---
title: "東京大学大学院　情報理工入試過去問（数学）解答・２０１８年　第三問"
tags: u-tokyo, exam
---

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$$
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# 東京大学大学院　情報理工入試過去問（数学）<br><span style="float: right;">解答・２０１８年　第三問</span>

+ こちらの解答は非公式に作成されており、大学院側に認められたものではありません。正確性についての保証は致しかねます。
+ 問題本文は研究科のウェブサイトから見ることができます。
+ 満足出来た解答でしたらコメントしていただけると幸いです。

## ２０１８年　第三問
### (1) Prove that $\begin{cases}n \equiv 0\pmod{2} &\rightarrow& \mathrm{Im}(z_n) = 0\\ n \equiv 1\pmod{2} &\rightarrow& \mathrm{Re}(z_n) = 0\end{cases}$
Let $\ell$ be the number of red card taken, and $n - \ell$ be the number of white card taken, with $0 \leq \ell \leq n$. So,
$$
\begin{align}
z_n &= i^{\ell}(-i)^{n - \ell} \cdot z_0 \\[1ex]
&= i^\ell [(-1)^{n - \ell}i^{n - \ell}] \cdot z_0 \\[1ex]
&= i^{\ell + (n - \ell)}(-1)^{n - \ell} \cdot z_0 \\[1ex]
&= i^n(-1)^{n - \ell} \cdot z_0
\end{align}
$$

If $n$ is even,
$$
i^n(-1)^{n - \ell} \cdot z_0 = (-1)^{n/2}(-1)^{n - \ell} \cdot z_0 = (-1)^{\frac{3}{2}n - \ell} \cdot z_0 = (-1)^{\frac{3}{2}n - \ell}
$$
Since $n$ is even, $(-1)^{\frac{3}{2}n - \ell} = (-1)^m$ for some $m \in \mathbb{N}$, or $m = 0$ when $n = \ell = 0$, resulting in $z_n = \pm 1$, thus $n \equiv 0\pmod{2} \rightarrow \mathrm{Im}(z_n) = 0$.

----

If $n$ is odd,
$$
i^n(-1)^{n - \ell} \cdot z_0 = i(-1)^{\frac{n - 1}{2}}(-1)^{n - \ell} \cdot z_0 = i(-1)^{\frac{3n - 1\;}{2} - \ell} \cdot z_0 = (-1)^{\frac{3n - 1\;}{2} - \ell}i
$$
Since $n$ is odd, $(-1)^{\frac{3n - 1\;}{2} - \ell} = (-1)^s$ where $s \in \mathbb{N}$, or $s = 0$ when $n = \ell = 1$, resulting in $z_n = \pm i$, thus $n \equiv 1\pmod{2} \rightarrow \mathrm{Re}(z_n) = 0$.

### (2) Find the recurrence equation for $\prob{z_n = 1} = P_n$ and $\prob{z_n = i} = Q_n$.
Since $z_0 = 1$, we can first find the following initial conditions:
$$
P_0 = 1, Q_0 = 0, P_1 = 0, Q_1 = \frac{2}{3}
$$

And the relation between $P_n$ and $Q_n$ being:
$$
P_n = \prob{z_n = 1} = \frac{1}{3}\prob{z_{n - 1} = i} + \frac{2}{3}\prob{z_{n - 1} = -i} = \frac{1}{3}Q_{n - 1} + \frac{2}{3}\prob{z_{n - 1} = -i}\\
Q_n = \prob{z_n = i} = \frac{2}{3}\prob{z_{n - 1} = 1} + \frac{1}{3}\prob{z_{n - 1} = -1} = \frac{2}{3}P_{n - 1} + \frac{1}{3}\prob{z_{n - 1} = -1}\\
$$

