01T00:19:50Z<p>LanceLongFei: /* Question 1 */ Removed line introducing unused variable r, replaced 'is equivalent to' with 'implies' for correctness and cleaned up brackets in solution 1</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]]. Even though it is 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 that a has a remainder of 6 and b has a remainder of 1 when divided by 14, what is the remainder of x when divided by 14, given x is an integer solution to <math>x^2-2ax+b=0</math>?<br />
<br />
'''Solution:'''<br />
<br />
This problem can be solved simply with [[wikipedia:Modular arithmetic|modular arithmetic]]:<br />
<br />
*<math>a \equiv 6 \pmod{14}</math>;<br />
*<math>b \equiv 1 \pmod{14}</math>.<br />
<br />
*<math>\displaystyle{x^2 - 2ax + b = 0}</math> implies<br />
<br />
*<math>x^2 - 12x + 1 \equiv 0 \pmod{14}</math><br />
<br />
*<math>x^2 + 2x + 1 \equiv 0 \pmod{14} </math> (because <math>-12 \equiv 2 \pmod{14}</math>)<br />
<br />
*<math>(x + 1)^2 \equiv 0 \pmod{14} </math><br />
<br />
*<math> x + 1 \equiv 0 \pmod{14} </math> (because 14 is square-free)<br />
<br />
*<math> x \equiv 13 \pmod{14}</math> (because <math> -1 \equiv 13 \pmod{14}</math>).<br />
<br />
'''Second Solution:'''<br />
<br />
Homura used in Episode 1 a basic approach with usual integer [[wikipedia:arithmetic|arithmetic]].<br />
<br />
Let <math> \displaystyle{x = 14q + r, a = 14s + 6, b = 14t + 1} </math>.<br />
<br />
Substitute into <math> \displaystyle{x^2 - 2ax + b = 0} </math> to get after some calculations<br />
<br />
<math> \displaystyle{x^2 - 2ax + b = 14c + (r+1)^2 = 0} \mbox{ with } \displaystyle{c = 14q^2 + 2q - 28qs - 2rs - 12q + t - r}</math><br />
<br />
14 divides 14c and 0, hence it also divides <math>(r+1)^2</math>. This implies that <math>r+1</math> 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://archive.foolz.us/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 <math>(1+n)^p - n^p - 1 \,</math> 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 />
<math> a^p \equiv a \pmod{p} \,</math><br />
<br />
Thus:<br />
<br />
<math>(1+n)^p - n^p - 1 \pmod{p} \,</math> is equivalent to<br />
<br />
<math>(1+n) - n - 1 \pmod{p} \,</math> is equivalent to<br />
<br />
<math>0 \pmod{p} \,</math><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 <math>\binom{p}{k}</math> 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 <math>\binom{p}{0} = \binom{p}{p} = 1</math>.<br />
<br />
<br />
It holds that <math>\binom{p}{1} = p </math>.<br />
Since p is a prime number, the factor p in <math>\binom{p}{k}</math> is not divisible by k, k-1,...,2 for 2 ≤ k ≤ p-1.<br />
Therefore, p divides <math>\binom{p}{k}</math> 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 />
===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 <math>(a^3 + a^2 - 1) - (a - 1)b = 0 \,</math><br />
<br />
'''Solution:'''<br />
[[file:Episode One Math S3.jpg|thumb|Episode one Math solution to question 3]]<br />
<br />
<math><br />
\begin{align}<br />
(&a^3 + a^2 - 1) - (a - 1)b \\<br />
& = (a^3 - 1) + (a^2 - 1) + 1 - (a - 1)b \\<br />
& = (a - 1)(a^2 + a + 1) + (a - 1)(a + 1) + 1 - (a - 1)b \\<br />
& = (a - 1)(a^2 + 2a - b + 2) + 1 = 0<br />
\end{align}<br />
</math><br />
<br />
Therefore,<br> <br />
<math>(a - 1)(a^2 + 2a - b + 2) = -1 \,</math><br />
<br />
a, b are integers, therefore the factors a - 1 and <math>a^2 + 2a - b + 2 \,</math> are integers too. Since the product is -1, one of the factors must be equal to -1, the other to 1.<br />
<br />
If <math>a - 1 = -1</math> then a = 0 and from <math>a^2 + 2a - b + 2 = 1 \,</math> we obtain b = 1.<br />
<br />
If <math>a - 1 = 1</math> then a = 2 and <math>a^2 + 2a - b + 2 = -1 \,</math> implies that b = 11.<br />
<br />
There are two integer solutions: <math>(a,b) = (0,1) \,</math> or <math>(a,b) = (2,11) \,</math>.<br />
