Discriminant function and paritition function - modular forms - algebra really

[binbagsss]

Homework Statement

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,

Homework Equations

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

The Attempt at a Solution

[/B]
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.

 

[fresh_42]

Have you tried to use Euler's formula to write ##\Delta(q)^{-1}##?

 

[binbagsss]

Have you tried to use Euler's formula to write ##\Delta(q)^{-1}##?

erm yeah I get ##1/\Delta = 1/q (\sum\limits^{\infty}_{n=0} p(n)q^{n} ) ^{24} ## ..

 

[binbagsss]

bump.

 

[binbagsss]

erm yeah I get ##1/\Delta = 1/q (\sum\limits^{\infty}_{n=0} p(n)q^{n} ) ^{24} ## ..

am i missing some properties of partition function or anything?

 

[fresh_42]

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).