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
Description
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)
\[
f^{(n)}(x) = (x^2+2x+2)e^x
\]
(4)
\[
f^{(n)}(x) = (-\frac{1}{2})^n(n-1)![\frac{1}{(x-1)^(n+1)}-\frac{1}{(x+!)^(n+1)}]
\]
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 the asked integral be denoted as \(I\), we have
\[
\begin{align}
I &= \int^{\infty}_{-\infty}\int^{\infty}_{-\infty}ax^2e^{-(ax^2+by^2)}\mathrm{d}x\mathrm{d}y + \int^{\infty}_{-\infty}\int^{\infty}_{-\infty}by^2e^{-(ax^2+by^2)}\mathrm{d}x\mathrm{d}y \\
&= \int^{\infty}_{-\infty}e^{-by^2}\mathrm{d}y\int^{\infty}_{-\infty}ax^2e^{-ax^2}\mathrm{d}x + \int^{\infty}_{-\infty}e^{-ax^2}\mathrm{d}x\int^{\infty}_{-\infty}by^2e^{-by^2}\mathrm{d}y
\end{align}
\]
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
\[
\int^{\infty}_{-\infty}x^2e^{-x^2}\mathrm{d}x = \frac{\sqrt{\pi}}{2}
\]
Insert the above 2 integrals, we have the answer:
\[
I = \frac{\pi}{\sqrt{ab}}
\]