关于适用mathematica化简多项式,求教简单快速的方法

2025-03-11 11:18:42
推荐回答(2个)
回答1:

化简多项式:

In[1]:= FullSimplify[x^3 - 6 x^2 + 11 x - 6]

Out[1]= (-3 + x) (-2 + x) (-1 + x)

In[2]:= FullSimplify[(x^10 - 1) (x^10 + 1)]

Out[2]= -1 + x^20

将双曲线表达式化简为指数形式:

In[1]:= FullSimplify[Cosh[x] - Sinh[x]]

Out[1]= E^-x

将指数表达式化简为三角形式:

In[1]:= FullSimplify[(1 + I) E^(-I x) + (1 - I) E^(I x)]

Out[1]= 2 (Cos[x] + Sin[x])

化简一个代数数:

In[1]:= FullSimplify[Sqrt[2] + Sqrt[3] - Sqrt[5 + 2 Sqrt[6]]]

Out[1]= 0

化简超越数:

In[1]:= FullSimplify[-I Log[(1 + 2 I)/Sqrt[5]]]

Out[1]= ArcTan[2]

In[2]:= FullSimplify[16 ArcTan[1/5] - 4 ArcTan[1/239]]

Out[2]= \[Pi]

化简包含特殊函数的表达式:

In[1]:= FullSimplify[ExpIntegralE[1 - n, x] x^n]

Out[1]= Gamma[n, x]

In[2]:= FullSimplify[Csc[Pi v] (BesselI[-v, z] - BesselI[v, z])/2]

Out[2]= BesselK[v, z]/\[Pi]

用假设化简表达式:

In[1]:= FullSimplify[ProductLog[x E^x], x >= -1]

Out[1]= x

In[2]:= FullSimplify[E^(EllipticF[x, 1]), -Pi/2 < x < Pi/2]

Out[2]= Sec[x] + Tan[x]

In[3]:= FullSimplify[EulerPhi[p^2] + p, Element[p, Primes]]

Out[3]= p^2

根据公理系统证明定理:

In[1]:= FullSimplify[f[f[b, a], a] == f[a, f[b, a]],
ForAll[{a, b}, f[a, b] == f[b, a]]]

Out[1]= True

In[2]:= FullSimplify[f[a, a] == f[a, b], ForAll[{a, b}, f[f[a, a], b] == a]]

Out[2]= True

任意表达式可以用作一个变量:

In[3]:= FullSimplify[
Subscript[a, 1]\[CirclePlus]Subscript[a, 1] ==
Subscript[a, 1]\[CirclePlus]Subscript[a, 2],
ForAll[{a, b}, (a\[CirclePlus]a)\[CirclePlus]b == a]]

Out[3]= True

在定理中,无定量的变量被当作常量处理:

In[4]:= FullSimplify[f[f[e, e], e] == e, ForAll[a, f[a, e] == a]]

Out[4]= True

假定左边的恒等性和逆的存在性,证明右边逆的存在:

In[5]:= FullSimplify[ForAll[x, Exists[y, g[x, y] == e]],
ForAll[{x, y, z},
g[x, g[y, z]] == g[g[x, y], z] && g[e, x] == x && g[inv[x], x] == e]]

Out[5]= True

多看看自带帮助(F1)

回答2:

In[2]:=y=3*x^3+a*b/c*x+2/.a*b/c->dOut[2]=2+dx+3x^3

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