<br />
'''Second Solution'''<br />
<br />
The question can also be solved by polynomial division.<br />
<br />
<math>(a^3 + a^2 - 1) - (a - 1)b = 0 \,</math><br />
<br />
<math> b = \frac{a^3 + a^2 - 1}{a - 1} = a^2 + 2a + 2 + \frac{1}{a - 1} \,</math><br />
<br />
As <math>b \in \Z</math> is an integer, <math> a^2 + 2a + 2 + \frac{1}{a - 1} \in \Z</math>, so <math> \frac{1}{a - 1} \in \Z</math><br />
<br />
As <math> 1 </math> has only two factors: <math>1</math> and <math>-1</math>, <br />
<br />
<math> a-1 = 1 \,</math> or <math> a-1 = -1 \,</math><br />
<br />
<math> a = 2 \,</math> or <math> a = 0 \,</math><br />
<br />
Substituting the above values into the equation to find the corresponding <math> b </math>, we have <br />
<br />
<math> (a, b) = (0, 1) \,</math> or <math> (2, 11) \,</math><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 <math> \displaystyle{F(1)+F(2)+F(3)+...+F(60)}</math>.<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 />
Notice that for any variables a & b:<br />
<br />
<math> \displaystyle{(a-b)(a+b) = a^2-b^2} </math>. <br />
<br />
Let <math> \displaystyle{a = \sqrt{2x + 1}} </math> and <math> \displaystyle{b = \sqrt{2x - 1}} </math>. <br />
<br />
Multiply <math>\displaystyle{F(x)}</math> by 1 or <math>\displaystyle{\frac{a-b}{a-b}}</math> which is equal to <math> \displaystyle{\frac{\sqrt{2x + 1} - \sqrt{2x - 1}}{\sqrt{2x + 1} - \sqrt{2x - 1}}} </math>. The denominator becomes:<br />
<br />
<math><br />
\begin{align}<br />
&(\sqrt{2x + 1} + \sqrt{2x - 1})(\sqrt{2x + 1} - \sqrt{2x - 1})\\<br />
&= (2x + 1) - (2x - 1)\\<br />
&= 2<br />
\end{align}<br />
</math><br />
<br />
The numerator becomes:<br />
<br />
<math><br />
\begin{align}<br />
&(4x + \sqrt{4x^2 - 1} ) (\sqrt{2x + 1} - \sqrt{2x - 1})\\ <br />
<br />
&=(4x + \sqrt{(2x)^2 - 1)}) (a - b)\\<br />
<br />
&=(4x + \sqrt{2x - 1} \sqrt{2x + 1} ) (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 />
\end{align}<br />
<br />
</math><br />
<br />
Therefore:<br />
<br />
<math> \displaystyle {F(x) = \frac{a^3 - b^3 }{2}} </math><br />
<br />
Note that when <math> \displaystyle{ x = 1, 2, 3, 4, \ldots, 60} </math>;<br />
<br />
<math> \displaystyle{ a^3 = 3^\frac{3}{2}, 5^\frac{3}{2}, 7^\frac{3}{2}, 9^\frac{3}{2}, \ldots, 119^\frac{3}{2}, 121^\frac{3}{2}} </math>;<br />
<br />
<math> \displaystyle{ b^3 = 1^\frac{3}{2}, 3^\frac{3}{2}, 5^\frac{3}{2}, 7^\frac{3}{2}, 9^\frac{3}{2}, \ldots, 119^\frac{3}{2} } </math>;<br />
<br />
Taking the sum over <math> \displaystyle{F(x)} </math> for <math> \displaystyle{x} </math> between 1 and 60, observe that the majority of the terms in <math> \displaystyle{a^3} </math> and <math> \displaystyle{b^3} </math> cancel out, leaving:<br />
<br />
<math><br />
\begin{align}<br />
\sum_{x=1}^{60}F(x) &= \frac{121^\frac{3}{2} - 1^\frac{3}{2}}{2}\\<br />
&=\frac{(11^2)^\frac{3}{2} - 1}{2}\\<br />
&= 665<br />
\end{align}<br />
</math><br />
<br />
Thus, sum of <math> \displaystyle{F(x)} </math> for <math> \displaystyle{x} </math> between 1 and 60 is 665. The solution can be generalized as equal to <math> \displaystyle{\frac{a(x_{max}^3) - b(x_{min}^3)}{2}} </math>. Where <math> \displaystyle{a} </math> and <math> \displaystyle{b} </math> are interpreted as functions of <math> \displaystyle{x} </math>.<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 [[wikipedia:probability|probability]] of a future event by analysis of data from past events of similar nature.<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]]''' Entrance Exam:<br />
<br />
===Problem===<br />
