# User:Titus III

yo mama

## Update, 11/5/2012

This is similar to the 11/3/2012 update. Let,

${\displaystyle m={\frac {12}{n-1}}}$

be a positive integer for some odd number n. Thus, there are only 4 possibilities: ${\displaystyle n=3,5,7,13}$. Given the standard Ramanujan theta functions ${\displaystyle \phi (q),\;\psi (q),\;f(-q)}$, then,

${\displaystyle \left({\frac {\phi (-q^{2})\phi (-q^{6n})}{\phi (-q^{6})\phi (-q^{2n})}}\right)^{m}\,+\,q^{3}\left({\frac {\psi (q)\psi (q^{3n})}{\psi (q^{3})\psi (q^{n})}}\right)^{m}-\,q^{3}\left({\frac {\psi (-q)\psi (-q^{3n})}{\psi (-q^{3})\psi (-q^{n})}}\right)^{m}-\,2mq^{2}\left({\frac {f(-q^{2})f(-q^{6n})}{f(-q^{6})f(-q^{2n})}}\right)^{m}=1}$

For example, for n = 5, hence m = 3, we have,

${\displaystyle \left({\frac {\phi (-q^{2})\phi (-q^{30})}{\phi (-q^{6})\phi (-q^{10})}}\right)^{3}\,+\,q^{3}\left({\frac {\psi (q)\psi (q^{15})}{\psi (q^{3})\psi (q^{5})}}\right)^{3}-\,q^{3}\left({\frac {\psi (-q)\psi (-q^{15})}{\psi (-q^{3})\psi (-q^{5})}}\right)^{3}-\,6q^{2}\left({\frac {f(-q^{2})f(-q^{30})}{f(-q^{6})f(-q^{10})}}\right)^{3}=1}$

and so on for the other three n.

--------------END--------------

## Update, 11/4/2012

Define,

${\displaystyle h_{k}=(-1)^{k-1}q^{k(3k-25)/50}\,{\frac {f(-q^{2k},-q^{25-2k})}{f(-q^{k},-q^{25-k})}},\;\;{\text{for}}\;k=1,...,12.}$

(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 ${\displaystyle j={\frac {\eta (\tau )}{\eta (25\tau )}}}$, analogous to the case for n = 13 discovered by Ramanujan. Since one pair is a constant, ${\displaystyle h_{5}h_{10}=-1}$, then the remaining five are the roots of a quintic (naturally enough) given by,

${\displaystyle 1+(10+10j+4j^{2}+j^{3})x+(5+15j+11j^{2}+5j^{3}+j^{4})x^{2}-2(5+5j+3j^{2}+j^{3})x^{3}+j^{2}x^{4}+x^{5}=0}$

with j as the eta quotient given above. The roots are then,

${\displaystyle [x_{1},\,x_{2},\,x_{3},\,x_{4},\,x_{5}]=[h_{1}h_{7},\,h_{2}h_{11},\,h_{3}h_{4},\,h_{6}h_{8},\,h_{9}h_{12}]}$

(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.)

--------------END--------------

## Update, 11/3/2012

Here is a generalization of one identity described by Berndt as "...fascinating, but with no direct proof" (Ramanujan's Notebooks III, p.322). Let,

${\displaystyle k={\frac {24}{n-1}}}$

be a positive integer for some odd number n. Of course, there are only 6 possibilities: ${\displaystyle n=3,5,7,9,13,25}$. Given the standard Ramanujan theta functions ${\displaystyle \phi (q),\;\psi (q),\;f(-q)}$, then it is proposed that,

