At the crossroads of datas: Using artificial intelligence for efficient and resilient planning in African cities

Cheikh Cisse; Cheikh Cisse

doi:10.3934/urs.2023019

Urban Resilience and Sustainability

2023, Volume 1, Issue 4: 309-313. doi: 10.3934/urs.2023019

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Editorial

At the crossroads of datas: Using artificial intelligence for efficient and resilient planning in African cities

Cheikh Cisse ^,

Urban Planner-Geographer, Freelance Consultant and affiliated Researcher at the African Studies Centre Leiden, Leiden University, Netherlands

Received: 25 December 2023 Revised: 29 December 2023 Accepted: 30 December 2023 Published: 31 December 2023

By collecting and conducting in-depth analysis of official statistical data, empirical evidence, and exploring integrated artificial intelligence (AI) approaches in diverse African urban landscapes, this article examines the integration of formal and informal data to bolster urban resilience in Africa. It underscores the significance of this diverse data in crafting effective urban planning strategies tailored to the needs of African city-dwellers. The AI-driven analysis, amalgamating these datasets, provides a comprehensive understanding of urban realities and community needs in African cities. Additionally, the article emphasizes how harnessing so-called informal data—often overlooked yet crucial—alongside formal data enables more contextual and adaptive planning aligned with local needs and aspirations. Highlighting the transformative potential of data-driven approaches, the article presents AI strategies as proactive tools to address the challenges faced by African cities, fostering resilience amidst socio-economic and environmental shifts. Ultimately, it advocates for the integration of diverse data sources in urban planning, utilizing AI as a catalyst, underlining the importance of a holistic approach that values both formal and informal data for sustainable and resilient urban development in Africa.

Keywords:

Citation: Cheikh Cisse. At the crossroads of datas: Using artificial intelligence for efficient and resilient planning in African cities[J]. Urban Resilience and Sustainability, 2023, 1(4): 309-313. doi: 10.3934/urs.2023019

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Abstract

1. Introduction and main results

If $h$ and $k$ are integers with $k > 0$ , the classical Dedekind sums $S(h, k)$ are defined as

$S(h,k) = \sum\limits_{a = 1}^{k}\left(\left(\frac{a}{k}\right)\right)\left(\left(\frac{ah}{k}\right) \right),$

where

$((x)) = \left\{\begin{array}{ll} x-[x]-\frac{1}{2}, &\hbox{if $x$ is not an integer; } \\ 0, &\hbox{if $x$ is an integer. } \end{array}\right.$

The various properties of $S(h, k)$ were investigated by many authors, one of which is reciprocity theorem (see Tom M. Apostol ^[1] or L. Carlitz ^[2]). That is, for all positive integers $h$ and $k$ with $(h, k) = 1$ , we have the identity

$S(h,k)+S(k,h) = \frac{h^{2}+k^{2}+1}{12hk}-\frac{1}{4}.$

Conrey et al. ^[3] studied the mean value distribution of $S(h, k)$ and deduced the important asymptotic formula

$\begin{eqnarray*} && \mathop{{\sum}'}^{k}_{h = 1}\left|S(h,k)\right|^{2m} = f_{m}(k)\left( \frac{k}{12} \right)^{2m}+O\left(\left(k^{\frac{9}{5}}+k^{2m-1+\frac{1}{m+1}} \right)\log^{3}k \right), \end{eqnarray*}$

where $\begin{eqnarray*} \mathop{{\sum}'}^{k}_{h = 1}\end{eqnarray*}$ denotes the summation over all $h$ such that $(h, k) = 1$ and

$\begin{eqnarray*} && \sum\limits_{n = 1}^{\infty}\frac{f_{m}(n)}{n^{s}} = 2\frac{\zeta^{2}(2m)}{\zeta(4m)}\cdot\frac{\zeta(s+4m-1)}{\zeta^{2}(s+2m)}\zeta(s). \end{eqnarray*}$

Moreover, X. L. He and W. P. Zhang ^[4] gave an interesting asymptotic formula for the Dedekind sums with a weight of Hurwitz zeta-function as follows:

$\begin{eqnarray*} && \mathop{{\sum}'}^{k}_{h = 1}\zeta^{2}\left(\frac{1}{2},\frac{h}{k}\right)S^{2}(h,k) = \frac{k^{3}}{144}\zeta(3)\prod\limits_{p|k}\left( 1-\frac{1}{p^{3}}\right)+O\left( k^{\frac{5}{2}}\exp\left(\frac{3\log k}{\log\log k} \right) \right). \end{eqnarray*}$

Other sums analogous to the Dedekind sums are the Hardy sums. Using the notation of Berndt and Goldberg ^[5], they defined

$S_{1}(h,k) = \sum\limits_{j = 1}^{k-1}(-1)^{j+1+\left[\frac{hj}{k}\right]},$

where $h$ and $k$ are integers with $k > 0$ .

