common sums

∑j∈J,k∈Kajbk=(∑j∈Jaj)(∑k∈Kbk)\sum_{j\in J,k\in K}a_{j}b_{k}=(\sum_{j\in J}a_j)(\sum_{k\in K}b_k)j∈J,k∈K∑ajbk=(j∈J∑aj)(k∈K∑bk)
mapping

∑j∈J∑k∈K(j)aj,k=∑k∈K′∑j∈J′(k)aj,k\sum_{j\in J}\sum_{k\in K(j)}a_{j,k}=\sum_{k\in K^{'}}\sum_{j\in J^{'}(k)}a_{j,k}j∈J∑k∈K(j)∑aj,k=k∈K′∑j∈J′(k)∑aj,k
[1⩽j⩽n][j⩽k⩽n]=[1⩽j⩽k⩽n]=[1⩽k⩽n][1⩽j⩽k][1\leqslant j\leqslant n][j\leqslant k\leqslant n]=[1\leqslant j\leqslant k\leqslant n]=[1\leqslant k\leqslant n][1\leqslant j\leqslant k][1⩽j⩽n][j⩽k⩽n]=[1⩽j⩽k⩽n]=[1⩽k⩽n][1⩽j⩽k]

∑j=1n∑k=jnaj,k=∑1⩽j⩽k⩽naj,k=∑k=1n∑j=1kaj,k\sum_{j=1}^{n}\sum_{k=j}^{n}a_{j,k}=\sum_{1\leqslant j\leqslant k\leqslant n}a_{j,k}=\sum_{k=1}^{n}\sum_{j=1}^{k}a_{j,k}j=1∑nk=j∑naj,k=1⩽j⩽k⩽n∑aj,k=k=1∑nj=1∑kaj,k

example 1

[a1a1a1a2a1a3...a1ana2a1a2a2a2a3...a2ana3a1a3a2a3a3...a3an⋮⋮⋮⋮ana1ana2ana3...anan]\left[ \begin{array}{ccccc} a_1a_1 & a_1a_2 & a_1a_3 & ... & a_1a_n \\ a_2a_1 & a_2a_2 & a_2a_3 & ... & a_2a_n \\ a_3a_1 & a_3a_2 & a_3a_3 & ... & a_3a_n \\ \vdots & \vdots & \vdots & & \vdots \\ a_na_1 & a_na_2 & a_na_3 & ... & a_na_n \end{array} \right]⎣⎢⎢⎢⎢⎢⎡a1a1a2a1a3a1⋮ana1a1a2a2a2a3a2⋮ana2a1a3a2a3a3a3⋮ana3............a1ana2ana3an⋮anan⎦⎥⎥⎥⎥⎥⎤

How about the sum of right upper triangle
S⊲=∑1⩽j⩽k⩽najakS_{\lhd}=\sum_{1\leqslant j\leqslant k\leqslant n}a_ja_kS⊲=1⩽j⩽k⩽n∑ajak

The question here is how to simplify the formula S⊲S_{\lhd}S⊲ S⊳+S⊲=∑1⩽j,k⩽najak+∑1⩽j=k⩽najak=(∑1⩽j⩽naj)2+∑i⩽k⩽nak2\left. \begin{aligned} S_{\rhd}+S_{\lhd} &=\sum_{1 \leqslant j,k\leqslant n}a_ja_k+\sum_{1\leqslant j=k\leqslant n}a_ja_k\\ &=(\sum_{1\leqslant j\leqslant n}a_j)^2+\sum_{i\leqslant k \leqslant n}a_k^2 \end{aligned} \right.S⊳+S⊲=1⩽j,k⩽n∑ajak+1⩽j=k⩽n∑ajak=(1⩽j⩽n∑aj)2+i⩽k⩽n∑ak2

from
∑j=1n∑k=jnaj,k=∑1⩽j⩽k⩽naj,k=∑k=1n∑j=1kaj,k\sum_{j=1}^{n}\sum_{k=j}^{n}a_{j,k}=\sum_{1\leqslant j\leqslant k\leqslant n}a_{j,k}=\sum_{k=1}^{n}\sum_{j=1}^{k}a_{j,k}∑j=1n∑k=jnaj,k=∑1⩽j⩽k⩽naj,k=∑k=1n∑j=1kaj,k,
we can get S⊳=S⊲S_{\rhd}=S_{\lhd}S⊳=S⊲

