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Cote
Cote 2011 年 4 月 18 日
[EDIT: 20110512 17:03 EDT - reformat - WDR]
I keep getting the error Undefined function or method 'Newton' for input arguments of type 'inline'. I'm doing newtons method and I can't figure out what that error means and what part of my code is wrong.
function x = Newton(f, fp, x, nmax, error)
x=1;
e=2.71828;
f(x)=inline('(x)*(5000)*(e^(-x)))-((100)+(0.73)*(5000)*(e^(-x))');
fp(x)=inline('-5000*(e^(-x))*(x-1.73)');
nmax = 10;
error = 1.0e-15;
x = Newton(f,fp,x,nmax,error)
fprintf('x(0) = %10g \n', x)
for n = 1:nmax
d = f(x)/fp(x);
x = x - d;
fprintf('x(%i) = %10g \n', n, x)
if abs(d) < error
fprintf('Converged! \n')
return
end
end
end

採用された回答

Paulo Silva
Paulo Silva 2011 年 4 月 18 日
There's no Newton function on MATLAB current path or any other path that MATLAB is aware, you get the same error for
whateverfunctionisayso(inline('x'))
This should be your script:
x=1;
f=inline('x*5000*exp(-x)-(100+0.73*5000*exp(-x))');
fp=inline('-5000*exp(-x)*(x-1.73)');
nmax = 10;
error = 1.0e-15;
x = Newton(f,fp,x,nmax,error)
fprintf('x(0) = %10g \n', x)
And this is your function:
function x = Newton(f, fp, x, nmax, error)
for n = 1:nmax
d = f(x)/fp(x);
x = x - d;
fprintf('x(%i) = %10g \n', n, x)
if abs(d) < error
fprintf('Converged! \n')
return
end
end
end
Save the function to the current path or another that MATLAB is aware of and run the script.
  16 件のコメント
Cote
Cote 2011 年 4 月 19 日
Nevermind I got it to work thanks for all your help!
Cote
Cote 2011 年 4 月 19 日
And that thank you is directed towards Paulo and Matt

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その他の回答 (1 件)

Matt Tearle
Matt Tearle 2011 年 4 月 18 日
That message generally means that your function file isn't on the MATLAB path. Is Newton.m in your current directory?
There are some other issues with your code, but they wouldn't cause that error.
  5 件のコメント
Paulo Silva
Paulo Silva 2011 年 4 月 19 日
Matt is also right about using function handles instead of inline expressions but I wouldn't bother much with it, all should work good with your current code.
Cote
Cote 2011 年 4 月 19 日
okay thank you

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