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Recently I came along the following problem concerning a lower bound on the difference of two series:

I want to show that for every $q \in [e^{-2},e^{-\frac{1}{2}}]$ we have

$$ f(q) := \sum_{n=0}^\infty (4n+1)q^{\left (\frac{4n+1}{2}\right)^2} - \sum_{n=1}^\infty (4n-1)q^{\left (\frac{4n-1}{2}\right)^2} \geq \frac{1}{10} $$

If I plot the function $f$ on the interval $[e^{-2},e^{-\frac{1}{2}}]$ I get the following: enter image description here

Hence, the minimum seems to be attained at $q=$ with $f(e^{-\frac{1}{2}}) \approx 0.113$. Does anyone has an idea or a simple approach how to show the above estimate?

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    $\begingroup$ Letting $\chi_4$ be the unique non-principal Dirichlet character modulo $4$, your $f$ may be written as $\frac{1}{2} \times \sum_{n \in \mathbb{Z}} \chi_4(n) n q^{n^2/4} = \sum_{n \ge 1} \chi_4(n) n q^{n^2/4}$, which suggests that the theory of modular forms might be relevant. Perhaps $f$ factorizes? $\endgroup$ Sep 4, 2021 at 20:17
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    $\begingroup$ @OfirGorodetsky It is $q^{1/4}\prod_{n\geqslant1}(1-q^{2n})^3$ $\endgroup$ Sep 5, 2021 at 6:01
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    $\begingroup$ In other words, it is $\eta(q^2)^3$ $\endgroup$ Sep 5, 2021 at 6:07

1 Answer 1

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Let $$s_q(N):=\sum_{n=0}^N f_q(n),\quad t_q(N):=\sum_{n=1}^N g_q(n),$$ where $$f_q(n):=(4n+1)q^{(4n+1)^2/4},\quad g_q(n):=(4n-1)q^{(4n-1)^2/4}.$$ We want to show that $$s_q(\infty)-t_q(\infty)\overset{\text{(?)}}\ge1/10 \tag{1}$$ for all $$q\in[e^{-2},e^{-1/2}]. \tag{2} $$

For such $q$, $g_q(n)$ is decreasing in $n\ge1$ and increasing in $q$, and hence $$t_q(\infty)-t_q(2) =\sum_{n=3}^\infty g_q(n) \\ <\int_2^\infty g_{e^{-1/2}}(u)\,du=e^{-49/8}.$$ So, for $q$ as in (2),
$$ \begin{aligned} &s_q(\infty)-t_q(\infty) \\ &>s_q(2)-t_q(2)-e^{-49/8} \\ &=h(q):=9 q^{81/4}-7 q^{49/4}+5 q^{25/4}-3 q^{9/4}+q^{1/4}-e^{-49/8} \\ &\ge h(e^{-1/2})>1/10; \end{aligned} \tag{3}$$ the penultimate inequality, $h(q)\ge h(e^{-1/2})$, in the above multiline display is easy to prove, since $h(q)$ is a simple polynomial in $q^{1/4}$ (see a proof below).

So, (1) indeed holds for all $q$ as in (2).


Proof of the inequality $h(q)\ge h(e^{-1/2})$ for $q$ as in (2): For $u\in[e^{-1/2},e^{-1/8}]$, let $$H(u):=h(u^4),\quad H_2(u):=\frac{H''(u)}{24u^7}, \quad H_3(u):=\frac{H_2'(u)}{80 u^{15}},$$ $$ H_4(u):=\frac{H_3'(u)}{168 u^{23}}=729 u^{32}-49\le H_4(e^{-1/8})<0.$$ So, $H_3$ is decreasing (on $[e^{-1/2},e^{-1/8}]$), to $H_3(e^{-1/8})>0$. So, $H_3>0$ and hence $H_2$ is increasing, from $H_2(e^{-1/2})<0$ to $H_2(e^{-1/8})>0$.

So, for some $c\in(e^{-1/2},e^{-1/8})$, $H$ is concave on $[e^{-1/2},c]$ and convex on $[c,e^{-1/8}]$. Also, $H(e^{-1/2})>H(e^{-1/8})$ and $H'(e^{-1/8})<0$.

So, $H(u)\ge H(e^{-1/8})$ for $u\in[e^{-1/2},e^{-1/8}]$ -- that is, $h(q)\ge h(e^{-1/2})$ for $q$ as in (2). $\quad\Box$

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  • $\begingroup$ @J.Swail : Are you satisfied with this answer? $\endgroup$ Sep 10, 2021 at 0:21
  • $\begingroup$ Yes, I like the answer very much. Thanks a lot!! $\endgroup$
    – J. Swail
    Sep 10, 2021 at 18:05

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