at last

S⊲=12[(∑1⩽j⩽naj)2+∑i⩽k⩽nak2]S_{\lhd} = \frac{1}{2}[(\sum_{1\leqslant j\leqslant n}a_j)^2+\sum_{i\leqslant k \leqslant n}a_k^2]S⊲=21[(1⩽j⩽n∑aj)2+i⩽k⩽n∑ak2]

example 2

S=∑1⩽j<k⩽n(ak−aj)(bk−bj)S=\sum_{1\leqslant j <k \leqslant n}(a_k-a_j)(b_k-b_j)S=1⩽j<k⩽n∑(ak−aj)(bk−bj)

The question here is how to simplify the formula of SSS. Let us look
at the fact below,

[1⩽j<k⩽n]+[1⩽k<j⩽k]=[1⩽j,k⩽n]−[1⩽j=k⩽n][1\leqslant j < k \leqslant n]+[1\leqslant k < j \leqslant k]=[1\leqslant j, k \leqslant n]-[1\leqslant j = k \leqslant n][1⩽j<k⩽n]+[1⩽k<j⩽k]=[1⩽j,k⩽n]−[1⩽j=k⩽n]

S=∑1⩽j<k⩽n(ak−aj)(bk−bj)=∑1⩽k<j⩽n(aj−ak)(bj−bk)S=\sum_{1\leqslant j <k \leqslant n}(a_k-a_j)(b_k-b_j)=\sum_{1\leqslant k <j \leqslant n}(a_j-a_k)(b_j-b_k)S=1⩽j<k⩽n∑(ak−aj)(bk−bj)=1⩽k<j⩽n∑(aj−ak)(bj−bk)

2S=∑1⩽j,k⩽n(ak−aj)(bk−bj)−∑1⩽j=k⩽n(ak−aj)(bk−bj)2S=\sum_{1\leqslant j,k\leqslant n}(a_k-a_j)(b_k-b_j)-\sum_{1\leqslant j = k \leqslant n}(a_k-a_j)(b_k-b_j)2S=1⩽j,k⩽n∑(ak−aj)(bk−bj)−1⩽j=k⩽n∑(ak−aj)(bk−bj)

We should aware that the second part of the left equation is zero.

2S=∑1⩽j,k⩽n(ak−aj)(bk−bj)2S=\sum_{1\leqslant j,k\leqslant n}(a_k-a_j)(b_k-b_j)2S=1⩽j,k⩽n∑(ak−aj)(bk−bj)

Let’s expand (ak−aj)(bk−bj)(a_k-a_j)(b_k-b_j)(ak−aj)(bk−bj) to akbk+ajbj−akbj−ajbka_kb_k+a_jb_j-a_kb_j-a_jb_kakbk+ajbj−akbj−ajbk.
It’s obvious that

∑1⩽j,k⩽nakbk=∑1⩽j,k⩽najbj,∑1⩽j,k⩽nakaj=∑1⩽j,k⩽najak\sum_{1\leqslant j,k\leqslant n}a_kb_k=\sum_{1\leqslant j,k\leqslant n}a_jb_j,\sum_{1\leqslant j,k\leqslant n}a_ka_j=\sum_{1\leqslant j,k\leqslant n}a_ja_k1⩽j,k⩽n∑akbk=1⩽j,k⩽n∑ajbj,1⩽j,k⩽n∑akaj=1⩽j,k⩽n∑ajak

We should get the formula below

2S=2∑1⩽j,k⩽nakbk−2∑1⩽j,k⩽najbk2S=2\sum_{1\leqslant j,k\leqslant n}a_kb_k-2\sum_{1\leqslant j,k\leqslant n}a_jb_k2S=21⩽j,k⩽n∑akbk−21⩽j,k⩽n∑ajbk