A number sequence <math> \displaystyle{ \{F(n)\} }</math> that can be defined as <math> \displaystyle{F(1) = 1, F(2) = 1}</math>, <math> \displaystyle{F(n+2) = F(n) + F(n+1)} </math> (where <math> \displaystyle{n \in \mathbb{N}} </math>) 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 <math> \displaystyle{ \{X(n)\} } </math> (where n is any natural number), in which each digit is either 0 or 1, set by the following rules:<br />
<br />
(i) <math> \displaystyle{X(1) = 1} </math><br />
<br />
(ii) We define <math> \displaystyle{X(n+1)} </math> as a natural number, which can be obtained by replacing the digits of <math> \displaystyle{X(n)} </math> with 1 if the digit is 0, and with 10 if the digit is 1.<br />
<br />
For example, <math> \displaystyle{X(1) = 1, X(2) = 10, X(3) = 101, X(4) = 10110, X(5) = 10110101, \ldots} </math> <br />
<br />
'' '''Anecdote''': <math> \displaystyle{X(n)} </math> 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 <math> \displaystyle{A(n)} </math>, defined as the number of digits of <math> \displaystyle{X(n)} </math>.<br />
<br />
(2) Find <math> \displaystyle{B(n)} </math>, defined as the numbers of times '01' appears in <math> \displaystyle{X(n)} </math>?<br />
For example, <math> \displaystyle{B(1) = 0, B(2) = 0, B(3) = 1, B(4) = 1, B(5) = 3, \ldots} </math><br />
<br />
<br />
===Solution===<br />
====Part 1====<br />
<br />
Let <math> \displaystyle{A(n)} </math> equal to the number of digits in X(n), which consists solely of 1s and 0s. Let's suppose <math> \displaystyle{x(n)} </math> is the number of 0s in <math> \displaystyle{X(n)} </math> at the n-th iteration (poor choice of variable by the student). Let's suppose <math> \displaystyle{y(n)} </math> is the number of 1s in <math> \displaystyle{X(n)} </math> in the n-th iteration, then <math> \displaystyle{x(n) + y(n) = A(n)} </math>. Since<br />
*Every time a 0 appears, it is replaced with 1 at the next iteration, contributing to a single 1 in <math> \displaystyle{X(n+1)} </math>.<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 <math>X(n+1)</math>.<br />
it follows that the number of 0s in the next iteration is equal to the number of 1s previously: <br />
<br />
<math> \displaystyle{x(n+1)= y(n)}</math> ;<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 />
<math>y(n+1)= \displaystyle{x(n)+y(n)} </math> .<br />
<br />
<br />
Next, prove that x(n) is a Fibonacci sequence, since we know that:<br />
<br />
*<math> \displaystyle{y(n+1) = x(n) + y(n)} </math> ;<br />
*<math> \displaystyle{x(n+2) = y(n+1)} </math> ;<br />
*<math> \displaystyle{x(n+1) = y(n)} </math> ;<br />
<br />
by substition we can show,<br />
<br />
<math><br />
\begin{align}<br />
&x(n+2)\\ <br />
&= y(n+1)\\<br />
&= x(n) + y(n)\\<br />
&= x(n) + x(n+1)<br />
\end{align}<br />
</math><br />
<br />
Hence, <math> \displaystyle{x(n+2) = x(n+1) + x(n)}</math>.<br />
<br />
Thus <math> \displaystyle{x(n)} </math> fits the definition of a Fibonacci sequence. Since <math> \displaystyle{x(n)} </math> is a Fibonacci sequence, it follows that <math> \displaystyle{y(n)} </math> is also a Fibonacci sequence (given that <math> \displaystyle{x(n+1) = y(n)}</math>). Therefore, since it has already been shown that:<br />
<br />
*<math> \displaystyle{x(n+2)=x(n+1)+x(n)} </math><br />
<br />
*<math> \displaystyle{y(n+2)=y(n+1)+y(n)} </math><br />
<br />
It follows that <math> \displaystyle {x(n+2)+y(n+2)=x(n+1)+y(n+1)+x(n)+y(n)} </math>.<br />
<br />
Recall that <math> \displaystyle{A(n)=x(n)+y(n)} </math>, therefore<br />
<br />
:<math> \displaystyle{A(n+2)=A(n+1)+A(n)} </math> .<br />
<br />
Since, <math> \displaystyle{X(1) = 1, X(2) = 10, X(3) = 101, X(4) = 10110, X(5) = 10110101, \ldots} </math><br />
<br />
It follows that <math> \displaystyle{A(1) = 1, A(2) = 2, A(3) = 3, A(4) = 5, A(5) = 8, \ldots} </math><br />
<br />
Recall by definition of the Fibonacci sequence: <math> \displaystyle{\{F(n)\} = \{1, 1, 2, 3, 5, 8, \ldots\}} </math>. Therefore <math> \displaystyle{A(n) = F(n+1)} </math>, 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 />
