Let $j$ be the Klein $j$invariant (from the theory of modular functions).
Does $j$ satisfy a differential equation of the form $j^\prime (z) = f(j(z),z)$ for
any rational function $f$?

$\begingroup$ Out of curiosity, why are you interested in such a question? (Lest it not be obvious, I mean absolutely no disrespect in asking. I'm merely interested in whether the (non)existence of such an equation would have interesting consequences.) $\endgroup$ – Nick Salter Feb 3 '11 at 23:35

$\begingroup$ (a) other interesting functions satisfy nice differential equations, but there's conspicuously no such equation for j in the usual books. (b) j does satisfy a thirdorder equation (as one of the other responders pointed out). (c) There is a nice computer graphic of j on Wikipedia, but googling for "C code for the jinvariant" doesn't produce much, and I doubt that using the qseries would be a good way to compute j, so maybe if it satisfied a nice differential equation that would be useful. $\endgroup$ – Michael Beeson Feb 4 '11 at 0:59

$\begingroup$ "I doubt that using the qseries would be a good way to compute j"  yes, you can compute the invariant with qseries; just repeatedly apply modular transformations until $\tau$ has large enough imaginary part (and thus a tiny enough $q$), after which you can now use theta functions. $\endgroup$ – J. M. is not a mathematician May 9 '11 at 2:57

$\begingroup$ In practice, once you've found an equivalent $\tau$ in the fundamental domain you might want to use not $q^{1} + 744 + 196884 q + \cdots$ but $E_4^3 / q \prod_n (1q^n)^{24}$ for which you need fewer terms for a good approximation. $\endgroup$ – Noam D. Elkies Sep 12 '11 at 17:49
No. Conceptually, the reason is that $j'(z)$ is a weakly holomorphic (= holomorphic except at the cusp at infinity, where it has a pole) modular form of weight $2$, so it cannot be expressed in terms of $j$ (weakly holomorphic modular form of weight $0$) and $z$ (not anywhere near being a modular form).
For a rigorous proof:
Note that $j(z+1) = j(z)$, so $j'(z+1) = j'(z)$.
Suppose that the $j$ invariant did satisfy a differential equation of your form. Then we'd have $f(j(z), z) = f(j(z+1), z+1) = f(j(z), z+1)$. Note that the functions $z$ and $j(z)$ are algebraically independent (this is just saying that $j(z)$ is a transcendental function). Hence the underlying twovariable rational function $f(x, y)$ satisfies $f(x, y) = f(x, y+1)$. This then easily implies that $f(x, y)$ must be independent of $y$, e.g. $f(x, y) = g(x)$ for some rational function $g$.
So our original differential equation must actually take the form $j'(z) = f(j(z))$. But the left hand side is a nonzero (weakly holomorphic) modular form of weight $2$ while the right hand side has weight 0, and a nonzero modular form has a unique weight, so this is impossible.
The jfunction satisfies a thirdorder differential equation. I've learned this from an old paper of Daniel Bertrand which I am having trouble locating right now. Maybe is this one:
MR0550281 (81i:10042) Bertrand, Daniel Propriétés arithmétiques des dérivées de la fonction modulaire $j(\tau )$. (French) Séminaire de Théorie des Nombres 1977–1978, Exp. No. 22, 4 pp., CNRS, Talence, 1978, 10F37 (10F35)

16$\begingroup$ Abstractly one can see this in the following way: let Q be the field of fractions of the ring of quasimodular forms for $\mathrm{SL}(2,\mathbf{Z})$. Then $j \in Q$, Q has transcendence degree three over $\mathbf{C}$ (generated by $E_2, E_4$ and $E_6$), and $Q$ is closed under differentiation since the derivative of a quasimodular form is quasimodular. Hence j and its first three derivatives are algebraically dependent, so j satisfies a nonlinear third degree differential equation. $\endgroup$ – Dan Petersen Feb 3 '11 at 21:29
(this is too long for a comment)
Here is the explicit equation of order three for the $q$expansion of $j$ multiplied by $q$. Keep in mind that this does not prove that there is no order one differential equation, so it is not an answer to the question.
n [x ]f(x): 3 4 4 3 5 2 , 2 5 (2x f(x)  6912x f(x) + 5971968x f(x) )f (x)  2x f(x) + 3 4 4 3 6912x f(x)  5971968x f(x) * ,,, f (x) + 3 4 4 3 5 2 ,, 2 ( 3x f(x) + 10368x f(x)  8957952x f(x) )f (x) + 2 4 3 3 4 2 , 5 (6x f(x)  20736x f(x) + 17915904x f(x) )f (x)  6x f(x) + 2 4 3 3 20736x f(x)  17915904x f(x) * ,, f (x) + 3 2 4 5 , 4 (x f(x)  1968x f(x) + 2654208x )f (x) + 2 3 3 2 4 , 3 ( 4x f(x) + 7872x f(x)  10616832x f(x))f (x) + 4 2 3 3 2 , 2 (5x f(x)  8352x f(x) + 12939264x f(x) )f (x) + 5 4 2 3 , 5 4 ( 2f(x) + 960x f(x)  4644864x f(x) )f (x) + 1488f(x)  331776x f(x) = 0 , 2 3 4 f(x)= 1 + 744x + 196884x + 21493760x + O(x )]

$\begingroup$ Please tell me how you computed that equation! $\endgroup$ – Michael Beeson Feb 4 '11 at 21:32

$\begingroup$ In FriCAS, create a list [1,744,196884,...] containing the first "few" (I used 6000, but much less should suffice) terms of the expansion, and enter guessADE(l, maxDerivative==3, maxPower==6, maxDegree==5, check==skip, debug==true), wait roughly 200 seconds. The settings for maxPower and maxDegree make the guessing much faster. If you have no clue (as I did) and guess without it can take an hour or so. $\endgroup$ – Martin Rubey Feb 5 '11 at 8:38
A thirdorder differential equation for $j(\tau)$ is gotten via the Schwarzian derivative. The result is (1.13) of the paper by Harnad: http://arxiv.org/abs/solvint/9902013