In 2014, H. Zhang and W. P. Zhang ^[6] obtained some beautiful identities involving $S_{1}(h, k)$ in the forms of

$\sum\limits_{m = 1}^{p-1}\sum\limits_{n = 1}^{p-1}K(m,p)K(n,p)S_{1}(2m\overline{n},p),$

$\sum\limits_{m = 1}^{p-1}\sum\limits_{n = 1}^{p-1}\left|K(m,p)\right|^{2}\left|K(n,p)\right|^{2}S_{1}(2m\overline{n},p),$

where $K(n, p)$ denotes the reduced form of the general Kloosterman sums attached to a Dirichlet character $\lambda$ modulo $k$ as

$\begin{eqnarray*} && K(r,l,\lambda;k) = \mathop{{\sum}'}^{k}_{a = 1}\lambda(a)e\left(\frac{ra+l\overline{a}}{k}\right), \end{eqnarray*}$

where $e(x) = e^{2 \pi i x}$ , $\overline{a}$ denotes the solution of the congruence $x \cdot a\equiv 1 \bmod k$ .

Recently, H. F. Zhang and T. P. Zhang ^[7] extended the results in ^[6] to a more general situation as

$\sum\limits_{m = 1}^{p-1}\sum\limits_{n = 1}^{p-1}K(m,s,\lambda;p)\overline{K(n,t,\lambda;p)}S_{1}(2m\overline{n},p),$

$\sum\limits_{m = 1}^{p-1}\sum\limits_{n = 1}^{p-1}\left|K(m,s,\lambda;p)\right|^{2}\left|K(n,t,\lambda;p)\right|^{2}S_{1}(2m\overline{n},p),$

where $\overline{K(n, t, \lambda; p)}$ denotes complex conjugate of $K(n, t, \lambda; p)$ .

Actually there are six forms of Hardy sums (see Berndt ^[8] and Goldberg ^[9]). A natural question is whether we can obtain similar results by replacing $S_{1}(h, k)$ with other forms of Hardy sums. Due to some technical reasons, for most of other forms of Hardy sums, the answer is no! Thanks to the important relationships among Hardy sums and Dedekind sums built by R. Sitaramachandrarao ^[10], we are lucky to find the only one $S_{3}(h, p)$ to replace, with

$\begin{eqnarray*} S_{3}(h,k) = \sum\limits_{j = 1}^{k}(-1)^{j}\left(\left(\frac{hj}{k}\right)\right). \end{eqnarray*}$

Our starting point relies heavily on the following in ^[10] as:

Proposition 1. Let $k$ be an odd positive integer, $h$ be an integer with $(h, k) = 1$ . Then

$\begin{eqnarray*} && S_{3}(h,k) = 2S(h,k)-4S(2h,k). \end{eqnarray*}$

Then applying the properties of Gauss sums and the mean square value of Dirichlet $L$ -functions, we have

Theorem 1. Let $p$ be an odd prime. Then for any Dirichlet character $\lambda \bmod p$ and any integer $s$ , $t$ with $(s, p) = (t, p) = 1$ , we have

$\begin{eqnarray*} && \sum\limits_{m = 1}^{p-1}\sum\limits_{n = 1}^{p-1}K(m,s,\lambda;p)\overline{K(n,t,\lambda;p)}S_{3}(m\overline{n},p) = \left\{ \begin{array}{ll} \frac{p-1}{2}, & \hbox{if $\overline{\lambda}\chi = \chi_{0}$}; \\ \frac{p(p-1)}{2}, & \hbox{if $\overline{\lambda}\chi\neq\chi_{0}$}, \end{array} \right. \end{eqnarray*}$

where $\chi$ is an odd Dirichlet character modulo $p$ and $\chi_{0}$ is the principal character modulo $p$ .

Theorem 2. Let $p$ be an odd prime with $p\equiv 1 \bmod 4$ . Then for any Dirichlet character $\lambda \bmod p$ and any integer $s$ , $t$ with $(s, p) = (t, p) = 1$ , we have

$\begin{eqnarray*} && \sum\limits_{m = 1}^{p-1}\sum\limits_{n = 1}^{p-1}\left|K(m,s,\lambda;p)\right|^{2}\left|K(n,t,\lambda;p)\right|^{2}S_{3}(m\overline{n},p)\\ &&\left\{ \begin{array}{ll} = \frac{p^{2}(p-1)}{2}, & \hbox{if $\overline{\lambda\chi}\neq\chi_{0},\ \overline{\lambda}\chi\neq\chi_{0}$};\\ = \frac{p(p-1)}{2}, & \hbox{if $\overline{\lambda\chi}\neq\chi_{0},\ \overline{\lambda}\chi = \chi_{0}$};\\ \leq p^{\frac{9}{2}}+\frac{1}{2}p^{4}-4p^{\frac{7}{2}}-p^{3}+5p^{\frac{5}{2}}+p^{2}-2p^{\frac{3}{2}}-\frac{1}{2}p, & \hbox{if $\overline{\lambda\chi} = \chi_{0},\ \overline{\lambda}\chi = \chi_{0}$}; \\ \leq p^{5}-3p^{4}+3p^{3}-\frac{1}{2}p^{2}-\frac{1}{2}p, & \hbox{if $\overline{\lambda\chi} = \chi_{0},\ \overline{\lambda}\chi\neq\chi_{0}$}. \end{array} \right. \end{eqnarray*}$

