Ramanujan's most beautiful identity

G. H. Hardy called the following equation Ramanujan’s “most beautiful identity.” For |q| < 1,

sum_{n=0}^infty p(5n+4) q^n = 5 prod_{n=1}^infty frac{(1 - q^{5n})^5}{(1 - q^n)^6}

If I understood it, I might say it’s beautiful, but for now I can only say it’s mysterious. Still, I explain what I can.

The function p on the left side is the partition function. For a positive integer argument n, p(n) is the number of ways one can write n as the sum of a non-decreasing sequence of positive integers.

The right side of the equation is an example of a q-series. Strictly speaking it’s a product, not a series, but it’s the kind of thing that goes under the general heading of q-series.

I hardly know anything about q-series, and they don’t seem very motivated. However, I keep running into them in unexpected places. They seem to be a common thread running through several things I’m vaguely familiar with and would like to understand better.

As mysterious as Ramanujan’s identity is, it’s not entirely unprecedented. In the eighteenth century, Euler proved that the generating function for partition numbers is a q-product:

sum_{n=0}^infty p(n) q^n = prod_{n=1}^infty frac{1}{(1 - q^n)}

So in discovering his most beautiful identity (and others) Ramanujan followed in Euler’s footsteps.

Reference: An Invitation to q-series

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5 comments on “Ramanujan's most beautiful identity
  1. Hei-Chi Chan, was a former professor of mine and he discovered a generalization of Ramanujan’s “most beautiful identity.” Here is a link to his paper: https://edocs.uis.edu/hchan1/www/Chan=CubicCFandMostBeautifulIdentityt=v03.pdf

  2. rhalbersma says:

    For a very readable proof of this identity, see:
    http://www.austms.org.au/Gazette/2005/Sep05/Hirschhorn.pdf

  3. PhilM says:

    Given his unorthodox background, I wonder if Ramanujan knew of Euler’s work. In any case, I am constantly amazed of these things that “real” mathematicians dream up!

  4. rhalbersma says:

    According to <a href="http://en.wikipedia.org/wiki/Srinivasa_Ramanujan&quot; Wikipedia (citing <a href="http://www.amazon.com/Ramanujan-Essays-Surveys-History-Mathematics/dp/0821826247&quot; Berndt & Rankin, Ramanujan independently rediscovered Euler’s identity.

  5. John says:

    Ramanujan wasn’t entirely self-taught. Hardy mentions some books as ones Ramanujan had definitely read, others he was unsure about.

    It’s plausible that Ramanujan rediscovered Euler’s identity. He was interested in partitions, and it’s natural to want to find the generating function of a sequence you’re interested in.