# Recurrences with some relation to Dedekind eta functions

Dedekind eta functions are know to satisfy certain difference equations/recurrence relations. The same is true for ratios of eta functions. Suppose some ratio of eta functions, say $A(q)$ satisfies the recurrence $$f(u, v) = (u - v^2)*(u^2 - v) + 2(u^2 + u*v + v^2)$$ by way of $f(A(q), b*A(q^2)) = 0$. In this case $A(q)$ is a polynomial where some of the terms are known.

As a particular example suppose some coefficients of the polynomial $A(q)$ are given by: $$A(q) = \frac{1}{q} + q - q^2 + q^3 + q^4 - q^7 + 2 \, q^8 + \cdots.$$ The question becomes: How can the Mathematica code be given such that the coefficients of a polynomial, $A(q)$, can be found that satisfies the recurrence $f(A(q), b*A(q^2)) = 0$ ?

In terms of a sample towards a working code:

f[u_, v_]:= (u - v^2)*(u^2 - v) + 2(u^2 + u*v + v^2);
0:= f[A[q], A[q^2]];
CoefficientList[Series[A[q], {q,0,30}], q]


what is missed, for me, is how to define the polynomial A[q_] to start the process.

• Please show the code that you have tried and describe what problems you are encountering. Jul 2 '18 at 4:55
• Is b a constant? Does the expression for A[q] show all its terms of order q^8 or lower - or might there be unknown terms like c q^5? Jul 3 '18 at 3:40
• @bbgodfrey $A(q)$ is an infinite series, $b$ is a constant. For this example I'm supposing $f(A(q), A(q^2))=0$ leads to the first few known values of $A(q)$. It could actually be of the form $f(A(q), A(q^6)/2) = 0$ or some other similar form. Jul 3 '18 at 3:49

The more general problem can be solved as follows. Let

a = Sum[d[i] q^i, {i, 0, 20}] + 1/q;


and determine b and the first eleven d.

Solve[Thread[CoefficientList[q^6 Series[f[a, b a /. q -> q^2], {q, 0, 5}]
// Normal, q] == 0], Flatten@{b, d, Array[d, 10]}] // Flatten

(* {b -> 1, d -> 0, d -> 1, d -> 1, d -> 1, d -> 1,
d -> 3, d -> 2, d -> 2, d -> 4, d -> 4, d -> 4} *)


Comparing a in the question, expanded to eighth order, shows that it is not consistent with the general solution.