Theorem 3. Let $p$ be an odd prime with $p\equiv 3 \bmod 8$ . Then for any Dirichlet character $\lambda \bmod p$ and any integer $s$ , $t$ with $(s, p) = (t, p) = 1$ , we have

$\begin{eqnarray*} && \sum\limits_{m = 1}^{p-1}\sum\limits_{n = 1}^{p-1}\left|K(m,s,\lambda;p)\right|^{2}\left|K(n,t,\lambda;p)\right|^{2}S_{3}(m\overline{n},p)\\ &&\left\{ \begin{array}{ll} = \frac{p^{2}(p-1)}{2}-6p^{2}h_{p}^{2}, & \hbox{if $\overline{\lambda\chi}\neq\chi_{0},\ \overline{\lambda}\chi\neq\chi_{0}$};\\ = \frac{p(p-1)}{2}-6ph_{p}^{2}, & \hbox{if $\overline{\lambda\chi}\neq\chi_{0},\ \overline{\lambda}\chi = \chi_{0}$};\\ \leq p^{\frac{9}{2}}+\frac{1}{2}p^{4}-4p^{\frac{7}{2}}-p^{3}+5p^{\frac{5}{2}}+p^{2}-2p^{\frac{3}{2}}-\frac{1}{2}p+6p^{2}h_{p}^{2}, & \hbox{if $\overline{\lambda\chi} = \chi_{0},\ \overline{\lambda}\chi = \chi_{0}$}; \\ \leq p^{5}-3p^{4}+3p^{3}-\frac{1}{2}p^{2}-\frac{1}{2}p+6p^{3}h_{p}^{2}, & \hbox{if $\overline{\lambda\chi} = \chi_{0},\ \overline{\lambda}\chi\neq\chi_{0}$}, \end{array} \right. \end{eqnarray*}$

where $h_{p}$ denotes the class number of the quadratic field $Q\left(\sqrt{-p}\right)$ .

Theorem 4. Let $p$ be an odd prime with $p\equiv 7 \bmod 8$ . Then for any Dirichlet character $\lambda \bmod p$ and any integer $s$ , $t$ with $(s, p) = (t, p) = 1$ , we have

$\begin{eqnarray*} && \sum\limits_{m = 1}^{p-1}\sum\limits_{n = 1}^{p-1}\left|K(m,s,\lambda;p)\right|^{2}\left|K(n,t,\lambda;p)\right|^{2}S_{3}(m\overline{n},p)\\ &&\left\{ \begin{array}{ll} = \frac{p^{2}(p-1)}{2}+2p^{2}h_{p}^{2}, & \hbox{if $\overline{\lambda\chi}\neq\chi_{0},\ \overline{\lambda}\chi\neq\chi_{0}$};\\ = \frac{p(p-1)}{2}+2ph_{p}^{2}, & \hbox{if $\overline{\lambda\chi}\neq\chi_{0},\ \overline{\lambda}\chi = \chi_{0}$};\\ \leq p^{\frac{9}{2}}+\frac{1}{2}p^{4}-4p^{\frac{7}{2}}-p^{3}+5p^{\frac{5}{2}}+p^{2}-2p^{\frac{3}{2}}-\frac{1}{2}p+2p^{2}h_{p}^{2}, & \hbox{if $\overline{\lambda\chi} = \chi_{0},\ \overline{\lambda}\chi = \chi_{0}$}; \\ \leq p^{5}-3p^{4}+3p^{3}-\frac{1}{2}p^{2}-\frac{1}{2}p+2p^{3}h_{p}^{2}, & \hbox{if $\overline{\lambda\chi} = \chi_{0},\ \overline{\lambda}\chi\neq\chi_{0}$}. \end{array} \right. \end{eqnarray*}$

Taking $\lambda = \lambda_{0}$ , $s = t = 1$ in Theorems 1–4, we immediately deduce the following results.

Corollary 1. Let $p$ be an odd prime. Then we have the identity

$\sum\limits_{m = 1}^{p-1}\sum\limits_{n = 1}^{p-1}K(m,p)K(n,p)S_{3}(m\overline{n},p) = \frac{p(p-1)}{2}.$

Corollary 2. Let $p$ be an odd prime. Then we have

$\sum\limits_{m = 1}^{p-1}\sum\limits_{n = 1}^{p-1}\left|K(m,p)\right|^{2}\left|K(n,p)\right|^{2}S_{3}(m\overline{n},p) = \left\{ \begin{array}{ll} \frac{p^{2}(p-1)}{2}, & \hbox{if $p\equiv 1 \bmod 4$}; \\ \frac{p^{2}(p-1)}{2}-6p^{2}h_{p}^{2}, & \hbox{if $p\equiv 3 \bmod 8$};\\ \frac{p^{2}(p-1)}{2}+2p^{2}h_{p}^{2}, & \hbox{if $p\equiv 7 \bmod 8$}.\\ \end{array} \right.$

2. Some Lemmas

To prove the Theorems, we need the following Lemmas.

