# Math Help - Values of derivatives

1. ## Values of derivatives

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be infinitely differentiable and suppose that
$f(\frac{1}{n})= \frac{n^{2}}{n^{2}+1}$
for all $n=1,2,3,...$. Determine the values of
$f^{'}(0), f^{''}(0), f^{'''}(0), f^{(4)}(0),...$.

2. Originally Posted by Markeur
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be infinitely differentiable and suppose that
$f(\frac{1}{n})= \frac{n^{2}}{n^{2}+1}$
for all $n=1,2,3,...$. Determine the values of
$f^{'}(0), f^{''}(0), f^{'''}(0), f^{(4)}(0),...$.
Note first that by continuity $\displaystyle f(0)=f\left(\lim \frac{1}{n}\right)=\lim f\left(\frac{1}{n}\right)=1$. Note then that since $f'(x)$ exists one has that if $x_n\to 0$ and $x_n\ne 0,\text{ }n\in\mathbb{N}$ then $\displaystyle f'(x)=\lim_{n\to\infty}\frac{f\left(x+x_n\right)-f(x)}{x_n}$. In particular $\displaystyle f'(0)=\lim_{n\to\infty}\frac{f\left(\frac{1}{n}\ri ght)-f(0)}{\frac{1}{n}}=\lim_{n\to\infty}\frac{\frac{n^ 2}{n^2+1}-1}{\frac{1}{n}}=0$. Generalize.

3. Originally Posted by Markeur
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be infinitely differentiable and suppose that
$f(\frac{1}{n})= \frac{n^{2}}{n^{2}+1}$
for all $n=1,2,3,...$. Determine the values of
$f^{'}(0), f^{''}(0), f^{'''}(0), f^{(4)}(0),...$.
Since $f\bigl(\frac1n\bigr) = \frac{n^{2}}{n^{2}+1} = \frac1{1+(1/n)^2}$, it follows that one such function would be $f(x) = \frac1{1+x^2} = (1+x^2)^{-1} = 1-x^2+x^4-x^6+\ldots$. Comparing that with the Taylor series $f(x) = 1+xf'(0)+\frac{x^2}{2!}f''(0)+\ldots$, you see that for this particular function $f^{(n)}(0) = 0$ if n is odd, and $f^{(2n)}(0) = (-1)^nn!$.

That tells you what you should expect the answer to be, but of course it isn't a proof. To set about proving the result, you will have to use techniques like those in Drexel28's comment. But knowing what to expect the answer to be can sometimes be a help in establishing it.