Evaluation of the limit $lim_qrightarrow 1 fracphi^5(q)_inftyphi(q^5)_infty$
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Given the Euler function $phi(q)=prod_n = 1^infty(1-q^n)$ which is a modular form where $q=exp(2pi i tau)$, $|q|lt1$
Then what is the limit $lim_qrightarrow 1fracphi^5(q)_inftyphi(q^5)_infty$ ?
calculus limits modular-forms theta-functions
asked Aug 16 at 6:25
Nicco
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up vote
2
down vote
favorite
Given the Euler function $phi(q)=prod_n = 1^infty(1-q^n)$ which is a modular form where $q=exp(2pi i tau)$, $|q|lt1$
Then what is the limit $lim_qrightarrow 1fracphi^5(q)_inftyphi(q^5)_infty$ ?
calculus limits modular-forms theta-functions
asked Aug 16 at 6:25
Nicco
1,073726
1
You can use the expression for Dedekind's eta function in terms of elliptic integral $K$ and modulus $k$. Thus we have $$q^1/24phi(q)=2^-1/6sqrtfrac2Kpik^1/12k'^1/3$$
– Paramanand Singh
Aug 16 at 6:55
@Paramanand Singh: that's actually a good idea
– Nicco
Aug 16 at 7:09
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Given the Euler function $phi(q)=prod_n = 1^infty(1-q^n)$ which is a modular form where $q=exp(2pi i tau)$, $|q|lt1$
Then what is the limit $lim_qrightarrow 1fracphi^5(q)_inftyphi(q^5)_infty$ ?
calculus limits modular-forms theta-functions
asked Aug 16 at 6:25
Nicco
1,073726
Given the Euler function $phi(q)=prod_n = 1^infty(1-q^n)$ which is a modular form where $q=exp(2pi i tau)$, $|q|lt1$
Then what is the limit $lim_qrightarrow 1fracphi^5(q)_inftyphi(q^5)_infty$ ?
calculus limits modular-forms theta-functions
asked Aug 16 at 6:25
Nicco
1,073726
edited Aug 17 at 0:42
asked Aug 16 at 6:25
Nicco
1,073726
asked Aug 16 at 6:25
Nicco
1,073726
asked Aug 16 at 6:25
Nicco
1,073726
1,073726
1
You can use the expression for Dedekind's eta function in terms of elliptic integral $K$ and modulus $k$. Thus we have $$q^1/24phi(q)=2^-1/6sqrtfrac2Kpik^1/12k'^1/3$$
– Paramanand Singh
Aug 16 at 6:55
@Paramanand Singh: that's actually a good idea
– Nicco
Aug 16 at 7:09
add a comment |Â
1
You can use the expression for Dedekind's eta function in terms of elliptic integral $K$ and modulus $k$. Thus we have $$q^1/24phi(q)=2^-1/6sqrtfrac2Kpik^1/12k'^1/3$$
– Paramanand Singh
Aug 16 at 6:55
@Paramanand Singh: that's actually a good idea
– Nicco
Aug 16 at 7:09
1
1
You can use the expression for Dedekind's eta function in terms of elliptic integral $K$ and modulus $k$. Thus we have $$q^1/24phi(q)=2^-1/6sqrtfrac2Kpik^1/12k'^1/3$$
– Paramanand Singh
Aug 16 at 6:55
You can use the expression for Dedekind's eta function in terms of elliptic integral $K$ and modulus $k$. Thus we have $$q^1/24phi(q)=2^-1/6sqrtfrac2Kpik^1/12k'^1/3$$
– Paramanand Singh
Aug 16 at 6:55
@Paramanand Singh: that's actually a good idea
– Nicco
Aug 16 at 7:09
@Paramanand Singh: that's actually a good idea
– Nicco
Aug 16 at 7:09
add a comment |Â
2 Answers
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When $qto 1^-$, $tau to 0$, hence $-1/tau to iinfty$,
$$beginalignedfracphi ^5(q)_infty phi (q^5)_infty = fraceta ^5(tau )eta (5tau ) &= fracleft( frac - 1tau i right)^5/2eta ^5(frac - 1tau )left( frac - 15tau i right)^1/2eta (frac - 15tau ) \ &= -fracsqrt 5 tau ^2frace^ - frac5pi i12tau prodlimits_n = 1^infty left( 1 - e^ - frac2pi nitau right)^5 e^ - fracpi i60tau prodlimits_n = 1^infty left( 1 - e^ - frac2pi ni5tau right) \ &sim -fracsqrt 5 tau ^2e^ - frac2pi i5tau to 0endaligned$$
Note that the two infinite products both $to 1$.