Lemma 1. Let $k > 2$ be an integer. Then for any integer $a$ with $(a, k) = 1,$ we have the identity

$S(a,k) = \frac{1}{\pi^2k}\sum\limits_{d\mid k}\frac{d^2}{\phi(d)}\mathop{\sum\limits_{\chi \bmod d}}_{\chi(-1) = -1}\chi(a)\left|L(1,\chi)\right|^2,$

where $L(1, \chi)$ denotes the Dirichlet $L$ -function corresponding to Dirichlet character $\chi$ mod $d$ .

Proof. See Lemma 2 of ^[11].

Lemma 2. Let $p$ be an odd prime, $s$ be any integer with $(s, p) = 1$ . Then for any non-principal character $\chi \bmod p$ and any Dirichlet character $\lambda \bmod p$ , we have

$\left|\sum\limits_{m = 1}^{p-1}\chi(m)K(m,s,\lambda;p)\right| = \left\{ \begin{array}{ll} p^{{\frac{1}{2}}}, & \hbox{if $\overline{\lambda}\chi = \chi_{0}$}; \\ p, & \hbox{if $\overline{\lambda}\chi\neq\chi_{0}$}. \end{array} \right.$

Proof. See Lemma 2 of reference ^[7].

Lemma 3. Let $p$ be an odd prime, $s$ be any integer with $(s, p) = 1$ . Then for any non-principal character $\chi \bmod p$ and any Dirichlet character $\lambda \bmod p$ , we have

$\begin{eqnarray*} && \left|\sum\limits_{m = 1}^{p-1} \chi(m)\left|K(m,s,\lambda;p)\right|^{2}\right| = \left\{ \begin{array}{ll} p\left|\tau\left(\overline{\chi}^{2}\right)\right|, & \hbox{if $\overline{\lambda\chi}\neq\chi_{0},\ \overline{\lambda}\chi\neq\chi_{0}$};\\ p^{\frac{1}{2}}\left|\tau\left(\overline{\chi}^{2}\right)\right|, & \hbox{if $\overline{\lambda\chi}\neq\chi_{0},\ \overline{\lambda}\chi = \chi_{0}$}; \\ p\left|\tau\left(\overline{\chi}^{2}\right)+(p-1)\right|, & \hbox{if $\overline{\lambda\chi} = \chi_{0},\ \overline{\lambda}\chi = \chi_{0}$}; \\ p\left|-\tau\left(\overline{\chi}^{2}\right)\tau\left(\overline{\lambda}\chi\right)+(p-1)\right|, & \hbox{if $\overline{\lambda\chi} = \chi_{0},\ \overline{\lambda}\chi\neq\chi_{0}$}, \end{array} \right. \end{eqnarray*}$

where $\tau(\chi) = \sum_{a = 1}^{p}\chi(a) e\left(\frac{a}{p}\right)$ denotes the classical Gauss sums.

Proof. See Lemma 1 of reference ^[7].

Lemma 4. Let $p$ be an odd prime, then we have

$\begin{eqnarray*} && \mathop{\sum\limits_{\chi \bmod p}}_{\chi(-1) = -1}\left|L(1,\chi)\right|^2 = \frac{\pi^2}{12}\cdot\frac{(p-1)^2(p-2)}{p^2} , \end{eqnarray*}$

$\begin{eqnarray*} && \mathop{\sum\limits_{\chi \bmod p}}_{\chi(-1) = -1}\chi(2)\cdot\left|L(1,\chi)\right|^2 = \frac{\pi^2}{24}\cdot\frac{(p-1)^2(p-5)}{p^2}. \end{eqnarray*}$

Proof. See Lemma 5 of reference ^[6].

3. Proof of Theorems

Now we come to prove our Theorems.