It can be seen that odd(n) = odd(X(n)) for all n. To prove this inductively: observe that odd(X(1)) = odd(1). Now let odd(X(n)) = odd(n), then if n is even, X(n) is even, thus X(n) ends with 0, which gets mapped to 1, making X(n+1) odd and so odd(X(n+1)) = odd(n+1). And if n is odd, then X(n) is odd, thus X(n) ends with 1, which gets mapped to 10, making X(n+1) even, and so odd(X(n+1)) = odd(n+1).<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(n) <br />
<br />
B(n) = F(n-1) - odd(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]]<br />
<br />
==Episode 11==<br />
<br />
[[File:Homura Walpurgis Ballistics Calculations.jpg|thumb|250px]]<br />
Pieces of papers floats by Homura as she faces [[Walpurgis Night]]. On it are calculations done by hand by [[Homura]]. Most likely, these are [[wikipedia:ballistic]] of the big guns. Two pairs of Xs and Ys on each of the four pieces of paper, making it 8 shots in all.<br />
<br />
== Movie 3: Rebellion ==<br />
[[File:Rebellion Math 1.jpg|thumb]]<br />
<br />
=== Problem A ===<br />
<br />
The first problem posed, (a), was to calculate <math>\int\frac{\arcsin^3x}{\sqrt{1-x^2}}\,dx</math><br />
<br />
We can perform the substitution <math>u=\arcsin x \longrightarrow \frac{du}{dx}=\frac{1}{\sqrt{1-x^2}}</math> and so rewrite the integral as <math>\int u^3\,du</math><br />
<br />
This can be easily solved via basic integration formulas: <math>\int u^3\,du=\frac{1}{4}u^4+C</math><br />
<br />
Undoing the substitution we get the solution: <math>\frac{1}{4}\arcsin^4x + C</math><br />
<br />
=== Problem B ===<br />
<br />
The second problem, (b), was to calculate <math>\int x\ln(x^2+y)\,dx</math><br />
In order to solve it we'll assume <math>\displaystyle y</math> to be a parameter independent from the value of <math>\displaystyle x</math> and <math>\displaystyle\ln x</math> to denote the natural logarithm of <math>\displaystyle x</math>.<br />
<br />
Performing the substitution <math>u=x^2+y\longrightarrow\frac{du}{dx}=2x</math> we can rewrite the integral as <math>\frac{1}{2}\int\ln u\,du</math><br />
<br />
This can be solved integrating by parts: <math>\frac{1}{2}\int \ln u\,du=\frac{1}{2}(u\ln u - \int1\,du)=\frac{1}{2}u(\ln u - 1)</math><br />
<br />
And the final answer in obtained undoing the substitution: <math>\frac{1}{2}(x^2+y)\ln((x^2+y)-1)+C</math><br />
<br />
=== Problem C ===<br />
<br />
One of the problems posed, (c), was to calculate <math>\int\frac{x^3+2x^2+10x}{x^2-x+1}\, dx</math>.<br />
<br />
The answer is to first simplify the function:<br />
<math>\frac{x^3+2x^2+10x}{x^2-x+1} = x + 3 + \frac{12x - 3}{x^2-x+1} = x + 3 + 6\frac{2x - 1}{x^2-x+1} + \frac{3}{x^2-x+1}</math>.<br />
Note that <math>x^2-x+1</math> has only simple complex zeros.<br />
<br />
It holds <math>\int x + 3 \, dx = \frac{1}{2}x^2 + 3x + C</math>. <br />
<br />
The general formula <math>\int\frac{f'(x)}{f(x)}\, dx = \ln|f(x)| dx + C</math> can be used to calculate <math>\int \frac{2x - 1}{x^2-x+1} \, dx = \ln(x^2-x+1) + C</math>.<br />
<br />
<br />
For the last term, <math>\int\frac{dx}{a^2 + x^2} = \frac{1}{a}\arctan \frac{x}{a} + C</math> can be used (with some additional calculations) to obtain the solution<br />
<br />
<math>\int\frac{x^3+2x^2+10x}{x^2-x+1}\, dx = \frac{1}{2}x^2 + 3x + 6\ln(x^2-x+1) + 2\sqrt{3} \arctan \frac{2x-1}{\sqrt{3}} + C</math></div>LanceLongFeihttps://wiki.puella-magi.net/w/index.php?title=Puella_Magi_Madoka_Magica_Side_Story:_Magia_Record&diff=89766Puella Magi Madoka Magica Side Story: Magia Record2020-03-28T22:06:12Z<p>LanceLongFei: /* Anime Adaptation */ Grammar improvements</p>
<hr />
