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How to simplify $$\frac{\sqrt[3]{\vartheta _3\left(0,e^{-5 \pi }\right)}}{\sqrt[12]{2} \sqrt[6]{\vartheta _2\left(0,e^{-5 \pi }\right) \vartheta _4\left(0,e^{-5 \pi }\right)}}$$

This is a Ramanujan's Class Invariant $G(25)$.

Class invariant is defined as $G(n)=(2k(e^{-\pi\sqrt{n}})k'(e^{-\pi\sqrt{n}}))^{-1/12}$, Where $k(q)$ is a Elliptic Modulus

k[q_] := (EllipticTheta[2, 0, q]/EllipticTheta[3, 0, q])^2

kc[q_] := (EllipticTheta[4, 0, q]/EllipticTheta[3, 0, q])^2

G[n_] := (2 k[E^(-Pi Sqrt[n])] kc[E^(-Pi Sqrt[n])])^(-1/12)

G[25] // N 

1.61803

I know that $G(25)$ is the golden ratio $$G(25)=\frac{1+\sqrt{5}}{2}$$ But "FullSimplify" doesn't work

G[25] // FullSimplify

Sorry for my English. Thanks!

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  • 1
    $\begingroup$ You can reduce this problem to showing that ModularLambda[5 I] == Root[1 - 414728 #1 + 414744 #1^2 - 32 #1^3 + 16 #1^4 &, 1]. I'm not sure how to get Mathematica to do this, however, as its modular functions are not aware of the special algebraic values. $\endgroup$ – J. M. will be back soon Feb 20 '16 at 3:26
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k[q_] := (EllipticTheta[2, 0, q]/EllipticTheta[3, 0, q])^2
kc[q_] := (EllipticTheta[4, 0, q]/EllipticTheta[3, 0, q])^2
G[n_] := (2 k[E^(-Pi Sqrt[n])] kc[E^(-Pi Sqrt[n])])^(-1/12)

TrueQ[G[25] == N@GoldenRatio]

(*True*)

If you want a more exact solution, you might want to check this out.

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    $\begingroup$ Try e.g. TrueQ[G[25] == N[GoldenRatio + 1/10^14]]. Even though you get true it doesn't answer the question. $\endgroup$ – Artes Feb 1 '16 at 8:18
  • $\begingroup$ That's right. But I will keep the answer for reference. $\endgroup$ – thedude Feb 1 '16 at 9:44

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