Find a minimum value of $\frac{f^{1000}(0)}{f^{999}(0)}$ The task: let $\mathbb A$ be a set of infinitely differentiable functions $f:(-1,1)\to R$, such that for every $n \in N$ following relation holds: $f(\frac 1n)=\frac{3n^2}{(5n+1)^2}$. Find minimum value of $\frac{f^{1000}(0)}{f^{999}(0)}$ on $\mathbb A$.
My solution is:
Since for every $n \in N$ $f(\frac 1n)=\frac{3n^2}{(5n+1)^2}$, then m-th derivative at point $\frac 1n$ is $f^m(\frac 1n)=\frac{5^{m-2}m!(-10n+m-1)3}{(5n+1)^{m+2}}$, and given that $f(0)=\lim_{n\to\infty}f(\frac 1n)$ we have following
$$\frac{f^{1000}(0)}{f^{999}(0)}=\lim_{n\to\infty}\frac{f^{1000}(\frac 1n)}{f^{999}(\frac 1n)}=\lim_{n\to\infty}\frac{\frac{5^{998}3(-10n+999)1000!}{(5n+1)^{1002}}}{\frac{-5^{997}3(-10n+998)999!}{(5n+1)^{1001}}}=\lim_{n\to\infty}\frac{-5(-10n+999)1000}{5(5n+1)(-10n+998)}=0$$.
Is my solution correct?
 A: The function
$$g(x)=\frac3{(5+x)^2}$$
shares the values with the given function $f(x)$ at the sequence of points $x=1/n$, $n$ a positive integer. This suggests that we should consider the difference
$$\phi(x)=g(x)-f(x).$$
We know that $\phi\in C^\infty(((-1,1))$ and $\phi(1/n)=0$ for all positive integers $n$.
Claim. For all $k=0,1,2,\ldots$, we have $D^k\phi(0)=0$.
Proof. Assume contrariwise that some derivative $\phi$ does not vanish at $x=0$. Let $m$ be the smallest non-negative integer such $D^m\phi(0)=A\neq0$.
By continuity of $D^m\phi(x)$ there exists an interval $I=[-\delta,\delta],\delta>0$ such that $|D^m\phi(x)|\ge|A|/2$ for all $x\in I$.
The degree $m-1$ Taylor polynomial of $\phi$ at zero vanishes by minimality of $m$. So by Taylor's theorem for all $x\in I$ we have
$$
\phi(x)=\frac{\phi^{(m)}(\xi)}{m!}x^m,
$$
where $\xi$ lies between $0$ and $x$. By our earlier consideration we know that
$|\phi^{(m)}(\xi)|\ge |A|/2$. It follows that for all $x\in I$ we have
$$
|\phi(x)|\ge \frac{|A||x|^m}{2 m!}.
$$
In particular, $\phi(x)\neq0$ for all $x\in I, x\neq0$. This is a contradiction.
So it follows that for all $f\in C^\infty((-1,1))$ with the prescribed values at the points $1/n, n=1,2,\ldots$, we have
$$
\frac{f^{(1000)}(0)}{f^{(999)}(0)}=\frac{g^{(1000)}(0)}{g^{(999)}(0)}.
$$
Leaving the rest to you.
A: For $\phi$ like in Jyrki's answer, using Rolle's theorem, you can easily prove by induction that for any $k\in\Bbb N,$ $\phi^{(k)}$ admits a decreasing sequence of zeros which converges to $0.$
Therefore, $$\forall k\in\Bbb N\quad\phi^{(k)}(0)=0.$$