+ Consider when $n$ is odd, $n - 1$ is even, from question (1) we know that:
$$
\mathrm{Re}(z_n) = 0 \rightarrow \prob{z_n = 1} = 0, \prob{z_n = -1} = 0 \rightarrow \prob{z_{n} = -i} = 1 - Q_{n}\\
\mathrm{Im}(z_{n - 1}) = 0 \rightarrow \prob{z_{n - 1} = i} = 0, \prob{z_{n - 1} = -i} = 0 \rightarrow \prob{z_{n - 1} = -1} = 1 - P_{n - 1}
$$
Thus the relation become:
$$
P_n = 0 \\
Q_n = \frac{2}{3}P_{n - 1} + \frac{1}{3}\prob{z_{n - 1} = -1} = \frac{2}{3}P_{n - 1} + \frac{1}{3}(1 - P_{n - 1}) = \frac{1}{3} + \frac{1}{3}P_{n - 1}
$$

+ Consider when $n$ is even, $n - 1$ is odd, from question (1) we know that:
$$
\mathrm{Im}(z_n) = 0 \rightarrow \prob{z_n = i} = 0, \prob{z_n = -i} = 0 \rightarrow \prob{z_{n - 1} = -1} = 1 - P_{n - 1}\\
\mathrm{Re}(z_{n - 1}) = 0 \rightarrow \prob{z_{n - 1} = 1} = 0, \prob{z_{n - 1} = -1} = 0 \rightarrow \prob{z_{n - 1} = -i} = 1 - Q_{n - 1}
$$
Thus the relation become:
$$
P_n = \frac{1}{3}Q_{n - 1} + \frac{2}{3}\prob{z_{n - 1} = -i} = \frac{1}{3}Q_{n - 1} + \frac{2}{3}(1 - Q_{n - 1}) = \frac{2}{3} - \frac{1}{3}Q_{n - 1} \\
Q_n = 0
$$

Then, for some even $n$ where $n \geq 0$, we get:
$$
P_{n + 2} = \frac{2}{3} - \frac{1}{3}Q_{n + 1} = \frac{2}{3} - \frac{1}{3}\bigg( \frac{1}{3} + \frac{1}{3}P_{n} \bigg) = \frac{5}{9} - \frac{1}{9}P_{n}
$$

And for some odd $n$ when $n \geq 1$, we get:
$$
Q_{n + 2} = \frac{1}{3} + \frac{1}{3}P_{n + 1} = \frac{1}{3} + \frac{1}{3}\bigg( \frac{2}{3} - \frac{1}{3}Q_{n} \bigg) = \frac{5}{9} - \frac{1}{9}Q_{n}
$$

Resulting in the recurrence equations:
$$
\begin{array}{c | c}
\begin{cases}
P_0 = 1 \\[1ex]
P_{n + 2} = \frac{5}{9} - \frac{1}{9}P_{n} &, n > 0, n \equiv 0 \pmod 2\\
P_n = 0 &, n \equiv 1 \pmod 2
\end{cases} &
\begin{cases}
Q_1 = \frac{2}{3} \\[1ex]
Q_n = 0 &, n \equiv 0 \pmod 2\\
Q_{n + 2} = \frac{5}{9} - \frac{1}{9}Q_{n} &, n > 1, n \equiv 1 \pmod 2
\end{cases}
\end{array}
$$

### (3) Find the probability for $\prob{z_n = 1}$, $\prob{z_n = -1}$, $\prob{z_n = i}$ and $\prob{z_n = -i}$.
:::info
In case someone did try to solve the following monstrosity using matrix diagonalization or some other magic, Heres the solution:
$$
\begin{cases}
P(z_n = 1) &= P_0 = 1 \\
P(z_n = i) &= Q_0 = 0 \\
P(z_n = -1) &= R_0 = 0 \\
P(z_n = -i) &= S_0 = 0 \\
P_{n + 1} &= \displaystyle\frac{1}{3}(Q_n + 2S_n) \\
Q_{n + 1} &= \displaystyle\frac{1}{3}(R_n + 2P_n) \\
R_{n + 1} &= \displaystyle\frac{1}{3}(S_n + 2Q_n) \\
S_{n + 1} &= \displaystyle\frac{1}{3}(P_n + 2R_n)
\end{cases} \Rightarrow
\begin{align}
\displaystyle \prob{z_n = 1} &= \frac{1}{4}\bigg( (-1)^n + (-\frac{i}{3})^n + (\frac{i}{3})^n + 1 \bigg) \\
\displaystyle \prob{z_n = 1} &= \frac{1}{4}\bigg(-(-1)^n + i(-\frac{i}{3})^n - i(\frac{i}{3})^n + 1 \bigg) \\
\displaystyle \prob{z_n = 1} &= \frac{1}{4}\bigg( (-1)^n - (-\frac{i}{3})^n - (\frac{i}{3})^n + 1 \bigg) \\
\displaystyle \prob{z_n = 1} &= \frac{1}{4}\bigg(-(-1)^n - i(-\frac{i}{3})^n + i(\frac{i}{3})^n + 1 \bigg)
\end{align}
$$
:::
First, we know that:
+ When $n$ is even, $\prob{z_n = i} = \prob{z_n = -i} = 0, \prob{z_n = -1} = 1 - \prob{z_n = 1}$
+ When $n$ is odd, $\prob{z_n = 1} = \prob{z_n = -1} = 0, \prob{z_n = -i} = 1 - \prob{z_n = i}$

