use f=@(x) as an argument of a function.m

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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?

Accepted Answer

per isakson
per isakson on 13 Mar 2015
Edited: per isakson on 13 Mar 2015
Replace
New_Raph(@f1,@df1,0,0.1)
by
New_Raph(f,df1,0,0.1)
f or f1 - a typo?
&nbsp
Addendum triggered by &nbsp 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 Comments
DDD
DDD on 13 Mar 2015
Edited: DDD on 13 Mar 2015
It did not work. I tryed making f1.m and df1.m and it solved but in this way it is much cleaner.
Stephen23
Stephen23 on 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?

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More Answers (1)

Matt J
Matt J on 13 Mar 2015
x= x + delta;

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