# Taylor approximation

• Mar 10th 2010, 02:30 AM
tim_mannire
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.
• Mar 10th 2010, 02:33 AM
Miss
Quote:

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?
• Mar 10th 2010, 02:38 AM
tim_mannire
yes, i believe taylor series and taylor approximation mean the same thing.
• Mar 10th 2010, 02:41 AM
Miss
centered at what?
• Mar 10th 2010, 05:19 AM
CaptainBlack
Quote:

Originally Posted by Miss
centered at what?

Assume MacLaurin (Taylor centred at 0)

CB
• Mar 10th 2010, 06:46 AM
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.
• Mar 10th 2010, 06:52 AM
Drexel28
Quote:

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.

$\displaystyle T_2(x;0)=f(0)+f'(0)x+\frac{f''(0)}{2}x^2$.

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

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

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

So $\displaystyle T_2(x;0)=\frac{-1}{2}x^2$
• Mar 10th 2010, 06:55 AM
Drexel28
Quote:

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 $\displaystyle \ln(\cos(x))=\ln\left(\sqrt{1-\sin^2(x)}\right)=\frac{1}{2}\ln\left(1-\sin^2(x)\right)$ and remembering that $\displaystyle \ln(1-x)=-x-\frac{x^2}{2}-\cdots$ we see that $\displaystyle \frac{1}{2}\ln\left(1-\sin^2(x)\right)=\frac{-1}{2}\left(\sin^2(x)+\frac{\sin^4(x)}{2}+\cdots\ri ght)$.

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