${\displaystyle {\frac {\phi ^{k}(-q^{2n})}{\phi ^{k}(-q^{2})}}+q^{3}\left({\frac {\psi ^{k}(q^{n})}{\psi ^{k}(q)}}-{\frac {\psi ^{k}(-q^{n})}{\psi ^{k}(-q)}}\right)-2kq^{2}\left({\frac {f^{k}(-q^{2n})}{f^{k}(-q^{2})}}\right)=1}$

and, as suggested by Michael Somos, by using a Fricke involution,

${\displaystyle {\frac {\phi ^{k}(-q^{2})}{\phi ^{k}(-q^{2n})}}+{\frac {1}{q^{3}}}\left({\frac {\psi ^{k}(q)}{\psi ^{k}(q^{n})}}-{\frac {\psi ^{k}(-q)}{\psi ^{k}(-q^{n})}}\right)-{\frac {2k}{q^{2}}}\left({\frac {f^{k}(-q^{2})}{f^{k}(-q^{2n})}}\right)=n^{k/2}}$

Thus, for n = 3, we have,

${\displaystyle {\frac {\phi ^{12}(-q^{6})}{\phi ^{12}(-q^{2})}}+q^{3}\left({\frac {\psi ^{12}(q^{3})}{\psi ^{12}(q)}}-{\frac {\psi ^{12}(-q^{3})}{\psi ^{12}(-q)}}\right)-24q^{2}\left({\frac {f^{12}(-q^{6})}{f^{12}(-q^{2})}}\right)=1}$

and,

${\displaystyle {\frac {\phi ^{12}(-q^{2})}{\phi ^{12}(-q^{6})}}+{\frac {1}{q^{3}}}\left({\frac {\psi ^{12}(q)}{\psi ^{12}(q^{3})}}-{\frac {\psi ^{12}(-q)}{\psi ^{12}(-q^{3})}}\right)-{\frac {24}{q^{2}}}\left({\frac {f^{12}(-q^{2})}{f^{12}(-q^{6})}}\right)=3^{6}}$

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--------------

## Update, 11/2/2012

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,

${\displaystyle h_{k}=(-1)^{k-1}q^{k(-7+3k)/14}\,{\frac {f(-q^{2k},-q^{7-2k})}{f(-q^{k},-q^{7-k})}}}$

hence,

${\displaystyle h_{1}={\frac {f(-q^{2},-q^{5})}{q^{2/7}f(-q,-q^{6})}},\;\;h_{2}={\frac {-f(-q^{3},-q^{4})}{q^{1/7}f(-q^{2},-q^{5})}},\;\;h_{3}={\frac {q^{3/7}f(-q,-q^{6})}{f(-q^{3},-q^{4})}}}$

(Kindly note that the even-index ${\displaystyle h_{k}}$ are negative.) Given the Dedekind eta function ${\displaystyle \eta (\tau )}$, let,

${\displaystyle j=\left({\frac {\eta (\tau )}{\eta (7\tau )}}\right)^{4}}$

then Ramanujan found that the 3 roots of the cubic,

${\displaystyle x^{3}-(57+14j+j^{2})x^{2}-(289+126j+19j^{2}+j^{3})x+1=0}$

are,

${\displaystyle x_{1}={h_{1}}^{7},\;\;x_{2}={h_{2}}^{7},\;\;x_{3}={h_{3}}^{7}}$

It turns out that for p = 13, then the 13th power of the analogous theta quotients are the 6 roots of a sextic.

II. Case p = 13

Define,

${\displaystyle h_{k}=(-1)^{k-1}q^{k(-13+3k)/26}\,{\frac {f(-q^{2k},-q^{13-2k})}{f(-q^{k},-q^{13-k})}}}$

hence,

${\displaystyle h_{1}={\frac {f(-q^{2},-q^{11})}{q^{5/13}f(-q,-q^{12})}},\;\;h_{2}={\frac {-f(-q^{4},-q^{9})}{q^{7/13}f(-q^{2},-q^{11})}},\;\;h_{3}={\frac {f(-q^{6},-q^{7})}{q^{6/13}f(-q^{3},-q^{10})}}}$

