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).