answered Aug 16 at 7:29
pisco
10k21336
@Paramanand Singh: do you agree with this answer?
– Nicco
Aug 16 at 7:39
2
@Nicco: using my approach via elliptic integrals also gives the result as $0$ but with more difficulty. The current answer gives a more elegant / direct approach. +1 to pisco.
– Paramanand Singh
Aug 16 at 8:14
I like this direct approach
– Nicco
Aug 17 at 0:22
add a comment |Â
up vote
2
down vote
Given the following identity $frac(q)^5_infty(q^5)_infty=1+5sum_n=1^inftyfrac(-1)^n nq^fracn(n+1)2prod_k=1^n-1Big(R_k+q^kBig)$ from my old post where $frac1R_n=R(q^n,q)$ is the Generalized Rogers Ramanujan continued fraction
If we consider the fact that $lim_xrightarrow 0+frac1R(e^-nx,e^-x)=phi$
where $phi=frac1+sqrt52$ is the golden ratio
Then we are led to $$lim_qrightarrow 1fracphi^5(q)_inftyphi(q^5)_infty=1+sum_n=1^inftyfrac5n (-1)^n (1+phi)^n$$
Whereby the infinite series on the RHS converges to zero.
answered Aug 17 at 0:07
Nicco
1,073726
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
When $qto 1^-$, $tau to 0$, hence $-1/tau to iinfty$,
$$beginalignedfracphi ^5(q)_infty phi (q^5)_infty = fraceta ^5(tau )eta (5tau ) &= fracleft( frac - 1tau i right)^5/2eta ^5(frac - 1tau )left( frac - 15tau i right)^1/2eta (frac - 15tau ) \ &= -fracsqrt 5 tau ^2frace^ - frac5pi i12tau prodlimits_n = 1^infty left( 1 - e^ - frac2pi nitau right)^5 e^ - fracpi i60tau prodlimits_n = 1^infty left( 1 - e^ - frac2pi ni5tau right) \ &sim -fracsqrt 5 tau ^2e^ - frac2pi i5tau to 0endaligned$$
Note that the two infinite products both $to 1$.
answered Aug 16 at 7:29
pisco
10k21336
@Paramanand Singh: do you agree with this answer?
– Nicco
Aug 16 at 7:39
2
@Nicco: using my approach via elliptic integrals also gives the result as $0$ but with more difficulty. The current answer gives a more elegant / direct approach. +1 to pisco.
– Paramanand Singh
Aug 16 at 8:14
I like this direct approach
– Nicco
Aug 17 at 0:22
add a comment |Â
up vote
3
down vote
When $qto 1^-$, $tau to 0$, hence $-1/tau to iinfty$,
$$beginalignedfracphi ^5(q)_infty phi (q^5)_infty = fraceta ^5(tau )eta (5tau ) &= fracleft( frac - 1tau i right)^5/2eta ^5(frac - 1tau )left( frac - 15tau i right)^1/2eta (frac - 15tau ) \ &= -fracsqrt 5 tau ^2frace^ - frac5pi i12tau prodlimits_n = 1^infty left( 1 - e^ - frac2pi nitau right)^5 e^ - fracpi i60tau prodlimits_n = 1^infty left( 1 - e^ - frac2pi ni5tau right) \ &sim -fracsqrt 5 tau ^2e^ - frac2pi i5tau to 0endaligned$$
Note that the two infinite products both $to 1$.
answered Aug 16 at 7:29
pisco
10k21336
@Paramanand Singh: do you agree with this answer?
