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Power Series over q

asked 2023-12-27 22:48:03 +0200

Tobi gravatar image

I need to compute the power series of $g^2(z) + (\theta^4(z))^2$, where $$g(z) := \frac{\eta^5(5z)}{\eta(z)},$$ where $\eta$ is the dedekind eta-function and $$\theta(z) = \sum_{n=0}^\infty q^{n^2}.$$ I am using the online Version and ChatGPT is only giving errors...

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Presumably there is a relationship between q and z?

John Palmieri gravatar imageJohn Palmieri ( 2023-12-28 00:14:27 +0200 )edit

Also, what is $g^2(z)$? Do you mean $g(z)^2$ or is it the second compositional power?

Max Alekseyev gravatar imageMax Alekseyev ( 2023-12-28 14:32:01 +0200 )edit

$g^2(z)$ means $g(z)^2$ and $q = e^{2\pi iz}$

Tobi gravatar imageTobi ( 2024-01-02 18:09:24 +0200 )edit

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answered 2024-01-01 15:32:24 +0200

dan_fulea gravatar image

I will use $\tau$ instead of $z$, and assume that we have the nome-convention: $$ q = q(\tau)=e^{2\pi i\; \tau}\ . $$ There is in sage already a (related) function for the expansion of the mentioned $\theta$-function,

sage: f = (1 + theta_qexp(150)) / 2
sage: f
1 + q + q^4 + q^9 + q^16 + q^25 + q^36 + q^49 + q^64 + q^81 + q^100 + q^121 + q^144 + O(q^150)

(The implemented theta_qexp corresponds to the "same" sum where $n$ runs in $\Bbb Z$.)

Now we can take our own power series ring in (an other) variable $q$, or use this one above. I will use "mine", so that a precision is also in there.

The $g$ is quickly computed, using $$ \eta(\tau)=q^{1/24}\prod_{n\ge 1}(1-q^n)\ , $$ We have, using some higher precision:

PREC = 500
R.<q> = PowerSeriesRing(QQ)
f = sum([q^(n^2) for n in range(floor(sqrt(PREC)))]) + O(q^PREC)
g = q * prod([ (1 - q^(5*n) + O(q^PREC))^5/(1 - q^n + O(q^PREC)) for n in [1..PREC] ])
h = g^2 + f^8

print(h + O(q^20))

The last print shows some first few terms, i see no "good" pattern...

1 + 8*q + 29*q^2 + 58*q^3 + 83*q^4 + 122*q^5 + 216*q^6 + 314*q^7 + 354*q^8 + 404*q^9 + 617*q^10 + 836*q^11 + 912*q^12 + 956*q^13 + 1332*q^14 + 1726*q^15 + 1851*q^16 + 1776*q^17 + 2309*q^18 + 2984*q^19 + O(q^20)
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So I should compute g by using

PREC = 500 R. = PowerSeriesRing(QQ) g =q* prod([ (1 - q^(5*n) + O(q^PREC))^5/(1 - q^(n) + O(q^PREC)) for n in [1..PREC] ])

print(g + O(q^20))

right? But somehow this isn't giving the correct series. I the paper I'm using, its given, that $$g(z) = q-q^2-q^4+q^5+4q^6+...$$

Tobi gravatar imageTobi ( 2024-01-02 18:00:39 +0200 )edit

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Asked: 2023-12-27 22:48:03 +0200

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Last updated: Jan 01