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

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.

 Risposta accettata

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

6 Commenti

I don't get it, when i run it on my MATLAB it stopped on the second iteration
It gives this error message:
Output argument "p" (and maybe others) not assigned during call to "fixpt".
Maybe you used a different value for "tol" in both cases ?
I'm not sure if that's the case, since different value of "tol" would still give the array of values..
It seems like my indentation is wrong, I have seen your code and fixed it. Thank you!

Accedi per commentare.

Più risposte (1)

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
Is this correct?

3 Commenti

No, what I expected is a list that contains the iterations. In this code, when I it in the form of function, i can't get those values.
If you copy-paste the code above in a new script, does it work?
Yeah it does work, it seems like my indentation is the problem. Sorry that I could only choose one answer. Anyways, Thank you and have a great day.

Accedi per commentare.

Prodotti

Release

R2021a

Richiesto:

il 6 Ott 2022

Commentato:

il 6 Ott 2022

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