Firstly, we prove Theorem 1. Applying Proposition $1$ and Lemma $1$ , we obtain

$\begin{eqnarray*} && \sum\limits_{m = 1}^{p-1}\sum\limits_{n = 1}^{p-1}K(m,s,\lambda;p)\overline{K(n,t,\lambda;p)}S_{3}(m\overline{n},p)\\ && = \frac{2p}{\pi^2(p-1)}\sum\limits_{\chi \bmod p \atop \chi(-1) = -1}\sum\limits_{m = 1}^{p-1}\chi(m) K(m,s,\lambda;p)\cdot\sum\limits_{n = 1}^{p-1}\chi(\overline{n})\overline{K(n,t,\lambda;p)}\cdot|L(1,\chi)|^2\\ &&\quad-\frac{4p}{\pi^2(p-1)}\sum\limits_{\chi \bmod p \atop \chi(-1) = -1}\chi(2)\sum\limits_{m = 1}^{p-1}\chi(m) K(m,s,\lambda;p)\cdot\sum\limits_{n = 1}^{p-1}\chi(\overline{n})\overline{K(n,t,\lambda;p)}\cdot|L(1,\chi)|^2\\ && = \frac{2p}{\pi^2(p-1)}\sum\limits_{\chi \bmod p \atop \chi(-1) = -1}\left|\sum\limits_{m = 1}^{p-1}\chi(m)K(m,s,\lambda;p)\right|^{2}\left|L(1,\chi)\right|^2\\ &&\quad-\frac{4p}{\pi^2(p-1)}\sum\limits_{\chi \bmod p \atop \chi(-1) = -1} \chi(2)\left|\sum\limits_{m = 1}^{p-1}\chi(m)K(m,s,\lambda;p)\right|^{2}\left|L(1,\chi)\right|^2. \end{eqnarray*}$

Then from Lemma $2$ and Lemma $4$ , if $\overline{\lambda}\chi = \chi_{0}$ , we have

$\begin{eqnarray*} && \sum\limits_{m = 1}^{p-1}\sum\limits_{n = 1}^{p-1}K(m,s,\lambda;p)\overline{K(n,t,\lambda;p)}S_{3}(m\overline{n},p)\\ && = \frac{2p^{2}}{\pi^2(p-1)}\sum\limits_{\chi \bmod p \atop \chi(-1) = -1}\left|L(1,\chi)\right|^2 -\frac{4p^{2}}{\pi^2(p-1)}\sum\limits_{\chi \bmod p \atop \chi(-1) = -1}\chi(2)\left|L(1,\chi)\right|^2\\ && = \frac{2p^{2}}{\pi^2(p-1)}\cdot\frac{\pi^{2}}{12}\frac{(p-1)^{2}(p-2)}{p^{2}} -\frac{4p^{2}}{\pi^2(p-1)}\cdot\frac{\pi^{2}}{24}\frac{(p-1)^{2}(p-5)}{p^{2}}\\ && = \frac{p-1}{2}. \end{eqnarray*}$

While if $\overline{\lambda}\chi\neq\chi_{0}$ , we have

$\begin{eqnarray*} && \sum\limits_{m = 1}^{p-1}\sum\limits_{n = 1}^{p-1}K(m,s,\lambda;p)\overline{K(n,t,\lambda;p)}S_{3}(m\overline{n},p)\\ && = \frac{2p^{3}}{\pi^2(p-1)}\sum\limits_{\chi \bmod p \atop \chi(-1) = -1}\left|L(1,\chi)\right|^2 -\frac{4p^{3}}{\pi^2(p-1)}\sum\limits_{\chi \bmod p \atop \chi(-1) = -1}\chi(2)\left|L(1,\chi)\right|^2\\ && = \frac{2p^{3}}{\pi^2(p-1)}\cdot\frac{\pi^{2}}{12}\cdot\frac{(p-1)^{2}(p-2)}{p^{2}} -\frac{4p^{3}}{\pi^2(p-1)}\cdot\frac{\pi^{2}}{24}\cdot\frac{(p-1)^{2}(p-5)}{p^{2}}\\ && = \frac{p(p-1)}{2}. \end{eqnarray*}$

This completes the proof of Theorem 1.

Then we prove Theorem 2. From Proposition $1$ and Lemma $1$ , we obtain

Since $p\equiv 1 \bmod 4$ , and notice that $\left|\tau\left(\overline{\chi}^{2}\right)\right| = \sqrt{p}$ . From Lemma $3$ and Lemma $4$ , if $\overline{\lambda\chi}\neq\chi_{0}$ , $\overline{\lambda}\chi\neq\chi_{0}$ , we have

Similarly, if $\overline{\lambda\chi}\neq\chi_{0}, \ \overline{\lambda}\chi = \chi_{0}$ , we have

$\begin{eqnarray*} && \sum\limits_{m = 1}^{p-1}\sum\limits_{n = 1}^{p-1}\left|K(m,s,\lambda;p)\right|^{2}\left|K(n,t,\lambda;p)\right|^{2}S_{3}(m\overline{n},p) = \frac{p(p-1)}{2}. \end{eqnarray*}$

If $\overline{\lambda\chi} = \chi_{0}, \ \overline{\lambda}\chi = \chi_{0}$ , we can obtain

$\begin{eqnarray*} && \left|\sum\limits_{m = 1}^{p-1}\chi(m)\left|K(m,s,\lambda;p)\right|^{2}\right|^{2}\\ && = p^{2}\left[ \left(Re\ \tau\left(\overline{\chi}^{2}\right)+(p-1)\right)^{2}+\left(Im\ \tau\left(\overline{\chi}^{2}\right)\right)^{2}\right]\\ && = p^{2}\left[p+2(p-1)Re\ \tau\left(\overline{\chi}^{2}\right)+(p-1)^{2}\right]. \end{eqnarray*}$

