## 2019年9月16日月曜日

### 数学 - Python - 解析学 - 級数 - 絶対収束と交代級数の収束 - 累乗(べき乗、有理数)

1. $\sum \left|\frac{{\left(-1\right)}^{n}}{{n}^{\frac{5}{2}}+n}\right|=\sum \frac{1}{{n}^{\frac{5}{2}}+n}$

よって絶対収束する。

ゆえに収束する。

コード

Python 3

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

print('18.')

n = symbols('n')
f = (-1) ** n / (n ** Rational(5, 2) + n)
s1 = summation(f, (n, 1, oo))
s2 = summation(abs(f), (n, 1, oo))
for o in [s1, s2]:
pprint(o)
print()

def g(m):
return sum([f.subs({n: n0}) for n0 in range(1, m)])

def h(m):
return sum([abs(f.subs({n: n0})) for n0 in range(1, m)])

p = plot(f, abs(f),
(n, 1, 11),
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('sample18.png')

ms = range(1, 11)
plt.plot(ms, [g(m) for m in ms],
ms, [h(m) for m in ms],
ms, [f.subs({n: m}) for m in ms])
plt.legend(['Σ (-1)^n * 1 / (n^(5/2) + n)',
'Σ |(-1)^n * 1 / (n^(5/2) + n)|',
'(-1)^n * 1 / (n^(5/2) + n)',
'|(-1)^n * 1 / (n^(5/2) + n)|'])
plt.savefig('sample18.png')


C:\Users\...>py sample18.py
18.
∞
_____
╲
╲         n
╲    (-1)
╲  ────────
╱   5/2
╱   n    + n
╱
╱
‾‾‾‾‾
n = 1

∞
____
╲
╲    -π⋅im(n) │   1    │
╲  ℯ        ⋅│────────│
╱            │ 5/2    │
╱             │n    + n│
╱
‾‾‾‾
n = 1

c:\Users\...>