How do i write the function for the taylor polynomial of cos(x) using for loops and no anonymous function

Pn is the taylor polynomial of the degree (at most) n centred at x0=0 for the function cos(x)

The function should take inputs a and n and return Pn evaluated at x=a

function Pn = taylor_cos(a,n)
x=a;
Pn=1;
for i=1:n;
addterm=((-1)^i)*(x^(2*i))/factorial(2*(i));
Pn=Pn+addterm;
end
% thats my version but for some reason its incorrect

[Wan Ji]

For what reason it is incorrect? for example, a large x is selected?

[B]

Well the matlab grader says ' the code is incorrect for some choices of a and n'. Not sure what that means

[Wan Ji]

Did you test the code modified by me?

[B]

yeah come up with the same error in the grader

[Wan Ji]

Would you like to show us the website of the grader, I have tried many of the cody questions in Matlab Community Cody. And always I get full marks

[B]

apologise i had inputed the modifed code incorrectly, its working now

Answer 1

[Wan Ji]

If a large x is selected, then use

function Pn = taylor_cos(a,n)
x=mod(abs(a),2*pi); %
if(x>=pi)
    x = 2*pi-x;
end
Pn=1;
for i=1:n
    addterm=((-1)^i)*(x^(2*i))/factorial(2*(i));
    Pn=Pn+addterm;
end

Answer 2

[John D'Errico]

We can see how well it works. Your code is below.

Now, let me test it out, in a symbolic form.

syms X
P8 = taylor_cos(X,4)

P8 = 

Is that a correct Taylor expansion for cos(X)?

Yes. In fact, it is.

taylor(cos(X),'order',10)

ans = 

Does it correctly extimate cos(X), for reasonably small X?

taylor_cos(0.1,4)

ans = 0.9950

cos(0.1)

ans = 0.9950

It seems to predict reasonably well.

My guess is you are hoping this will predict correctly for large values of X. And that is of course not true.

taylor_cos(20,10)

ans = 2.0966e+07

Yes, that is garbage. Perhaps you need to do some reading about convergence of infinite series. Even if this series will converge in theory, after a vast number of terms, this does not mean you will expect it to converge in practice, using computation with a finite number of decimal digits.

function Pn = taylor_cos(a,n)
x=a;
Pn=1;
for i=1:n;
addterm=((-1)^i)*(x^(2*i))/factorial(2*(i));
Pn=Pn+addterm;
end
end