Value of the J Invariant at $frac{1+sqrt{-163}}{2}$












4












$begingroup$


For a while I've wanted to be able to show why $e^{pisqrt{163}}approx 744+640320^3$, but I have no idea how to show that $j(frac{1+sqrt{-163}}{2})=-640320^3$.
I considered using the fact that $mid eta(frac{1+sqrt{-163})}{2})mid^{4}=frac{1}{326pi}prod_{n=1}^{162}Gamma({frac{n}{163}} )^{(frac{n}{163})}$ (exponent is the legendre symbol) but I'd also need $eta(sqrt{-163})$ and also I have no idea how the simplification would work out. I've seen values of J being related to roots of polynomials and was wondering if that would be the easiest way to compute it. If anyone could explain methods that could be used to compute it I'd appreciate it.






special-functions closed-form modular-forms







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Adjoining $alpha = j(frac{1+sqrt{-163}}{2})$ to $K = mathbb{Q}(sqrt{-163})$ gives its Hilbert class field, which is itself. Hence $alphain K$, since $alpha$ is real and an algebraic integer, $alpha in mathbb{Z}$, so you can just numerically compute $alpha$ to a desired accuracy.
    $endgroup$
    – pisco
    Nov 26 '18 at 2:53








  • 1




    $begingroup$
    @pisco Thanks for your answer! Is the fact that J at that number is an integer the only way to show the value is -640320^3? Are there any other ways to directly get that value without using numerical computation?
    $endgroup$
    – uhhhhidk
    Nov 26 '18 at 4:39










  • $begingroup$
    You can see this arxiv.org/abs/0807.2976
    $endgroup$
    – Nikos Bagis
    Nov 27 '18 at 21:09






  • 1




    $begingroup$
    The j invariant is related to the Ramanujan class invariant $G$ via $j=-dfrac{27G^{48}}{(G^{24}-4)^3}$ but the calculation of $G$ and further computation of $j$ is difficult at least via pen and paper.
    $endgroup$
    – Paramanand Singh
    Nov 28 '18 at 6:08










  • $begingroup$
    @ParamanandSingh is that $G=G_{163}$?
    $endgroup$
    – uhhhhidk
    Dec 3 '18 at 4:23
















4












$begingroup$


For a while I've wanted to be able to show why $e^{pisqrt{163}}approx 744+640320^3$, but I have no idea how to show that $j(frac{1+sqrt{-163}}{2})=-640320^3$.
I considered using the fact that $mid eta(frac{1+sqrt{-163})}{2})mid^{4}=frac{1}{326pi}prod_{n=1}^{162}Gamma({frac{n}{163}} )^{(frac{n}{163})}$ (exponent is the legendre symbol) but I'd also need $eta(sqrt{-163})$ and also I have no idea how the simplification would work out. I've seen values of J being related to roots of polynomials and was wondering if that would be the easiest way to compute it. If anyone could explain methods that could be used to compute it I'd appreciate it.






special-functions closed-form modular-forms







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Adjoining $alpha = j(frac{1+sqrt{-163}}{2})$ to $K = mathbb{Q}(sqrt{-163})$ gives its Hilbert class field, which is itself. Hence $alphain K$, since $alpha$ is real and an algebraic integer, $alpha in mathbb{Z}$, so you can just numerically compute $alpha$ to a desired accuracy.
    $endgroup$
    – pisco
    Nov 26 '18 at 2:53








  • 1




    $begingroup$
    @pisco Thanks for your answer! Is the fact that J at that number is an integer the only way to show the value is -640320^3? Are there any other ways to directly get that value without using numerical computation?
    $endgroup$
    – uhhhhidk
    Nov 26 '18 at 4:39










  • $begingroup$
    You can see this arxiv.org/abs/0807.2976
    $endgroup$
    – Nikos Bagis
    Nov 27 '18 at 21:09






  • 1




    $begingroup$
    The j invariant is related to the Ramanujan class invariant $G$ via $j=-dfrac{27G^{48}}{(G^{24}-4)^3}$ but the calculation of $G$ and further computation of $j$ is difficult at least via pen and paper.
    $endgroup$
    – Paramanand Singh
    Nov 28 '18 at 6:08










  • $begingroup$
    @ParamanandSingh is that $G=G_{163}$?
    $endgroup$
    – uhhhhidk
    Dec 3 '18 at 4:23














4












4








4


1



$begingroup$


For a while I've wanted to be able to show why $e^{pisqrt{163}}approx 744+640320^3$, but I have no idea how to show that $j(frac{1+sqrt{-163}}{2})=-640320^3$.
I considered using the fact that $mid eta(frac{1+sqrt{-163})}{2})mid^{4}=frac{1}{326pi}prod_{n=1}^{162}Gamma({frac{n}{163}} )^{(frac{n}{163})}$ (exponent is the legendre symbol) but I'd also need $eta(sqrt{-163})$ and also I have no idea how the simplification would work out. I've seen values of J being related to roots of polynomials and was wondering if that would be the easiest way to compute it. If anyone could explain methods that could be used to compute it I'd appreciate it.






special-functions closed-form modular-forms







share|cite|improve this question











$endgroup$




For a while I've wanted to be able to show why $e^{pisqrt{163}}approx 744+640320^3$, but I have no idea how to show that $j(frac{1+sqrt{-163}}{2})=-640320^3$.
I considered using the fact that $mid eta(frac{1+sqrt{-163})}{2})mid^{4}=frac{1}{326pi}prod_{n=1}^{162}Gamma({frac{n}{163}} )^{(frac{n}{163})}$ (exponent is the legendre symbol) but I'd also need $eta(sqrt{-163})$ and also I have no idea how the simplification would work out. I've seen values of J being related to roots of polynomials and was wondering if that would be the easiest way to compute it. If anyone could explain methods that could be used to compute it I'd appreciate it.