Let’s simplify the right part of the upper equation
- ∑1⩽j,k⩽nakbk=n∑i⩽k⩽nakbk\sum_{1\leqslant j,k\leqslant n}a_kb_k = n \sum_{i\leqslant k\leqslant n}a_kb_k1⩽j,k⩽n∑akbk=ni⩽k⩽n∑akbk
- ∑1⩽j,k⩽najbk=(∑1⩽k⩽nak)(∑1⩽k⩽nbk)\sum_{1\leqslant j,k\leqslant n}a_jb_k = (\sum_{1\leqslant k\leqslant n}a_k)(\sum_{1\leqslant k\leqslant n}b_k)1⩽j,k⩽n∑ajbk=(1⩽k⩽n∑ak)(1⩽k⩽n∑bk)
Finally,

2S=2n∑i⩽k⩽nakbk−2(∑1⩽k⩽nak)(∑1⩽k⩽nbk)2S=2n \sum_{i\leqslant k\leqslant n}a_kb_k-2(\sum_{1\leqslant k\leqslant n}a_k)(\sum_{1\leqslant k\leqslant n}b_k)2S=2ni⩽k⩽n∑akbk−2(1⩽k⩽n∑ak)(1⩽k⩽n∑bk)

Rewrite the upper equation

(∑1⩽k⩽nak)(∑1⩽k⩽nbk)=n∑k=1nakbk−∑1⩽j<k⩽n(ak−aj)(bk−bj)(\sum_{1\leqslant k\leqslant n}a_k)(\sum_{1\leqslant k\leqslant n}b_k) = n \sum_{k=1}^{n}a_kb_k-\sum_{1\leqslant j<k\leqslant n}(a_k-a_j)(b_k-b_j)(1⩽k⩽n∑ak)(1⩽k⩽n∑bk)=nk=1∑nakbk−1⩽j<k⩽n∑(ak−aj)(bk−bj)

Chebyshev’s monotonic inequlities

We can go back to equation

(∑1⩽k⩽nak)(∑1⩽k⩽nbk)=n∑k=1nakbk−∑1⩽j<k⩽n(ak−aj)(bk−bj)(\sum_{1\leqslant k\leqslant n}a_k)(\sum_{1\leqslant k\leqslant n}b_k) = n \sum_{k=1}^{n}a_kb_k-\sum_{1\leqslant j<k\leqslant n}(a_k-a_j)(b_k-b_j)(1⩽k⩽n∑ak)(1⩽k⩽n∑bk)=nk=1∑nakbk−1⩽j<k⩽n∑(ak−aj)(bk−bj)
If a1⩽a2⩽...⩽ana_1\leqslant a_2\leqslant ...\leqslant a_na1⩽a2⩽...⩽an and
b1⩽b2⩽...⩽bnb_1\leqslant b_2\leqslant ... \leqslant b_nb1⩽b2⩽...⩽bn, we can get

(∑1⩽k⩽nak)(∑1⩽k⩽nbk)⩽n∑k=1nakbk(\sum_{1\leqslant k\leqslant n}a_k)(\sum_{1\leqslant k\leqslant n}b_k)\leqslant n \sum_{k=1}^{n}a_kb_k(1⩽k⩽n∑ak)(1⩽k⩽n∑bk)⩽nk=1∑nakbk

If a1⩽a2⩽...⩽ana_1\leqslant a_2\leqslant ...\leqslant a_na1⩽a2⩽...⩽an and
b1⩾b2⩾...⩾bnb_1\geqslant b_2\geqslant ... \geqslant b_nb1⩾b2⩾...⩾bn, we can get

(∑1⩽k⩽nak)(∑1⩽k⩽nbk)⩽n∑k=1nakbk(\sum_{1\leqslant k\leqslant n}a_k)(\sum_{1\leqslant k\leqslant n}b_k)\leqslant n \sum_{k=1}^{n}a_kb_k(1⩽k⩽n∑ak)(1⩽k⩽n∑bk)⩽nk=1∑nakbk

The original Chebyshev’s monotonic inequalities

[(∫abf(x)dx][∫abg(x)dx]⩽(b−a)∫abf(x)g(x)dx[(\int_{a}^{b}{f(x)dx}][\int_{a}^{b}{g(x)dx}]\leqslant (b-a)\int_{a}^{b}{f(x)g(x)dx}[(∫abf(x)dx][∫abg(x)dx]⩽(b−a)∫abf(x)g(x)dx

the condition is that f(x)f(x)f(x) and g(x)g(x)g(x) are monotonic nodecreasing
functions.

