Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $n$: $$f^{(n)}(0)=a_n,\quad f^{(n)}(1)=b_n$$

share|improve this question
3  
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 '13 at 14:49
3  
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 '13 at 14:51
    
For analytic functions the answer is no in general, no matter whether they are complex or real valued. –  AndreasT Feb 6 '13 at 14:51
    
Real or complex analytic, the answer is no. Take $a_n=(n!)^2$ and think about the radius of convergence of $\sum \frac{a_n}{n!}x^n$. –  1015 Feb 6 '13 at 14:51
    
$C^\infty$ need not be analytic. –  user59671 Feb 6 '13 at 14:58

2 Answers 2

up vote 4 down vote accepted

For smooth functions, this is a version of Whitney's extension theorem.

For analytic functions, the answer is 'no' as already stated in the comments. It's not even true for one point.

share|improve this answer
    
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 '13 at 15:40
    
I don't understand what Whitney says. So such $C^\infty$ function exists?! –  user59671 Feb 6 '13 at 18:37
1  
@CutieKrait: Yes. –  mrf Feb 6 '13 at 20:01

Essentially a duplicate of

Order of growth of derivatives at given x

For a cutoff function $g(x)$ equal to $1$ near $0$ and vanishing outside $(-1/2,1/2)$, take $\sum (\frac{a_n}{n!}x^n g(nx) + \frac{b_n}{n!}(x-1)^n g(n(x-1))$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.