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Partition function, discriminant function modular forms, really just algebra

1. The problem statement, all variables and given/known data

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

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

2. Relevant equations

Euler's result that :

n=0p(n)qn=Πn=1(1qn)1 \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

n=0p(n)qn \sum\limits^{\infty}_{n=0} p(n)q^{n}

raised to a negative power, as raising to +24+24 looks like your going to get something like (1qn)24(1-q^{n})^{-24}..

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

Δ(q)=qΠn=1(1qn)24=qΠn=1(1qn)25Πn=1(1qn)=(Πn=1(1qn)25)qn=1p(n)qn \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.
Original post by xfootiecrazeesarax
1. The problem statement, all variables and given/known data

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

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

2. Relevant equations

Euler's result that :

n=0p(n)qn=Πn=1(1qn)1 \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

n=0p(n)qn \sum\limits^{\infty}_{n=0} p(n)q^{n}

raised to a negative power, as raising to +24+24 looks like your going to get something like (1qn)24(1-q^{n})^{-24}..

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

Δ(q)=qΠn=1(1qn)24=qΠn=1(1qn)25Πn=1(1qn)=(Πn=1(1qn)25)qn=1p(n)qn \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.

I'm not *too* familiar with the quantities involved here but there seem to be a few issues - for example, what is tt? I'm also unsure about the q in the denominator of the final answer, as well as the lack of tt-dependence.

Nonetheless, purely using the definitions you've given, Euler's result implies:

Πn=1(1qn)=(Πn=1(1qn)1)1=(n=0p(n)qn)1\displaystyle\Pi _{n=1}^{\infty}(1-q^n) = \left( \displaystyle\Pi _{n=1}^{\infty}(1-q^n)^{-1}\right) ^{-1} = \left(\displaystyle\sum_{n=0}^{ \infty }p(n)q^n \right)^{-1}

And raising both sides to the power of 24 yields,

Πn=1(1qn)24=(n=0p(n)qn)24\displaystyle\Pi _{n=1}^{\infty}(1-q^n)^{24} = \left(\displaystyle\sum_{n=0}^{ \infty }p(n)q^n \right)^{-24}

Hence, by direct substitution into the discriminant function as written, we have

Δ(q)=q/(n=0p(n)qn)24\Delta (q) = q/\left(\displaystyle\sum_{n=0}^{ \infty }p(n)q^n \right)^{24}

Which is almost the answer you've provided, except the argument of Δ\Delta is qq and the free qq factor on the RHS appears in the numerator instead of the denominator. So either I'm missing some further properties of these special functions, or there's something wrong with the answer you're chasing.

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