Actually

If a1⩽a2⩽...⩽ana_1\leqslant a_2\leqslant ...\leqslant a_na1⩽a2⩽...⩽an, the
∑k=1nakbk\sum_{k=1}^{n}a_kb_k∑k=1nakbk will get the largest value when
b1⩽b2⩽..⩽bnb_1\leqslant b_2\leqslant ..\leqslant b_nb1⩽b2⩽..⩽bn, and the
∑k=1nakbk\sum_{k=1}^{n}a_kb_k∑k=1nakbk will get the smallest value when
b1⩾b2⩾...⩾bnb_1\geqslant b_2\geqslant ...\geqslant b_nb1⩾b2⩾...⩾bn

How to proof this simple fact: (Hint1: Draw pictures, Hint2:
Recursive method)

Mapping

f:J→Kf:J \rightarrow Kf:J→K

∑j∈Jaf(j)=∑k∈Kak#f−1(k),f−1(k)={j∣f(j)=k}\sum_{j\in J}a_{f(j)}=\sum_{k\in K}a_k \#f^{-1}(k), f^{-1}(k)=\{j|f(j)=k\}j∈J∑af(j)=k∈K∑ak#f−1(k),f−1(k)={j∣f(j)=k}

#\## is how many j maps to the same k

example:

Sn=∑1⩽j<k⩽n1k−jS_n=\sum_{1\leqslant j<k\leqslant n}\frac{1}{k-j}Sn=1⩽j<k⩽n∑k−j1

method 1 Let’s accumulate jjj first.

Sn=∑1⩽k⩽n∑1⩽j<k1k−j=∑1⩽k⩽n∑1⩽k−j<k1j=∑1⩽k⩽n∑0<j⩽k−11j=∑1⩽k⩽nHk−1=∑0⩽k<nHk\left. \begin{aligned} S_n&=\sum_{1\leqslant k\leqslant n}\sum_{1\leqslant j< k} \frac{1}{k-j}\\ &=\sum_{1\leqslant k\leqslant n} \sum_{1\leqslant k-j<k}\frac{1}{j}\\ &=\sum_{1\leqslant k\leqslant n} \sum_{0<j\leqslant k-1}\frac{1}{j}\\ &=\sum_{1\leqslant k\leqslant n} H_{k-1}\\ &=\sum_{0\leqslant k<n}H_k \end{aligned} \right.Sn=1⩽k⩽n∑1⩽j<k∑k−j1=1⩽k⩽n∑1⩽k−j<k∑j1=1⩽k⩽n∑0<j⩽k−1∑j1=1⩽k⩽n∑Hk−1=0⩽k<n∑Hk

We cannot get closed form at last.
method 2

Let’s accumulate k first.

Sn=∑0⩽j<nHjS_n=\sum_{0\leqslant j<n}H_jSn=∑0⩽j<nHj

It is the same as method 1
method 3

Can we map k to k-j?

Sn=∑1⩽j<k⩽n1k−j=∑1⩽j<k+j⩽n1k=∑1⩽k⩽n∑1⩽j⩽n−k1k=∑1⩽k⩽nn−kk=∑1⩽k⩽nnk−∑1⩽k⩽n=n∑1⩽k⩽n1k−n=nHn−n\left. \begin{aligned} S_n&=\sum_{1\leqslant j<k\leqslant n}\frac{1}{k-j}\\ &=\sum_{1\leqslant j<k+j\leqslant n}\frac{1}{k}\\ &=\sum_{1\leqslant k\leqslant n}\sum_{1\leqslant j\leqslant n-k}\frac{1}{k}\\ &=\sum_{1\leqslant k\leqslant n}\frac{n-k}{k}\\ &=\sum_{1\leqslant k\leqslant n}\frac{n}{k}-\sum_{1\leqslant k\leqslant n}\\ &=n \sum_{1\leqslant k\leqslant n}\frac{1}{k}-n=nH_n-n \end{aligned} \right.Sn=1⩽j<k⩽n∑k−j1=1⩽j<k+j⩽n∑k1=1⩽k⩽n∑1⩽j⩽n−k∑k1=1⩽k⩽n∑kn−k=1⩽k⩽n∑kn−1⩽k⩽n∑=n1⩽k⩽n∑k1−n=nHn−n

