## 2019年8月28日水曜日

### 数学 - Python - 解析学 - 級数 - 積分による判定法 - 対数関数の累乗(べき乗)、累乗(べき乗)、逆数、収束、部分積分方

1. $\begin{array}{l}\int \frac{1}{x{\left(\mathrm{log}x\right)}^{1+\epsilon }}\mathrm{dx}\\ =\frac{\mathrm{log}x}{{\left(\mathrm{log}x\right)}^{1+\epsilon }}-\int \left(\mathrm{log}x\right)·\frac{-1}{{\left(\mathrm{log}x\right)}^{2\left(1+\epsilon \right)}}\left(1+\epsilon \right){\left(\mathrm{log}x\right)}^{\epsilon }·\frac{1}{x}\mathrm{dx}\\ =\frac{1}{{\left(\mathrm{log}x\right)}^{\epsilon }}+\left(1+\epsilon \right)\int \frac{1}{x{\left(\mathrm{log}x\right)}^{1+\epsilon }}\mathrm{dx}\\ -\epsilon \int \frac{1}{x{\left(\mathrm{log}x\right)}^{1+\epsilon }}\mathrm{dx}=\frac{1}{{\left(\mathrm{log}x\right)}^{\epsilon }}\\ \int \frac{1}{x{\left(\mathrm{log}x\right)}^{1+\epsilon }}=-\frac{1}{\epsilon {\left(\mathrm{log}x\right)}^{\epsilon }}\end{array}$

よって、

$\begin{array}{l}\underset{b\to \infty }{\mathrm{lim}}{\int }_{2}^{b}\frac{1}{x{\left(\mathrm{log}x\right)}^{1+\epsilon }}\mathrm{dx}\\ =\underset{b\to \infty }{\mathrm{lim}}\left(-\frac{1}{\epsilon }{\left[\frac{1}{{\left(\mathrm{log}x\right)}^{\epsilon }}\right]}_{2}^{b}\right)\\ =-\frac{1}{\epsilon }\underset{b\to \infty }{\mathrm{lim}}\left(\frac{1}{{\left(\mathrm{log}b\right)}^{\epsilon }}-\frac{1}{{\left(\mathrm{log}2\right)}^{\epsilon }}\right)\\ =\frac{1}{\epsilon {\left(\mathrm{log}2\right)}^{\epsilon }}\end{array}$

ゆえに、 問題の無限級数は収束する。

コード

Python 3

#!/usr/bin/env python3
from sympy import pprint, symbols, summation, oo, Integral, plot, log
import matplotlib.pyplot as plt

print('13.')

n = symbols('n')
epsilon = 2
f = 1 / (n * log(n) ** (1 + epsilon))
s = summation(f, (n, 2, oo))
I = Integral(f, (n, 2, oo))

for o in [s, I, I.doit()]:
pprint(o)
print()

p = plot(f,
(n, 2, 12),
legend=True,
show=False)
colors = ['red', 'green', 'blue', 'brown', 'orange',
'purple', 'pink', 'gray', 'skyblue', 'yellow']

for s, color in zip(p, colors):
s.line_color = color

p.show()
p.save('sample13.png')

def g(m):
return sum([f.subs({n: k}) for k in range(2, m)])

ms = range(2, 12)
plt.plot(ms, [g(m) for m in ms])
plt.legend(['Σ 1 / n(log n) ^ (1+ε)', '1 / n(log n)^(1+ε)'])
plt.savefig('sample13.png')


C:\Users\...>py sample12.py
13.
∞
____
╲
╲       1
╲  ─────────
╱       3
╱   n⋅log (n)
╱
‾‾‾‾
n = 2

∞
⌠
⎮     1
⎮ ───────── dn
⎮      3
⎮ n⋅log (n)
⌡
2

1
─────────
2
2⋅log (2)

c:\Users\...>