mathematica 化简举个简单的例子,比如:Y = (a+b+1)^2,y1 = a^2 + b^2,y2 = 2ab,显然我们知道:Y = y1 + y2 + 2a + 2b + 1,但是如何用Mathematica化简给出这样的结果呢?我试了FullSimplify[ (a + b + 1)^2,Assumptions -> y1 == a

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mathematica 化简举个简单的例子,比如:Y = (a+b+1)^2,y1 = a^2 + b^2,y2 = 2ab,显然我们知道:Y = y1 + y2 + 2a + 2b + 1,但是如何用Mathematica化简给出这样的结果呢?我试了FullSimplify[ (a + b + 1)^2,Assumptions -> y1 == a
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mathematica 化简举个简单的例子,比如:Y = (a+b+1)^2,y1 = a^2 + b^2,y2 = 2ab,显然我们知道:Y = y1 + y2 + 2a + 2b + 1,但是如何用Mathematica化简给出这样的结果呢?我试了FullSimplify[ (a + b + 1)^2,Assumptions -> y1 == a
mathematica 化简
举个简单的例子,比如:
Y = (a+b+1)^2,
y1 = a^2 + b^2,
y2 = 2ab,
显然我们知道:Y = y1 + y2 + 2a + 2b + 1,
但是如何用Mathematica化简给出这样的结果呢?
我试了FullSimplify[ (a + b + 1)^2,Assumptions -> y1 == a^2 + b^2 && y2 == 2*a*b ] 不行.

mathematica 化简举个简单的例子,比如:Y = (a+b+1)^2,y1 = a^2 + b^2,y2 = 2ab,显然我们知道:Y = y1 + y2 + 2a + 2b + 1,但是如何用Mathematica化简给出这样的结果呢?我试了FullSimplify[ (a + b + 1)^2,Assumptions -> y1 == a
ClearAll["`*"]; Y = (a + b + 1)^2;
Expand@Y /.a^2 + b^2 -> y1 /.(2 a b -> y2)
不要给y1,y2定义,直接替换就行.