Three methods have three kinds of direction of sum. (upper and
down),(left and right) and (diagonal). (k=1k=2k=3k=4j=1111213j=21112j=311j=4)\left( \begin{array}{ccccc} & k=1 & k=2 & k=3 & k=4 \\ j=1 & & \frac{1}{1} & \frac{1}{2} & \frac{1}{3} \\ j=2 & & & \frac{1}{1} & \frac{1}{2} \\ j=3 & & & & \frac{1}{1} \\ j=4 & & & & \end{array} \right)⎝⎜⎜⎜⎜⎛j=1j=2j=3j=4k=1k=211k=32111k=4312111⎠⎟⎟⎟⎟⎞

具体数学--(各种基本和式)相关推荐

数学知识：和式的处理及二项式系数的运用
数学知识:和式的处理及二项式系数的运用〇这段时间看了一部分<具体数学>上的内容,正如第一章中所说,这本书的主要目的是: 说明不具备超人洞察力的人如何求解问题也就是说,这本书主要讲述的 ...
【做题】SRM701 Div1 Hard - FibonacciStringSum——数学和式＆矩阵快速幂
原文链接 https://www.cnblogs.com/cly-none/p/SRM701Div1C.html 题意:定义"Fibonacci string"为没有连续1的01串 ...
数学基础：和式极限（可爱因子理解）连续，极限定义：导数：微积分：推荐数学电影
目录和式极限(可爱因子理解) 连续,极限定义: 导数: 微积分:
【学习笔记】和式（《具体数学》第二章）
干货！神经网络原来是这样和数学挂钩的
近几年,有几个被媒体大肆报道的事件,如下表所示. 如上所示,深度学习作为人工智能的一种具有代表性的实现方法,取得了很大的成功.那么,深度学习究竟是什么技术呢?深度学习里的"学习"是 ...
干货！神经网络原来是这样和数学挂钩的 // 深度学习的数学
https://www.toutiao.com/a6688175496821211662/ 近几年,有几个被媒体大肆报道的事件,如下表所示. 如上所示,深度学习作为人工智能的一种具有代表性的实现方法, ...
神经网络原来是这样和数学挂钩的
来源:遇见数学近几年,有几个被媒体大肆报道的事件,如下表所示. 如上所示,深度学习作为人工智能的一种具有代表性的实现方法,取得了很大的成功.那么,深度学习究竟是什么技术呢?深度学习里的"学 ...
python求分段函数值_高中数学知识点整理（2）——函数概念及基本初等函数篇（上）...
大家好!我是高考数学易老师,今天是我来知乎的第二天,今天更新函数概念及基本初等函数知识点.如果有任何关于高中数学知识点,可随时询问呢. 函数 1. 函数与映射(1) 函数的概念设是两个非空的数集, ...
常见数学符号：等号、不等号、算术运算符号、几何符号、三角函数、指数、对数、微分、积分符号、集合符号、逻辑符号
一般数学符号 1. 有序符号符号意义 a1, a2, - , an 数列:序列 2. 等号和不等号符号意义符号意义符号意义＝等于 -.∝ 正比于 << ...
像个字段相减绝对值_【高考数学】33个知识点+66个易混点大整合
乐学数韵(ID/抖音:Vlxsy8 视频号/B站:乐学数韵) 教研.解题.资源 Q群: 314559613 ,1078982440 (2群) 相关链接: 数学干 ...

具体数学--(各种基本和式)

common sums

Chebyshev’s monotonic inequlities

Actually

Mapping

具体数学--(各种基本和式)相关推荐

最新文章

热门文章