Solving equation with two variables, and giving the result for one as a function of the other

I have an equation with two variables (Alpha, and Ze_grad), i wont an equation that locks like Alpha= f(Ze_grad), but i cant do that!

mfreqs=[200 5000];
syms Alpha  Ze_grad
Zm = 1./((2i*pi*mfreqs).^Alpha);
Zm_grad = (imag(Zm(2))-imag(Zm(1)))/(real(Zm(2))-real(Zm(1)));
equ= Zm_grad - Ze_grad == 0 ;
alpha = solve(equ, Alpha) 

Warning: Unable to find explicit solution. For options, see help.

 
alpha =
 
Empty sym: 0-by-1
 

[Bakr Abdelgaliel]

Any Help!

Answer 1

[Torsten]

Alpha0 = 2.0;
Ze_grad = 3.0;
Alpha = fsolve(@(x)fun(x,Ze_grad),Alpha0)

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

Alpha = 1.2048

fun(Alpha,Ze_grad)

ans = 1.1813e-13

function res = fun(Alpha,Ze_grad)
  mfreqs=[200 5000];
  Zm = 1./((2i*pi*mfreqs).^Alpha);
  Zm_grad = (imag(Zm(2))-imag(Zm(1)))/(real(Zm(2))-real(Zm(1)));
  res = abs(Zm_grad - Ze_grad);
end

[Bakr Abdelgaliel]

Thanks for your help, but I need a symbolic equation. I need to have a resukt in a form of Alpha=f(Ze_grad).

Somthing like Sa=f(x) for this equation:

syms a b c x
eqn = a*x^2 + b*x + c == 0;

Sa = solve(eqn,a)

Sa = 

[Torsten]

But you read what was returned from "solve", didn't you ?

Unable to find explicit solution.

So "solve" was not able to solve your equation for "Alpha".

So within MATLAB, only a numerical solution seems possible.

[Bakr Abdelgaliel]

Thats is my main question!

Is there a way to get what I need using another function (because SOLVE cannot help)?

Alsso, i know that 0<=Alpha<=1. (if this may helps)

[Torsten]

"solve" is the symbolic MATLAB solver. If it doesn't succeed, there is no other symbolic solver in MATLAB.

You might want to try MATHEMATICA, Octave, Maple ...

Can you really assume Alpha to be real-valued and 0<=Alpha<=1 for arbitrary (maybe complex) values of Ze_grad ? I doubt it. My example from above already shows that this is not true.

[Bakr Abdelgaliel]

Actually, I know the value of Ze_grad (which is a real number), and i can calculate Alpha by knowing the value of Ze_grad using Solve or vpasolve around for 0<=Alpha<=1.

But i need to make this operations many many times in my code and its computational time is really high specially due to using syms.

So i am wondering if i could have a direct equation in the form of Alpha=f(Ze_grad) to directly calculate Alpha?

[Torsten]

You can save a (Alpha,Ze_grad) matrix obtained from

Ze_grad = f(Alpha)

and use interp1 to interpolate the inverse function:

mfreqs=[200 5000];
Alpha = 0.001:0.001:1-0.001;
for i = 1:numel(Alpha)
  Zm = 1./((2i*pi*mfreqs).^Alpha(i));
  Zm_grad(i) = (imag(Zm(2))-imag(Zm(1)))./(real(Zm(2))-real(Zm(1)));
end
plot(Alpha,Zm_grad)

Zm_grad_interp = (Zm_grad(20)+Zm_grad(21))/2;
Alpha_interp = interp1(Zm_grad,Alpha,Zm_grad_interp)

Alpha_interp = 0.0205

[Bakr Abdelgaliel]

Thanks alot for your effort and time,.

But i think Binary search will be mor time effecient than that.

[Torsten]

Depends on your problem. If mfreqs is the same for many Alpha-calulations, I think the interpolation will be quite fast compared to binary search. If mfreqs often changes and Zm_grad has to be built anew, you might be correct.

[Bakr Abdelgaliel]

I will Try👍.