开发者

solving an equation for the square of the variable?

开发者 https://www.devze.com 2023-03-08 20:05 出处:网络
Given a expression (polynomial, or any equation in general) such as a s^2+b = 0 I want to solve for s^2, to get s^2 = -b/a. We all know that one can\'t just write

Given a expression (polynomial, or any equation in general) such as

a s^2+b = 0

I want to solve for s^2, to get s^2 = -b/a. We all know that one can't just write

Solve[eq==0,s^2]

because s^2 is not a 'variable开发者_运维百科'. only s is a 'variable'. So what I do is

eq  = a s^2+b;
sol = First@Solve[eq==0/.s^2->z,z];
z/.sol

-(b/a)

I was wondering if there is a way to do the above, without the intermediate variable substitution? I tried many commands, but no success (reduce, collect, eliminate, factor. etc...).

thanks --Nasser


One way is to solve for s and then square it...

eq=a s^2+b;
sol=#^2 &@ (s/.Solve[eq==0,s])//DeleteDuplicates

Out[1]= {-(b/a)}


You could use the Notation package, but it leads to other issues. So here is your original equation:

In[1]:= Solve[b + a s^2 == 0, s^2]
During evaluation of In[1]:= Solve::ivar: s^2 is not a valid variable. >>
Out[1]= Solve[b + a s^2 == 0, s^2]

Now Symbolize s^2 so that the normal Mathematica evaluator treats it like any other symbol

In[2]:= Needs["Notation`"]
In[3]:= Symbolize[ParsedBoxWrapper[SuperscriptBox["s", "2"]]]

In[4]:= Solve[b + a s^2 == 0, s^2]

Out[4]= {{s^2 -> -(b/a)}}

The problem is that s^2 really is treated as just another symbol, eg

In[6]:= Sqrt[s^2] // PowerExpand
Out[6]= Sqrt[s^2]

A work around is to replace s^2 with s*s, since Symbolize only acts on user inputed expressions (ie at the level of interpreting inputted Box structures)

In[7]:= Sqrt[s^2] /. s^2 -> s s // PowerExpand
Out[7]= s
0

精彩评论

暂无评论...
验证码 换一张
取 消