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京都大学 情報学研究科 知能情報学専攻 2023年8月実施 情報学基礎 F1-2

Author

Isidore, Casablanca

Description

設問1

以下の積分を求めよ。計算過程も明示すること。

(1) \(\int_0^{\infty} \frac{1}{(x^2 + 1)^2} \text{d}x\)

(2) \(D = \left \{ (x, y) \mid x^2 + \frac{y^2}{4} \leq 1 \right \}\) としたときに、

\[ \iint_D x^2 y^2 \text{d}x \text{d}y \]

設問2

以下の問いに答えよ。計算過程も明示すること。

(1) \(\log_e (1.02)\) の小数第 \(7\) 位を四捨五入し、小数第 \(6\) 位まで求めよ。

(2) \(x > 0\) に対して、次の不等式が成り立つことを示せ。

\[ x - \frac{x^2}{2} < \log_e (1 + x) < 1 - \frac{x^2}{2} + \frac{x^3}{3} \]

設問3

\(3x^2 + 2y^2 + z^2 = 1\) の条件の下で、\(xyz\) の最大値と最小値を求めよ。

Kai

設問1

(1)

Perform the substitution \(x = \tan \theta\), the answer is \(\pi\)

(2)

Perform the substitution \(x = r \cos \theta, y = 2r \sin \theta\), the Jacobian determinant is \(2r\) and the answer is \(\frac{1}{3}\pi\)

設問2

(1)

Perform a Taylor series expansion of the function \(f(x) = \log_e(x+ \triangle x)\), then insert \(x=1, \triangle x = 0.02\). Calculating until the 5th term could lead to a result \(0.01980256\), and then round it to \(0.019803\)

(2)

(solution by Isidore)

Perform the same expansion as (1) will directly prove

\[ x-\frac{x^2}{2} < \log_e(1+x) \]

Let \(f(x) = 1-\frac{x^2}{2}+\frac{x^3}{3} - \log_e(1+x)\), then its derivative is

\[ f'(x) = \frac{x^3-x-1}{x+1} \]

We use Newton-Raphson's method to calculate the root of \(x^3-x-1 = 0\).

Starting at \(x_0 = 1.5\), the value of \(x_1\) can be calculated as following

\[ x_{1} = x_0 - \frac{x_0^3-x_0-1}{3x_0^2-1} \approx 1.3478 \]

Therefore, insert \(x=1.3478\),

\[ \min \{f\} \approx f(1.3478) = 0.05 > 0 \]

Q.E.D

(solution by Casablanca)

Easy to see that we only need to prove that:

\[ 1+x < e^{1- \frac{x^2}{2}+ \frac{x^3}{3}} \]

Let \(f(x) = e^{1- \frac{x^2}{2}+ \frac{x^3}{3}}\). Then we have \(f(x) \geq f(1) = e^{\frac 56}\).

  • (i) for \(x \in (0, 1], x+1 \leq 2 < e^{\frac 56} \leq f(x)\) \(\quad\) (note: \(e^x > 1+ x + \frac{x^2}{2}, e^{\frac 56} > \frac{157}{72}>2\) )
  • (ii) for \(x \in (1, + \infty)\), \(f'(x) = (x^2 - x)f(x) > 0\), \(f(x)\) increases
  • and we consider the point \((\frac{3}{2},e)\) on \((x,f(x))\), \(f'(\frac 32) = \frac 34 e\).

Let

\[ F(x) = f(x) - (\frac 34 e(x - \frac 32) + e) = f(x) - \frac 34 e + \frac e8, \]

where

\[ x \in (1, +\infty) \]

and we know that: \(F(\frac 32) = 0, F'(x) = (x^2-x)f(x) - \frac 34 e\), it's obvious that \(F'(x)\) increases for \(x > 1\), and \(F'(\frac 32) = 0\).

Hence \(F(x) \geq F(\frac 32 ) = 0\) for \(x > 1\).

Thus

\[ f(x) > \frac 34 ex - \frac e8, x \in (1,+\infty) \]

And consider the intersection \((x_0,y_0)\) of \(y = x+1\) and \(y = \frac 34 ex - \frac e8\), we have

\[ x_0 = \frac{8+e}{6e} < 1 \]

thus for \(x > x_0, \frac 34 ex - \frac e8 > x+1\), hence for \(x > 1\),

\[ f(x) > \frac 34 ex - \frac e8 > x+1 \]

and from (i) and (ii), finally ,we know that \(f(x) > 1+x\)

設問3

Perform the Lagrange multipliers method, we get

\[ L(x,y,z;\lambda) = xyz - \lambda(3x^2+2y^2+z^2-1) \]

Calculate its derivative with the roots, the answer is

\[ \min \{xyz\} = -\frac{1}{18}\sqrt{2}, \quad \max \{xyz\} = \frac{1}{18}\sqrt{2} \]