So we have

$\begin{eqnarray*} && \sum\limits_{m = 1}^{p-1}\sum\limits_{n = 1}^{p-1}\left|K(m,s,\lambda;p)\right|^{2}\left|K(n,t,\lambda;p)\right|^{2}S_{3}(m\overline{n},p)\\ && = \frac{2p^{3}}{\pi^2(p-1)}\sum\limits_{\chi \bmod p \atop \chi(-1) = -1}\left[p+2(p-1)Re\ \tau\left(\overline{\chi}^{2}\right)+(p-1)^{2}\right]\left|L(1,\chi)\right|^2\\ &&\quad-\frac{4p^{3}}{\pi^2(p-1)}\sum\limits_{\chi \bmod p \atop \chi(-1) = -1}\chi(2)\left[p+2(p-1)Re\ \tau\left(\overline{\chi}^{2}\right)+(p-1)^{2}\right]\left|L(1,\chi)\right|^2\\ && = \frac{2p^{3}[p+(p-1)^{2}]}{\pi^2(p-1)}\sum\limits_{\chi \bmod p \atop \chi(-1) = -1}\left|L(1,\chi)\right|^2+\frac{4p^{3}}{\pi^2}\sum\limits_{\chi \bmod p \atop \chi(-1) = -1}Re\ \left(\tau\left(\overline{\chi}^{2}\right)\right)\left|L(1,\chi)\right|^2\\ &&\quad -\frac{4p^{3}[p+(p-1)^{2}]}{\pi^2(p-1)} \sum\limits_{\chi \bmod p \atop \chi(-1) = -1}\chi(2)\left|L(1,\chi)\right|^2-\frac{8p^{3}}{\pi^2}\sum\limits_{\chi \bmod p \atop \chi(-1) = -1}\chi(2)Re\ \left(\tau\left(\overline{\chi}^{2}\right)\right)\left|L(1,\chi)\right|^2\\ && = \frac{p(p-1)(p^{2}-p+1)}{2}+\frac{4p^{3}}{\pi^{2}}\sum\limits_{\chi \bmod p \atop \chi(-1) = -1} Re\ \left(\tau \left(\overline{\chi}^{2}\right)\right) \left|L(1,\chi)\right|^{2}\\ &&\quad -\frac{8p^{3}}{\pi^{2}}\sum\limits_{\chi \bmod p \atop \chi(-1) = -1}\chi(2) Re\ \left(\tau\left(\overline{\chi}^{2}\right)\right)\left|L(1,\chi)\right|^{2}. \end{eqnarray*}$

Noting that $x\leq |x|$ holds for any real number $x$ , we have

$\begin{eqnarray*} && \sum\limits_{m = 1}^{p-1}\sum\limits_{n = 1}^{p-1}\left|K(m,s,\lambda;p)\right|^{2}\left|K(n,t,\lambda;p)\right|^{2}S_{3}(m\overline{n},p)\\ &&\leq\left|\sum\limits_{m = 1}^{p-1}\sum\limits_{n = 1}^{p-1}\left|K(m,s,\lambda;p)\right|^{2}\left|K(n,t,\lambda;p)\right|^{2}S_{3}(m\overline{n},p)\right|\\ &&\leq\frac{p(p-1)(p^{2}-p+1)}{2}+\frac{4p^{\frac{7}{2}}}{\pi^{2}}\sum\limits_{\chi \bmod p \atop \chi(-1) = -1}\left|L(1,\chi)\right|^{2}+\frac{8p^{\frac{7}{2}}}{\pi^{2}}\sum\limits_{\chi \bmod p \atop \chi(-1) = -1}\left|L(1,\chi)\right|^{2}\\ && = \frac{p(p-1)(p^{2}-p+1)}{2}+\frac{4p^{\frac{7}{2}}}{\pi^{2}}\cdot\frac{\pi^{2}}{12}\cdot\frac{(p-1)^{2}(p-2)}{p^{2}}+ \frac{8p^{\frac{7}{2}}}{\pi^{2}}\cdot\frac{\pi^{2}}{12}\cdot\frac{(p-1)^{2}(p-2)}{p^{2}}\\ && = p^{\frac{9}{2}}+\frac{1}{2}p^{4}-4p^{\frac{7}{2}}-p^{3}+5p^{\frac{5}{2}}+p^{2}-2p^{\frac{3}{2}}-\frac{1}{2}p. \end{eqnarray*}$

Similarly, if $\overline{\lambda\chi} = \chi_{0}, \ \overline{\lambda}\chi\neq\chi_{0}$ , we have

This completes the proof of Theorem 2.