special-functions closed-form modular-forms




special-functions closed-form modular-forms






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 27 '18 at 15:09









nospoon

4,5411432




4,5411432










asked Nov 26 '18 at 2:32









uhhhhidkuhhhhidk

764




764








  • 1




    $begingroup$
    Adjoining $alpha = j(frac{1+sqrt{-163}}{2})$ to $K = mathbb{Q}(sqrt{-163})$ gives its Hilbert class field, which is itself. Hence $alphain K$, since $alpha$ is real and an algebraic integer, $alpha in mathbb{Z}$, so you can just numerically compute $alpha$ to a desired accuracy.
    $endgroup$
    – pisco
    Nov 26 '18 at 2:53








  • 1




    $begingroup$
    @pisco Thanks for your answer! Is the fact that J at that number is an integer the only way to show the value is -640320^3? Are there any other ways to directly get that value without using numerical computation?
    $endgroup$
    – uhhhhidk
    Nov 26 '18 at 4:39










  • $begingroup$
    You can see this arxiv.org/abs/0807.2976
    $endgroup$
    – Nikos Bagis
    Nov 27 '18 at 21:09






  • 1




    $begingroup$
    The j invariant is related to the Ramanujan class invariant $G$ via $j=-dfrac{27G^{48}}{(G^{24}-4)^3}$ but the calculation of $G$ and further computation of $j$ is difficult at least via pen and paper.
    $endgroup$
    – Paramanand Singh
    Nov 28 '18 at 6:08










  • $begingroup$
    @ParamanandSingh is that $G=G_{163}$?
    $endgroup$
    – uhhhhidk
    Dec 3 '18 at 4:23














  • 1




    $begingroup$
    Adjoining $alpha = j(frac{1+sqrt{-163}}{2})$ to $K = mathbb{Q}(sqrt{-163})$ gives its Hilbert class field, which is itself. Hence $alphain K$, since $alpha$ is real and an algebraic integer, $alpha in mathbb{Z}$, so you can just numerically compute $alpha$ to a desired accuracy.
    $endgroup$
    – pisco
    Nov 26 '18 at 2:53








  • 1




    $begingroup$
    @pisco Thanks for your answer! Is the fact that J at that number is an integer the only way to show the value is -640320^3? Are there any other ways to directly get that value without using numerical computation?
    $endgroup$
    – uhhhhidk
    Nov 26 '18 at 4:39










  • $begingroup$
    You can see this arxiv.org/abs/0807.2976
    $endgroup$
    – Nikos Bagis
    Nov 27 '18 at 21:09






  • 1




    $begingroup$
    The j invariant is related to the Ramanujan class invariant $G$ via $j=-dfrac{27G^{48}}{(G^{24}-4)^3}$ but the calculation of $G$ and further computation of $j$ is difficult at least via pen and paper.
    $endgroup$
    – Paramanand Singh
    Nov 28 '18 at 6:08










  • $begingroup$
    @ParamanandSingh is that $G=G_{163}$?
    $endgroup$
    – uhhhhidk
    Dec 3 '18 at 4:23








1




1




$begingroup$
Adjoining $alpha = j(frac{1+sqrt{-163}}{2})$ to $K = mathbb{Q}(sqrt{-163})$ gives its Hilbert class field, which is itself. Hence $alphain K$, since $alpha$ is real and an algebraic integer, $alpha in mathbb{Z}$, so you can just numerically compute $alpha$ to a desired accuracy.
$endgroup$
– pisco
Nov 26 '18 at 2:53






$begingroup$
Adjoining $alpha = j(frac{1+sqrt{-163}}{2})$ to $K = mathbb{Q}(sqrt{-163})$ gives its Hilbert class field, which is itself. Hence $alphain K$, since $alpha$ is real and an algebraic integer, $alpha in mathbb{Z}$, so you can just numerically compute $alpha$ to a desired accuracy.
$endgroup$
– pisco
Nov 26 '18 at 2:53






1




1




$begingroup$
@pisco Thanks for your answer! Is the fact that J at that number is an integer the only way to show the value is -640320^3? Are there any other ways to directly get that value without using numerical computation?
$endgroup$
– uhhhhidk
Nov 26 '18 at 4:39




$begingroup$
@pisco Thanks for your answer! Is the fact that J at that number is an integer the only way to show the value is -640320^3? Are there any other ways to directly get that value without using numerical computation?
$endgroup$
– uhhhhidk
Nov 26 '18 at 4:39












$begingroup$
You can see this arxiv.org/abs/0807.2976
$endgroup$
– Nikos Bagis
Nov 27 '18 at 21:09




$begingroup$
You can see this arxiv.org/abs/0807.2976
$endgroup$
– Nikos Bagis
Nov 27 '18 at 21:09




1




1




$begingroup$
The j invariant is related to the Ramanujan class invariant $G$ via $j=-dfrac{27G^{48}}{(G^{24}-4)^3}$ but the calculation of $G$ and further computation of $j$ is difficult at least via pen and paper.
$endgroup$
– Paramanand Singh
Nov 28 '18 at 6:08




$begingroup$
The j invariant is related to the Ramanujan class invariant $G$ via $j=-dfrac{27G^{48}}{(G^{24}-4)^3}$ but the calculation of $G$ and further computation of $j$ is difficult at least via pen and paper.
$endgroup$
– Paramanand Singh
Nov 28 '18 at 6:08












$begingroup$
@ParamanandSingh is that $G=G_{163}$?
$endgroup$
– uhhhhidk
Dec 3 '18 at 4:23




$begingroup$
@ParamanandSingh is that $G=G_{163}$?
$endgroup$
– uhhhhidk
Dec 3 '18 at 4:23










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