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 fix

docs/kakomonn/kyoto_university/informatics/ist_202208_kiso_f1_2.md

@@ -7,12 +7,25 @@ tags:
 # 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](https://github.com/heacsing)
+[Isidore](https://github.com/heacsing), 祭音Myyura
 
 ## **Description**
-<figure style="text-align:center;">
-  <img src="https://s2.loli.net/2024/06/27/fSDUupmWIkTFJr5.png" width="640"/>
-</figure>
+### 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
@@ -31,13 +44,19 @@ $$
 #### (3)
 
 $$
-f^{(n)}(x) = (x^2+2x+2)e^x
+\begin{aligned}
+f^{(n)}(x) &= \sum_{k=0}^{n}\binom{n}{k}(x^{2})^{(k)}(e^{x})^{(n-k)} \\
+&= \sum_{k=0}^{2}\binom{n}{k}(x^{2})^{(k)}(e^{x})^{(n-k)} \\
+&= \binom{n}{0}(x^{2})^{(0)}(e^{x})^{(n)}+\binom{n}{1}(x^{2})^{(1)}(e^{x})^{(n-1)}+\binom{n}{2}(x^{2})^{(2)}(e^{x})^{(n-2)}\\
+&= x^{2}e^{x}+2nxe^{x}+n(n-1)e^{x} \\
+&= \left(x^{2}+2nx+n(n-1)\right)e^{x}
+\end{aligned}
 $$
 
 #### (4)
 
 $$
-f^{(n)}(x) = (-\frac{1}{2})^n(n-1)![\frac{1}{(x-1)^(n+1)}-\frac{1}{(x+!)^(n+1)}]
+f^{(n)}(x) = \frac{n!}{2}(-1)^{n}\left\{(x-1)^{-n-1}-(x+1)^{-n-1}\right\}
 $$
 
 ### 設問2
@@ -61,30 +80,26 @@ $$
 $$
 
 #### (2)
-
-Let the asked integral be denoted as $I$, we have
+Let $u=\sqrt{a}x,v=\sqrt{b}y$.
 
 $$
-\begin{align}
-    I &= \int^{\infty}_{-\infty}\int^{\infty}_{-\infty}ax^2e^{-(ax^2+by^2)}\mathrm{d}x\mathrm{d}y + \int^{\infty}_{-\infty}\int^{\infty}_{-\infty}by^2e^{-(ax^2+by^2)}\mathrm{d}x\mathrm{d}y \\
-    &= \int^{\infty}_{-\infty}e^{-by^2}\mathrm{d}y\int^{\infty}_{-\infty}ax^2e^{-ax^2}\mathrm{d}x + \int^{\infty}_{-\infty}e^{-ax^2}\mathrm{d}x\int^{\infty}_{-\infty}by^2e^{-by^2}\mathrm{d}y
-\end{align}
+\begin{aligned}
+\int_{0}^{\infty}\int_{0}^{\infty}(ax^{2}+by^{2})e^{-(ax^{2}+by^{2})}dxdy
+&= \frac{1}{\sqrt{ab}}\int_{0}^{\infty}\int_{0}^{\infty}(u^{2}+v^{2})e^{-(u^{2}+v^{2})}dudv\\
+&= \frac{2}{\sqrt{ab}}\int_{0}^{\infty}\int_{0}^{\infty}u^{2}e^{-(u^{2}+v^{2})}dudv\\
+&= \frac{2}{\sqrt{ab}}\int_{0}^{\infty}u^{2}e^{-u^{2}}du\int_{0}^{\infty}e^{-v^{2}}dv\\
+&= \frac{\sqrt{\pi}}{\sqrt{ab}}\int_{0}^{\infty}u^{2}e^{-u^{2}}du
+\end{aligned}
 $$
 
-By (1) we have,
-
-$$
-\int^{\infty}_{-\infty}e^{-x^2}\mathrm{d}x = \sqrt{\pi}
-$$
-
-Perform the integration by parts ($e^{-x^2} = (x)'e^{-x^2}$) to the above integral, we have
-
 $$
-\int^{\infty}_{-\infty}x^2e^{-x^2}\mathrm{d}x = \frac{\sqrt{\pi}}{2}
+\begin{aligned}
+\int_{0}^{\infty}u^{2}e^{-u^{2}}du &= \int_{0}^{\infty}u\cdot (ue^{-u^{2}})du\\
+&= \int_{0}^{\infty}u\cdot\left(\frac{e^{-u^{2}}}{-2}\right)^{\prime}du\\
+&= \left[\frac{ue^{-u^{2}}}{-2}\right]_{0}^{\infty}+\frac{1}{2}\int_{0}^{\infty}e^{-u^{2}}du\\
+&= \frac{1}{4}\int_{-\infty}^{\infty}e^{-u^{2}}du \\
+&= \frac{\sqrt{\pi}}{4}
+\end{aligned}
 $$
 
-Insert the above 2 integrals, we have the answer:
-
-$$
-I = \frac{\pi}{\sqrt{ab}}
-$$
+Hence the result is $\frac{\pi}{4\sqrt{ab}}$.