Next we turn to prove Theorem 3. Since $p\equiv 3 \bmod 4$ , note that $\left(\frac{-1}{p}\right) = \chi_{2}(-1) = -1$ , $L(1, \chi_{2}) = \frac{\pi\cdot h_{p}}{\sqrt{p}}$ , and $\tau\left(\overline{\chi}^{2}_{2}\right) = -1$ . From Lemma $3$ and Lemma $4$ , if $\overline{\lambda\chi}\neq\chi_{0}, \ \overline{\lambda}\chi\neq\chi_{0}$ , we have

Similarly, if $\overline{\lambda\chi}\neq\chi_{0}, \ \overline{\lambda}\chi = \chi_{0}$ , we have

If $\overline{\lambda\chi} = \chi_{0}, \ \overline{\lambda}\chi = \chi_{0}$ , we can obtain

$\begin{eqnarray*} && p^{2}\left|\tau\left(\overline{\chi}^{2}_{2}\right)+(p-1)\right|^{2} = p^{2}\left[1+2(p-1)Re\ \tau\left(\overline{\chi}^{2}_{2}\right)+(p-1)^{2}\right]. \end{eqnarray*}$

So we have

$\begin{eqnarray*} && \sum\limits_{m = 1}^{p-1}\sum\limits_{n = 1}^{p-1}\left|K(m,s,\lambda;p)\right|^{2}\left|K(n,t,\lambda;p)\right|^{2}S_{3}(m\overline{n},p)\\ && = \frac{2p^{3}}{\pi^2(p-1)}\sum\limits_{\chi \bmod p \atop \chi(-1) = -1}\left[p+2(p-1)Re\ \tau\left(\overline{\chi}^{2}\right)+(p-1)^{2}\right]\left|L(1,\chi)\right|^2\\ &&\quad-\frac{2p^{3}}{\pi^2(p-1)}\left[p+2(p-1)Re\ \tau\left(\overline{\chi}^{2}_{2}\right)+(p-1)^{2}\right]\left|L(1,\chi_{2})\right|^2\\ &&\quad+\frac{2p^{3}}{\pi^2(p-1)}\left[1+2(p-1)Re\ \tau\left(\overline{\chi}^{2}_{2}\right)+(p-1)^{2}\right]\left|L(1,\chi_{2})\right|^2\\ &&\quad-\frac{4p^{3}}{\pi^2(p-1)}\sum\limits_{\chi \bmod p \atop \chi(-1) = -1}\chi(2)\left[p+2(p-1)Re\ \tau\left(\overline{\chi}^{2}\right)+(p-1)^{2}\right]\left|L(1,\chi)\right|^2\\ &&\quad+\frac{4p^{3}}{\pi^2(p-1)}\left[p+2(p-1)Re\ \tau\left(\overline{\chi}^{2}_{2}\right)+(p-1)^{2}\right]\chi_{2}(2)\left|L(1,\chi_{2})\right|^2\\ &&\quad-\frac{4p^{3}}{\pi^2(p-1)}\left[1+2(p-1)Re\ \tau\left(\overline{\chi}^{2}_{2}\right)+(p-1)^{2}\right]\chi_{2}(2)\left|L(1,\chi_{2})\right|^2\\ && = \frac{p(p-1)(p^{2}-p+1)}{2}+\frac{4p^{3}}{\pi^{2}}\sum\limits_{\chi \bmod p \atop \chi(-1) = -1}Re\ \left(\tau\left(\overline{\chi}^{2}\right)\right)\left|L(1,\chi)\right|^{2}\\ &&\quad-\frac{8p^{3}}{\pi^{2}}\sum\limits_{\chi \bmod p \atop \chi(-1) = -1}\chi(2) Re\ \left(\tau\left(\overline{\chi}^{2}\right)\right)\left|L(1,\chi)\right|^{2}-2p^{2}h_{p}^{2}+4p^{2}h_{p}^{2}\left(\frac{2}{p}\right). \end{eqnarray*}$

Similarly, if $\overline{\lambda\chi} = \chi_{0}, \ \overline{\lambda}\chi\neq\chi_{0}$ , we have

$\begin{eqnarray*} && \sum\limits_{m = 1}^{p-1}\sum\limits_{n = 1}^{p-1}|K(m,s,\lambda;p)|^{2}|K(n,t,\lambda;p)|^{2}S_{3}(m\overline{n},p)\\ && = \frac{p(p-1)(2p^{2}-2p+1)}{2}-\frac{4p^{3}}{\pi^{2}}\sum\limits_{\chi \bmod p \atop \chi(-1) = -1}Re\ \left(\tau(\overline{\chi}^{2})\tau(\overline{\lambda}\chi)\right)|L(1,\chi)|^{2} \\ &&\quad+\frac{8p^{3}}{\pi^{2}}\sum\limits_{\chi \bmod p \atop \chi(-1) = -1}\chi(2) Re\ \left(\tau(\overline{\chi}^{2})\tau(\overline{\lambda}\chi)\right)|L(1,\chi)|^{2}-2p^{3}h_{p}^{2}+4p^{3}h_{p}^{2}\left(\frac{2}{p}\right). \end{eqnarray*}$

