Solving Equation

6 次查看（过去 30 天）
Mark
Mark 2011-2-11
I'm trying to solve a equation, and I'm having some issues...
syms Xell Xelr Xerl Xerr s q; s = solve('(Xell-Xelr)/(Xerl-Xerr) = -(((s-1)*(s-(1+2/q)))/((s+1)*(s+(1+2/q))))', s);
How do i use the results and enter values in for the syms (i.e. Xell Xelr Xerl Xerr q) to solve for s?
Thanks! Mark

请先登录，再回答此问题。

采纳的回答

Oleg Komarov
Oleg Komarov 2011-2-11
Not sure the best method, convert to function handles:
syms Xell Xelr Xerl Xerr s q;
s = solve('(Xell-Xelr)/(Xerl-Xerr) = -(((s-1)*(s-(1+2/q)))/((s+1)*(s+(1+2/q))))', s)
s1 = str2func(['@(Xell, Xelr, Xerl, Xerr, q)' char(s(1))]);
s2 = str2func(['@(Xell, Xelr, Xerl, Xerr, q)' char(s(2))]);
% Evaluate at the desired values
s1(1,1,3,1,2)
s2(1,1,4,1,2)
Oleg

更多回答（3 个）

Mark
Mark 2011-2-11
Thanks Oleg, but I hit a snag...
If you solve the following code: syms Xell Xelr Xerl Xerr s q; s = solve('(Xell-Xelr)/(Xerl-Xerr) = -(((s-1)*(s-(1+2/q)))/((s+1)*(s+(1+2/q))))', s)
you get this: (1/(2*(Q*Xell - Q*Xelr + Q*Xerl - Q*Xerr)))*(2*Xell - 2*Q*Xell - 2*Xelr + 2*Q*Xelr - 2*Xerl + 2*Q*Xerl + 2*Xerr - 2*Q*Xerr - sqrt(-4*(Q*Xell - Q*Xelr + Q*Xerl - Q*Xerr)*(-2*Xell + Q*Xell + 2*Xelr - Q*Xelr - 2*Xerl + Q*Xerl + 2*Xerr - Q*Xerr) + (-2*Xell + 2*Q*Xell + 2*Xelr - 2*Q*Xelr + 2*Xerl - 2*Q*Xerl - 2*Xerr + 2*Q*Xerr)^2));
When I run your code, I get the following: (Xelr-Xell+Xerl-Xerr-Q*Xell+Q*Xelr+Q*Xerl-Q*Xerr+(4*Q^2*Xell*Xerr-4*Q^2*Xell*Xerl+4*Q^2*Xelr*Xerl-4*Q^2*Xelr*Xerr-8*Q*Xell*Xerl+8*Q*Xell*Xerr+8*Q*Xelr*Xerl-8*Q*Xelr*Xerr+Xell^2-2*Xell*Xelr-2*Xell*Xerl+2*Xell*Xerr+Xelr^2+2*Xelr*Xerl-2*Xelr*Xerr+Xerl^2-2*Xerl*Xerr+Xerr^2)^(1/2))/(Q*Xell-Q*Xelr+Q*Xerl-Q*Xerr)
I'm get a real number for my equation, but an imaginary for yours.... Any suggestions?
Thanks for the help!
  1 个评论
Oleg Komarov
Oleg Komarov 2011-2-11
I'm running your part of the code first and then my part of the code and don't get the difference you say.
Clear all and run again the script, which I'll complete with your part.

请先登录，再进行评论。

Mark
Mark 2011-2-11
Nope, I get the same issue. If I run the following code, I get the below code:
s = solve('(Xell-Xelr)/(Xerl-Xerr) = -(((s-1)*(s-(1+2/q)))/((s+1)*(s+(1+2/q))))', s)
sneg = (1/(2*(Q*Xell - Q*Xelr + Q*Xerl - Q*Xerr)))*(2*Xell - 2*Q*Xell - 2*Xelr + 2*Q*Xelr - 2*Xerl + 2*Q*Xerl + 2*Xerr - 2*Q*Xerr - sqrt(-4*(Q*Xell - Q*Xelr + Q*Xerl - Q*Xerr)*(-2*Xell + Q*Xell + 2*Xelr - Q*Xelr - 2*Xerl + Q*Xerl + 2*Xerr - Q*Xerr) + (-2*Xell + 2*Q*Xell + 2*Xelr - 2*Q*Xelr + 2*Xerl - 2*Q*Xerl - 2*Xerr + 2*Q*Xerr)^2));
spos = (1/(2*(Q*Xell - Q*Xelr + Q*Xerl - Q*Xerr)))*(2*Xell - 2*Q*Xell - 2*Xelr + 2*Q*Xelr - 2*Xerl + 2*Q*Xerl + 2*Xerr - 2*Q*Xerr + sqrt(-4*(Q*Xell - Q*Xelr + Q*Xerl - Q*Xerr)*(-2*Xell + Q*Xell + 2*Xelr - Q*Xelr - 2*Xerl + Q*Xerl + 2*Xerr - Q*Xerr) + (-2*Xell + 2*Q*Xell + 2*Xelr - 2*Q*Xelr + 2*Xerl - 2*Q*Xerl - 2*Xerr + 2*Q*Xerr)^2));
I'm getting the following for your s1 and s2
(Xelr-Xell+Xerl-Xerr-Q*Xell+Q*Xelr+Q*Xerl-Q*Xerr+(4*Q^2*Xell*Xerr-4*Q^2*Xell*Xerl+4*Q^2*Xelr*Xerl-4*Q^2*Xelr*Xerr-8*Q*Xell*Xerl+8*Q*Xell*Xerr+8*Q*Xelr*Xerl-8*Q*Xelr*Xerr+Xell^2-2*Xell*Xelr-2*Xell*Xerl+2*Xell*Xerr+Xelr^2+2*Xelr*Xerl-2*Xelr*Xerr+Xerl^2-2*Xerl*Xerr+Xerr^2)^(1/2))/(Q*Xell-Q*Xelr+Q*Xerl-Q*Xerr)
-(Xell-Xelr-Xerl+Xerr+Q*Xell-Q*Xelr-Q*Xerl+Q*Xerr+(4*Q^2*Xell*Xerr-4*Q^2*Xell*Xerl+4*Q^2*Xelr*Xerl-4*Q^2*Xelr*Xerr-8*Q*Xell*Xerl+8*Q*Xell*Xerr+8*Q*Xelr*Xerl-8*Q*Xelr*Xerr+Xell^2-2*Xell*Xelr-2*Xell*Xerl+2*Xell*Xerr+Xelr^2+2*Xelr*Xerl-2*Xelr*Xerr+Xerl^2-2*Xerl*Xerr+Xerr^2)^(1/2))/(Q*Xell-Q*Xelr+Q*Xerl-Q*Xerr)
Different then what I get using the solver.
  1 个评论
Oleg Komarov
Oleg Komarov 2011-2-11
How can you get sneg = ... if you're solving s = solve...? Where does sneg come from? (btw I corrected again my original post)

请先登录，再进行评论。

Walter Roberson
Walter Roberson 2011-2-11
You are going to get two solutions returned. You need to take that in to account. You are effectively solving a quadratic, so two solutions.
for K=1:length(s); funcs{K} = matlabFunction(s(K)); end
Now funcs{K} will be a function handle that you can pass the values in to and get solutions for s. Please see the matlabFunction documentation for information as to what parameter order the generated functions will expect.

类别

在 Help Center 和 File Exchange 中查找有关 Mathematics 的更多信息

标签

产品

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by