1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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

    User Avatar
    2017 Award

    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

    User Avatar
    2017 Award

    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).
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Loading...