Let $P_{2k}, \;k = 0, 1, 2 \ldots$ form another sequence, sequence $A_0 = P_0, A_1 = P_2, A_2 = P_4 \ldots$, then we get recurrence equation $\begin{cases}A_0 = 1 \\ A_{k + 1} = \frac{5}{9} - \frac{1}{9}A_{k}\end{cases}$, which the characteristic equation is:
$$
\alpha = -\frac{1}{9}\alpha + \frac{5}{9} \Rightarrow \alpha = \frac{1}{2}
$$

Inducing that we can rewrite the recurrence equation as:
$$
\begin{align}
A_{k + 1} - \frac{1}{2} = -\frac{1}{9}(A_k - \frac{1}{2}) &\Rightarrow A_{k + 1} - \frac{1}{2} = \bigg(-\frac{1}{9}\bigg)^{k+1}\bigg( A_0 - \frac{1}{2} \bigg) = \frac{1}{2}\bigg(-\frac{1}{9}\bigg)^{k+1} \\
&\Rightarrow A_{k} = \frac{1}{2}\bigg(-\frac{1}{9}\bigg)^{k} + \frac{1}{2}
\end{align}
$$

So the formula of $\prob{z_n = 1}$ and $\prob{z_n = -1}$ becomes:
$$
\prob{z_n = 1} = P_n = \begin{cases}
\displaystyle \frac{1}{2} + \frac{1}{2}\bigg( -\frac{1}{9} \bigg)^\frac{n}{2} &, n \equiv 0 \pmod 2 \\
0 &, n \equiv 1 \pmod 2
\end{cases} \\
\prob{z_n = -1} = \begin{cases}
\displaystyle \frac{1}{2} - \frac{1}{2}\bigg( -\frac{1}{9} \bigg)^\frac{n}{2} &, n \equiv 0 \pmod 2 \\
0 &, n \equiv 1 \pmod 2
\end{cases}
$$

----

Let $Q_{2t + 1}, \;t = 0, 1, 2 \ldots$ form another sequence $B_0 = Q_1, B_1 = Q_3, B_2 = Q_5, \ldots$, then we can get recurrence equation $\begin{cases}B_0 = \frac{2}{3} \\ B_{t + 1} = \frac{5}{9} - \frac{1}{9}B_{t}\end{cases}$, which the characteristic equation is:
$$
\beta = -\frac{1}{9}\beta + \frac{5}{9} \Rightarrow \beta = \frac{1}{2}
$$
Inducing we can rewrite the recurrence equation as:
$$
\begin{align}
B_{t + 1} - \frac{1}{2} = -\frac{1}{9}(B_t - \frac{1}{2}) &\Rightarrow B_{t + 1} - \frac{1}{2} = \bigg(-\frac{1}{9}\bigg)^{t+1}\bigg( B_0 - \frac{1}{2} \bigg) = \frac{1}{6}\bigg(-\frac{1}{9}\bigg)^{t+1} \\
&\Rightarrow B_{t} = \frac{1}{6}\bigg(-\frac{1}{9}\bigg)^{t} + \frac{1}{2}
\end{align}
$$

