学習環境
代数への出発
(新装版 数学入門シリーズ) (松坂
和夫 (著)、岩波書店 )の第7章(不等式)の練習問題14.の解答を求めてみる。
a
3
+
b
3
-
a
2
b
-
a
b
2
=
(
a
+
b
)
3
-
3
a
2
b
-
3
a
b
2
-
a
2
b
-
a
b
2
=
(
a
+
b
)
3
-
4
a
b
(
a
+
b
)
=
(
a
+
b
)
(
(
a
+
b
)
2
-
4
a
b
)
=
(
a
+
b
)
(
a
2
+
b
2
-
2
a
b
)
=
(
a
+
b
)
(
a
-
b
)
2
≥
0
a
3
+
b
3
≥
a
2
b
+
a
b
2
2
(
a
5
+
b
5
)
-
(
a
2
+
b
2
)
(
a
3
+
b
3
)
=
2
(
a
5
+
b
5
)
-
(
a
5
+
a
2
b
3
+
a
3
b
2
+
b
5
)
=
a
5
+
b
5
-
a
2
b
3
-
a
3
b
2
=
(
a
+
b
)
5
-
5
a
4
b
-
10
a
3
b
2
-
10
a
2
b
3
-
5
a
b
4
-
a
2
b
3
-
a
3
b
2
=
(
a
+
b
)
5
-
5
a
4
b
-
11
a
3
b
2
-
11
a
2
b
3
-
5
a
b
4
=
(
a
+
b
)
5
-
a
b
(
5
a
3
+
11
a
2
b
+
11
a
b
2
+
5
b
3
)
=
(
a
+
b
)
5
-
a
b
(
5
a
3
+
15
a
2
b
+
15
a
b
2
+
5
b
2
-
4
a
2
b
-
4
a
b
2
)
=
(
a
+
b
)
5
-
a
b
(
5
(
a
+
b
)
3
-
4
a
b
(
a
+
b
)
)
=
(
a
+
b
)
(
(
a
+
b
)
4
-
a
b
(
5
(
a
+
b
)
2
-
4
a
b
)
)
=
(
a
+
b
)
(
a
4
+
4
a
3
b
+
6
a
2
b
2
+
4
a
b
3
+
b
4
-
a
b
(
5
a
2
+
6
a
b
+
5
b
2
)
)
=
(
a
+
b
)
(
a
4
-
a
3
b
-
a
b
3
+
b
4
)
=
(
a
+
b
)
(
a
3
(
a
-
b
)
-
b
3
(
a
-
b
)
)
=
(
a
+
b
)
(
a
-
b
)
(
a
3
-
b
3
)
=
(
a
+
b
)
(
a
-
b
)
2
(
a
2
+
a
b
+
b
2
)
≥
0
2
(
a
5
+
b
5
)
≥
(
a
2
+
b
2
)
(
a
3
+
b
3
)
コード
表示/非表示
#!/usr/bin/env python3
from sympy.plotting import plot3d
from sympy.abc import a, b
print ('14.' )
p = plot3d(
a ** 3 + b ** 3 ,
a ** 2 + b + a * b ** 2 ,
(a, 0 , 5 ),
(b, 0 , 5 ),
show= False
)
p. xlabel = a
p. ylabel = b
p. save('sample14_1.png' )
p = plot3d(
2 * (a ** 5 + b ** 5 ),
(a ** 2 + b ** 2 ) * (a ** 3 + b ** 3 ),
(a, 0 , 5 ),
(b, 0 , 5 ),
show= False
)
p. xlabel = a
p. ylabel = b
p. save('sample14_2.png' )
p. show()
入出力結果(Zsh、PowerShell、Terminal、Jupyter(IPython))
表示/非表示
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