- #1

binbagsss

- 1,278

- 11

## 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

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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.