京都大学 情報学研究科 知能情報学専攻 2022年8月実施 情報学基礎 F1-2

Author

Isidore, 祭音Myyura

Description

設問1

以下の関数の \(x\) に関する \(n\) 階導関数を求めよ。ただし \(a\) は実数、かつ \(a>0\)、\(a \neq 1\) である。

• (1) \(\log_e x\)
• (2) \(a^x\)
• (3) \(x^2e^x\)
• (4) \(\frac{1}{x^2-1}\)

設問2

\(z = f(x,y)\), \(x = e^u \cos v\), \(y = e^u \sin v\) とする。\(\frac{\partial^2 z}{\partial u^2} + \frac{\partial^2 z}{\partial v^2}\) を \(x, y, \frac{\partial^2 z}{\partial x^2}, \frac{\partial^2 z}{\partial y^2}\) で表せ。

設問3

以下の積分を求めよ。計算過程を明示すること。

• (1) \(\int^{\infty}_{-\infty} e^{-x^2} dx\)
• (2) \(\int_0^{\infty} \int_0^{\infty} (ax^2 + by^2)e^{-(ax^2 + by^2)} dxdy\)、但し、\(a>0\) かつ \(b>0\) とし、(1) の結果を用いてよい。

Kai

設問1

(1)

\[ f^{(n)}(x) = (-1)^{n+1}\frac{(n-1)!}{x^n} \]

(2)

\[ f^{(n)}(x) = a^x(\log_e a)^n \]

(3)

\[ \begin{aligned} f^{(n)}(x) &= \sum_{k=0}^{n}\binom{n}{k}(x^{2})^{(k)}(e^{x})^{(n-k)} \\ &= \sum_{k=0}^{2}\binom{n}{k}(x^{2})^{(k)}(e^{x})^{(n-k)} \\ &= \binom{n}{0}(x^{2})^{(0)}(e^{x})^{(n)}+\binom{n}{1}(x^{2})^{(1)}(e^{x})^{(n-1)}+\binom{n}{2}(x^{2})^{(2)}(e^{x})^{(n-2)}\\ &= x^{2}e^{x}+2nxe^{x}+n(n-1)e^{x} \\ &= \left(x^{2}+2nx+n(n-1)\right)e^{x} \end{aligned} \]

(4)

\[ f^{(n)}(x) = \frac{n!}{2}(-1)^{n}\left\{(x-1)^{-n-1}-(x+1)^{-n-1}\right\} \]

設問2

\[ \frac{\partial^2 z}{\partial u^2} + \frac{\partial^2 z}{\partial v^2} = (\frac{\partial^2 z}{\partial x^2} + \frac{\partial^2 z}{\partial y^2})(x^2 + y^2) \]

設問3

(1)

Perform the substitution \(x^2 = u\), we have

\[ \int^{\infty}_{-\infty}e^{-x^2}\mathrm{d}x = 2\int^{\infty}_{0}e^{-x^2}\mathrm{d}x = \int^{\infty}_{0}u^{-\frac{1}{2}}e^{-u}\mathrm{d}u \]

By the properties of Gamma Function, the above integral equals

\[ \int^{\infty}_{0}u^{\frac{1}{2}-1}e^{-u}\mathrm{d}u = \Gamma(\frac{1}{2}) = \sqrt{\pi} \]

(2)

Let \(u=\sqrt{a}x,v=\sqrt{b}y\).

\[ \begin{aligned} \int_{0}^{\infty}\int_{0}^{\infty}(ax^{2}+by^{2})e^{-(ax^{2}+by^{2})}dxdy &= \frac{1}{\sqrt{ab}}\int_{0}^{\infty}\int_{0}^{\infty}(u^{2}+v^{2})e^{-(u^{2}+v^{2})}dudv\\ &= \frac{2}{\sqrt{ab}}\int_{0}^{\infty}\int_{0}^{\infty}u^{2}e^{-(u^{2}+v^{2})}dudv\\ &= \frac{2}{\sqrt{ab}}\int_{0}^{\infty}u^{2}e^{-u^{2}}du\int_{0}^{\infty}e^{-v^{2}}dv\\ &= \frac{\sqrt{\pi}}{\sqrt{ab}}\int_{0}^{\infty}u^{2}e^{-u^{2}}du \end{aligned} \]

\[ \begin{aligned} \int_{0}^{\infty}u^{2}e^{-u^{2}}du &= \int_{0}^{\infty}u\cdot (ue^{-u^{2}})du\\ &= \int_{0}^{\infty}u\cdot\left(\frac{e^{-u^{2}}}{-2}\right)^{\prime}du\\ &= \left[\frac{ue^{-u^{2}}}{-2}\right]_{0}^{\infty}+\frac{1}{2}\int_{0}^{\infty}e^{-u^{2}}du\\ &= \frac{1}{4}\int_{-\infty}^{\infty}e^{-u^{2}}du \\ &= \frac{\sqrt{\pi}}{4} \end{aligned} \]

Hence the result is \(\frac{\pi}{4\sqrt{ab}}\).