User:Titus III

Update, 5/7/2022
I. Rogers-Selberg Mod 7 Identities

$$A(q)= \frac{(q^3,q^4,q^7;q^7)_\infty}{(q^2;q^2)_\infty}= \frac{f(-q^3,-q^4)}{f(-q^2,-q^4)}= \prod_{n=1}^\infty\frac{(1-q^{7n})(1-q^{7n-3})(1-q^{7n-4})}{(1-q^{2n})}$$

$$B(q)= \frac{(q^2,q^5,q^7;q^7)_\infty}{(q^2;q^2)_\infty}= \frac{f(-q^2,-q^5)}{f(-q^2,-q^4)}= \prod_{n=1}^\infty\frac{(1-q^{7n})(1-q^{7n-2})(1-q^{7n-5})}{(1-q^{2n})}$$

$$C(q)= \frac{(q,q^6,q^7;q^7)_\infty}{(q^2;q^2)_\infty}\,= \frac{f(-q,-q^6)}{f(-q^2,-q^4)}\,= \prod_{n=1}^\infty\frac{(1-q^{7n})(1-q^{7n-1})(1-q^{7n-6})}{(1-q^{2n})}$$

'''II. Bailey Mod 9 Identities'''

$$A(q)= \frac{(q^4,q^5,q^9;q^9)_\infty}{(q^3;q^3)_\infty}= \frac{f(-q^4,-q^5)}{f(-q^3,-q^6)}= \prod_{n=1}^\infty\frac{(1-q^{9n})(1-q^{9n-4})(1-q^{9n-5})}{(1-q^{3n})}$$

$$B(q)= \frac{(q^2,q^7,q^9;q^9)_\infty}{(q^3;q^3)_\infty}= \frac{f(-q^2,-q^7)}{f(-q^3,-q^6)}= \prod_{n=1}^\infty\frac{(1-q^{9n})(1-q^{9n-2})(1-q^{9n-7})}{(1-q^{3n})}$$

$$C(q)= \frac{(q,q^8,q^9;q^9)_\infty}{(q^3;q^3)_\infty}\,= \frac{f(-q,-q^8)}{f(-q^3,-q^6)}\,= \prod_{n=1}^\infty\frac{(1-q^{9n})(1-q^{9n-1})(1-q^{9n-8})}{(1-q^{3n})}$$

'''III. Rogers Mod 14 Identities'''

$$A(q)= \frac{(q^6,q^8,q^{14};q^{14})_\infty}{(q;q)_\infty}\,= \frac{f(-q^6,-q^8)}{f(-q,-q^2)}\,= \prod_{n=1}^\infty\frac{(1-q^{14n})(1-q^{14n-6})(1-q^{14n-8})}{(1-q^{n})}$$

$$B(q)= \frac{(q^4,q^{10},q^{14};q^{14})_\infty}{(q;q)_\infty}= \frac{f(-q^4,-q^{10})}{f(-q,-q^2)}= \prod_{n=1}^\infty\frac{(1-q^{14n})(1-q^{14n-4})(1-q^{14n-10})}{(1-q^{n})}$$

$$C(q)= \frac{(q^2,q^{12},q^{14};q^{14})_\infty}{(q;q)_\infty}= \frac{f(-q^2,-q^{12})}{f(-q,-q^2)}= \prod_{n=1}^\infty\frac{(1-q^{14n})(1-q^{14n-2})(1-q^{14n-12})}{(1-q^{n})}$$

'''IV. Dyson Mod 27 Identities'''

$$A(q)= \frac{(q^{12},q^{15},q^{27};q^{27})_\infty}{(q;q)_\infty}= \frac{f(-q^{12},-q^{15})}{f(-q,-q^2)}= \prod_{n=1}^\infty\frac{(1-q^{27n})(1-q^{27n-12})(1-q^{27n-15})}{(1-q^{n})}$$

$$B(q)= \frac{(q^{9},q^{18},q^{27};q^{27})_\infty}{(q;q)_\infty}= \frac{f(-q^{9},-q^{18})}{f(-q,-q^2)}= \prod_{n=1}^\infty\frac{(1-q^{27n})(1-q^{27n-9})(1-q^{27n-18})}{(1-q^{n})}$$

