7
$\begingroup$

In a paper titled "RAMANUJAN’S UNPUBLISHED MANUSCRIPT ON THE PARTITION AND TAU FUNCTIONS WITH PROOFS AND COMMENTARY" by Bruce C. Berndt and Ken Ono, it is mentioned that Ramanujan derived the formula \begin{align}\sum_{n = 0}^{\infty}p(25n + 24)q^{n} &= 5^{2}\cdot 63\frac{(q^{5};q^{5})_{\infty}^{6}}{(q;q)_{\infty}^{7}} + 5^{5}\cdot 52q\frac{(q^{5};q^{5})_{\infty}^{12}}{(q;q)_{\infty}^{13}}\notag\\ &\,\,\,\,+\,5^{7}\cdot 63q^{2}\frac{(q^{5};q^{5})_{\infty}^{18}}{(q;q)_{\infty}^{19}} + 5^{10}\cdot 6q^{3}\frac{(q^{5};q^{5})_{\infty}^{24}}{(q;q)_{\infty}^{25}}\notag\\ &\,\,\,\,+\,5^{12}q^{4}\frac{(q^{5};q^{5})_{\infty}^{30}}{(q;q)_{\infty}^{31}}\tag{1}\end{align} via the use of the formulas $$\frac{(q;q)_{\infty}^{6}}{(q^{5};q^{5})_{\infty}^{6}} = A^{5} - 11q + q^{2}B^{5}\tag{2}$$ and

$\dfrac{(q^{5};q^{5})_{\infty}}{(q^{1/5};q^{1/5})_{\infty}}$ $=\dfrac{(A^{4} + 3Bq) + q^{1/5}(A^{3} + 2B^{2}q) + q^{2/5}(2A^{2} + B^{3}q) + q^{3/5}(3A + B^{4}q) + 5q^{4/5}}{A^{5} - 11q + q^{2}B^{5}}\text{ (3)}$

and $$\sum_{n = 0}^{\infty}p(5n + 4)q^{n} = 5\frac{(q^{5};q^{5})_{\infty}^{5}}{(q;q)_{\infty}^{6}}\tag{4}$$

This is done by replacing $q$ by $q^{1/5}$ in $(4)$ and writing the formula as $$\sum_{n = 0}^{\infty}p(5n + 4)q^{n/5} = \frac{5}{(q;q)_{\infty}}\left(\frac{(q^{5};q^{5})_{\infty}}{(q^{1/5};q^{1/5})_{\infty}}\right)^{6}\frac{(q;q)_{\infty}^{6}}{(q^{5};q^{5})_{\infty}^{6}}$$ This effectively requires us to raise equation $(3)$ to $6^{\text{th}}$ power and then take the coefficient of $q^{4/5}$. I wonder how this leads to a beautiful result like equation $(1)$. If we use multinomial theorem to calculate $6^{\text{th}}$ power then it leads to a huge number of terms containing $q^{4/5}$ (42 terms and each term a polynomial in $A, B, q$) and I don't know if this can be done via hand calculation.

Is there is any other method to find the coefficient of $q^{4/5}$ or some alternative way to derive the complicated formula $(1)$?

Update: Here $A, B$ are given by $$A = \frac{(q^{2};q^{5})_{\infty}(q^{3};q^{5})_{\infty}}{(q;q^{5})_{\infty}(q^{4};q^{5})_{\infty}},\,\, AB = -1$$

$\endgroup$
10
$\begingroup$

Hirschhorn and Hunt have published a note M. D. Hirschhorn; D. C. Hunt, A simple proof of the Ramanujan conjecture for powers of 5, J. Reine Angew. Math., 326 (1981), 1-17., where they give a proof (what they call "a simple proof") for the formula for the generating function of $p(5^{\alpha}n+\delta_n)$, with $\alpha\ge 1$ and $\delta_n$ being the reciprocal modulo $5^{\alpha}$ of 24; see Theorem $1.4$.

$\endgroup$
  • $\begingroup$ Thanks Dietrich! This seems to be exactly what I needed. It seems to be much simpler and systematic in terms of calculation. $\endgroup$ – Paramanand Singh Sep 4 '14 at 11:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.