use f=@(x) as an argument of a function.m
78 visualizzazioni (ultimi 30 giorni)
Mostra commenti meno recenti
I have the function
function [ k ] = New_Raph( f,df,x,tol)
x_old =x;
x=1000;
while abs(x_old-x) > tol
x_old = x;
delta=-f(x)/df(x);
x= x-delta;
end
k=x;
end
And i want to calculate in a new .m file:
f=@(x) x.^5-0.4475.*x.^4-3.1.*x.^3+3.085.*x.^2-0.962.*x+0.0943;
df1=@(x) 5.*x.^4-1.79.*x.^3-9.3.*x^2+6.17.*x-0.962;
New_Raph(@f1,@df1,0,0.1)
What is wrong?
0 Commenti
Risposta accettata
per isakson
il 13 Mar 2015
Modificato: per isakson
il 13 Mar 2015
Replace
New_Raph(@f1,@df1,0,0.1)
by
New_Raph(f,df1,0,0.1)
f or f1 - a typo?
 
Addendum triggered by   It did not work
Try this example
>> my_sin = @(x) sin(x)
my_sin =
@(x)sin(x)
>> arrayfun( my_sin, pi/6*[1,2,3] )
ans =
0.5000 0.8660 1.0000
>> arrayfun( @sin, pi/6*[1,2,3] )
ans =
0.5000 0.8660 1.0000
and
>> arrayfun( @my_sin, pi/6*[1,2,3] )
Error: "my_sin" was previously used as a variable, conflicting with its use
here as the name of a function or command.
2 Commenti
Stephen23
il 13 Mar 2015
"It did not work" does not tell us what happened when you tried it: was there an error message, unexpected values or something else?
It should work, according to the information that you have given us. Can you show us your exact code?
Più risposte (1)
Vedere anche
Categorie
Scopri di più su Matrix Indexing in Help Center e File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!