Newton Method Matlab code

Question

Malcolm Kevin 2021-1-25

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/725988-newton-method-matlab-code

回答： Vishesh 2022-10-3

function approximation = newtonsMethod( fnc, x0, delta )

maxnumIter = 100;

n = 0; % Initialize iteration counter.

syms fp; % derivative

fp = diff(fnc);

c = x0;

while ( (abs(subs(fnc,c))>tol) & (n < maxnumIter) )

c = double(c - subs(fnc,xcur)/subs(fp,c));

n = n+1;

end

if (n<100)

c(end);

disp('num_iter f(x) fprime(x) ')

disp('________________________________________________________________________________________')

end

for i=1:n

fprintf('%d \t %20f \t %20f\n',i ,fnc (c(i)),fp (c(i)))

end

if( abs(subs(fnc,c))> delta)

disp(['Warning: Tolerance not met after ' num2str(maxnumIter) ' iterations.']);

end

approximation = c;

Above is a code I attempted for newton method. I was a bit confused about how to print out a table of values for iteration number, x, f(x), f'(x). I tried to add a "if" statement below "while" and use display, but I got error like this

Array indices must be positive integers or logical values.

Error in sym/subsref (line 870)

R_tilde = builtin('subsref',L_tilde,Idx);

Error in newtonsMethod (line 32)

fprintf('%d \t %20f \t %20f\n',i ,fnc(c(i)),fp(c(i)))

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Answer 1

Vishesh 2022-10-3

0
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https://ww2.mathworks.cn/matlabcentral/answers/725988-newton-method-matlab-code#answer_1065540

在 MATLAB Online 中打开

In the above code, you haven't stored the "c" values anywhere and you are trying to access the "c" values in "fprintf('%d \t %20f \t %20f\n',i ,fnc(c(i)),fp(c(i)))".
In your code, "c" is just a variable but you are trying to access it as an array.
Below is the modified code where i am storing the "c" values (which is "x" values) in "x_values".

function approximation = newtonsMethod( fnc, x0, delta )
maxnumIter = 1000;
n = 0;  % Initialize iteration counter.
syms fp;   % derivative
fp = diff(fnc);
c = x0;
x_values=[x0];
while ( (abs(subs(fnc,c))>delta)  &&  (n < maxnumIter) ) 
        c =  double(c - subs(fnc,c)/subs(fp,c));
        x_values=[x_values c];
        n = n+1;
end
if (n<100)
    c(end);
    disp('num_iter          f(x)           fprime(x)             ')
    disp('________________________________________________________________________________________')
    for i=1:n
      fprintf('%d \t %20f \t %20f\n',i ,fnc (x_values(i)),fp (x_values(i)))
    end
end
if( abs(subs(fnc,c))> delta)
        disp(['Warning:  Tolerance not met after ' num2str(maxnumIter) ' iterations.']);
end
approximation = c;
end