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There are many identities in which on one side we have a summation series and on the other side we have a product series. For e.g. the very famous rogers-ramanujan identity ∑_(n=0)^∞▒q^(n^2 )/〖(q;q)〗n =∏(n=1)^∞▒1/((1-q^(5n-1))(1-q^(5n-4))), here 〖(q;q)〗_n is known as Pochhammer symbols which is defined in inbuilt mathematica. Now I have learnt that it is possible to convert a sum series, as the L.H.S. of above e.g., to an equivalent product series using some mathematica code. can anyone plz help in making that mathematica code..

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    $\begingroup$ Please give things properly formatted either in $\LaTeX$ or (preferred) in Mathematica. The jumble of symbols there is really not very helpful - it's in fact ambiguous. $\endgroup$ – Patrick Stevens Sep 7 '15 at 7:34
  • $\begingroup$ Sir I want to convert a sum series (which is power series in q) to equivalent product series such that both have the same power series expansion. For e.g. we have such identity known as Roger-Ramanujan identitiy which is \begin{eqnarray}\label{c1} \sum_{\lambda=0}^{\infty}\frac{q^{\lambda^2}}{(q;q)_{\lambda}}=\prod_{\lambda=1}^{\infty}\frac{1}{(1-q^{5\lambda-1})(1-q^{5\lambda-4})},\nonumber \end{eqnarray} $\endgroup$ – Megha Goyal Sep 7 '15 at 8:20
  • $\begingroup$ Now I want to convert \sum_{\lambda=0}^{\infty}\frac{q^{\lambda^2+2\lambda}}{(q;q)_{\lambda}} to an equivalent product series using mathematica... can you please help me in making some mathematica module to convert this sum series to equivalent product series. $\endgroup$ – Megha Goyal Sep 7 '15 at 8:24
  • $\begingroup$ Greetings! Make the most of Mma.SE and take the tour. Help us to help you, write an excellent question. Edit if improvable, show due diligence, give brief context, include minimum working examples of code and data in formatted form. As you receive give back, vote and answer questions, keep the site useful, be kind, correct mistakes and share what you have learned. $\endgroup$ – rhermans Sep 7 '15 at 11:35
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THIS IS NOT AN ANSWER BUT RATHER AN EXTENDED COMMENT

Mathematica can evaluate both the product and sum in closed form

f1[q_] = Product[1/((1 - q^(5 k - 1)) (1 - q^(5 k - 4))), {k, 1, Infinity}]

(* ((1 - 1/q^4)*(-1 + q))/
   (q*QPochhammer[1/q^4, q^5]*
      QPochhammer[1/q, q^5]) *)

Limit[f1[q], q -> 0]

(* 1 *)

f2[q_] = Sum[q^(k^2)/QPochhammer[q, q, k], {k, 0, Infinity}]

(* 1/(QPochhammer[q, q^5]*
      QPochhammer[q^4, q^5]) *)

f2[0]

(* 1 *)

While Mathematica does not recognize these as equal

Assuming[{0 < q < 1}, f1[q] == f2[q] // FullSimplify]

(* ((1 - 1/q^4)*(-1 + q))/
     (q*QPochhammer[1/q^4, q^5]*
        QPochhammer[1/q, q^5]) == 
   1/(QPochhammer[q, q^5]*
        QPochhammer[q^4, q^5]) *)

Mathematica can show that the series expansions are equal to any arbitrary degree

With[{n = 150}, (Series[f1[q], {q, 0, n}]) === (Series[f2[q], {q, 0, n}])]

(* True *)

With[{n = 150}, Series[f1[q] - f2[q], {q, 0, n}]]

(* SeriesData[q, 0, {}, 151, 151, 1] *)

And that the functions are numerically equal for any specific values of q

With[{delta = .0001`20}, 
 And @@ Table[f1[q] == f2[q], {q, delta, 1 - delta, delta}]]

(* True *)
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  • $\begingroup$ thnx for your explaination.. but there is a code in mathematica which people use to convert a sum series to a product series. Can anybody plz help me. $\endgroup$ – Megha Goyal Sep 8 '15 at 2:13

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