# How to use secant method to solve two equations

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I would like to solve these two equations (with the help of using anonymous functions) using the secant method.

460cos(theta)*t + 210.6*t-25082 = 0

460sin(theta)*t -6248 = 0

I am trying to solve for theta and t.

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### Answers (2)

MICHAEL MUTWIRI
on 23 Jun 2021

clc;clear;close all

% define the function as anonymous

f = @(x,t) [460.*cos(x).*t+210.6.*t-25082;460.*sin(x).*t-6248]; % x = theta

theta_0 =0;% Initial guess for Theta

t_0 = 1; % Inital guess for t

es = 1e-1; % percentage tolerance es = (new-old)/new * 100%

maxit = 60; % Maximum number of iterations

dh = 1e-3; % step size to used in finite differences method to compute the

% Jacobian

Initial_guess = [theta_0; t_0]; % column vector

[x,f,ea,iter]=secantNon_linear(f,Initial_guess,es,maxit,dh);

your_function_Solution=x

function_at_the_found_solution = f

disp('Solution does not converge')

%% TEST THE CODE USING A FUNCTION WITH KNOW SOLUTION

disp(' ')

disp(' ')

disp(' ')

disp('*****************************************************************')

disp('Testing SECANT method with functions with known solution')

disp('fsolve built-in function is used to confirm the solution')

disp('of the test function')

fprintf('Test functions\n')

fprintf(' : f1 = x1.^2+x2.^2-5\n')

fprintf(' : f2 = x2+1-x1.^2\n')

% Here is a test for the code to just confirm the code works

func = @(x1,x2) [x1.^2+x2.^2-5;x2+1-x1.^2];

x0 = [1.2;1.2];

es = 5;

dh = 1e-3;

maxit = 50;

[x,f,ea,iter] = secantNon_linear(func,x0,es,maxit,dh);

secant__TEST_solution = x'

% define the function again in way fsolve can use it: x1 = t(1); x2 = t(2)

func_for_fsolve = @(t) [t(1).^2+t(2).^2-5;t(2)+1-t(1).^2];

fsolve_CONFIRM_solution = fsolve(func_for_fsolve,x0')

%% THE SECANT CODE

function [x,f,ea,iter]=secantNon_linear(fun_xy,Initial_guess,tolerance,maxit,dh)

iter = 0;

x=Initial_guess;

while (1)

f=fun_xy(x(1),x(2));

% Jacobian matrix computed using the finite differences method

dfdx = (fun_xy(x(1)+dh,x(2))-fun_xy(x(1),x(2)))./dh;

dfdy = (fun_xy(x(1),x(2)+dh)-fun_xy(x(1),x(2)))./dh;

J = [dfdx dfdy];

dx=J\f;

x=x-dx;

iter = iter + 1;

ea=100*max(abs(dx./x));

if iter>=maxit||ea<=tolerance, break, end

end

end

##### 1 Comment

MICHAEL MUTWIRI
on 23 Jun 2021

John D'Errico
on 2 Dec 2018

The secant method does not have a simple extension into multiple dimensions, although I am sure one could cobble something up. Far better however is to simply use tools that ARE designed for multiple variables, such as Newton-Raphson. Better yet of course, is to NOT write your own code to solve nonlinear equations. Never write your own numerical code to do something you do not fully understand, especially when professionally written code is available. (And if you do want to use the secant method here, then it is also clear you do not indeed fully understand the issues.)

Instead, use fsolve (from the optimization toolbox) or solve/vpasolve (from the symbolic toolbox), or lacking those TBs, you could even use fminsearch.

And if you cannot figure out how to implement it with fminsearch, then it is trivially simple to just solve the second equation for t, then substitute into the first equation.

t = 6248/(460*sin(theta))

When you eliminate t in the first equation, you now have a simple equation, solvable using fzero. That would find a solution at theta around 0.35502 (radians). Then just recover the value of t. Of course, there are infinitely may solutions.

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