${\displaystyle h_{4}={\frac {-f(-q^{5},-q^{8})}{q^{2/13}f(-q^{4},-q^{9})}},\;\;h_{5}={\frac {q^{5/13}f(-q^{3},-q^{10})}{f(-q^{5},-q^{8})}},\;\;h_{6}={\frac {-q^{15/13}f(-q,-q^{12})}{f(-q^{6},-q^{7})}}}$

(As before, the even-index ${\displaystyle h_{k}}$ are negative.) Let,

${\displaystyle j=\left({\frac {\eta (\tau )}{\eta (13\tau )}}\right)^{2}}$

Ramanujan discovered that the 3 roots of the cubic,

${\displaystyle j={\frac {y^{3}+y^{2}-4y+1}{y(1-y)}}}$

are,

${\displaystyle y_{1}=h_{1}h_{5},\;\;y_{2}=h_{2}h_{3},\;\;y_{3}=h_{4}h_{6}}$

However, if a modular equation can be found between, say, ${\displaystyle h_{1},h_{5}}$, then we can eliminate ${\displaystyle h_{5}}$, and have an equation solely in ${\displaystyle h_{1}}$ and ${\displaystyle j}$. After some effort using Mathematica's integer relations algorithm, I found that, let,

${\displaystyle u=h_{1},v=h_{5},x=h_{1}h_{5}}$

then,

{\displaystyle {\begin{aligned}u^{2}v^{2}(uv-1)^{5}(u^{13}+v^{13})&=1-7x+19x^{2}-27x^{3}+35x^{4}-68x^{5}+100x^{6}-84x^{7}\\&+69x^{8}-86x^{9}+70x^{10}-29x^{11}+11x^{12}-4x^{13}+x^{14}\end{aligned}}}

(The same relation exists between the other pairs.) Eliminating ${\displaystyle h_{5}}$, one gets the sextic in ${\displaystyle z=h_{1}^{13}}$,

${\displaystyle z^{6}+P_{1}z^{5}+P_{2}z^{4}+P_{3}z^{3}+P_{4}z^{2}+P_{5}z-1=0}$

where the ${\displaystyle P_{i}}$ are polynomials in the eta quotient ${\displaystyle j}$ of degrees 7, 13, 18, 20, 15, respectively. Explicitly, the first one is,

{\displaystyle {\begin{aligned}P_{1}&=-(h_{1}^{13}+h_{2}^{13}+h_{3}^{13}+h_{4}^{13}+h_{5}^{13}+h_{6}^{13})\\&=j^{7}+13j^{6}+91j^{5}+377j^{4}+962j^{3}+1040j^{2}-845j-4083\end{aligned}}}

though the others are too tedious to write down. One can then ascertain that the 6 roots of the sextic are in fact ${\displaystyle z_{i}=h_{i}^{13}}$.

(P.S. I do not see this sextic, nor the modular relation between the ${\displaystyle h_{i}}$, 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 -----------

## Update, 9/23/2012

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 ${\displaystyle \eta (\tau )=q^{1/24}f(-q)}$, for convenience I'll use ${\displaystyle \eta (\tau )}$ instead.

I. p = 2

For comparison, Ramanujan found modular relations between ${\displaystyle P'={\tfrac {f(-q)}{f(-q^{n})}}}$ and ${\displaystyle Q'={\tfrac {f(-q^{2})}{f(-q^{2n})}}}$ for n = 3, 5, 7, 9, 13, 25. For example, he found,

1. Define ${\displaystyle P={\tfrac {\eta (t)}{\eta (13t)}},\;Q={\tfrac {\eta (2t)}{\eta (26t)}},\;V_{k}={\big (}{\tfrac {Q}{P}}{\big )}^{k}+{\big (}{\tfrac {P}{Q}}{\big )}^{k}}$, then,

${\displaystyle PQ+{\tfrac {13}{PQ}}=V_{3}-4V_{1}}$

II. p = 11

But there are also relations between ${\displaystyle P'={\tfrac {f(-q)}{f(-q^{n})}}}$ and ${\displaystyle Q'={\tfrac {f(-q^{11})}{f(-q^{11n})}}}$ for n = 2, 3, (and 5, 7, 13?) with the first two as,