$$C(q)= \frac{(q^{6},q^{21},q^{27};q^{27})_\infty}{(q;q)_\infty}\,= \frac{f(-q^{6},-q^{21})}{f(-q,-q^2)}\,= \prod_{n=1}^\infty\frac{(1-q^{27n})(1-q^{27n-6})(1-q^{27n-21})}{(1-q^{n})}$$

$$D(q)= \frac{(q^{3},q^{24},q^{27};q^{27})_\infty}{(q;q)_\infty}\,= \frac{f(-q^{3},-q^{24})}{f(-q,-q^2)}\,= \prod_{n=1}^\infty\frac{(1-q^{27n})(1-q^{27n-3})(1-q^{27n-24})}{(1-q^{n})}$$

Update, 11/5/2012
This is similar to the 11/3/2012 update. Let,

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

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

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

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

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

and so on for the other three n.

--END--

Update, 11/4/2012
Define,

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

(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 $$j=\frac{\eta(\tau)}{\eta(25\tau)}$$, analogous to the case for n = 13 discovered by Ramanujan. Since one pair is a constant, $$h_5 h_{10} = -1$$, then the remaining five are the roots of a quintic (naturally enough) given by,

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

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

$$[x_1,\, x_2,\, x_3,\, x_4,\, x_5] = [h_1 h_7,\, h_2 h_{11},\, h_3 h_4,\, h_6 h_8,\, h_9 h_{12}]$$

(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.)

Update April 28, 2022: After almost 10 years, I finally found the answer to my "note". This is just the EMMA LEHMER QUINTIC!

--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,

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

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

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

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

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

Thus, for n = 3, we have,

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

and,

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

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,

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

hence,

$$h_1 = \frac{f(-q^2,-q^5)}{q^{2/7}f(-q,-q^6)} = \frac{1}{q^{2/7}}\prod_{n=1}^\infty \frac{(1-q^{7n-2})(1-q^{7n-5})}{(1-q^{7n-1})(1-q^{7n-6})}$$

$$h_2 = \frac{-f(-q^3,-q^4)}{q^{1/7}f(-q^2,-q^5)} = \frac{-1}{q^{1/7}}\prod_{n=1}^\infty \frac{(1-q^{7n-3})(1-q^{7n-4})}{(1-q^{7n-2})(1-q^{7n-5})}$$

$$h_3 = \frac{q^{3/7}f(-q,-q^6)}{f(-q^3,-q^4)} = \frac{1}{q^{-3/7}}\prod_{n=1}^\infty \frac{(1-q^{7n-1})(1-q^{7n-6})}{(1-q^{7n-3})(1-q^{7n-4})}$$

functions highly analogous to the Rogers–Ramanujan continued fraction. (Kindly note that the even-index $$h_k$$ is negative.) Given the Dedekind eta function $$\eta(\tau)$$, let,

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

then Ramanujan found that the 3 roots of the cubic,

$$x^3-(57+14c+c^2)x^2-(289+126c+19c^2+c^3)x+1 = 0$$

are,

$$x_1 = {h_1}^7,\;\;x_2 = {h_2}^7,\;\;x_3 = {h_3}^7$$

Note also another use for the eta quotient $$c$$ is,

$$\begin{align} &j(\tau) = \frac{(y^2 + 5y + 1)^3(y^2 + 13y + 49)}{y}\\ &j(\tau)-1728 = \frac{(y^4+14y^3+63y^2+70y-7)^2}{y} \end{align}$$

where $$y = \frac{7^2}{c}.$$

'''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,

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

hence,

$$h_1 = \frac{f(-q^2,-q^{11})}{q^{5/13}f(-q,-q^{12})},\;\;h_2 = \frac{-f(-q^4,-q^9)}{q^{7/13}f(-q^2,-q^{11})},\;\;h_3 = \frac{f(-q^6,-q^7)}{q^{6/13}f(-q^3,-q^{10})}$$

$$h_4 = \frac{-f(-q^5,-q^8)}{q^{2/13}f(-q^4,-q^9)},\;\;h_5 = \frac{q^{5/13}f(-q^3,-q^{10})}{f(-q^5,-q^8)},\;\;h_6 = \frac{-q^{15/13}f(-q,-q^{12})}{f(-q^6,-q^7)}$$

(As before, the even-index $$h_k$$ are negative.) Let,

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

Ramanujan discovered that the 3 roots of the cubic,

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

are,

$$y_1 = h_1h_5,\;\;y_2 = h_2h_3,\;\;y_3 = h_4h_6$$

Note: Incidentally, $$y^3+y^2-4y+1=0$$ is a well-known equation for $$y = 2\cos(2\pi/13)+2\cos(10\pi/13)$$ and similar roots.