– Nicco
Aug 16 at 7:39
2
@Nicco: using my approach via elliptic integrals also gives the result as $0$ but with more difficulty. The current answer gives a more elegant / direct approach. +1 to pisco.
– Paramanand Singh
Aug 16 at 8:14
I like this direct approach
– Nicco
Aug 17 at 0:22
add a comment |Â
up vote
3
down vote
up vote
3
down vote
When $qto 1^-$, $tau to 0$, hence $-1/tau to iinfty$,
$$beginalignedfracphi ^5(q)_infty phi (q^5)_infty = fraceta ^5(tau )eta (5tau ) &= fracleft( frac - 1tau i right)^5/2eta ^5(frac - 1tau )left( frac - 15tau i right)^1/2eta (frac - 15tau ) \ &= -fracsqrt 5 tau ^2frace^ - frac5pi i12tau prodlimits_n = 1^infty left( 1 - e^ - frac2pi nitau right)^5 e^ - fracpi i60tau prodlimits_n = 1^infty left( 1 - e^ - frac2pi ni5tau right) \ &sim -fracsqrt 5 tau ^2e^ - frac2pi i5tau to 0endaligned$$
Note that the two infinite products both $to 1$.
answered Aug 16 at 7:29
pisco
10k21336
When $qto 1^-$, $tau to 0$, hence $-1/tau to iinfty$,
$$beginalignedfracphi ^5(q)_infty phi (q^5)_infty = fraceta ^5(tau )eta (5tau ) &= fracleft( frac - 1tau i right)^5/2eta ^5(frac - 1tau )left( frac - 15tau i right)^1/2eta (frac - 15tau ) \ &= -fracsqrt 5 tau ^2frace^ - frac5pi i12tau prodlimits_n = 1^infty left( 1 - e^ - frac2pi nitau right)^5 e^ - fracpi i60tau prodlimits_n = 1^infty left( 1 - e^ - frac2pi ni5tau right) \ &sim -fracsqrt 5 tau ^2e^ - frac2pi i5tau to 0endaligned$$
Note that the two infinite products both $to 1$.
answered Aug 16 at 7:29
pisco
10k21336
answered Aug 16 at 7:29
pisco
10k21336
answered Aug 16 at 7:29
pisco
10k21336
answered Aug 16 at 7:29
pisco
10k21336
10k21336
@Paramanand Singh: do you agree with this answer?
– Nicco
Aug 16 at 7:39
2
@Nicco: using my approach via elliptic integrals also gives the result as $0$ but with more difficulty. The current answer gives a more elegant / direct approach. +1 to pisco.
– Paramanand Singh
Aug 16 at 8:14
I like this direct approach
– Nicco
Aug 17 at 0:22
add a comment |Â
@Paramanand Singh: do you agree with this answer?
– Nicco
Aug 16 at 7:39
2
@Nicco: using my approach via elliptic integrals also gives the result as $0$ but with more difficulty. The current answer gives a more elegant / direct approach. +1 to pisco.
– Paramanand Singh
Aug 16 at 8:14
I like this direct approach
– Nicco
Aug 17 at 0:22
@Paramanand Singh: do you agree with this answer?
– Nicco
Aug 16 at 7:39
@Paramanand Singh: do you agree with this answer?
– Nicco
Aug 16 at 7:39
2
2
@Nicco: using my approach via elliptic integrals also gives the result as $0$ but with more difficulty. The current answer gives a more elegant / direct approach. +1 to pisco.
– Paramanand Singh
Aug 16 at 8:14
@Nicco: using my approach via elliptic integrals also gives the result as $0$ but with more difficulty. The current answer gives a more elegant / direct approach. +1 to pisco.