1. Define ${\displaystyle P={\tfrac {\eta (t)}{\eta (2t)}},\;Q={\tfrac {\eta (11t)}{\eta (22t)}},\;R_{k}=(PQ)^{k}+({\tfrac {2}{PQ}})^{k}}$, then,

${\displaystyle R_{5}+11R_{3}+44R_{1}={\big (}{\tfrac {Q}{P}}{\big )}^{6}-{\big (}{\tfrac {P}{Q}}{\big )}^{6}}$

2. Define ${\displaystyle P={\tfrac {\eta (t)}{\eta (3t)}},\;Q={\tfrac {\eta (11t)}{\eta (33t)}},\;S_{k}=(PQ)^{k}+({\tfrac {3}{PQ}})^{k}}$, then,

${\displaystyle S_{5}+11(S_{4}+6S_{3}+23S_{2}+63S_{1}+126)={\big (}{\tfrac {Q}{P}}{\big )}^{6}+{\big (}{\tfrac {P}{Q}}{\big )}^{6}}$

3. For comparison, define ${\displaystyle P={\tfrac {\eta (t)}{\eta (33t)}},\;Q={\tfrac {\eta (3t)}{\eta (11t)}},\;T_{k}=({\tfrac {P}{Q}})^{k}-({\tfrac {3Q}{P}})^{k}}$, then,

${\displaystyle T_{5}+2T_{4}+3T_{3}-8T_{2}+9T_{1}=(PQ)^{3}-({\tfrac {11}{PQ}})^{3}}$

No.3 is equivalent to Somos' level 33 identity.

- - - - - - - - - - - - - - - - - - - - -

III. p = 13

Likewise, there are modular relations between ${\displaystyle P'={\tfrac {f(-q)}{f(-q^{n})}}}$ and ${\displaystyle Q'={\tfrac {f(-q^{13})}{f(-q^{13n})}}}$ for n = 2, 3, 5, 7, with the first two being,

1. Define ${\displaystyle P={\tfrac {\eta (t)}{\eta (2t)}},\;Q={\tfrac {\eta (13t)}{\eta (26t)}},\;U_{k}={\big (}{\tfrac {Q}{P}}{\big )}^{k}-{\big (}{\tfrac {P}{Q}}{\big )}^{k}}$, then,

${\displaystyle U_{7}-13U_{5}+52U_{3}-78U_{1}=(PQ)^{6}+{\big (}{\tfrac {2}{PQ}})^{6}}$

2. Define ${\displaystyle P={\tfrac {\eta (t)}{\eta (3t)}},\;Q={\tfrac {\eta (13t)}{\eta (39t)}},\;V_{k}={\big (}{\tfrac {Q}{P}}{\big )}^{k}+{\big (}{\tfrac {P}{Q}}{\big )}^{k}}$, then,

${\displaystyle V_{7}+13(-V_{6}+4V_{5}-2V_{4}-19V_{3}+28V_{2}+17V_{1}-50)=(PQ)^{6}+{\big (}{\tfrac {3}{PQ}})^{6}}$

I do not know if there is a p = 17 identity similar to the ones above.

--------------END--------------

## Update, 9/16/2012

Given the Dedekind eta function ${\displaystyle \eta (\tau )}$. Let p be a prime and define ${\displaystyle m=(p-1)/2}$,

1. Let p be a prime of form ${\displaystyle p=12v+5}$. Then for ${\displaystyle n=2,4,8,14}$:

${\displaystyle \sum _{k=0}^{p-1}{\Big (}e^{\pi imk/12}\eta {\big (}{\tfrac {\tau +mk}{p}}{\big )}{\Big )}^{n}=-{\big (}{\sqrt {p}}\,\eta (p\tau ){\big )}^{n}}$

2. Let p be a prime of form ${\displaystyle p=12v+11}$. Then for ${\displaystyle n=2,6,10,14}$:

${\displaystyle \sum _{k=0}^{p-1}{\Big (}e^{\pi imk/12}\eta {\big (}{\tfrac {\tau +mk}{p}}{\big )}{\Big )}^{n}={\big (}{\sqrt {p}}\,\eta (p\tau ){\big )}^{n}}$

Are these two multi-grade identities true?

