# Thread: Taylor approximation

1. ## Taylor approximation

Find the second Taylor approximation for the following

g(x)=ln(cos)(x))

I think I have to use the chain rule, but i'm not sure how. I really need help.

2. Originally Posted by tim_mannire
Find the second Taylor approximation for the following

g(x)=ln(cos)(x))

I think I have to use the chain rule, but i'm not sure how. I really need help.
Do you mean taylor series?

3. yes, i believe taylor series and taylor approximation mean the same thing.

4. centered at what?

5. Originally Posted by Miss
centered at what?
Assume MacLaurin (Taylor centred at 0)

CB

6. Wouldn't it just be -ln(x^2/2!)?

How I got my answer was just by identifying that cos(x) = 1 - x^2/2!, then just take the ln of each nth. This left me with 0 - ln(x^2/2!).

Please keep in my that I could be wrong, fore I am learning Taylor/MacLaurin series at this moment.

7. Originally Posted by tim_mannire
Find the second Taylor approximation for the following

g(x)=ln(cos)(x))

I think I have to use the chain rule, but i'm not sure how. I really need help.
$T_2(x;0)=f(0)+f'(0)x+\frac{f''(0)}{2}x^2$.

So, $f(0)=\ln(1)=0$

$f'(0)\frac{-\sin(0)}{\cos(0)}=0$

$f''(0)=\frac{-1}{\cos^2(0)}=-1$.

So $T_2(x;0)=\frac{-1}{2}x^2$

8. Originally Posted by Brandong954
Wouldn't it just be -ln(x^2/2!)?

How I got my answer was just by identifying that cos(x) = 1 - x^2/2!, then just take the ln of each nth. This left me with 0 - ln(x^2/2!).

Please keep in my that I could be wrong, fore I am learning Taylor/MacLaurin series at this moment.
There is a way to doing this. For a small enough neighborhood around zero $\ln(\cos(x))=\ln\left(\sqrt{1-\sin^2(x)}\right)=\frac{1}{2}\ln\left(1-\sin^2(x)\right)$ and remembering that $\ln(1-x)=-x-\frac{x^2}{2}-\cdots$ we see that $\frac{1}{2}\ln\left(1-\sin^2(x)\right)=\frac{-1}{2}\left(\sin^2(x)+\frac{\sin^4(x)}{2}+\cdots\ri ght)$.

But, as to your question, since when does $\ln(a+b)=\ln(a)+\ln(b)$?