What is wrong with the function code when it can work perfectly without function code

Question

Kenichi 2022-10-6

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评论： Torsten 2022-10-6

Hi There!

I am having trouble on my code where I made the code in function [input] = f(output) doesn't give me the expected answer but a normal script do.

function [k,p,err,P] = fixpt(g,po,tol,maxi)
% g is the function input as function handle
% po is the first guess
% tol is the tolerance
% maxi is the maximum iteration
% K is the number of iteration
% p is the sequence {pn}
% error is the absolute error
P(1) = po
    for k = 2:maxi
        P(k) = g(P(k-1));
        err = abs(P(k) - P(k-1));
        relerr = (err/abs(P(k)));
        p = P(k);
        if abs(err)<tol || abs(relerr)<tol
            return
        end
    end
    if k == maxi
        fprintf('Iteration exceed maximum.')
    end
end

If I do run it:

fixpt(@(x) sin(x)-1,-1,0.001,100)

It doesn't give me the the answer I expected whereas:

g = @(x) sin(x)-1
maxi = 100
tol = 0.00001
po = -1
P(1) = po
for k = 2:maxi
        P(k) = g(P(k-1));
        err = abs(P(k) - P(k-1));
        relerr = (err/abs(P(k)));
        p = P(k);
        if abs(err)<tol || abs(relerr)<tol
            return
        end
    end
if k == maxi
    fprintf('Iteration exceed maximum.')
end

This does.

Im using R2021a. Thank you.

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Answer 1

Torsten 2022-10-6

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https://ww2.mathworks.cn/matlabcentral/answers/1818735-what-is-wrong-with-the-function-code-when-it-can-work-perfectly-without-function-code#answer_1067960

编辑：Torsten 2022-10-6

在 MATLAB Online 中打开

Works for me.

[k,p,err,P] = fixpt(@(x) sin(x)-1,-1,0.00001,100)
k = 12
p = -1.9346
err = 1.0380e-05
P = 1×12
   -1.0000   -1.8415   -1.9636   -1.9238   -1.9383   -1.9332   -1.9350   -1.9344   -1.9346   -1.9345   -1.9346   -1.9346
function [k,p,err,P] = fixpt(g,po,tol,maxi)
% g is the function input as function handle
% po is the first guess
% tol is the tolerance
% maxi is the maximum iteration
% K is the number of iteration
% p is the sequence {pn}
% error is the absolute error
P(1) = po;
for k = 2:maxi
    P(k) = g(P(k-1));
    err = abs(P(k) - P(k-1));
    relerr = (err/abs(P(k)));
    p = P(k);
    if abs(err)<tol || abs(relerr)<tol
        return
    end
end
if k == maxi
    fprintf('Iteration exceed maximum.')
end
end

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Kenichi 2022-10-6

It seems like my indentation is wrong, I have seen your code and fixed it. Thank you!

Torsten 2022-10-6

Maybe you should restart MATLAB.

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Answer 2

Davide Masiello 2022-10-6

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编辑：Davide Masiello 2022-10-6

在 MATLAB Online 中打开

The value of for tol differs of two orders of magnitude in the examples you gave (i.e., you used 0.001 in the function call, 0.00001 in the functionless script).

[k,p,err,P] = fixpt(@(x) sin(x)-1,-1,0.00001,100)
P = -1
k = 12
p = -1.9346
err = 1.0380e-05
P = 1×12
   -1.0000   -1.8415   -1.9636   -1.9238   -1.9383   -1.9332   -1.9350   -1.9344   -1.9346   -1.9345   -1.9346   -1.9346
function [k,p,err,P] = fixpt(g,po,tol,maxi)
% g is the function input as function handle
% po is the first guess
% tol is the tolerance
% maxi is the maximum iteration
% K is the number of iteration
% p is the sequence {pn}
% error is the absolute error
P(1) = po
    for k = 2:maxi
        P(k) = g(P(k-1));
        err = abs(P(k) - P(k-1));
        relerr = (err/abs(P(k)));
        p = P(k);
        if abs(err)<tol || abs(relerr)<tol
            return
        end
    end
    if k == maxi
        fprintf('Iteration exceed maximum.')
    end
end