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

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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

##### 6 Comments

Wan Ji
on 2 Sep 2021

### Accepted Answer

Wan Ji
on 2 Sep 2021

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

##### 0 Comments

### More Answers (1)

John D'Errico
on 2 Sep 2021

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)

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

Yes. In fact, it is.

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

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

taylor_cos(0.1,4)

cos(0.1)

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)

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

##### 0 Comments

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