Transforming a sum to another one.

[MathematicalPhysicist]

In the next paper:
http://www.math.psu.edu/vstein/alg/antheory/preprint/andrews/lew4.pdf

In page 7, in (21) I don't understand how does changing from q to -q gives the RHS in (5) in page 2, obviously changing gives the LHS in (5), but I don't see how it gives the RHS in (5).

Any tips?

I tried dissecting to sums that passes through all n which are divisble by 2 and all sums that aren't, and noticing that 6n+1=3(2n)+1 , 6n+4=3(2n+1)+1, 6n+2=3(2n)+2, 6n+5=3(2n+1)+2.
but I didn't get far with it.

 

[LCKurtz]

I can't speak for others, but I'm guessing your lack of responses so far is because instead of taking the time to just post the equations in question, you expect your readers to download a pdf and chase down the references.

 

[MathematicalPhysicist]

No problem.

I have the next sum:
$$1-2(\sum_{n=0}^{\infty} \frac{q^{6n+5}}{1-q^{6n+5}}-\frac{q^{6n+1}}{1-q^{6n+1}}+\frac{q^{6n+2}}{1+q^{6n+2}}-\frac{q^{6n+4}}{1+q^{6n+4}})$$
and this sum ought to be the next sum:
$1+2(\sum_{n=0}^{\infty}\frac{q^{3n+1}}{1-q^{3n+1}} -\frac{q^{3n+2}}{1-q^{3n+2}})+4(\sum_{n=0}^{\infty}\frac{q^{12n+4}}{1-q^{12n+4}}-\frac{q^{12n+8}}{1-q^{12n+8}})$

 

[Stephen Tashi]

Is the second equation missing a bracket after the 12n + 4? Should it be:

$1+2(\sum_{n=0}^{\infty}\frac{q^{3n+1}}{1-q^{3n+1}} -\frac{q^{3n+2}}{1-q^{3n+2}})+4(\sum_{n=0}^{\infty}\frac{q^{12n+4}}{1-q^{12n+4}}-\frac{q^{12n+8}}{1-q^{12n+8}})$

 

[MathematicalPhysicist]

Yes, Stephen.

 

[Stephen Tashi]

The paper mentions that $q$ is a complex number with modulus < 1.

I think you have accidentally reversed two terms in your first equation.

Is the problem this:

Define
$$S(q) = 1-2(\sum_{n=0}^{\infty} \frac{q^{6n+1}}{1-q^{6n+1}}-\frac{q^{6n+5}}{1-q^{6n+5}}+\frac{q^{6n+2}}{1+q^{6n+2}}-\frac{q^{6n+4}}{1+q^{6n+4}})$$

Show
$S(-q) = 1+2(\sum_{n=0}^{\infty}\frac{q^{3n+1}}{1-q^{3n+1}} -\frac{q^{3n+2}}{1-q^{3n+2}})+4(\sum_{n=0}^{\infty}\frac{q^{12n+4}}{1-q^{12n+4}}-\frac{q^{12n+8}}{1-q^{12n+8}})$

 

[MathematicalPhysicist]

Yes this is the problem, thought I think I have given S(-q) in the first line of latex.
I mean if you plug -q instead of q in S(q) you get:
$$1-2(\sum_{n=0}^{\infty} \frac{q^{6n+5}}{1-q^{6n+5}}-\frac{q^{6n+1}}{1-q^{6n+1}}+\frac{q^{6n+2}}{1+q^{6n+2}}-\frac{q^{6n+4}}{1+q^{6n+4}})$$

 

[MathematicalPhysicist]

I mean we have $$S(q)=1-2(\sum_{n=0}^{\infty} -\frac{q^{6n+5}}{1+q^{6n+5}}+\frac{q^{6n+1}}{1+q^{6n+1}}+\frac{q^{6n+2}}{1+q^{6n+2}}-\frac{q^{6n+4}}{1+q^{6n+4}})$$
I plugged -q instead of q and got the above in post 7.

Thanks for your help.

 

[Stephen Tashi]

Maybe it was simpler to let people read the paper!

Fixing my mistake in the signs, is the problem:

$q$ is a complex number with modulus < 1.


Define
$$S(q) = 1-2(\sum_{n=0}^{\infty} \frac{q^{6n+1}}{1+q^{6n+1}}-\frac{q^{6n+5}}{1+q^{6n+5}}+\frac{q^{6n+2}}{1+q^{6n+2}}-\frac{q^{6n+4}}{1+q^{6n+4}})$$

Show
$S(-q) = 1+2(\sum_{n=0}^{\infty}\frac{q^{3n+1}}{1-q^{3n+1}} -\frac{q^{3n+2}}{1-q^{3n+2}})+4(\sum_{n=0}^{\infty}\frac{q^{12n+4}}{1-q^{12n+4}}-\frac{q^{12n+8}}{1-q^{12n+8}})$

 

[MathematicalPhysicist]

Yes.

 

[Stephen Tashi]

My only idea this morning is to try a more elaborate version of your original approach.

Define
$$S(q) = 1-2(\sum_{n=0}^{\infty} \frac{q^{6n+1}}{1+q^{6n+1}}-\frac{q^{6n+5}}{1+q^{6n+5}}+\frac{q^{6n+2}}{1+q^{6n+2}}-\frac{q^{6n+4}}{1+q^{6n+4}})$$

Let $R(q) = S(-q)$
As you pointed out in your earlier post,

$$R(q) = 1-2(\sum_{n=0}^{\infty} \frac{q^{6n+5}}{1-q^{6n+5}}-\frac{q^{6n+1}}{1-q^{6n+1}}+\frac{q^{6n+2}}{1+q^{6n+2}}-\frac{q^{6n+4}}{1+q^{6n+4}})$$

Let

$T(q) = 1+2(\sum_{n=0}^{\infty}\frac{q^{3n+1}}{1-q^{3n+1}} -\frac{q^{3n+2}}{1-q^{3n+2}})+4(\sum_{n=0}^{\infty}\frac{q^{12n+4}}{1-q^{12n+4}}-\frac{q^{12n+8}}{1-q^{12n+8}})$

For a given positive integer v, we can try to analyze whether the terms involving $q^v$ are the same in R(q) and T(q). This involves various cases.

For example, suppose v is odd. Then in R(q), v appears when there are nonnegative integer solutions for n in either of the equations
6n+1 = v
6n + 5 = v
but if there is a solution to 6n+1 = v, there can't be a solution to 6n+5 = v. So we have two sub-cases to consider. Consider the first case: v = 6n + 1

Where can a v of that form appear in T(q)? It would have to appear with an exponent m so that 3m + 1 = 6n + 1. So it would appear when m = 2n. So in both R(q) and T(q), in this case and sub-case, the term involving the exponent v is $2 \frac {q^v}{1- q^v}$

...then we would have to consider all the other cases and sub cases.

I don't really like this approach because it gets into number theory and it doesn't reveal (to me) how one would discover the formula for T(q).

 

[MathematicalPhysicist]

Ok, I solved it, it was rather easy when I think of it, and I am shmuck for not seeing it before.