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Discriminant function and paritition function - modular forms - algebra really

  1. Nov 20, 2016 #1
    1. The problem statement, all variables and given/known data

    I am wanting to show that ##\Delta (t) = 1/q (\sum\limits^{\infty}_{n=0} p(n)q^{n})^{24} ##

    where ##\Delta (q) = q \Pi^{\infty}_{n=1} (1-q^{n})^{24} ## is the discriminant function
    and ##p(n)## is the partition function,


    2. Relevant equations

    Euler's result that : ## \sum\limits^{\infty}_{n=0} p(n)q^{n} = \Pi^{\infty}_{n=1} (1-q^{n})^{-1} ##



    3. The attempt at a solution

    To be honest , I'm probably doing something really stupid, but at first sight, I would have thought we need
    ## \sum\limits^{\infty}_{n=0} p(n)q^{n} ## raised to a negative power, as raising to ##+24## looks like your going to get something like ##(1-q^{n})^{-24}##...

    I've had a little play and get the following...

    ## \Delta (q) = q \Pi^{\infty}_{n=1} (1-q^{n})^{24} = \frac{q \Pi^{\infty}_{n=1}(1-q^{n})^{25}}{\Pi^{\infty}_{n=1}(1-q^{n}})=(\Pi^{\infty}_{n=1} (1-q^{n})^{25}) q \sum\limits^{\infty}_{n=1} p(n) q^{n} ##

    (don't know whether it's in the right direction or where to turn next..)

    Many thanks in advance.
     
  2. jcsd
  3. Nov 20, 2016 #2

    fresh_42

    Staff: Mentor

    Have you tried to use Euler's formula to write ##\Delta(q)^{-1}##?
     
  4. Nov 21, 2016 #3
    erm yeah I get ##1/\Delta = 1/q (\sum\limits^{\infty}_{n=0} p(n)q^{n} ) ^{24} ## ..
     
  5. Nov 23, 2016 #4
  6. Nov 27, 2016 #5
    am i missing some properties of partition function or anything?
     
  7. Nov 27, 2016 #6

    fresh_42

    Staff: Mentor

    Let me list what I've found out, and what not.

    We first have Dedekind's ##\eta-##function defined by ##\eta(z)=q^{\frac{1}{24}}\cdot \prod_{n \geq 1} (1-q^n)## where ##q=\exp(2 \pi i z )## and ##\mathcal{Im}(z) > 0##. This defines our modular discriminant by ##\Delta(z)=c\cdot \eta^{24}(z)## with ##c := (2\pi)^{12}##. So this results in
    $$ \Delta(z) = c\cdot \eta^{24}(z) = c\cdot q \cdot \prod_{n \geq 1}(1-q^n)^{24}\; , \;\left(c := (2\pi)^{12}\; , \;q=\exp(2\pi i z)\; , \;\mathcal{Im}(z) > 0\right)$$
    Then we have Ramanujan's ##\tau-##function ##\tau: \mathbb{N}\rightarrow\mathbb{Z}## defined by
    $$\sum_{n=1}^{\infty}\tau(n)q^n=q\cdot \prod_{n \geq 1}(1-q^n)^{24} = \eta(z)^{24} = c^{-1} \Delta(z)$$.

    Euler's identity is ##\sum_{k=0}^{\infty}p(k)x^k = \prod_{n \geq 1}(1-x^n)^{-1}## with the partition function ##p(k): \mathbb{N}\rightarrow\mathbb{N}##.

    Unfortunately, I haven't found (or dug any deeper if you like) whether Euler's identity holds for our special choice ##x=q## with positive imaginary part of ##z##. Also the sums aren't over the same range ##(k \geq 0\; , \;n\geq 1)## and involve two different functions ##\tau## and ##p##. But I'm no specialist in number theory (and have only a book with basics).
     
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