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

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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

Torsten
Torsten il 6 Ott 2022
Modificato: Torsten il 6 Ott 2022
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
Kenichi
Kenichi il 6 Ott 2022
It seems like my indentation is wrong, I have seen your code and fixed it. Thank you!

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Più risposte (1)

Davide Masiello
Davide Masiello il 6 Ott 2022
Modificato: Davide Masiello il 6 Ott 2022
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
Kenichi
Kenichi il 6 Ott 2022
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.

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