– Paramanand Singh
Aug 16 at 8:14
I like this direct approach
– Nicco
Aug 17 at 0:22
I like this direct approach
– Nicco
Aug 17 at 0:22
add a comment |Â
up vote
2
down vote
Given the following identity $frac(q)^5_infty(q^5)_infty=1+5sum_n=1^inftyfrac(-1)^n nq^fracn(n+1)2prod_k=1^n-1Big(R_k+q^kBig)$ from my old post where $frac1R_n=R(q^n,q)$ is the Generalized Rogers Ramanujan continued fraction
If we consider the fact that $lim_xrightarrow 0+frac1R(e^-nx,e^-x)=phi$
where $phi=frac1+sqrt52$ is the golden ratio
Then we are led to $$lim_qrightarrow 1fracphi^5(q)_inftyphi(q^5)_infty=1+sum_n=1^inftyfrac5n (-1)^n (1+phi)^n$$
Whereby the infinite series on the RHS converges to zero.
answered Aug 17 at 0:07
Nicco
1,073726
add a comment |Â
up vote
2
down vote
Given the following identity $frac(q)^5_infty(q^5)_infty=1+5sum_n=1^inftyfrac(-1)^n nq^fracn(n+1)2prod_k=1^n-1Big(R_k+q^kBig)$ from my old post where $frac1R_n=R(q^n,q)$ is the Generalized Rogers Ramanujan continued fraction
If we consider the fact that $lim_xrightarrow 0+frac1R(e^-nx,e^-x)=phi$
where $phi=frac1+sqrt52$ is the golden ratio
Then we are led to $$lim_qrightarrow 1fracphi^5(q)_inftyphi(q^5)_infty=1+sum_n=1^inftyfrac5n (-1)^n (1+phi)^n$$
Whereby the infinite series on the RHS converges to zero.
answered Aug 17 at 0:07
Nicco
1,073726
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Given the following identity $frac(q)^5_infty(q^5)_infty=1+5sum_n=1^inftyfrac(-1)^n nq^fracn(n+1)2prod_k=1^n-1Big(R_k+q^kBig)$ from my old post where $frac1R_n=R(q^n,q)$ is the Generalized Rogers Ramanujan continued fraction
If we consider the fact that $lim_xrightarrow 0+frac1R(e^-nx,e^-x)=phi$
where $phi=frac1+sqrt52$ is the golden ratio
Then we are led to $$lim_qrightarrow 1fracphi^5(q)_inftyphi(q^5)_infty=1+sum_n=1^inftyfrac5n (-1)^n (1+phi)^n$$
Whereby the infinite series on the RHS converges to zero.
answered Aug 17 at 0:07
Nicco
1,073726
Given the following identity $frac(q)^5_infty(q^5)_infty=1+5sum_n=1^inftyfrac(-1)^n nq^fracn(n+1)2prod_k=1^n-1Big(R_k+q^kBig)$ from my old post where $frac1R_n=R(q^n,q)$ is the Generalized Rogers Ramanujan continued fraction
If we consider the fact that $lim_xrightarrow 0+frac1R(e^-nx,e^-x)=phi$
where $phi=frac1+sqrt52$ is the golden ratio
Then we are led to $$lim_qrightarrow 1fracphi^5(q)_inftyphi(q^5)_infty=1+sum_n=1^inftyfrac5n (-1)^n (1+phi)^n$$
Whereby the infinite series on the RHS converges to zero.
answered Aug 17 at 0:07
Nicco
1,073726
edited Aug 17 at 0:19
answered Aug 17 at 0:07
Nicco
1,073726
answered Aug 17 at 0:07
Nicco
1,073726
answered Aug 17 at 0:07
Nicco
1,073726
1,073726
add a comment |Â
add a comment |Â
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1
You can use the expression for Dedekind's eta function in terms of elliptic integral $K$ and modulus $k$. Thus we have $$q^1/24phi(q)=2^-1/6sqrtfrac2Kpik^1/12k'^1/3$$
– Paramanand Singh
Aug 16 at 6:55
@Paramanand Singh: that's actually a good idea
– Nicco
Aug 16 at 7:09