Let f(x)= ln (1+x^2). Find f^15(0)
I know the answer is 0 but why?
Hello,
We know that $\displaystyle \ln(1+x)=-\sum_{n\geq 1}\frac{x^n}{n}$
So here, $\displaystyle f(x)=-\sum_{n\geq 1}\frac{x^{2n}}{n}$
So you can notice that the factors of the odd powers of x are 0.
But we also know that $\displaystyle f(x)=\sum_{n\geq 0} \frac{f^{(n)}(0)}{n!} \cdot x^n=f(0)+\sum_{n\geq 1} \frac{f^{(n)}(0)}{n!} \cdot x^n=\sum_{n\geq 1} \frac{f^{(n)}(0)}{n!} \cdot x^n$
So by identification of the 15-th term
$\displaystyle \frac{f^{(15)}(0)}{15!}=0$
And hence $\displaystyle f^{(15)}(0)=0$