6
$\begingroup$

It is a well studied problem to take a function $f$ expressed (usually expressed as a solution to a differential equation w/ some initial conditions) and ask if it has an "elementary closed form". The Risch "algorithm" allows one to do this for just integration and generally the field of Differential Galois Theory allows one to look at the more general case of differential equations.

I wanted to consider a slightly higher-order problem: given a function as a power series, is there any polynomial differential equation of finite differential order $P(f,f',f'',....f^{(k)})$ which this $f$ satisfies, without necessarily finding such a polynomial differential equation?

My concrete motivation for the problem was if the theta function $\theta(z) = \sum_{n=0}^{\infty} z^{n^2}$ can be expressed as the solution to some polynomial Diff-Eq.

I suspect the answer is no, (in fact probably no for any lacunary series)

But I don't know of any theory/tools/framework for answering problems like this.

Has anyone seen something similar?

Update:

It seems people have thought about this before: An example of a series that is not differentially algebraic?

It also seems that consensus is that lacunary functions are transcendental.

$\endgroup$
6
  • 16
    $\begingroup$ Jacobi showed in 1847 that $\sum_{n\geq 0}z^{n^2}$ does satisfy a polynomial differential equation. The equation appears for instance on page 282 of Enumerative Combinatorics, vol. 2. It is open, for instance, whether $\sum_{n\geq 0}z^{n^3}$ satisfies a polynomial differential equation, but most likely it doesn't. $\endgroup$ Jan 15, 2022 at 16:01
  • 1
    $\begingroup$ That is unbelievable! I am legitimately shocked $\endgroup$ Jan 15, 2022 at 16:02
  • 2
    $\begingroup$ Look for the topic "holonomic function" en.wikipedia.org/wiki/Holonomic_function $\endgroup$ Feb 6, 2022 at 23:55
  • 2
    $\begingroup$ @SidharthGhoshal: not to my knowledge, except in a trivial way. E.g., let $g(z)=f(z)$, regarded as a known function. Then for instance $f'(z)=g'(z)$ (or even $f(z)=g(z)$). This shows that you have to be a little careful about what kind of ODE's are allowed. $\endgroup$ Nov 12, 2023 at 1:48
  • 1
    $\begingroup$ Related ... "Otto Hölder proved in 1887 that the gamma function at least does not satisfy any algebraic differential equation" en.wikipedia.org/wiki/Gamma_function#History $\endgroup$ Nov 13, 2023 at 0:44

2 Answers 2

6
$\begingroup$

I had not seen this post before, but although the differential equation may be complicated, the result is easy: any modular form on $\Gamma_0(4)$ (but this is true much more generally), here $\theta(\tau)$, is a solution of a third degree differential equation. A nonconstructive proof in this special case is as follows: the ring of quasi-modular forms (i.e., including $E_2$) of $\Gamma_0(4)$ is $\Bbb C[E_2,\theta,F_2]$ where $F_2=\sum_{n\ge1,\ n\text{ odd}}\sigma_1(n)q^n$, and this ring is stable by differentiation $D=qd/dq$ and of transcendence degree $3$, so for any $F$ in the ring, $F$, $DF$, $D^2F$, and $D^3F$ must be algebraically dependent.

$\endgroup$
4
$\begingroup$

For the lazy reader I have wrapped Richard Stanley's very helpful comment into an answer. Richard Stanley's Enumerative Combinatorics Volume 2 on Page 282 reviews Jacobis Result that

$$ y = \sum_{n=-\infty}^{\infty} x^{n^2}$$

Obeys $$ \left( y^2z_3 - 15yz_1 z_2 + 30 z_1^3 \right)^2 + 32(yz_2 -3z_1^2)^3 - y^{10} (yz_2 -3z_1^2)^2 = 0 $$

Where

$$ z_1 = xy' \\ z_2 = xy' + x^2 y'' \\ z_3 = xy' + 3x^2 y'' + x^3 y''' $$

Where Jacobi's Original paper is : K. G. J. Jacobi, J. Reine Angew. Math. ( = Crelle’s Journal.) 36 (1847), 97-112

I tried reading Jacobi's original paper but unfortunately it is in German which I cannot read and in google translate he does say something to the effect of "this identity cannot be verified without going to the 14th order..." suggesting that his proof which doesn't involve series level manipulations was at the time the ONLY way to verify the identity (it was physically too much effort to expand and combine terms at the series level to check it (although the fact he had some number like 14 to measure how hard it was might mean Jacobi did actually go ahead and check)) but using Mathematica today hopefully this can be verified in the naive way much more easily.

$\endgroup$
3
  • 1
    $\begingroup$ If we work mod $x^4$ then $y=1+2x$ and $y'=2$ and $y''=y'''=0$ so $z_1=z_2=z_3=2x$ and the left hand side of your claimed identity is $480x^3\neq 0$ so presumably something has been transcribed incorrectly. I do not have my copy of Enumerative Combinatorics to hand so I cannot check it. $\endgroup$ Nov 12, 2023 at 18:06
  • 2
    $\begingroup$ Yes I had a "$+$" on the $15$ when it should have been a "$-$". Good catch $\endgroup$ Nov 12, 2023 at 18:15
  • 4
    $\begingroup$ With that correction, I verified the identity in Maple mod $x^{10001}$, for what that's worth. $\endgroup$ Nov 12, 2023 at 18:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.