I. Rogers-Selberg Mod 7 Identities
II. Bailey Mod 9 Identities
III. Rogers Mod 14 Identities
IV. Dyson Mod 27 Identities
This is similar to the 11/3/2012 update. Let,
be a positive integer for some odd number n. Thus, there are only 4 possibilities:
. Given the standard Ramanujan theta functions
, then,
For example, for n = 5, hence m = 3, we have,
and so on for the other three n.
--------------END--------------
Define,
(Note, as usual, that the even-index h_k have a negative sign.) It turns out that appropriate pairs of h_k are roots of a polynomial whose coefficients are in
, analogous to the case for n = 13 discovered by Ramanujan. Since one pair is a constant,
, then the remaining five are the roots of a quintic (naturally enough) given by,
with j as the eta quotient given above. The roots are then,
(Note: The quintic looks vaguely familiar to me, it seems I've come across it before, or something similar in the course of my research into the Rogers-Ramanujan continued fraction, but I cannot recall precisely in what context.)
Update April 28, 2022: After almost 10 years, I finally found the answer to my "note". This is just the EMMA LEHMER QUINTIC!
--------------END--------------
Here is a generalization of one identity described by Berndt as "...fascinating, but with no direct proof" (Ramanujan's Notebooks III, p.322). Let,
be a positive integer for some odd number n. Of course, there are only 6 possibilities:
. Given the standard Ramanujan theta functions
, then it is proposed that,
and, as suggested by Michael Somos, by using a Fricke involution,
Thus, for n = 3, we have,
and,
and so on for the other five n, with the case n = 7 given in page 322 which was the inspiration for this generalization. I have no rigorous proof for this "family", but one can easily see via Mathematica that the proposed equality indeed holds true for hundreds of decimal places.
--------------END--------------
It seems Ramanujan missed certain aspects of theta quotients at p = 13.
I. Case p = 7
To illustrate, define the quotients for p = 7 as,
hence,
functions highly analogous to the Rogers–Ramanujan continued fraction. (Kindly note that the even-index
is negative.) Given the Dedekind eta function
, let,
then Ramanujan found that the 3 roots of the cubic,
are,
Note also another use for the eta quotient
is,
where
II. Case p = 13
It turns out that for p = 13, then the 13th power of the analogous theta quotients are the 6 roots of a sextic.
Define,
hence,
(As before, the even-index
are negative.) Let,
Ramanujan discovered that the 3 roots of the cubic,
are,
Note: Incidentally,
is a well-known equation for
and similar roots.
However, if a modular equation can be found between, say,
, then we can eliminate
, and have an equation solely in
and
. After some effort using Mathematica's integer relations algorithm, I found that, let,
then,
(The same relation exists between the other pairs.) Eliminating
, one gets the sextic in
,
where the
are polynomials in the eta quotient
of degrees 7, 13, 18, 20, 15, respectively. Explicitly, the first one is,
though the others are too tedious to write down. One can then ascertain that the 6 roots of the sextic are in fact
.
(P.S. I do not see this sextic, nor the modular relation between the
, in Ramanujan's Notebooks III, Entry 8, page 372, which discusses the theta quotients for p = 13. But I find it satisfying that the results for p = 7 can be extended to the next "lacunary" prime p = 13.)
----------- End -----------
In Ramanujan's Notebook IV, (entries 51-72, p.207-237), there are 23 of Ramanujan's P-Q modular equations. However, it seems he missed the prime orders p = 11,13. Since the Dedekind eta function is
, for convenience I'll use
instead.
I. p = 2
For comparison, Ramanujan found modular relations between
and
for n = 3, 5, 7, 9, 13, 25. For example, he found,
1. Define
, then,
![{\displaystyle PQ+{\tfrac {13}{PQ}}=V_{3}-4V_{1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4780eb181f2b5b0b3062abd8dae71f5be041b613)
II. p = 11
But there are also relations between
and
for n = 2, 3, (and 5, 7, 13?) with the first two as,
1. Define
, then,
![{\displaystyle R_{5}+11R_{3}+44R_{1}={\big (}{\tfrac {Q}{P}}{\big )}^{6}-{\big (}{\tfrac {P}{Q}}{\big )}^{6}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ee94655e01a0ec6ec107d21c985cbccef9d262e2)
2. Define
, then,
![{\displaystyle S_{5}+11(S_{4}+6S_{3}+23S_{2}+63S_{1}+126)={\big (}{\tfrac {Q}{P}}{\big )}^{6}+{\big (}{\tfrac {P}{Q}}{\big )}^{6}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e1e08a6f9724e5d30a936f4f954ad04baa7a4db1)
3. For comparison, define
, then,
![{\displaystyle T_{5}+2T_{4}+3T_{3}-8T_{2}+9T_{1}=(PQ)^{3}-({\tfrac {11}{PQ}})^{3}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ad1deddefb6fac954c3c8cd552ee0e4979157f2b)
No.3 is equivalent to Somos' level 33 identity.
- - - - - - - - - - - - - - - - - - - - -
III. p = 13
Likewise, there are modular relations between
and
for n = 2, 3, 5, 7, with the first two being,
1. Define
, then,
![{\displaystyle U_{7}-13U_{5}+52U_{3}-78U_{1}=(PQ)^{6}+{\big (}{\tfrac {2}{PQ}})^{6}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1e59aa0e68488e9c667a5754c6c67033c4136156)
2. Define
, then,
![{\displaystyle V_{7}+13(-V_{6}+4V_{5}-2V_{4}-19V_{3}+28V_{2}+17V_{1}-50)=(PQ)^{6}+{\big (}{\tfrac {3}{PQ}})^{6}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/adb8f5b31bb93e3c9df6bbfae668b609bf2ba64b)
I do not know if there is a p = 17 identity similar to the ones above.
