Welcome to MathOverflow

Question

3

4

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.

edited Dec 26 2010 at 18:03

asked Dec 26 2010 at 13:38

Unknown
138●1●9

1

Note that my question here is not restricted to algebraic differential equations. – Unknown Dec 26 2010 at 14:07

2

$x=f$ 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 2010 at 14:53

1

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 2010 at 17:42

1

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 2010 at 1:23

1

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 2012 at 9:37

show 5 more comments

infact · Answer 1 · 2010-12-27 01:10:09Z UTC

2

A power series $\small f(z) = \sum_{n=0}^\infty c_n(z-x)^n$ has coefficients $c_n$ that are related to the derivatives by $\small f^{(n)}(x) = c_nn!$ with which you may write $\small g(f,f',f'',...,f^{(n)}) = 0$ and obtain a function that relates the derivatives to one another.

answered Dec 27 2010 at 1:10

infact
29●4

1

If the power series is in $z$ with center $x$, then the equation you propose is not a differential equation in $z$. Rather it is an equation of the center and might not yield, upon solution, the power series we started with. The OP seems to suggest a different problem. – To be cont'd Dec 27 2010 at 11:24

Gerry Myerson · Answer 2 · 2010-12-26 15:39:40Z UTC

1

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.

answered Dec 26 2010 at 15:39

Gerry Myerson
15.6k●1●47●87

	@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 2010 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 2010 at 16:00
	The question asks "can we find, or prove that it exists"... – Mariano Suárez-Alvarez Dec 26 2010 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 2010 at 2:40

To be cont'd · Answer 3 · 2010-12-27 11:39:38Z UTC

If the problem as it stands has been solved, then Wikipedia would not hint that the problem of determining a non-algebraic differential equation characterizing the Gamma function is open. So, my current bet is that nothing is known about characterizations of power series in terms of DE's beyond the theory of algebraic DE's.

From power series to differential equations

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

3 Answers

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.