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Power Series over q

asked 1 year ago

Tobi gravatar image

I need to compute the power series of g2(z)+(θ4(z))2, where g(z):=η5(5z)η(z), where η is the dedekind eta-function and θ(z)=n=0qn2. 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 ( 1 year ago )

Also, what is g2(z)? Do you mean g(z)2 or is it the second compositional power?

Max Alekseyev gravatar imageMax Alekseyev ( 1 year ago )

g2(z) means g(z)2 and q=e2πiz

Tobi gravatar imageTobi ( 1 year ago )

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answered 1 year ago

dan_fulea gravatar image

I will use τ instead of z, and assume that we have the nome-convention: q=q(τ)=e2πiτ . There is in sage already a (related) function for the expansion of the mentioned θ-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 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 η(τ)=q1/24n1(1qn) , 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)=qq2q4+q5+4q6+...

Tobi gravatar imageTobi ( 1 year ago )

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Asked: 1 year ago

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Last updated: Jan 01 '24