<div>[[File:Magia record.PNG|500px|thumb|right]]<blockquote>Guided by strange power, magical girls gather in this town. <br \ >There is a place where a magical girl can stay as she is. <br \ > They're fighting with a new power from witches.</blockquote><br />
<br />
[http://magireco.com/ {{nihongo|'''''Puella Magi Madoka Magica Side Story: Magia Record'''''|マギアレコード 魔法少女まどか☆マギカ外伝|Magia Recōdo: Mahō Shōjo Madoka Magika Gaiden|}}], also known as {{nihongo|'''''MagiReco'''''|マギレコ|}} for short, is a smartphone-exclusive role-playing game distributed by Aniplex and developed by F4samurai which was released on August 22, 2017. Originally slated for Mid-May 2017, the game was pushed back for further development. The game is available on iOS and Android. It follows the story of Iroha Tamaki, a magical girl who arrives at [[Kamihama City]] to find her missing younger sister.<br />
<br />
The theme song of the first part of the story is called {{nihongo|"[[Kakawari]]"|かかわり||"Connection"}}, and the theme song of the sencd part is called {{nihongo|"[[Utsuroi]]"|うつろい||"Change"}}; both were written and composed by Sho Watanabe and sung by TrySail, a group consisting of Momo Asakura, Sora Amamiya, and Shina Natsukawa, the voice actors for Iroha, Yachiyo, and Tsuruno respectively.<br />
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*Read the <u>[[Magia Record Getting Started | Getting Started]]</u> page to learn how to download and play the game.<br />
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==Summary==<br />
It was first announced in 2011 that a Gaiden work of Puella Magi Madoka☆Magica" would be developed. On September 22-25, 2016 the game was officially announced at the SHAFT 40th anniversary event "Madogatari"[http://www.4gamer.net/games/358/G035840/20160930121/] . Atsuhiro Iwakami, the producer of the original anime [[Manga Time Kirara Magica Vol.30|said]] "Since it seems that it will take some more time until the anime's new work, the idea that I want to make a title that "Madoka Magica" is active will work." <br />
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Ume Aoki was responsible for the design of the original character that serves as the game's protagonist, as well as designing more than 10 new and original characters for the game. Gekidan Inu Curry is also responsible for the designs of the new witches. In addition to this, Shaft has produced the opening animation and magical girl transformation scenes.<br />
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:The '''[[Magia Record Story]]''' page includes translations of some parts of the game's story. Magia Record has a total of six story modes:<br />
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:*'''[[Magia Record Main Story|Main Story]]''', which details [[Iroha Tamaki|Iroha's]] quest to find her sister amid the mysteries of Kamihama City. Part One consists of ten chapters.<br />
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:*'''[[Magia Record Another Story|Another Story]]''', which focuses on what other magical girls were doing during the Main Story. There are currently nine chapters (starting with chapter 2).<br />
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:*'''[[Magia Record Mirrors Story|Mirrors]]''', the PvP mode. The story of Mirrors is that Iroha and her companions are exploring the barrier of the Mirror Witch. Mirrors is divided into levels which are unlocked by playing the mode. Unlocking levels displays new story scenes. <br />
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:*Each character has their own '''Side Story''' that is three chapters long. (Some characters have a fourth chapter with no story that is used to unlock their Doppel.)<br />
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::Certain characters have '''Costume Stories''' which are basically side stories tied to limited edition costumes.<br />