So the formula of $\prob{z_n = i}$ and $\prob{z_n = -i}$ becomes:
$$
\prob{z_n = i} = Q_n = \begin{cases}
0 &, n \equiv 0 \\
\displaystyle \frac{1}{2} + \frac{1}{6}\bigg( -\frac{1}{9} \bigg)^{\frac{n - 1}{2}} &, n \equiv 1
\end{cases} \\
\prob{z_n = -i} = \begin{cases}
0 &, n \equiv 0 \\
\displaystyle \frac{1}{2} - \frac{1}{6}\bigg( -\frac{1}{9} \bigg)^{\frac{n - 1}{2}} &, n \equiv 1
\end{cases}
$$

----

Concluding the result:
$$
\prob{z_n = 1} = P_n = \begin{cases}
\displaystyle \frac{1}{2} + \frac{1}{2}\bigg( -\frac{1}{9} \bigg)^\frac{n}{2} &, n \equiv 0 \pmod 2 \\
0 &, n \equiv 1 \pmod 2
\end{cases} \\
\prob{z_n = i} = Q_n = \begin{cases}
0 &, n \equiv 0 \pmod 2 \\
\displaystyle \frac{1}{2} + \frac{1}{6}\bigg( -\frac{1}{9} \bigg)^{\frac{n - 1}{2}} &, n \equiv \pmod 2
\end{cases}
$$

### (4) Show that $\excp{z_n} = \displaystyle \Big(\frac{i}{3}\Big)^n$
$$
\excp{z_n} = \overbrace{1 \cdot \prob{z_n = 1} - 1 \cdot \prob{z_n = -1}}^{=0\text{ if } n \text{ is odd}} + \overbrace{i \cdot \prob{z_n = i} - i \cdot \prob{z_n = -i}}^{=0\text{ if } n \text{ is even}}
$$

+ If $n$ is even, we get:
$$
\begin{align}
\excp{z_n} &= \prob{z_n = 1} - \prob{z_n = -1} + 0 \\[1ex]
&= \displaystyle \frac{1}{2} + \frac{1}{2}\Big( -\frac{1}{9} \Big)^{\frac{n}{2}} - \Big[\frac{1}{2} - \frac{1}{2}\Big( -\frac{1}{9} \Big)^{\frac{n}{2}}\Big] \\[1ex]
&= \Big( -\frac{1}{9} \Big)^{\frac{n}{2}} \\[1ex]
&= (-1)^{\frac{n}{2}}\Big(\frac{1}{9}\Big)^{\frac{n}{2}} \\[1ex]
&= i^n \cdot \Big( \frac{1}{3} \Big)^n = \Big(\frac{i}{3}\Big)^n
\end{align}
$$

+ If $n$ is odd, we get
$$
\begin{align}
\excp{z_n} &= 0 + i\prob{z_n = i} - i\prob{z_n = -i} \\[1ex]
&= i\Big[ \frac{1}{2} + \frac{1}{6}\bigg( -\frac{1}{9} \bigg)^{\frac{n - 1}{2}} \Big] - i\Big[ \frac{1}{2} - \frac{1}{6}\bigg( -\frac{1}{9} \bigg)^{\frac{n - 1}{2}} \Big] \\[1ex]
&= i \Big[ \frac{1}{3}\Big( -\frac{1}{9} \Big)^{\frac{n - 1}{2}} \Big] \\[1ex]
&= \frac{i}{3}\Big[ (-1)^{\frac{n - 1}{2}}\Big( \frac{1}{9} \Big)^{\frac{n - 1}{2}} \Big] \\[1ex]
&= \frac{i}{3} \cdot i^{n - 1} \cdot \Big(\frac{1}{3}\Big)^{n - 1} \\[1ex]
&= \frac{i}{3} \cdot \Big( \frac{i}{3} \Big)^{n - 1} = \Big( \frac{i}{3} \Big)^{n}
\end{align}
$$

### (5) Find $\prob{z_n = w_n}$
We can see $z_n$ as a unit vector rotating on the complex plane, doing $90 ^{\circ}$ jumps.

+ In every step, $z_n$ has a $\frac{2}{3}$ chance to go counter-clockwisely, and a $\frac{1}{3}$ chance to jump clockwisely.
+ In every step, $w_n$ has a $\frac{2}{3}$ chance to go clockwisely, and a $\frac{1}{3}$ chance to jump counter-clockwisely.