--------------END--------------

## Update, 9/11/2012

In "An Identity for the Dedekind eta function involving two independent complex variables", given two complex numbers ${\displaystyle a,b}$ with imaginary part > 0, Berndt and Hart gave the identity,

${\displaystyle \eta ^{3}{\big (}{\tfrac {a}{3}}{\big )}\eta ^{3}{\big (}{\tfrac {b}{3}}{\big )}+i\eta ^{3}{\big (}{\tfrac {a+1}{3}}{\big )}\eta ^{3}{\big (}{\tfrac {b+1}{3}}{\big )}-\eta ^{3}{\big (}{\tfrac {a+2}{3}}{\big )}\eta ^{3}{\big (}{\tfrac {b+2}{3}}{\big )}=3^{3}\eta ^{3}(3a)\eta ^{3}(3b)}$

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,

${\displaystyle \sum _{k=0}^{p-1}e^{2\pi ik/4}\eta ^{3}{\big (}{\tfrac {a+k}{p}}{\big )}\eta ^{3}{\big (}{\tfrac {b+k}{p}}{\big )}=p^{3}\eta ^{3}(pa)\eta ^{3}(pb)}$

where p is ANY PRIME of form ${\displaystyle p=4n-1}$, with the Hart-Berndt identity simply the case ${\displaystyle p=3}$. 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--------------

## Update, 4/10/2009

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,

${\displaystyle {\Big (}{\frac {a}{b}}{\Big )}\pi ^{k}=\sum _{n=1}^{\infty }{\frac {1}{n^{k}}}{\Big (}{\frac {2^{k-1}}{q^{n}-1}}-{\frac {2^{k-1}+1}{q^{2n}-1}}+{\frac {1}{q^{4n}-1}}{\Big )}}$

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 ${\displaystyle 2^{2}*3^{2}*5}$
7 13 ${\displaystyle 3^{4}*5^{2}*7}$
11 32072 ${\displaystyle 3^{5}*5^{3}*7^{2}*11*13}$
15 219824 ${\displaystyle 3^{8}*5^{3}*7^{2}*13*17}$

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

${\displaystyle r\pi ^{k}=\sum _{n=1}^{\infty }{\frac {1}{n^{k}}}{\Big (}{\frac {2^{k-1}}{q^{n}-1}}-{\frac {2^{k-1}+1}{q^{2n}-1}}+{\frac {1}{q^{4n}-1}}{\Big )}}$

Where r is a rational number defined by,

${\displaystyle r=2^{4m-5}\sum _{j=0}^{2m}{\frac {(-1)^{j}(16^{m}+4-4^{j+1})B[2j]B[4m-2j]}{(2j)!(4m-2j)!}}}$

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,

${\displaystyle r\pi ^{k}=\sum _{n=1}^{\infty }{\frac {1}{n^{k}}}{\Big (}{\frac {2^{6m+1}+(-16)^{m}}{q^{n}-1}}-{\frac {2^{6m+1}+(-16)^{m}+(-1)^{m}}{q^{2n}-1}}+{\frac {(-1)^{m}}{q^{4n}-1}}{\Big )}}$

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:

1. ^ Plouffe, Simon. "Indentities inspired by Ramanujan's Notebooks (part 2)" (PDF). Retrieved 2009-4-10. Check date values in: |access-date= (help)
2. ^ Vepstas, Linas. "On Plouffe's Ramanujan Identities" (PDF). Retrieved 2009-5-18. Check date values in: |access-date= (help)