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:*'''[[Magia Record Events|Events]]''', which are one-time stories with different types of gameplay.<br />
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==Gameplay==<br />
Iroha Tamaki is the main character of the story. The gameplay follows a "Main Story", which is the focus of the game, a "Side Story" which is a timeline different from the Main Story, and "Character Story" where the background of each character is further developed. It features the magical girl cast from the anime and four original magical girls created for the game . The game will also feature other characters from the spinoff mangas such as Oriko and Kirika. The game uses character designs, animation, and music from the anime as well as incorporating new designs and songs for the dialogue scenes. Characters are superdeformed or "chibi" in the fight scenes. Players can also team up in a move called 'Connect' to defeat powerful enemies. The game will also feature a new gameplay element called "Doppel" in which magical girls can summon their witch forms and use new powers in combat.<br />
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*For details on how the game works, see the <u>[[Magia Record Gameplay]]</u> page.<br />
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===A battle against other players unfolds in the [[Mirror Witch]]'s Barrier===<br />
Within the mirror witch's barrier, a battle with the other player's reflections will unfold. Pass through several of the "mirror's layers" and aim for the center of the barrier. Unlock more mirror battles by advancing through the story quests.<br />
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The magical girls you fight in PvP are mere copies, not real magical girls. The mirror witch can make copies of magical girls who enter her barrier. The more time a girl spends in the barrier, the better the quality of the copy. Initially you are limited to a certain number of Battle Points which you use to initiate combat in the Mirror Witch's Barrier. As time passes, your Battle Points will replenish.<br />
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===Doppel===<br />
[[File:Doppel witch.PNG|300px|thumb|right]]<br />
According to the official website ''"In Kamihama City, some magical girls activate the power of "Doppel". This is the ability to activate and freely exercise the power of the inner witch when the soul gem gets cloudy, only the awakened magical girl will have it!"''<br />
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In general, a magical girl changes into a witch when her soul gem, which is the main source of magical power that is created as a result of contracting, becomes cloudy. But in the world of Kamihama City the "Doppel" power is invoked instead. A magical girl's appearance changes into that of their witch form but will still leave her magical girl form intact resulting in a hybrid of their magical girl and witch forms. When doppel is activated, the magical girl is able to use powerful techniques they would not normally be able to use.<br />
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==Characters==<br />
''See [[Magia Record Characters]]''<br />
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The game introduces a truly staggering amount of new magical girls. Twenty-two entirely new characters were playable on the game's launch, and other magical girls have been added as the game continues. Most of these characters are (currently) irrelevant to the main plot, though each playable magical girl gets their own special side-story.<br />