However, if a modular equation can be found between, say, $$h_1,h_5$$, then we can eliminate $$h_5$$, and have an equation solely in $$h_1$$ and $$d$$. After some effort using Mathematica's integer relations algorithm, I found that, let,

$$u = h_1,\, v = h_5,\, x = h_1h_5$$

then,

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

(The same relation exists between the other pairs.) Eliminating $$h_5$$, one gets the sextic in $$z = h_1^{13}$$,

$$z^6+P_1z^5+P_2z^4+P_3z^3+P_4z^2+P_5z-1 = 0$$

where the $$P_i$$ are polynomials in the eta quotient $$d$$ of degrees 7, 13, 18, 20, 15, respectively. Explicitly, the first one is,

$$\begin{align}P_1 &= -(h_1^{13}+h_2^{13}+h_3^{13}+h_4^{13}+h_5^{13}+h_6^{13})\\ &= d^7+13d^6+91d^5+377d^4+962d^3+1040d^2-845d-4083\end{align}$$

though the others are too tedious to write down. One can then ascertain that the 6 roots of the sextic are in fact $$z_i = h_i^{13}$$.

(P.S. I do not see this sextic, nor the modular relation between the $$h_i$$, 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 $$\eta(\tau) = q^{1/24}f(-q)$$, for convenience I'll use $$\eta(\tau)$$ instead.

I. p = 2

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

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


 * $$PQ+\tfrac{13}{PQ} = V_3-4V_1$$

II. p = 11

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

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


 * $$R_5+11R_3+44R_1 = \big(\tfrac{Q}{P}\big)^6-\big(\tfrac{P}{Q}\big)^6$$

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


 * $$S_5+11(S_4+6S_3+23S_2+63S_1+126) = \big(\tfrac{Q}{P}\big)^6+\big(\tfrac{P}{Q}\big)^6$$

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


 * $$T_5+2T_4+3T_3-8T_2+9T_1 = (PQ)^3-(\tfrac{11}{PQ})^3$$

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

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

III. p = 13

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

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


 * $$U_7-13U_5+52U_3-78U_1 = (PQ)^6+\big(\tfrac{2}{PQ})^6$$

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


 * $$V_7+13(-V_6+4V_5-2V_4-19V_3+28V_2+17V_1-50) = (PQ)^6+\big(\tfrac{3}{PQ})^6$$

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 $$\eta(\tau)$$. Let p be a prime and define $$m = (p-1)/2$$,

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


 * $$\sum_{k=0}^{p-1} \Big(e^{\pi i m k/12} \eta\big(\tfrac{\tau+m k}{p}\big)\Big)^n = -\big(\sqrt{p}\,\eta(p\tau)\big)^n $$

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


 * $$\sum_{k=0}^{p-1} \Big(e^{\pi i m k/12} \eta\big(\tfrac{\tau+m k}{p}\big)\Big)^n = \big(\sqrt{p}\,\eta(p\tau)\big)^n $$

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 $$a,b$$ with imaginary part > 0, Berndt and Hart gave the identity,


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

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,


 * $$\sum_{k=0}^{p-1} e^{2\pi i k/4}\eta^3\big(\tfrac{a+k}{p}\big)\eta^3\big(\tfrac{b+k}{p}\big) = p^3\eta^3(p a)\eta^3(p b)$$

where p is ANY PRIME of form $$p = 4n-1$$, with the Hart-Berndt identity simply the case $$p=3$$. 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. (April 10, 2009)

Let q = eπ and k be of the form 4m+3. Then it is true that,


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

where a,b are integral. (The denominator b turns out to be a highly factorable number.)

For the first few k, we have:

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.

Let q = eπ and k = 4m-1 (note this minor change), then


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

Where r is a rational number defined by,


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

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,


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

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: