Find the sum of the series and determine the radius and interval of convergence

**twilightstr** · May 14th 2009, 11:22 AM

Let f(x)= ln (1+x^2). Find f^15(0)
I know the answer is 0 but why?

**Moo** · May 14th 2009, 11:35 AM

Hello,

Originally Posted by twilightstr

Let f(x)= ln (1+x^2). Find f^15(0)
I know the answer is 0 but why?

We know that $\ln(1+x)=-\sum_{n\geq 1}\frac{x^n}{n}$

So here, $f(x)=-\sum_{n\geq 1}\frac{x^{2n}}{n}$
So you can notice that the factors of the odd powers of x are 0.

But we also know that $f(x)=\sum_{n\geq 0} \frac{f^{(n)}(0)}{n!} \cdot x^n=f(0)+\sum_{n\geq 1} \frac{f^{(n)}(0)}{n!} \cdot x^n=\sum_{n\geq 1} \frac{f^{(n)}(0)}{n!} \cdot x^n$

So by identification of the 15-th term
$\frac{f^{(15)}(0)}{15!}=0$

And hence $f^{(15)}(0)=0$

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