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As stated earlier, Ume Aoki was hired for the original character designs of several new magical girls that will feature in the game. In addition, other magical girls from the spin-off series such as [[Manga Time Kirara Magica]] series are also playable. The main characters are:<br />
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* '''[[Iroha Tamaki]]'''<br />
The main character of the game. Iroha became a magical girl to cure her little sister, Ui's illness. When Ui went missing she cam to Kamihama City to search for her. She is cheerful and earnest, but she pays too much attention to other people's needs. "Wait for me, Ui. Your big sister will find you."<br />
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* '''[[Madoka Kaname]]'''<br />
A magical girl with tremendous potential who works in Mitakihara City. She was fighting with a witch with her teammates, but she came to Kamihama City to look for Homura Akemi who suddenly disappeared. "Help is coming, Homura!"<br />
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* '''[[Yachiyo Nanami]]'''<br />
A magical girl who has been battling Witches in Kamihama City for seven years. She works alone and doesn't associate with others. She documents all of Kamihama City's rumors. "If you've learned your lesson, just stop. Life as a magical girl is no picnic."<br />
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* '''[[Felicia Mitsuki]]'''<br />
A mercenary magical girl who goes berserk if she sees a Witch. Felicia is selfish and lazy and always demands attention, but she is second to none in battle. "I'll rid the world of Witches!"<br />
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* '''[[Tsuruno Yui]]'''<br />
A foolhardy Magical Girl whose goal is to become the most powerful ever. She respects and loves her grandfather and her birthplace, which has a long, storied history. She is proud of the Chinese restaurant that her family has been running since her great-grandfather's time. “These are words I love: We won’t know if we don’t give a try! Leaving things to chance isn’t good? If we stopped before that, nothing would happen!”<br />
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* '''[[Sana Futaba]]'''<br />
A pessimistic Magical Girl who lacks confidence. She is submissive to people who need her, and if she's fond of them, she will quietly stand by them. "What… was I born for? I need to change, don't I?"<br />
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* '''[[Homura Akemi]]'''<br />
A timid magical girl who wished to become a being that could protect Madoka. A second year in Mitakihara Middle School. She is repeating the same time frame over and over to try and change the future where Madoka transforms into a witch. When she got a chance, she came to Kamihama City in hopes to find a way to change the future. Her appearance is that of the braided Homura from [[Episode 10]]. "With this, Madoka will be saved."<br />
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* '''[[Momoko Togame]]'''<br />
Like an older-sis magical girl, she always helps beginners when they’re in trouble. As a middle child of brothers in family, her character is frankly boyish. However, the truth is she loves adorable things like plush dolls. And because of that feeling, she’s also a fan of female idols. “Nothing to worry about, just leave it to this sis!”<br />
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==Magia☆Report==<br />
An original series by PAPA started on the game's official website on March 10, 2017. PAPA had previously serialized "The Veranda of Madoka" at Manga Time Kirara ☆ Magica. The comic updates every 3 to 4 days and explains the original animation and comments on the gameplay while exchanging gags. The original comic can be found [https://magireco.com/comic/01.html here] and translations are available [https://gamepress.gg/magiarecord/other/magiareport-season-1-archive here].<br />
