Value of the J Invariant at $frac{1+sqrt{-163}}{2}$
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
edited Nov 27 at 15:09
nospoon
4,5261432
asked Nov 26 at 2:32
uhhhhidk
564
|
show 2 more comments
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
edited Nov 27 at 15:09
nospoon
4,5261432
asked Nov 26 at 2:32
uhhhhidk
564
1
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.
– pisco
Nov 26 at 2:53
1
@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?
– uhhhhidk
Nov 26 at 4:39
You can see this arxiv.org/abs/0807.2976
– Nikos Bagis
Nov 27 at 21:09
1
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.
– Paramanand Singh
Nov 28 at 6:08
@ParamanandSingh is that $G=G_{163}$?
– uhhhhidk
Dec 3 at 4:23
|
show 2 more comments
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
edited Nov 27 at 15:09
nospoon
4,5261432
asked Nov 26 at 2:32
uhhhhidk
564
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
edited Nov 27 at 15:09
nospoon
4,5261432
asked Nov 26 at 2:32
uhhhhidk
564
edited Nov 27 at 15:09
nospoon
4,5261432
asked Nov 26 at 2:32
uhhhhidk
564
edited Nov 27 at 15:09
nospoon
4,5261432
edited Nov 27 at 15:09
nospoon
4,5261432
edited Nov 27 at 15:09
nospoon
4,5261432
4,5261432
asked Nov 26 at 2:32
uhhhhidk
564
asked Nov 26 at 2:32
uhhhhidk
564
asked Nov 26 at 2:32
uhhhhidk
564
564
1
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.
– pisco
Nov 26 at 2:53
1
@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?
– uhhhhidk
Nov 26 at 4:39
You can see this arxiv.org/abs/0807.2976
– Nikos Bagis
Nov 27 at 21:09
1
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.
– Paramanand Singh
Nov 28 at 6:08
@ParamanandSingh is that $G=G_{163}$?
– uhhhhidk
Dec 3 at 4:23
|
show 2 more comments
1
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.
– pisco
Nov 26 at 2:53
1
@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?
– uhhhhidk
Nov 26 at 4:39
You can see this arxiv.org/abs/0807.2976
– Nikos Bagis
Nov 27 at 21:09
1
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.
– Paramanand Singh
Nov 28 at 6:08
@ParamanandSingh is that $G=G_{163}$?
– uhhhhidk
Dec 3 at 4:23
1
1
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.
– pisco
Nov 26 at 2:53
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.
– pisco
Nov 26 at 2:53
1
1
@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?
– uhhhhidk
Nov 26 at 4:39
@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?
– uhhhhidk
Nov 26 at 4:39
You can see this arxiv.org/abs/0807.2976
– Nikos Bagis
Nov 27 at 21:09
You can see this arxiv.org/abs/0807.2976
– Nikos Bagis
Nov 27 at 21:09
1
1
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.
– Paramanand Singh
Nov 28 at 6:08
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.
– Paramanand Singh
Nov 28 at 6:08
@ParamanandSingh is that $G=G_{163}$?
– uhhhhidk
Dec 3 at 4:23
@ParamanandSingh is that $G=G_{163}$?
– uhhhhidk
Dec 3 at 4:23
|
show 2 more comments
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1
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.
– pisco
Nov 26 at 2:53
1
@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?
– uhhhhidk
Nov 26 at 4:39
You can see this arxiv.org/abs/0807.2976
– Nikos Bagis
Nov 27 at 21:09
1
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.
– Paramanand Singh
Nov 28 at 6:08
@ParamanandSingh is that $G=G_{163}$?
– uhhhhidk
Dec 3 at 4:23