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The MacLaurin series for $\frac{k}{1+k^2}$ is $=\sum_{n=0}^\infty (-1)^n k^{2n+1}$

I'm asked to find $f^5(0)$, so I substitute n=2 and find out it is $k^5$.

To find $f^6(0)$, I know the answer is $0$, but I'm not sure how to explain this.

Lastly, to conclude the value of $f^n(0)$, the series only picks up terms with odd powers on k's. Is this conclusion good enough?

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If you differentiate the infinite sum $5$ times and take the value at $x=0$, the only relevant term is the constant. The term with $k^5$ in the sum is $k^5$. Differntiating $5$ times gives $120$.

So we have $f^5(0)=120$.

Alternatively, you can consider $$f(k)=\sum_{m=0}^\infty \frac{k^m}{m!}f^m(0)$$

The term with $m=5$ is thereofore $\frac{k^5}{5!}f^5(0)$. Since the infinite sum has coefficient $1$ at the power $k^5$, we have $\frac{f^5(0)}{5!}=1$, which immediately gives $f^5(0)=120$.

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