Which we can see that $w_n$ is just the opposite of $z_n$. Hence, the probability of $z_n$, $w_n$ respectively on each spot become:

<span style="position: relative; left: -165px;">

$$
\begin{array}{ccccc | ccccc}
\prob{z_n = 1} &=& P_n &=&
\begin{cases}
\displaystyle \frac{1}{2} + \frac{1}{2}\bigg( -\frac{1}{9} \bigg)^\frac{\;n\;}{2} &, n \equiv 0 \pmod 2 \\
0 &, n \equiv 1 \pmod 2
\end{cases} &
\prob{w_n = 1} &=& P_n &=&
\begin{cases}
\displaystyle \frac{1}{2} + \frac{1}{2}\bigg( -\frac{1}{9} \bigg)^\frac{\;n\;}{2} &, n \equiv 0 \pmod 2 \\
0 &, n \equiv 1 \pmod 2
\end{cases} \\[1ex]
\prob{z_n = -1} &&&=&
\begin{cases}
\displaystyle \frac{1}{2} - \frac{1}{2}\bigg( -\frac{1}{9} \bigg)^\frac{\;n\;}{2} &, n \equiv 0 \pmod 2 \\
0 &, n \equiv 1 \pmod 2
\end{cases} &
\prob{w_n = -1} &&&=&
\begin{cases}
\displaystyle \frac{1}{2} - \frac{1}{2}\bigg( -\frac{1}{9} \bigg)^\frac{\;n\;}{2} &, n \equiv 0 \pmod 2 \\
0 &, n \equiv 1 \pmod 2
\end{cases} \\[1ex]
\prob{z_n = i} &=& Q_n &=&
\begin{cases}
0 &, n \equiv 0 \pmod 2 \\
\displaystyle \frac{1}{2} + \frac{1}{6}\bigg( -\frac{1}{9} \bigg)^{\frac{n - 1}{2}} &, n \equiv 1 \pmod 2
\end{cases} &
\prob{w_n = i} &&&=&
\begin{cases}
0 &, n \equiv 0 \pmod 2 \\
\displaystyle \frac{1}{2} - \frac{1}{6}\bigg( -\frac{1}{9} \bigg)^{\frac{n - 1}{2}} &, n \equiv 1 \pmod 2
\end{cases} \\[1ex]
\prob{z_n = -i} &&&=&
\begin{cases}
0 &, n \equiv 0 \pmod 2 \\
\displaystyle \frac{1}{2} - \frac{1}{6}\bigg( -\frac{1}{9} \bigg)^{\frac{n - 1}{2}} &, n \equiv 1 \pmod 2
\end{cases} &
\prob{w_n = -i} &=& Q_n &=&
\begin{cases}
0 &, n \equiv 0 \pmod 2 \\
\displaystyle \frac{1}{2} + \frac{1}{6}\bigg( -\frac{1}{9} \bigg)^{\frac{n - 1}{2}} &, n \equiv 1 \pmod 2
\end{cases} \\[1ex]
\end{array}
$$

</span>

And since $z_n$ and $w_n$ are independent, we can get:
+ For when $n$ is even,
$$
\begin{align}
\prob{z_n = w_n} &= \prob{z_n = 1}\prob{w_n = 1} + \prob{z_n = -1}\prob{w_n = -1} + 0 + 0 \\[1ex]
&= \Big[ \frac{1}{2} + \frac{1}{2}\Big( -\frac{1}{9} \Big)^{\frac{\,n\,}{2}} \Big]^2 + \Big[ \frac{1}{2} - \frac{1}{2}\Big( -\frac{1}{9} \Big)^{\frac{\,n\,}{2}} \Big]^2 \\[1ex]
&= 2\Big[ \Big( \frac{1}{2} \Big)^2 + \Big( \frac{1}{2}\Big( -\frac{1}{9} \Big)^\frac{\,n\,}{2} \Big)^2 \Big] \\[1ex]
&= 2\Big[ \frac{1}{4} + \frac{1}{4}\Big( -\frac{1}{9} \Big)^n \Big] \\[1ex]
&= \frac{1}{2} + \frac{1}{2}\Big( -\frac{1}{9} \Big)^n
\end{align}
$$

