京都大学 情報学研究科 知能情報学専攻 2024年8月実施 情報学基礎 F1-2

标签：

• Kyoto-University
• Differentiation
• Limit
• Triple-Integral
• Infinite-Series

Author

itsuitsuki

Description (English)

In the questions below, $\log x$ denotes the natural logarithm of $x$, and $e$ denotes Napier's constant (the base of the natural logarithm).

Q.1

Answer the following questions. Derivations must be clearly shown.

(1) Let $n$ be a positive integer. Compute the $n$-th derivative of $f(x) = x^2e^{-x}$.

(2) Compute the following limit.

$$\lim_{x \to \frac{\pi}{2}-0} \log(\tan x)^{\cos x}$$

Q.2

Compute the volume common to a sphere $x^2 + y^2 + z^2 \leq 36$ and a cylinder $x^2 + y^2 \leq 9$, $-\infty < z < \infty$. Derivation must be clearly shown.

Q.3

Answer the following questions.

(1) Show that the inequality

$$0 < e - \sum_{k=0}^{n} \frac{1}{k!} < \frac{3}{2(n+1)!}$$

holds for any positive integer $n$.

(2) Show that $e$ is an irrational number using the inequality in (1).

Kai

Q.1

(1)

Since

$$f(x)=x^2e^{-x},\quad f'(x)=2xe^{-x}-x^2e^{-x},\\ f''(x)=2e^{-x}-4xe^{-x}+x^2e^{-x},f'''(x)=-6e^{-x}+6xe^{-x}-x^2e^{-x},$$

let the $n$-th derivative be

$$f^{(n)}(x)=a_ne^{-x}+b_nxe^{-x}+(-1)^nx^2e^{-x},$$

where

$$a_0=a_1=0,\quad b_0=0,b_1=2.$$

Then

$$f^{(n+1)}(x)=-a_ne^{-x}+b_ne^{-x}-b_nxe^{-x}+2\cdot (-1)^n xe^{-x}+(-1)^{n+1}x^2e^{-x}$$

and therefore

$$a_{n+1}=b_n-a_n,\quad b_{n+1}=2(-1)^n-b_n\implies b_n=2(-1)^{n-1}n$$

so

$$a_{n+1}=2(-1)^{n-1}n-a_n=2(-1)^{n+1}n-a_n\implies {a_{n+1}\over (-1)^{n+1}}=2n+{a_n\over (-1)^n}\\\implies {a_n\over (-1)^n}=0+2(0+1+\dots+(n-1))=n^2-n\implies a_n=(-1)^n (n^2-n)$$

Hence

$$f^{(n)}(x)=(-1)^n(n^2-n)e^{-x}-2(-1)^nxe^{-x}+(-1)^nx^2e^{-x}.$$

(2)

詳しく書いてくれる方が居って頂ければ幸いです。自分が時間不足なので暫く略します。 $u=\sec x$ とする。

Q.2

詳しく書いてくれる方が居って頂ければ幸いです。自分が時間不足なので暫く略します。

$(288-108\sqrt 3)\pi$

cylinder coordinates $(x=r\cos\theta,y=r\sin\theta,z=z)$ and $V=\int_{範囲}\int_0^{2\pi}\int_0^{R=6} r\mathrm dr\mathrm d\theta\mathrm dz$

Q.3

(1)

Since $\forall k\in\Z^+,\quad \frac1{k!}>0$,

$$\sum_{k=0}^n {1\over k!} < \sum_{k=0}^{+\infty} {1\over k!} = e, \forall n=1,2,\dots$$

we have $0<e-\sum_{k=0}^n\frac1{k!}$.

And since

$$\begin{aligned} e-\sum_{k=0}^n \frac1{k!} &= {1\over (n+1)!} + {1\over (n+2)!} + {1\over (n+3)!} + \dots \\ & = {1\over (n+1)!} + {1\over (n+1)!(n+2)} + {1\over (n+1)!(n+2)(n+3)} + \dots \\ & < {1\over (n+1)!}\left[ 1+{1\over n+2} + {1\over (n+2)^2}+\dots \right] \\ & = {1\over (n+1)!}\cdot {1\over {n+1\over n+2}} \\ & = {n+2\over n+1}\cdot \frac1{(n+1)!}, \end{aligned}$$

and since

$$\begin{aligned} {n+2\over n+1}\le \frac32, \forall n=1,2,\dots, \end{aligned}$$

we can conclude

$$e-\sum_{k=0}^n\frac1{k!}<{3\over 2(n+1)!}.$$

(2)

Assume $e$ is rational, setting it $e=p/q$ where $p,q\in\Z^+$ and $\gcd(p,q)=1$.

Then

$$0<{p\over q}-\sum_{k=0}^n\frac1{k!}<{3\over 2(n+1)!}$$

where we set $n=q$, and multiply these with $q!$,

$$0<p(q-1)!-\sum_{k=0}^q \frac{q!}{k!}<{3\over 2(q+1)}<1,\quad q=1,2,\dots$$

with $p(q-1)!-\sum_{k=0}^q \frac{q!}{k!}$ should be integer but it is in $(0,1)$ by the inequality chain above, thus causing contradiction.

Hence no such $q$ exists, making $e$ irrational.