Iterative solution for a non-linear equation

I am looking for a solution to the problem:

ag_p*Gamma_p - Sa = 0 with Gamma_p = A+B*ag_p.

A, B and Sa are constants.

This can be solved by iterations starting with a certain guess value for ag_p. I tried to use the following code:

Sa = 2.7956;
A = 2.24;
B = 0.5547;

f = @(ag_p) ag_p*Gamma_p - Sa;
Gamma_p = A+B*ag_p;
z = fzero(@(ag_p) f(ag_p), 1E-3);

My code cant be run because the Gamma_p is a function of ag_p which is unknown. What would be the efficient way to tell the program to start with a guess value for ag_p ?

Answer 1

[Walter Roberson]

format long g
Sa = 2.7956;
A = 2.24;
B = 0.5547;

Gamma_p = @(ag_p) A+B*ag_p;
f = @(ag_p) ag_p*Gamma_p(ag_p) - Sa;
z = fzero(@(ag_p) f(ag_p), 1E-3)

z = 
          1.00026869288617

f(z)

ans = 
     -4.44089209850063e-16

[Raihan Rahmat Rabi]

Thank you Walter. Your help is appreciated.

[Raihan Rahmat Rabi]

Can you suggest another method, where the problem in question is solved by providing guess values to ag_p, let's say from [0...inf], and Gamma_p is updated based on the guess values of ag_p and the iterations are performed until the solution is found. I need to have a numerical value for Gamma_p in each iteration because I will be comparing its value with another constant parameter to define some "if" conditions before the fzero function is written.

Thank you,

[Walter Roberson]

https://en.wikipedia.org/wiki/Bisection_method

https://en.wikipedia.org/wiki/Newton%27s_method

Answer 2

[Chunru]

Sa = 2.7956;
A = 2.24;
B = 0.5547;
% Change the original code to the following.  Otherwise not working.
f = @(ag_p) ag_p*(A+B*ag_p) - Sa;
%Gamma_p = A+B*ag_p;
z = fzero(@(ag_p) f(ag_p), 1E-3)    % 1E-3 is the initial value

z = 1.0003

z0 = 0.99;      % a closer initial value
z = fzero(@(ag_p) f(ag_p), z0)

z = 1.0003