# Prove that $\sum_{k=1}^{\infty} (-1)^k q^{k^2} \geq \frac{2}{\pi}\ \sum_{k=1}^{\infty} \frac{(-1)^k}{2k-1}{q}^{\frac{ (2k-1)^2}{4}}$ for $0<q<1$ (updated bounty)

The title explains it. Prove that :

$\sum_{k=1}^{\infty} {(-1)^k\rm e}^{-\hat{t} k^2} \geq \frac{2}{\pi}\ \sum_{k=1}^{\infty} \frac{(-1)^k}{2k-1}{\rm e}^{\frac{-\hat{t} (2k-1)^2}{4}}$ for $\hat{t}>0$

For some background on the problem, you can see my recent paper:

I am an engineer and not a mathematician. However, it bugged me (and a reviewer of the paper!) that I wasn't able to prove that Equation 13 is always greater than Equation 15.

Graphically (Figure 5), it seems to be true that the flux sensor result $\bar{m}_{Flux}\geq\bar{m}_{Conc}$ for a given $\bar{t}$, but I was not able to prove it analytically! Would love to know how it's done. I imagine it will be something related to the fact that exponentials are always greater than zero. Maybe some clever stuff about the limit as $\bar{t}\to 0$ and taking the derivative? The difference function $\bar{m}_{Flux}-\bar{m}_{Conc}$ seems to be a single-peaked function. I'm not sure if that helps!

Note, in the paper I use $\bar{t} = \frac{t D}{L^2}$ whereas above I have simplified it further and use $\hat{t} = \frac{t D \pi^2}{ L^2}$

you can substitute ${\rm e}^{-t} = q$, for $t>0$, or $0<q<1$ (as I did in title of post vs body of text)

Note also that the left hand side is a Jacobi elliptic functions, and the right hand side sort of looks like an expansion of ArcTan. I'm not sure if that helps at all...

I recently was awarded $10 for answering a question correctly, so I have put that towards the bounty! • Erdos 0 Can you compare D with L? Could we assume that D is greater that L? The result should be true under this assumption, but if D is significantly smaller that L this may not be true. Is that compatible with your intuition about this problem? • Bob Lansdorp +1 D is the diffusion coefficient (units of cm^2/s) and L is a length (units of cm). So your question doesn't make sense. I non-dimensionalized the problem using$\bar{t} = t D/L^2$in the paper, and$\bar{t} = t D/(\Pi^2 L^2)$above. So if I translate what you're saying, it's basically, "is it true when \bar{t} is large and small?" and the answer is "yes I believe it is true for all$\bar{t}>0$". • Mathe +1 I believe I have solved the problem! I had already spent a good chunk of time on it, when you shared this problem last time. Would you consider increasing the bounty to$50?

• Mathe
+1

Oh, nevermind, I made a mistake. It's not done yet.

• Bob Lansdorp
0

Sounds promising! Thank you for spending your time on this problem. I have also spent a lot of time on it!

• M F H
0

you should avoid using index "i" (also "l" = lowercase L) and prefer other choices (k, m, n, ...), especially in the exponential most mathematicians will always see "i" as the imaginary unit at a first glance ...

• Bob Lansdorp
0

thank you and good point. updated from i to k. too bad l didn't change it in time for paper to be published! and yes i just used lowercase L there to be punny :)

• M F H
0

We can rewrite the inequality as $\pi \sum (-1)^k \, q^k^2 > \sum (-1)^k / (k-1/2) \, q^(k-1/2)^2$, where q = exp(-t) is an arbitrary number between 0 and 1. The L.H.S. can be written in terms of the elliptic theta function/constant $( \pi / 2 ) ( \theta_4 (q) - 1 )$. (That's more or less the definition: \theta_4 (q) := 1+ 2 sum (-1)^k q^k².) Maybe a possible approach is to check what the first few terms look like, and approximate/estimate later terms (k > 5 ?) by dropping the 1/2 ?

• Bob Lansdorp
0

yes LHS can be easily rewritten as elliptic theta. how does that relate to RHS though? when you say "terms" are you referring to some series expansion of elliptic theta that ends up looking like RHS? i like the q=exp(-t) substitution though! maybe a step in the right direction

• M F H
0

No I meant the first terms of the sum, i.e. what you get for k = 1,2,3,4, say. Given that you have k² in the exponent of q9<1, later terms are very likely "negligible" or more precisely very easily bounded from above.

• M F H
0

I think one might try to separate even and odd indices. The odd indices 2k-1 from the LHS might be rearranged with terms from the RHS also having exponents (2k-1)... I didn't yet have the time to develop that.

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