Iterative solution for a non-linear equation

2021 7 23
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更新时间：2021 7 24
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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 ?

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Walter Roberson
Walter Roberson 2021-7-23
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Ran in:
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

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Raihan Rahmat Rabi
Raihan Rahmat Rabi 2021-7-23
Thank you Walter. Your help is appreciated.
Raihan Rahmat Rabi
Raihan Rahmat Rabi 2021-7-23
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,

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Chunru
Chunru 2021-7-23
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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

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