How can I code this Newton-Raphson and Secant Methods?

Question

Umut Berkcan Karahan 2020-5-10

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https://ww2.mathworks.cn/matlabcentral/answers/524434-how-can-i-code-this-newton-raphson-and-secant-methods

评论： Muhammad Raihan Ekaputra Idrisatria 2020-11-2

Determine the root of the given equation (3*x*e^x)-1=0 for x ∈ [0,1] using Newton-Raphson and Secant methods. Stop calculation after seven iterations. I've coded a code but I'm not sure it is right or not...

%f is non-linear function.
%fder is derivation of f.
%x0 is initial value of iteration.
%e is tolerance of f function.
%d is first condition tolerance value.
%y is error amount.
%max is maximum iteration number.
syms x
x0 = 0;
x1 = 1;
d = 0.000001;
e = 0.00001;
max = 7;
f = @(x) 3*x*exp(x)-1;
fder = @(x) 3*(exp(x)+exp(x))*x;
    for iteration = 0:max
        t = fder(x0); %to check the derivation is zero or not.
        if abs(t)<e
            disp('Derivation value is very close to zero, algorithm stops')
            break
        end
        x1 = x0-(f(x0)/fder(x0));
        error = abs(x1-x0);
        odderror = 2*error/(abs(x1)+d);
        x0 = x1;
        y = f(x0);
        if(error<d)||(odderror<d)||(abs(y)<e)
            break
        end
    end
    disp('Wanted root value')
    x1
    disp('Error amount')
    y
    disp('Iteration number')
    iteration

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Muhammad Raihan Ekaputra Idrisatria 2020-10-29

Use this function that made by me https://www.mathworks.com/matlabcentral/fileexchange/81758-newton-raphson-secant-method

I mixed Newton-Raphson and backward euler to solve the derrivation using numeric method

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Answer 1

David Hill 2020-5-10

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在 MATLAB Online 中打开

The basic code is as follows. You can do other checks as desired.Your code looked ok, except you don't need symbolic variables and your fder function was missing a ')'.

x=0;
f = @(x) 3*x.*exp(x)-1;
fder = @(x) 3*(exp(x)+exp(x).*x);
for k=1:7%newton method
    x=x-f(x)/fder(x);
end
y=[0,1];
for k=1:7%secant method
    z=y(2);
    y(2)=y(2)-f(y(2))*diff(y)/diff(f(y));
    y(1)=z;
end