[已答复] 分别代入隐函数中不同的参数,进行solve求解时得不到结果

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as1025255 发表于 2020-11-26 11:24:43



如上述公式,我想要实现:代入一系列的A值,解得一系列的x值,并且绘制两者的曲线图。


我尝试先给出一个A值(例如pi),直接代入到表达式里求解,可以成功得到结果:


x=solve('x=((cos(pi)/(1+x))^2)*(sin(pi))*(((1+x)^2)-((cos(pi))^2)-1)^0.5','x')


回车得到结果x=0






然后我通过搜索,找到一个方法来实现代入一系列的A值进行求解,但是得不到想要的结果:


比如:


A=[pi/2,pi]


x=subs(solve('x=((cos(A)/(1+x))^2)*(sin(A))*(((1+x)^2)-((cos(A))^2)-1)^0.5','x'))



回车得到的结果是x=z1,我不明白z1是什么意思




我很奇怪,尝试了一个简单的例子,是可以成功算出来的:


A=[1,2,3]


x=subs(solve('x=A-2*x','x'))



回车得到结果为:x =[ 1/3, 2/3, 1]


恳请各位帮我指出问题所在,谢谢!

2 条回复


as1025255 发表于 2020-11-26 11:26:45
>> A=[pi/2,pi]

x=subs(solve('x=((cos(A)/(1+x))^2)*(sin(A))*(((1+x)^2)-((cos(A))^2)-1)^0.5','x'))

A =

    1.5708    3.1416

警告: Support of strings that are not valid variable names or define a number will be removed in a future release. To create symbolic
expressions, first create symbolic variables and then use operations on them.
> In sym>convertExpression (line 1536)
  In sym>convertChar (line 1441)
  In sym>tomupad (line 1198)
  In sym (line 177)
  In solve>getEqns (line 405)
  In solve (line 225)
警告: Do not specify equations and variables as character strings. Instead, create symbolic variables with syms.
> In solve>getEqns (line 445)
  In solve (line 225)
警告: The solutions are parameterized by the symbols: z1.
To include parameters and conditions in the solution, specify the 'ReturnConditions' option.
> In solve>warnIfParams (line 500)
  In solve (line 356)
警告: The solutions are valid under the following conditions: cos(A)^6*sin(A)^2 + 4*z1^3 + 6*z1^4 + 4*z1^5 + z1^6 ==
z1^2*(cos(A)^4*sin(A)^2 - 1) + 2*z1*cos(A)^4*sin(A)^2 & (z1 == 0 | signIm((z1*(z1 + 1)^2*1i)/(cos(A)^2*sin(A))) == 1) & z1 ~= -1.
To include parameters and conditions in the solution, specify the 'ReturnConditions' option.
> In solve>warnIfParams (line 507)
  In solve (line 356)

x =

z1

maple1314168 发表于 2020-11-27 09:46:44
as1025255 发表于 2020-11-26 11:26
>> A=

x=subs(solve('x=((cos(A)/(1+x))^2)*(sin(A))*(((1+x)^2)-((cos(A))^2)-1)^0.5','x'))

用新的写法吧!
syms A x
subs(solve(x-A),A,[2 3])

你以为随便的方程都可以解?
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