2019年12月26日木曜日

学習環境

解析入門(上) (松坂和夫 数学入門シリーズ 4) (松坂 和夫(著)、岩波書店)の第8章(積分の計算)、8.1(不定積分の計算)、問題4の解答を求めてみる。



    1. ddxsinm+1xcosn·1xm+n+n-1m+nsinmxcosn-2xdx=m+1sinmxcosnx+sinm+1xn-1cosn-2x-sinxm+n+n-1sinmxcosn-2xm+n=m+1sinmxcosnx-n-1sinn+2xcosn-2x+n-1sinmxcosn-2xm+n=sinmxcosn-2xm+1cos2x-n-1sin2x+n-1m+n=sinmxcosn-2xmcos2x-nsin2x+nm+n=sinmxcosn-2xmcos2x-n1-cos2x+nm+n=sinmxcosnx

      よって、

      Im,n=sinm+1xcosn-1xm+n+n-1m+nIm,n-2

      が成り立つ。

      ddx-sinm-1xcosn+1xm+n+m-1m+nIm-2,n=-m-1sinm-2xcosn+2x+sinmxn+1cosnxm+n+m-1sinm-2xcosnxm+n=sinm-2xcosnx-m-1cos2x+n+1sin2x+m-1m+n=sinm-2xcosnx-m1-sin2x+nsin2x+mm+n=sinmxcosnxIm,n=-sinm-1xcosn+1xn1+n+m-1m+nIm-2,n

    2. ddx-sinm+1xcosn+1xn+1+m+n+2n+1Im,n+2=-m+1sinmxcosn+2x+n+1sinm+2xcosnxn+1+m+n+2sinmxcosn+2xn+1=sinmxcosnx-m+1cos2x+n+1sin2x+m+n+2cos2xn+1=sinmxcosnxn+1cos2x+sin2xn+1=sinnxcosnx

      (証明終)


    3. ddxsinm+1xcosn+1xm+1+m+n+2m+1Im+2,n=m+1sinmxcosn+2x-n+1sinm+2xcosnxm+1+m+n+2sinn+2xcosnxm+1=sinmxcosnxm+1cos2x-n+1sin2x+m+n+2sin2xm+1=sinmxcosnxm+1cos2x+m+1sin2xm+1=sinmxcosnx

      (証明終)

コード

#!/usr/bin/env python3
from sympy import pprint, symbols, Integral, sin, cos, plot

print('4.')

x = symbols('x')
m, n = symbols('m, n', integer=True)
f = sin(x) ** m * cos(x) ** n

fs = [f.subs({m: m0, n: n0})
      for m0 in range(-1, 2)
      for n0 in range(-1, 3)]

for g in fs:
    I = Integral(g, x)
    for o in [I, I.doit()]:
        pprint(o)
        print()

p = plot(*fs,
         (x, -5, 5),
         ylim=(-5, 5),
         legend=False,
         show=False)
colors = ['red', 'green', 'blue', 'brown', 'orange',
          'purple', 'pink', 'gray', 'skyblue', 'yellow']

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

for o in zip(fs, colors):
    pprint(o)

p.show()
p.save('sample4.png')

入出力結果(Zsh、PowerShell、Terminal、Jupyter(IPython))

% ./sample4.py
4.
⌠                 
⎮       1         
⎮ ───────────── dx
⎮ sin(x)⋅cos(x)   
⌡                 

     ⎛   2       ⎞              
  log⎝sin (x) - 1⎠              
- ──────────────── + log(sin(x))
         2                      

⌠          
⎮   1      
⎮ ────── dx
⎮ sin(x)   
⌡          

log(cos(x) - 1)   log(cos(x) + 1)
─────────────── - ───────────────
       2                 2       

⌠          
⎮ cos(x)   
⎮ ────── dx
⎮ sin(x)   
⌡          

log(sin(x))

⌠           
⎮    2      
⎮ cos (x)   
⎮ ─────── dx
⎮  sin(x)   
⌡           

log(cos(x) - 1)   log(cos(x) + 1)         
─────────────── - ─────────────── + cos(x)
       2                 2                

⌠          
⎮   1      
⎮ ────── dx
⎮ cos(x)   
⌡          

  log(sin(x) - 1)   log(sin(x) + 1)
- ─────────────── + ───────────────
         2                 2       

⌠     
⎮ 1 dx
⌡     

x

⌠          
⎮ cos(x) dx
⌡          

sin(x)

⌠           
⎮    2      
⎮ cos (x) dx
⌡           

x   sin(x)⋅cos(x)
─ + ─────────────
2         2      

⌠          
⎮ sin(x)   
⎮ ────── dx
⎮ cos(x)   
⌡          

-log(cos(x))

⌠          
⎮ sin(x) dx
⌡          

-cos(x)

⌠                 
⎮ sin(x)⋅cos(x) dx
⌡                 

   2   
sin (x)
───────
   2   

⌠                  
⎮           2      
⎮ sin(x)⋅cos (x) dx
⌡                  

    3    
-cos (x) 
─────────
    3    

⎛      1           ⎞
⎜─────────────, red⎟
⎝sin(x)⋅cos(x)     ⎠
⎛  1          ⎞
⎜──────, green⎟
⎝sin(x)       ⎠
⎛cos(x)      ⎞
⎜──────, blue⎟
⎝sin(x)      ⎠
⎛   2          ⎞
⎜cos (x)       ⎟
⎜───────, brown⎟
⎝ sin(x)       ⎠
⎛  1           ⎞
⎜──────, orange⎟
⎝cos(x)        ⎠
(1, purple)
(cos(x), pink)
⎛   2         ⎞
⎝cos (x), gray⎠
⎛sin(x)         ⎞
⎜──────, skyblue⎟
⎝cos(x)         ⎠
(sin(x), yellow)
%

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