Is this decimal radial-energy identity a known cotangent/Dedekind-sum identity?
问题内容
Let $b\ge 2$. Partition $[0,1)$ into the $b$ equal half-open intervals
$$ I_j=\left[\frac{j}{b},\frac{j+1}{b}\right), \qquad 0\le j\le b-1. $$
Define the same-bin indicator
$$ H_b(x,y)= \begin{cases} 1, & x,y\text{ lie in the same }I_j,\\ 0, & \text{otherwise}, \end{cases} $$
and the centered function
$$ F_b(x,y)=H_b(x,y)-\frac1b. $$
I use the Fourier convention
$$ \widehat{F_b}(m,n) = \int_0^1\int_0^1 F_b(x,y)e^{-2\pi i(mx+ny)}\,dx\,dy. $$
For $t\ne 0$, put
$$ \alpha_t=\frac{1-e^{-2\pi i t/b}}{2\pi i t}, $$
and set $\alpha_0=1/b$. Since $F_b$ is centered, $\widehat{F_b}(0,0)=0$. For $(m,n)\ne(0,0)$, a direct computation appears to give
$$ \widehat{F_b}(m,n) = \begin{cases} b,\alpha_m\alpha_n, & m+n\equiv 0\pmod b,\\ 0, & \text{otherwise}. \end{cases} $$
For a primitive rational direction $(a,c)$, with $c>0$, $a\ne 0$, and $\gcd(a,c)=1$, define
$$ R_b(a,c)= \sum_{\ell\ne 0}\widehat{F_b}(\ell a,\ell c). $$
The support condition is
$$ b\mid \ell(a+c). $$
Let
$$ g=\gcd(b,a+c), \qquad h\equiv a\pmod g, \qquad 0\le h<g. $$
Then $\ell=(b/g)k$, and the numerator reduces to a sine-square term. Using
$$ \sum_{k\ne0}\frac{\sin^2(\pi kh/g)}{k^2} = \frac{\pi^2h(g-h)}{g^2}, $$
I obtain
$$ R_b(a,c) = -\frac{h(g-h)}{bac}. $$
Now define the radial energy by
$$ E_b= \sum_{\substack{c>0,\\ a\in\mathbb Z\setminus\{0\}\\ \gcd(a,c)=1}} |R_b(a,c)|^2. $$
Combining the two signs $a=\pm m$ gives
$$ E_b = \frac1{b^2} \sum_{\substack{m,c\ge 1\\ \gcd(m,c)=1}} \frac{Q_b(m,c)}{m^2c^2}, $$
where
$$ Q_b(m,c)=\Theta_b(m,c)+\Theta_b(-m,c), $$
with
$$ \Theta_b(a,c)=h_{a,c}^2\bigl(g_{a,c}-h_{a,c}\bigr)^2, $$
$$ g_{a,c}=\gcd(b,a+c), \qquad h_{a,c}\equiv a\pmod {g_{a,c}}, \qquad 0\le h_{a,c}<g_{a,c}. $$
For $b=10$, this gives a periodic weight $Q_{10}$ modulo $10$. With rows indexed by $r=m\bmod 10$ and columns by $s=c\bmod 10$, I get
$$ \begin{array}{c|rrrrrrrrrr} r\backslash s&0&1&2&3&4&5&6&7&8&9\\ \hline 0&0&0&0&0&0&0&0&0&0&0\\ 1&0&82&0&2&16&2&16&2&0&82\\ 2&0&0&256&36&0&0&0&36&256&0\\ 3&0&2&36&442&0&2&0&442&36&2\\ 4&0&16&0&0&576&0&576&0&0&16\\ 5&0&2&0&2&0&1250&0&2&0&2\\ 6&0&16&0&0&576&0&576&0&0&16\\ 7&0&2&36&442&0&2&0&442&36&2\\ 8&0&0&256&36&0&0&0&36&256&0\\ 9&0&82&0&2&16&2&16&2&0&82 \end{array} $$
Thus
$$ E_{10} = \frac1{100} \sum_{\substack{m,c\ge 1\\ \gcd(m,c)=1}} \frac{Q_{10}(m,c)}{m^2c^2}. $$
Equivalently, if
$$ Z_{10}(r,s) = \sum_{\substack{m,c\ge1\\ \gcd(m,c)=1\\ m\equiv r\pmod{10}\\ c\equiv s\pmod{10}}} \frac1{m^2c^2}, $$
then
$$ E_{10} = \frac1{100} \sum_{r=0}^9\sum_{s=0}^9 Q_{10}(r,s)Z_{10}(r,s). $$
My residue-class calculation gives a principal part and a quadratic-character part modulo $5$. Let $\chi_5(n)=\left(\frac{n}{5}\right)$. The calculation reduces to
$$ E_{10}=S_0\Lambda_0+S_\chi\Lambda_\chi, $$
where
$$ S_0= \frac{1}{\zeta(4)(1-2^{-4})(1-5^{-4})}, $$
$$ S_\chi= \frac{1}{L(4,\chi_5)(1+2^{-4})}, $$
and
$$ \Lambda_0=\frac{3783\pi^4}{125000}, \qquad \Lambda_\chi=-\frac{51}{25}L(4,\chi_5). $$
Using
$$ \zeta(4)=\frac{\pi^4}{90} $$
and
$$ L(4,\chi_5)=\frac{8\sqrt5,\pi^4}{1875}, $$
this gives
$$ S_0\Lambda_0=\frac{291}{100}, \qquad S_\chi\Lambda_\chi=-\frac{48}{25}, $$
and therefore
$$ E_{10} = \frac{291}{100}-\frac{48}{25} = \frac{99}{100}. $$
My questions are:
- Is the derivation of the radial coefficient
$$ R_b(a,c)=-\frac{h(g-h)}{bac} $$
correct, especially regarding the zero-frequency convention and the summation over $\ell\ne0$?
- Is the residue-class evaluation leading to
$$ E_{10}=\frac{99}{100} $$
correct as written?
- Is this identity already known as a special case of a Dedekind cotangent sum, Vasyunin-type cotangent sum, Bettin-Conrey sum, Estermann zeta identity, or another standard finite character/zeta evaluation?
I am not claiming novelty. I am trying to determine whether this is a known cotangent-sum specialization, a modest reformulation of a standard identity, or whether there is a gap in the derivation above. I am also not claiming the more general formula $E_b=1-b^{-2}$, nor any finite-prime modular-inverse discrepancy limit.
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