Anonymus Function from Input function

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

Answer 1

[Star Strider]

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);

[Jose Trevino]

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]

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]

Thank you very much! I got it

[Star Strider]

My pleasure!

If my Answer solved your problem, please Accept it.