Need to find the Taylor polynomial P2(x) of 2nd degree at 0 of the following function: $$ f(x)= \frac12 \ln \left(\frac{1+x}{1-x} \right), \quad x \in (-1,1) $$
What i've done so far: $$ 1/2(ln((1+x)/(1-x)) = 1/2(ln(1+x)-ln(1-x)) $$ taylor series first we need the f(0) which in this case is 0 as ln(1)=0
And because the function is odd, we don't need to derivative two times (I think). Literally stuck here, don't know what to do now
Note that you can simplify to write $$ f(x) = \frac{\ln(1+x) - \ln(1-x)}{2} $$ and therefore compute $$ f'(x) = \frac12 \left[\frac{1}{1+x} + \frac{1}{1-x}\right] = \frac{(1+x)^{-1} + (1-x)^{-1}}{2} $$ and $$ f''(x) = \frac{-(1+x)^{-2} + (1-x)^{-2}}{2} $$ Therefore, $f(0) = 0, f'(0) = 1$ and $f''(0) = 0$.
Taylor's formula says that $$ f(x) = \sum_{n=0}^\infty f^{(n)}(x)\frac{(x-a)^n}{n!}, $$ and applying this for $a=0$ and $n \le 2$ we get $$ f(x) \approx f(0) + xf'(0) + f''(0) \frac{x^2}{2} = x. $$