--------------END--------------
Given the Dedekind eta function
. Let p be a prime and define
,
1. Let p be a prime of form
. Then for
:
![{\displaystyle \sum _{k=0}^{p-1}{\Big (}e^{\pi imk/12}\eta {\big (}{\tfrac {\tau +mk}{p}}{\big )}{\Big )}^{n}=-{\big (}{\sqrt {p}}\,\eta (p\tau ){\big )}^{n}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/22b56af596c420d238c08be9f7f0ae8aa3273aea)
2. Let p be a prime of form
. Then for
:
![{\displaystyle \sum _{k=0}^{p-1}{\Big (}e^{\pi imk/12}\eta {\big (}{\tfrac {\tau +mk}{p}}{\big )}{\Big )}^{n}={\big (}{\sqrt {p}}\,\eta (p\tau ){\big )}^{n}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/417a02ac299445700f36d7c55eec143f138572e0)
Are these two multi-grade identities true?
--------------END--------------
In "An Identity for the Dedekind eta function involving two independent complex variables", given two complex numbers
with imaginary part > 0, Berndt and Hart gave the identity,
![{\displaystyle \eta ^{3}{\big (}{\tfrac {a}{3}}{\big )}\eta ^{3}{\big (}{\tfrac {b}{3}}{\big )}+i\eta ^{3}{\big (}{\tfrac {a+1}{3}}{\big )}\eta ^{3}{\big (}{\tfrac {b+1}{3}}{\big )}-\eta ^{3}{\big (}{\tfrac {a+2}{3}}{\big )}\eta ^{3}{\big (}{\tfrac {b+2}{3}}{\big )}=3^{3}\eta ^{3}(3a)\eta ^{3}(3b)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/79349d7b49c43b224b28e19bded84dbf062e112d)
and remarked that they, "...know of no other examples of a similar type." However, it seems the above is just the smallest member of an infinite family of cubes of the Dedekind eta function,
![{\displaystyle \sum _{k=0}^{p-1}e^{2\pi ik/4}\eta ^{3}{\big (}{\tfrac {a+k}{p}}{\big )}\eta ^{3}{\big (}{\tfrac {b+k}{p}}{\big )}=p^{3}\eta ^{3}(pa)\eta ^{3}(pb)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d5f5f832e26b923bd7a11161666e11360024224e)
where p is ANY PRIME of form
, with the Hart-Berndt identity simply the case
. It is easy to test the family using Mathematica and see that it holds for hundreds of decimal digits, but I have no proof that it is generally true.
--------------END--------------
Conjecture 1. Based on Simon Plouffe's work on pi.[1] (April 10, 2009)
Let q = eπ and k be of the form 4m+3. Then it is true that,
![{\displaystyle {\Big (}{\frac {a}{b}}{\Big )}\pi ^{k}=\sum _{n=1}^{\infty }{\frac {1}{n^{k}}}{\Big (}{\frac {2^{k-1}}{q^{n}-1}}-{\frac {2^{k-1}+1}{q^{2n}-1}}+{\frac {1}{q^{4n}-1}}{\Big )}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0f966a51af5e15b25142b2460cf9fbe3a293dd8c)
where a,b are integral. (The denominator b turns out to be a highly factorable number.)
For the first few k, we have:
k |
a |
b
|
3 |
1 |
|
7 |
13 |
|
11 |
32072 |
|
15 |
219824 |
|
and so on. Anyone knows how to prove this conjecture?
UPDATE (May 18, 2009)
Turns out there is a closed-form formula for (a/b). This is based on Theorem 6.7 (page 11) of Linas Vepstas' On Plouffe's Ramanujan Identities.[2]
Let q = eπ and k = 4m-1 (note this minor change), then
![{\displaystyle r\pi ^{k}=\sum _{n=1}^{\infty }{\frac {1}{n^{k}}}{\Big (}{\frac {2^{k-1}}{q^{n}-1}}-{\frac {2^{k-1}+1}{q^{2n}-1}}+{\frac {1}{q^{4n}-1}}{\Big )}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/90575d71110dca62246688a745ad58b8a6d13ab1)
Where r is a rational number defined by,
![{\displaystyle r=2^{4m-5}\sum _{j=0}^{2m}{\frac {(-1)^{j}(16^{m}+4-4^{j+1})B[2j]B[4m-2j]}{(2j)!(4m-2j)!}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b25b6c23b1e3f57d0b6b292f7bb7cb06594be53a)
and B[w] is a Bernoulli number.
Conjecture 2. Still based on Plouffe's work on pi but now involves powers k = 4m+1. (May 19, 2009)
Let q = eπ and k be of the form 4m+1. Then it is true that,
![{\displaystyle r\pi ^{k}=\sum _{n=1}^{\infty }{\frac {1}{n^{k}}}{\Big (}{\frac {2^{6m+1}+(-16)^{m}}{q^{n}-1}}-{\frac {2^{6m+1}+(-16)^{m}+(-1)^{m}}{q^{2n}-1}}+{\frac {(-1)^{m}}{q^{4n}-1}}{\Big )}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1340c77d937b8962932c86f4f53e76cf11ddd49c)
where r is a rational number.
For the first few k = {1,5,9,13,...} we have:
r = {1/24, 1/63, 164/13365, 76192/9823275,...}
and so on. Is there a closed-form formula for r when k = 4m+1?
References: