Question

Please use python or matlab. The function e^x −100x^2 =0 has three true solutions.Use secant method to locate the solutions with tolerance 10^(−10).

Answer 1

%Matlab code for Secant method
clear all
close all
%Function for which root have to find
f=@(x) exp(x)-100.*x.^2;
fprintf('For the function f(x)= ')
disp(f)

%finding root using Secant Method
e=10^-10;   %error convergence
max=50000;   %Maximum iterations

a=[-10 -5];    %initial guess
%Root using Secant method
[root]=secant_method(f, a, e, max);
fprintf('Finding root using Secant method\n')
fprintf('\tThe root for this function for initial guess [%d %d] is %f\n\n',a(1),a(2),root)

a=[2 10];    %initial guess
%Root using Secant method
[root]=secant_method(f, a, e, max);

fprintf('\tThe root for this function for initial guess [%d %d] is %f\n\n',a(1),a(2),root)

a=[1 2];    %initial guess
%Root using Secant method
[root]=secant_method(f, a, e, max);

fprintf('\tThe root for this function for initial guess [%d %d] is %f\n\n',a(1),a(2),root)

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%Matlab function for Secant Method
function [root]=secant_method(f, a, e, max)

    %a= vector of two initial guess
    %e= error tollerence
    %max= maximum number of iterations

    %x0 and x1 values for two initial guess
    x0=a(1); x1=a(2);
    %checking wheather function values for two initial guesses are same
    if f(x0)==f(x1)
        msg = 'Function values are identical at initial guesses.';
        error(msg);
    end
    %loop for Secant iterations
    k=0;
    for i=1:max
        k=k+1;
        xx=double(x1-(f(x1)*abs((x1-x0)/(f(x1)-f(x0)))));
        x0=x1;
        x1=xx;
        if(abs(xx-x0)<=e)
            break
        end

    end
    root=xx;
    %If maximum iteration reached
    if i==max
        fprintf('Has reached maximum iteration steps allowed.\n')
    end
end

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