## 2019年4月14日日曜日

### 数学 - Python - 解析学 - 数 - 数列と級数 - 数列(不等式、帰納法)

1. $\begin{array}{l}{\left(1+h\right)}^{n}-\left(1+nh+\frac{n\left(n-1\right)}{2}{h}^{2}\right)\\ ={\left(1+h\right)}^{n-1}\left(1+h\right)-\left(1+\left(n-1\right)h+h+\frac{\left(n-1\right)\left(n-2+2\right)}{2}{h}^{2}\right)\\ ={\left(1+h\right)}^{n-1}+h{\left(1+h\right)}^{n-1}-\\ \left(1+\left(n-1\right)h+h+\frac{\left(n-1\right)\left(n-2\right)}{2}{h}^{2}+\left(n-1\right){h}^{2}\right)\\ ={\left(1+h\right)}^{n-1}-\left(1+\left(n-1\right)h+\frac{\left(n-1\right)\left(n-2\right)}{2}{h}^{2}\right)+\\ h{\left(1+h\right)}^{n-1}-\left(h-\left(n-1\right){h}^{2}\right)\\ \ge h{\left(1+h\right)}^{n-1}-h+\left(n-1\right){h}^{2}\\ \ge h\left(1+\left(n-1\right)h\right)-h+\left(n-1\right){h}^{2}\\ =h+\left(n-1\right){h}^{2}-h+\left(n-1\right){h}^{2}\\ =0\end{array}$

よって、 帰納法により、

${\left(1+h\right)}^{n}\ge 1+nh+\frac{n\left(n-1\right)}{2}{h}^{2}$

（証明終）

コード

Python 3

#!/usr/bin/env python3
from sympy import symbols,  pprint, I, Rational, sqrt
from sympy.plotting import plot3d

print('3.')

h, n = symbols('h, n')
p = plot3d((1 + h) ** n - (1 + n * h + n * (n - 1) / 2 * h ** 2),
(h, 0.1, 5),
(n, 2, 10),
show=False)
p.xlabel = h
p.ylabel = n

p.show()
p.save('sample3.png')


C:\Users\...>py sample3.py
3.

C:\Users\...>