院試問題　東大　複雑理工学専攻　平成31年　専門基礎科目

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院試問題解答作成基本方針

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複雑理工学専攻

1

単純な計算問題なので省略.

2

線形代数の基本的な問題です.

(1)
This is obvious.

 (1)    \begin{align*} A = \left( \begin{matrix} 0 & 1 \\ 1 & 1 \\ \end{matrix} \right).  \end{align*}


(2)
For \lambda \in \mathbb{C}, we have

 (2)    \begin{align*} \det (\lambda - A) = \lambda^2 - \lambda - 1.  \end{align*}

Therefore \det (\lambda - A) = 0 \Leftrightarrow \lambda = (1 \pm \sqrt{5})/2, and we obtain

 (3)    \begin{align*} \lambda_\pm = \frac{1 \pm \sqrt{5}}{2}.  \end{align*}


(3)
Since it holds that

 (4)    \begin{align*} (\lambda_\pm - A) = \left( \begin{matrix} \lambda_\pm & -1 \\ -1 & - \lambda_\mp \\ \end{matrix} \right),  \end{align*}

an eigenvector u_\pm for the eigenvalue \lambda_\pm is

 (5)    \begin{align*} u_\pm = \left( \begin{matrix} 1 \\ \lambda_\pm \\ \end{matrix} \right). \end{align*}


Hence we obtain

 (6)    \begin{align*} P^{-1}AP = \left( \begin{matrix} \lambda_+ & 0 \\ 0 & \lambda_- \\ \end{matrix} \right), \end{align*}

where P = (u_+ ,u_-).
Then it follows that

 (7)    \begin{align*} P^{-1}A^nP = \left( \begin{matrix} \lambda_+^n & 0 \\ 0 & \lambda_-^n \\ \end{matrix} \right), \end{align*}

and we obtain the desired result.

(4)
Since we have

 (8)    \begin{align*} \left( \begin{matrix} x_n \\ x_{n+1} \\ \end{matrix} \right) = A^{n-1} \left( \begin{matrix} x_1 \\ x_2 \\ \end{matrix} \right), \end{align*}

we can easily check that

 (9)    \begin{align*} x_n = \frac{1}{\sqrt{5}}\left( \frac{1}{\lambda_+^2} + \frac{1}{\lambda_+} \right)\lambda_+^n + \frac{1}{\sqrt{5}}\left( \frac{1}{\lambda_-^2} + \frac{1}{\lambda_-} \right)\lambda_-^n = \frac{1}{\sqrt{5}}(\lambda_+^n - \lambda_-^n).  \end{align*}

Here we used the fact \lambda_\pm^2 - \lambda_\pm -1 = 0.

(5)
From the problem (3) and (4), we have

 (10)    \begin{align*} A^n     = \left( \begin{matrix} x_{n-1} & x_n \\ x_n & x_{n+1} \\ \end{matrix} \right). \end{align*}

Hence we obtain

 (11)    \begin{align*} y_n = a x_{n-2} + b x_{n-1}. \end{align*}


(6)
It is obvious that D_2 = 2.
By considering the cofactor expansion of D_n with respect to the first column, we have

 (12)    \begin{align*} D_n &= D_{n-1} + \det \left( \begin{matrix} 1 & 0 & \ldots & 0 \\ -1 &&& \\ 0 &&& \\ \vdots &&\text{\Huge $D_{n-2}$}& \\ 0 &&&  \end{matrix} \right) \\ &= D_{n-1} + D_{n-2}. \end{align*}

Hence from the problem (5), we obtain

 (13)    \begin{align*} D_n = x_{n-2} + 2x_{n-1}. \end{align*}

3

フーリエ変換の問題です. (4)以外は単純な計算問題です. (4)はs(\omega/2)^3の逆フーリエ変換を直接計算できるかと思いましたが, 計算がうまくいかなかったため, L^2におけるフーリエ変換のユニタリ性(特に単射性)を用いました. 少し大げさな解き方かもしれません.

(1)
We denote the Fourier transform of g as G.

