# From power series to differential equations

I was wondering whether for any convergent real power series (or a Frobenius series) we can find (or prove that there exists) a corresponding differential equation that characterizes it. I am aware of Hölder's theorem. So, in effect I am looking for results in these lines but, of course , for real analytic functions. (Generally, my question reads: Can real analytic functions be characterized by differential equations?)

P.S.: I feel these statements are rather vague. But I am eager to hear your comments/answers.

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Note that my question here is not restricted to algebraic differential equations. –  Unknown Dec 26 '10 at 14:07
$x=f$ is the unique solution to $x'=(f'/f)x, x(0)=f(0)$ if $f(0)\ne 0$. This is of course stupid, but it shows shows that you need to make your question a bit more precise. –  Felipe Voloch Dec 26 '10 at 14:53
Thanks. I reworked it exactly as you explained(even with the same variable t) before commenting but wondered why you chose to do so with the $x$'s and $f$'s in that cognitively dissonant way. –  Unknown Dec 26 '10 at 17:42
Maybe an algebraist thinks $x$ is an unknown to be solved for, while an analyst thinks $x$ is the independent variable of a function. –  Gerald Edgar Dec 27 '10 at 1:23
If you consider invertible functions as in oeis.org/A145271, then the autonomous diff. eqn. in that entry applies to characterize f. –  Tom Copeland Feb 1 '12 at 9:37
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I think there are more power series than there are differential equations you can write down, in the same sense that there are more real numbers than there are names for real numbers.

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@Gerry: not really: for every (formal!) power series $f$ there is an ODE $u'(x)=f(x)$, so there are at least as many ODEs as series. –  Mariano Suárez-Alvarez Dec 26 '10 at 15:44
@Mariano, yes, but: if you propose to write the right side out as an infinite power series, then you can't actually write it down, while if you just call the right side $f$, then eventually you run out of names for power series, without running out of power series. –  Gerry Myerson Dec 26 '10 at 16:00
The question asks "can we find, or prove that it exists"... –  Mariano Suárez-Alvarez Dec 26 '10 at 16:05
@Mariano, so, if the question had been whether for any real number $\alpha$ we can find, or prove there exists, a linear polynomial that characterizes it, you would say, yes, $x-\alpha$, and I would say that's cheating, because you generally can't write $\alpha$. I take your point; but I suspect OP had in mind an equation that didn't make such blatant use of the given power series, but rather used only, say, rational numbers and common functions. So I think it's up to OP to clarify intent. –  Gerry Myerson Dec 27 '10 at 2:40