Anonymus Function from Input function

2016 2 21
1 个回答
更新时间：2016 2 22
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0 个投票

Hi, I am writing a code that takes an input function (f) and then the code does some operations with it. It was using inline function but I prefer to use anonymous function. It was set as f1=inline(f); In order to use the anonymous instead of inline, I changed it to f1 = @(x)[f]; However, after declaring it like this, the code stops working correctly. I don't know what I'm doing wrong.
Thank you

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回答（1 个）

Star Strider
Star Strider 2016-2-21

0 个投票

you have to call the inline function as a function.
This works:
f = inline('cos(x) .* sin(x)');
f1 = @(x) f(x);
q = f1(pi/4);
Better is to just do:
f1 = @(x) cos(x) .* sin(x);

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Jose Trevino
Jose Trevino 2016-2-21
Thank you for your answer. I understand the procedure. However, I don't know how to apply it to my code:
function [ r, error ] = newton( f, df, x0, tol, N )
f = @(x)f;
df = @(x)df;
r(1) = x0 - (f(x0)/df(x0));
error(1) = abs(r(1)-x0);
k = 2;
while (error(k-1) >= tol) && (k <= N)
r(k) = r(k-1) - (f(r(k-1))/df(r(k-1)));
error(k) = abs(r(k)-r(k-1));
k = k+1;
end
end
I can't make it run as it should but if I swap to f =inline(f); df = inline (df); it Works. How can I make it work with the anonymous function instead of inline
Thank you again
Star Strider
Star Strider 2016-2-22
You need to rename and restate the secondary functions, and then change the other references to them in your code:
F = @(x)f(x);
DF = @(x)df(x);
Since MATLAB is case-sensitive, it won’t get confused.
However, you really don’t need to do all that. Something like this will work as well:
f = inline('sin(x) .* cos(x)');
deriv = @(f,x) (f(x+1E-8) -f(x))./1E-8; % Simple Numerical Derivative
x = linspace(0, 2*pi);
figure(1)
plot(x, f(x))
hold on
plot(x, deriv(f,x))
hold off
grid
This is just an illustration. If your functions are function files, you will have to pass them as function handles:
function [ r, error ] = newton( @f, @df, x0, tol, N )
Note the preceding ‘@’ sign, creating the function handles for ‘f’ and ‘df’.
Jose Trevino
Jose Trevino 2016-2-22
Thank you very much! I got it
Star Strider
Star Strider 2016-2-22
My pleasure!
If my Answer solved your problem, please Accept it.

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