数学 - Pytho - 代数学 - 整数 - 数学的帰納法と除法の定理(二項定理、組み合わせ、二項係数)

代数系入門

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代数系入門 (松坂 和夫(著)、岩波書店)の第1章(整数)、2(数学的帰納法と除法の定理)、問題5.を取り組んでみる。

5. 
   
   $\begin{array}{}\underset{r=0}{\overset{1}{\sum \; <!--\; n-ary\; summation\; -->}}(\begin{array}{c}1\\ r\end{array})\end{array}{x}^1-r_{y}r=(\begin{array}{c}1\\ 0\end{array})x^1{y}^0+(\begin{array}{c}1\\ 1\end{array})x^0{y}^1\\ =x+y$

   $\begin{array}{}{\left(x+y\right)}^n\\ =\left(x+y\right){\left(x+y\right)}^{n-1}\\ =\left(x+y\right)\underset{r=0}{\overset{n-1}{\sum \; <!--\; n-ary\; summation\; -->}}(\begin{array}{c}n-1\\ r\end{array})\end{array}{x}^{\left(n-1\right)-r}{\left(y\right)}^r=\underset{r=0}{\overset{n-1}{\sum \; <!--\; n-ary\; summation\; -->}}(\begin{array}{c}n-1\\ r\end{array})x^{n-r}{\left(y\right)}^r+\underset{r=0}{\overset{n-1}{\sum \; <!--\; n-ary\; summation\; -->}}(\begin{array}{c}n-1\\ r\end{array})x^{\left(n-1\right)-r}{\left(y\right)}^{r+1}\\ =\underset{r=0}{\overset{n-1}{\sum \; <!--\; n-ary\; summation\; -->}}(\begin{array}{c}n-1\\ r\end{array})x^{n-r}{\left(y\right)}^r+\underset{r=1}{\overset{n}{\sum \; <!--\; n-ary\; summation\; -->}}(\begin{array}{c}n-1\\ r-1\end{array})x^{n-r}{\left(y\right)}^r\\ =(\begin{array}{c}n-1\\ 0\end{array})x^n+\underset{r=1}{\overset{n-1}{\sum \; <!--\; n-ary\; summation\; -->}}(\begin{array}{c}n-1\\ r\end{array})x^{n-r}{\left(y\right)}^r+\underset{r=1}{\overset{n-1}{\sum \; <!--\; n-ary\; summation\; -->}}(\begin{array}{c}n-1\\ r-1\end{array})x^{n-r}{\left(y\right)}^r+(\begin{array}{c}n-1\\ n-1\end{array})y^n\\ =x^n+\underset{r=1}{\overset{n-1}{\sum \; <!--\; n-ary\; summation\; -->}}\left((\begin{array}{c}n-1\\ r\end{array})+(\begin{array}{c}n-1\\ r-1\end{array})\right)x^{n-r}{\left(y\right)}^r+y^n$

   ここで

   $\begin{array}{}\left(n_r-1\right)+(\begin{array}{c}n-1\\ r-1\end{array})\end{array}=\frac{\left(n-1\right)!}{r!\left(n-1-r\right)!}+\frac{\left(n-1\right)!}{\left(r-1\right)!\left(n-1-r+1\right)!}\\ =\frac{\left(n-1\right)!}{r!\left(n-r-1\right)!}+\frac{\left(n-1\right)!}{\left(r-1\right)!\left(n-r\right)!}\\ =\frac{\left(\left(n-r\right)+r\right)\left(n-1\right)!}{r!\left(n-r\right)!}\\ =\frac{n!}{r!\left(n-r\right)!}\\ =(\begin{array}{c}n\\ r\end{array})$

   よって

   $\begin{array}{}{\left(x+y\right)}^n\\ =x^n+\underset{r=1}{\overset{n-1}{\sum \; <!--\; n-ary\; summation\; -->}}(\begin{array}{c}n\\ r\end{array})\end{array}{x}^{n-r}{\left(y\right)}^r+y^n=\underset{r=0}{\overset{n}{\sum \; <!--\; n-ary\; summation\; -->}}(\begin{array}{c}n\\ r\end{array})x^{n-r}{\left(y\right)}^r$

   ゆえに帰納法により、すべての正の整数に対して問題の等式が成り立つ。（証明終）

コード(Emacs)

Python 3

#!/usr/bin/env python3
from sympy import pprint, symbols, factorial, summation
import random

print('5.')
x, y = symbols('x, y', real=True)
n = symbols('n', integer=True, positive=True)
r = symbols('r', integer=True)


def comb(n, r):
    return factorial(n) / (factorial(r) * factorial(n - r))

expr1 = (x + y) ** n
expr2 = summation(comb(n, r) * (x ** (n - r)) * (y ** r), (r, 0, n))

for t in [expr1, expr2, expr1 == expr2]:
    pprint(t)
    print()

for _ in range(5):
    expr = (expr1 - expr2).subs({x: random.random()
                                 * 10, y: random.random() * 10})
    print(expr)
    for n0 in range(1, 6):
        pprint(float(expr.subs({n: n0})))
    print()

入出力結果(Terminal, Jupyter(IPython))

$ ./sample5.py
5.
       n
(x + y) 

   ⎛⎧           n                  ⎞
   ⎜⎪    ⎛    y⎞            │y│    ⎟
   ⎜⎪    ⎜1 + ─⎟        for ─── ≤ 1⎟
   ⎜⎪    ⎝    x⎠            │x│    ⎟
   ⎜⎪                              ⎟
   ⎜⎪  n                           ⎟
 n ⎜⎪ ____                         ⎟
x ⋅⎜⎨ ╲                            ⎟
   ⎜⎪  ╲     -r  r                 ⎟
   ⎜⎪   ╲   x  ⋅y ⋅n!              ⎟
   ⎜⎪   ╱  ───────────   otherwise ⎟
   ⎜⎪  ╱   r!⋅(n - r)!             ⎟
   ⎜⎪ ╱                            ⎟
   ⎜⎪ ‾‾‾‾                         ⎟
   ⎝⎩r = 0                         ⎠

False

0
0.0
0.0
0.0
0.0
0.0

0
0.0
0.0
0.0
0.0
0.0

0
0.0
0.0
0.0
0.0
0.0

13.0597580799118**n - 5.76745887013144**n*Sum(5.76745887013144**(-r)*7.29229920978039**r*factorial(n)/(factorial(r)*factorial(n - r)), (r, 0, n))
-5.944494821485156e-16
1.366347006568968e-14
1.223615482318982e-13
-6.6597079572399804e-12
-1.6640724589167732e-11

-1.90692947958286**n*Sum(1.90692947958286**(-r)*2.78454558430684**r*factorial(n)/(factorial(r)*factorial(n - r)), (r, 0, n)) + 4.6914750638897**n
7.749650795345911e-16
2.2012324034062304e-15
2.198690027880009e-14
1.7931365695109826e-13
9.157060898030184e-13

$