How to find the approximation of the solution to an equation with an infinite series?

Hi, I am very new to matlab and I am trying to solve an equation with an infinite serie. The equation looks like this:

The variable is \alpha and the rest are all parameters.So this thing looks ugly, and almost clearely it has no closed form solution, but I can somehow gurantee that the thing on the left converges by choosing the right parameters. What I want to do is to choose numbers for all parameters and solve for \alpha. I think it would be ok if I replace infinity with a large number. But I have been trying this for a long time and still haven't find out how to do this. I would really appreciate it if someone can give me a hint about how to do this. (Even better if you can provide a code sample). Thanks!

Answer 1

[Bruno Luong]

N = 50; % number of sum
t = (1:N);
delta = 0.6;
R = 2;
c = 5;
M0 = 2;
Rc = R/c;
rhs = c/(M0*R);
sfun = @(beta) sum(((beta*delta).^(t-1).*(1-beta*delta))./(Rc-(Rc-M0)*beta.^t).^2);
beta = fzero(@(beta) sfun(beta)-rhs, 0);
alpha = 1-beta

[Jiahao Chen]

Hi Bruno, thanks for your answer! The code worked but as I change the parameters, sometimes the following error comes out:

Exiting fzero: aborting search for an interval containing a sign change because NaN or Inf function value encountered during search. (Function value at -5.12 is NaN.) Check function or try again with a different starting value.

I guess this is because under the chosen parameter there is no solution. But I am just wondering if there is a way to specify a starting point of searching for the solution? I am only concerned about the solutions that are between 0 and 1, so if the function goes to infinity somewhere else should not matter.

[Bruno Luong]

I don't know enough the equation to answer.

But you can first plot the function

>> ezplot(@(beta) sfun(beta)-rhs,[0 1]);

to see if it cross 0, and approximately where.

Answer 2

[Torsten]

function main
  c = ...;
  M0 = ...;
  R = ...;
  delta = ...;
  N = ...;
  x0 = ...;
  sol = fzero(@(x)fun(x,c,M0,R,delta,N),x0)
end

function res = fun(x,c,M0,R,delta,N)
  res = 0.0;
  for i = 1:N
    res = res + ((1-x)*delta))^(i-1)/(R/c-(R/c-M0)*(1-x)^i)^2
  end
  res = res - c/(M0*R)/(1-(1-x)*delta))
end

[Jiahao Chen]

Hi Torsten, thanks for your answer! But it seems that there is an error when I run the code:

 function main
 ↑
Error: Function definitions are not permitted in this context.

And I am a little bit confused about what is x0. Is that the starting point where the program start to look for a solution (change of sign)?