+ For when $n$ is odd,
$$
\begin{align}
\prob{z_n = w_n} &= 0 + 0 + \prob{z_n = i}\prob{w_n = i} + \prob{z_n = -i}\prob{w_n = -i} \\[1ex]
&= 2 \cdot \Big[ \frac{1}{2} + \frac{1}{6}\Big( -\frac{1}{9} \Big)^{\frac{\,n-1\,}{2}} \Big]\Big[ \frac{1}{2} - \frac{1}{6}\Big( -\frac{1}{9} \Big)^{\frac{\,n-1\,}{2}} \Big] \\[1ex]
&= 2 \cdot \Big[ \Big( \frac{1}{2} \Big)^2 - \Big( \frac{1}{6}\Big( -\frac{1}{9} \Big)^{\frac{\,n-1\,}{2}} \Big)^2 \Big] \\[1ex]
&= 2 \cdot \Big[ \frac{1}{4} - \frac{1}{6}\Big( -\frac{1}{9}\Big)^{n - 1} \Big] \\[1ex]
&= \frac{1}{2} - \frac{1}{3}\Big( -\frac{1}{9}\Big)^{n - 1} \\[1ex]
&= \frac{1}{2} + 3\Big( -\frac{1}{9}\Big)^{n}
\end{align}
$$

### (6) Find $\excp{z_n + w_n}$.
$$
\excp{z_n + w_n} = \excp{z_n} + \excp{w_n}
$$

+ For $\excp{z_n}$, use the result of (4). $\excp{z_n} = \displaystyle\Big( \frac{i}{3} \Big)^n$ for all $n$.
+ For $\excp{w_n}$:
$$
\excp{w_n} = \excp{w_n} = \overbrace{1 \cdot \prob{w_n = 1} - 1 \cdot \prob{w_n = -1}}^{=0\text{ if } n \text{ is odd}} + \overbrace{i \cdot \prob{w_n = i} - i \cdot \prob{w_n = -i}}^{=0\text{ if } n \text{ is even}}
$$
- If $n$ is even:
$$
\excp{w_n} = \prob{w_n = 1} - \prob{w_n = -1} + 0 = \Big( \frac{i}{3} \Big)^n
$$
- If $n$ is odd:
$$
\begin{align}
\excp{w_n} &= 0 + i\prob{w_n = i} - i\prob{w_n = -i} \\[1ex]
&= i\Big[ \frac{1}{2} - \frac{1}{6}\bigg( -\frac{1}{9} \bigg)^{\frac{n - 1}{2}} \Big] - i\Big[ \frac{1}{2} + \frac{1}{6}\bigg( -\frac{1}{9} \bigg)^{\frac{n - 1}{2}} \Big] \\[1ex]
&= i\Big[ -\frac{2}{6} \bigg( -\frac{1}{9} \bigg)^{\frac{n - 1}{2}} \Big] \\[1ex]
&= -\frac{i}{3} (-1)^{\frac{n - 1}{2}} \Big( \frac{1}{9} \Big)^{\frac{n - 1}{2}} \\[1ex]
&= -\frac{i}{3} i^{n - 1} \Big( \frac{1}{3} \Big)^{n - 1} \\[1ex]
&= - \Big( \frac{i}{3} \Big) \Big( \frac{i}{3} \Big)^{n - 1} = -\Big(\frac{i}{3} \Big)^{n}
\end{align}
$$

Thus,
$$
\excp{z_n + w_n} = \begin{cases}
2\Big( \frac{i}{3} \Big)^n &, n \equiv 0 \pmod 2 \\[1ex]
0 &, n \equiv 1 \pmod 2
\end{cases}
$$

### (7) Find $\excp{z_nw_n}$.
Since $z_n$ and $w_n$ are independent, $\excp{z_nw_n} = \excp{z_n}\excp{w_n}$. Thus using the result from (6),

$$
\excp{z_nw_n} = \begin{cases}
\Big( \frac{i}{3} \Big)^{2n} &, n \equiv 0 \pmod 2 \\[1ex]
-\Big( \frac{i}{3} \Big)^{2n} &, n \equiv 1 \pmod 2
\end{cases}
$$