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.
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 $\displaystyle \ln(a+b)=\ln(a)+\ln(b)$?