return a function without e.g. f(5)

I have never used Matlab before so it really is difficult for me as I now have a task in math course that says to interpolate a function with Lagrange.

My problem has nothing to do with the interpolation itself, more a question if something is possible and if so, how.

E.g. I say

f = @(x) x^3;
and p = @(x) f(2)*x

Matlab display p exactly like above but i would want it to display it as

p = @(x) 8*x

as is wouldn't make much sense to have the result as a polynomial with f(...) as values.

so i hope someone can help me with that.

Answer 1

[Walter Roberson]

If you have the symbolic toolbox,

   syms x
   p = matlabFunction(f(2)*x, x);

You did well, by the way, to realize that it was a matter of what was displayed rather than a matter of how it would execute.

[Marvin van Stegen]

thanks for your answer! that helped me a great bunch. i think it is kinda obvious that I tested if the execution would work dispite the display issue, maybe that's just me though.

i have another question for you, if you don't mind. while i can now display the polynomial in a way that is acceptable, I lack in another area.

p = @(x) (x+2)*(x-1) or something else, but you get the idea. is there another function built-in that displays that as x^2 + x -2?

i have the hinch that i'm doing the task in a way that is way too complicated :(

[Walter Roberson]

syms x
p = matlabFunction( expand( (x+2)*(x-1) ), x);

[Marvin van Stegen]

wow, that is easier than expected. can i ask you more specific things like how i can reduce the error of the interpolation?

currently my code looks like this:

 function p =  makeLagrangePoly(f,a,b,n)

 syms x

 X = zeros(1,n+1); 
 Y = zeros(1,n+1); 
 L = zeros(1,n+1); 

 x_k = a + k(b-a)/n
 for k = 0:n
    X(1,k+1) = a + (k/n)*(b-a);
    Y(1,k+1) = f(a + (k/n)*(b-a));
 end

 t = @(x) 0; 

 for k = 1:n+1
     l = @(x) 1;
    for j = 1:n+1
        if k ~= j
            l = @(x) (l(x).*(x-X(1,j))/((k-j)*(b-a)/n)); 
        end
    end
    t = @(x) (t(x) + Y(1,k).*l(x)); 
 end
 p = expand(t(x));

do you know a way to make the approximation better?

I realized by now, that i don't really need L, so i will delete that part.

[Thorsten]

Walter has answers two of your questions; I think it would be better to accept the answer and ask a new question in the forum with a new title for the new question.