# finding a function with predetermined $f^{(n)}$s at $0$ and $1$.

Let $a_n,b_n$ be 2 arbitrary sequences of real numbers. Is there any $C^\infty$ function $f:\mathbb{R}\to\mathbb{R}$ such that for any $n$: $$f^{(n)}(0)=a_n$$

$$f^{(n)}(1)=b_n$$

How about complex analytic functions? (I think the answer is no for complex functions)

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A complex analytic function is entirely determined by all its derivatives in a point. Therefore the sequences $\{a_n\}_n$ and $\{b_n\}_n$ depend mutually one on the other in case $f$ is required to be analytic. – AndreasT Feb 6 at 14:49
In the case of $C^\infty$-functions, the $a_n$ and $b_n$ can be chosen completely independently, because you can glue $C^\infty$-functions together with a partition of unity. So your question is reduced to the problem of finding $f$ such that $f^{(n)}(0)=a_n$ for all $n$. – Daan Michiels Feb 6 at 14:51
For analytic functions the answer is no in general, no matter whether they are complex or real valued. – AndreasT Feb 6 at 14:51
Real or complex analytic, the answer is no. Take $a_n=(n!)^2$ and think about the radius of convergence of $\sum \frac{a_n}{n!}x^n$. – julien Feb 6 at 14:51
$C^\infty$ need not be analytic. – CutieKrait Feb 6 at 14:58

Adding a bit to the last point on analytic functions, if we ignore $(b_n)$, a necessary but insufficient condition is $\limsup\sqrt[n]{|a_n|}<\infty$. A sufficient but unnecessary condition is $\limsup\sqrt[n]{|a_n|}=0$, if we're only talking about real analytic functions. In the complex analytic case where $f:\mathbb C\to\mathbb C$ is entire, the last condition would also be necessary. – Jonas Meyer Feb 6 at 15:40
I don't understand what Whitney says. So such $C^\infty$ function exists?! – CutieKrait Feb 6 at 18:37