While Loop for Percent Error Not Ending

Hello! This is the first time i am posting a question so i apologize if i format it incorrectly.

I am trying to create a while loop that will determine how many iterations are needed for the percent error between the taylor series for sin and the matlab sin() function to be below 2%. The code i currently have just continues to run without an end in sight. Here's what i have:

x=sym('x')
x=input('Enter a scalar value for x.')
n=0:100000;
e=100000;
k=0;
while e>2

      for i=1:length(n)
      y(i)=((-1).^n(i))*((x.^(2*n(i))+1)/factorial((2*n(i))+1));
      k=k+n(i); 
      u=sin(x);
      e=((abs(u-k)/abs(k))*100);
      %i=n(i);
      end

end
fprintf('Percent Error:%d',e)

For reference i am using the generalized version of the Taylor series that is attached.

Thank you in advance!

Answer 1

[Mischa Kim]

Madi, how about

x = input('Enter a scalar value for x: ');
n = 0;
e = 3;
k = 0;
while e>2
    dk = (-1)^n*x^(2*n + 1)/factorial(2*n + 1);
    k  = k + dk;
    u  = sin(x);
    e  = ((abs(u-k)/abs(k))*100);
    n  = n + 1;
    fprintf('Terms: %d, percent error: %3.2d\n', n, e)
end

No need for the sym command. The entire computation is done numerically.

[Madi Macias]

Oh wow, that makes so much more sense. Thank you very much!