Combining the fact that

$\begin{eqnarray*} && \left(\frac{2}{p}\right) = (-1)^{\frac{p^{2}-1}{8}} = \left\{ \begin{array}{ll} 1, & \hbox{if $p\equiv\pm 1 \bmod 8$}; \\ -1, & \hbox{if $p\equiv\pm 3 \bmod 8$}, \end{array} \right. \end{eqnarray*}$

we deduce that if $p\equiv 3 \bmod 8$ , then

$\begin{eqnarray*} && \sum\limits_{m = 1}^{p-1}\sum\limits_{n = 1}^{p-1}\left|K(m,s,\lambda;p)\right|^{2}\left|K(n,t,\lambda;p)\right|^{2}S_{3}(m\overline{n},p)\\ && = \left\{ \begin{array}{ll} \frac{p^{2}(p-1)}{2}-6p^{2}h_{p}^{2}, & \hbox{if $\overline{\lambda\chi}\neq\chi_{0},\ \overline{\lambda}\chi\neq\chi_{0}$};\\ \frac{p(p-1)}{2}-6ph_{p}^{2}, & \hbox{if $\overline{\lambda\chi}\neq\chi_{0},\ \overline{\lambda}\chi = \chi_{0}$}. \end{array} \right. \end{eqnarray*}$

If $\overline{\lambda\chi} = \chi_{0}, \ \overline{\lambda}\chi = \chi_{0}$ , we have

Then

$\begin{eqnarray*} && \sum\limits_{m = 1}^{p-1}\sum\limits_{n = 1}^{p-1}\left|K(m,s,\lambda;p)\right|^{2}\left|K(n,t,\lambda;p)\right|^{2}S_{3}(m\overline{n},p)\\ &&\leq \frac{p(p-1)(p^{2}-p+1)}{2}+\frac{4p^{3}}{\pi^{2}}\left|\sum\limits_{\chi \bmod p \atop \chi(-1) = -1}Re\ \left(\tau\left(\overline{\chi}^{2}\right)\right)\left|L(1,\chi)\right|^{2} \right|\\ &&\quad+\frac{8p^{3}}{\pi^{2}}\left|\sum\limits_{\chi \bmod p \atop \chi(-1) = -1}\chi(2) Re\ \left(\tau\left(\overline{\chi}^{2}\right)\right)\left|L(1,\chi)\right|^{2}\right|+6p^{2}h_{p}^{2}\\ && = p^{\frac{9}{2}}+\frac{1}{2}p^{4}-4p^{\frac{7}{2}}-p^{3}+5p^{\frac{5}{2}}+p^{2}-2p^{\frac{3}{2}}-\frac{1}{2}p+6p^{2}h_{p}^{2}. \end{eqnarray*}$

Similarly, if $\overline{\lambda\chi} = \chi_{0}, \ \overline{\lambda}\chi\neq\chi_{0}$ , we have

$\begin{eqnarray*} && \sum\limits_{m = 1}^{p-1}\sum\limits_{n = 1}^{p-1}|K(m,s,\lambda;p)|^{2}|K(n,t,\lambda;p)|^{2}S_{3}(m\overline{n},p) \leq p^{5}-3p^{4}+3p^{3}-\frac{1}{2}p^{2}-\frac{1}{2}p+6p^{3}h_{p}^{2}. \end{eqnarray*}$

This completes the proof of Theorem 3.

Theorem 4 can be derived by the same method. This completes the proof of our Theorems.

4. Conclusions

In this paper, we obtain some exact computational formulas or upper bounds for hybrid mean value involving Hardy sums and Kloosterman sums (both classical Kloosterman sums and general Kloosterman sums) by applying the properties of Gauss sums and the mean value of Dirichlet $L$ -function. But in some cases, unluckily, it is difficult to get the exact formula. So how to get the exact formula in all cases remains open.

Acknowledgments

The authors want to show their great thanks to the anonymous referee for his/her helpful comments and suggestions.

This work is supported by the National Natural Science Foundation of China (No. 11871317, 11926325, 11926321) and the Fundamental Research Funds for the Central Universities (No. GK201802011).

Conflict of interest

The authors declare that there are no conflicts of interest regarding the publication of this paper.

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沈阳化工大学材料科学与工程学院沈阳 110142

Urban Resilience and Sustainability

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Urban Resilience and Sustainability

At the crossroads of datas: Using artificial intelligence for efficient and resilient planning in African cities

Related Papers:

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1. Introduction and main results

2. Some Lemmas

3. Proof of Theorems

4. Conclusions

Acknowledgments

Conflict of interest

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Urban Resilience and Sustainability

At the crossroads of datas: Using artificial intelligence for efficient and resilient planning in African cities

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1. Introduction and main results

2. Some Lemmas

3. Proof of Theorems

4. Conclusions

Acknowledgments

Conflict of interest

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