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.

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- Mar 10th 2010, 02:30 AMtim_mannireTaylor 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 AMMiss
- Mar 10th 2010, 02:38 AMtim_mannire
yes, i believe taylor series and taylor approximation mean the same thing.

- Mar 10th 2010, 02:41 AMMiss
centered at what?

- Mar 10th 2010, 05:19 AMCaptainBlack
- Mar 10th 2010, 06:46 AMBrandong954
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 AMDrexel28
- Mar 10th 2010, 06:55 AMDrexel28
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 $\displaystyle \ln(a+b)=\ln(a)+\ln(b)$?