(14)   \begin{align*} G(\omega) &= \int_{-\infty}^{\infty}g(t)\mathrm{e}^{-\i \omega t}dt = 2\int_{0}^{1/2} \cos \omega t dt = \frac{2 \sin (\omega /2)}{\omega} = s(\omega/2). \end{align*}



(2)

(15)   \begin{align*} h(t) &= \int_{-\infty}^{\infty}g(\tau)g(t - \tau)d\tau \\ &= \int_{-1/2}^{1/2}g(t - \tau)d\tau \\ &=\int_{t - 1/2}^{t + 1/2}g(\tau)d\tau \\ &= (1 - |t|)1\{|t| \leq 1\}. \end{align*}



(3)
From the problem (1) and (2). we have from Fubini’s theorem

(16)   \begin{align*} \int_{-\infty}^{\infty}k(t)\mathrm{e}^{-\i \omega t}dt &= \int_{-\infty}^{\infty}(g\ast g)(t)\mathrm{e}^{-\i \omega t}dt \\ &= \left(\int_{-\infty}^{\infty}g(t)\mathrm{e}^{-\i \omega t}dt\right)^2 \\ &= s(\omega/2)^2. \end{align*}



(4)
It is not difficult to check that the Fourier transform of g\ast g \ast g is s(\omega/2)^3.
Since the function s(\omega/2)^2 is square-integrable on \mathbb{R}, we have from the unitarity of the Fourier transform l = g \ast g \ast g.

4

(2)の冒頭で示す離散の場合の部分積分公式を使うと計算が楽です.

(1)

(17)   \begin{align*} \mathrm{Pr}[X \geq a] = \sum_{i = a}^{4}2^{-i} + 2^{-4} = 2^{-a+1}(1 - 2^{-5+a}) + 2^{-4} = 2^{-a + 1}. \end{align*}



(2)
Note that for every function f: \{1,2,\cdots,k\} \rightarrow \mathbb{R} and random variable W, we have

(18)   \begin{align*} \sum_{i=1}^{k}f(i)\mathrm{Pr}[W = i] &= \sum_{i=1}^{k}f(i)(\mathrm{Pr}[W \geq i] - \mathrm{Pr}[W \geq i + 1]) \label{} \\ &= \sum_{i=1}^{k}f(i)\mathrm{Pr}[W \geq i] - \sum_{i=1}^{k-1}f(i)\mathrm{Pr}[W \geq i + 1] \label{} \\ &= \sum_{i=1}^{k}f(i)\mathrm{Pr}[W \geq i] - \sum_{i=2}^{k}f(i-1)\mathrm{Pr}[W \geq i] \label{} \\ &= f(1) + \sum_{i=2}^{k}(f(i) - f(i-1))\mathrm{Pr}[W \geq i]. \end{align*}


Hence we obtain

(19)   \begin{align*} E[Z] &= \sum_{i=1}^{5}2^{i}\mathrm{Pr}[X = i] = 2 + \sum_{i = 2}^{5}2^{i-1}\mathrm{Pr}[X \geq i] = 2 + \sum_{i = 2}^{5}1 = 6, \\ E[Z^2] &= 4 + \sum_{i = 2}^{5}3 \cdot 4^{i-1}\cdot 2^{-i + 1} = 4 + 3 \sum_{i = 2}^{5}2^{i-1} = 94, \\ \mathrm{Var}(Z) &= 94 - 36 = 58. \end{align*}



(3)
From the independence of {X_n}_n, we have

(20)   \begin{align*} \mathrm{Pr}[Y_n \geq a] = \mathrm{Pr}[X_i \geq a \ \text{for all} 1 \leq i \leq n] = \mathrm{Pr}[X \geq a]^n = 2^{n(-a + 1)}. \end{align*}



(4)
From the formula (18), we obtain

(21)   \begin{align*} E[Y_n] = 1 + \sum_{i = 2}^{k} \mathrm{Pr}[Y_n \geq i] = \frac{2^n - 2^{-n(k-1)}}{2^n -1}. \end{align*}

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