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==Stage Play==<br />
A stage play adaptation will be run in Japan [https://www.animenewsnetwork.com/news/2018-07-20/magia-record-puella-magi-madoka-magica-side-story-stage-play-reveals-cast/.134486 from August 24th through September 9th]<br />
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According to secondhand accounts, the stage play retells the first five Chapters of Magia Record's Main Story, with several alterations to the plot. For example, Tsuruno replaces Momoko in Chapter 1.<br />
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==Manga Adaptation==<br />
[https://www.animenewsnetwork.com/news/2018-07-24/magia-record-puella-magi-madoka-magica-side-story-game-gets-manga/.134626 A manga adaptation is serialized] in ''Manga Time Kirara Forward'', which started August 24th. The artist is Fujino Fuji.<br />
* See also [[Manga Time Kirara Forward Magia Record Serialization]]<br />
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==Anime Adaptation==<br />
An [[Magia Record Anime|anime]] adaptation [https://anime.magireco.com/ was announced] on September 1st, and it began its emission on January 4, 2020.<br />
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==Law of Cyclimpics==<br />
On April 1, 2019 a side bonus app was released on the google app store. Called "Law of Cyclimpics" ({{Lang|ja|円環の理大運動会}}), it features 3 mini games with a Health-Sports Day theme and a roster of various characters that can be unlocked by gaining new personal high scores or by sharing your high scores with others on twitter. The first mini game is called "Run away from despair! Heart pounding action" ({{Lang|ja|絶望から逃げろ! ドキドキがけっこ}}) and features a magical girl running a race away from a witch. The second game is called "Delivery to space! Kyubey epidemic" ({{Lang|ja|宇宙まで届け!キュゥべえ投げ}}) and has a magical girl in a hammer throw comptetition, complete with Kyubey hammer. The final game is called "Defeat despair! Stand against the witch and topple her over" ({{Lang|ja|絶望に抗え!誓り立つ魔女棒倒し}}) and has a magical girl attempt to destroy a witch and her familiars within 30 seconds or less, causing more damage to the witch as more familiars are destroyed. A competition was also announced with 3 random winners chosen from the pool of people posting their high scores on twitter. The winners will receive a "Madocancer" statue like the one featured in Anime Japan 2019.<br />
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* [[Magireco April Fools App|Gallery]]<br />
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==Galleries==<br />
* [[Gallery:Magia Record Game|Game Gallery]]<br />
** [[Magical Girl Stickers]]<br />
* [[Gallery:Magia Record Merchandise|Merchandise]]<br />
* [[Gallery:Magia Record Advertisements|Advertisements]]<br />
* [[Gallery:Magia Record Events 2018|Events 2018]]<br />
* [[Gallery:Magia Record Events 2019|Events 2019]]<br />
* [[Gallery:Magia Record Events 2020|Events 2020]]<br />
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==External links==<br />
*[http://magireco.com Official Magia Record Website - Japanese]<br />
*[https://magiarecord-en.com/ Official Magia Record Website - English]<br />
*[https://anime.magireco.com/ Official Magia Record anime Website]<br />
*[https://twitter.com/magireco Official Twitter]<br />
*[https://twitter.com/MagiaRecordEN Official English Twitter]<br />
*[https://magireco.gamerch.com/ Japanese Gamerch wiki]<br />
*[https://gamewith.jp/magireco/ Japanese GameWith wiki]<br />
*[http://appmedia.jp/magireco Japanese Appmedia wiki]<br />
*[http://muketsu.info/magireco/index.html Japanese Muketsu]<br />
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