Here, the operations of combinatorics and discrete optimization as well as group operations are mainly gathered.

Combinatorics

• An Essay

  • This is my Essay during the 2025 Spring Combinatorics course in Nankai.

Discrete Optimization

• An Essay

  • This is my Essay during the 2025 Spring Discrete Optimization course in Nankai.

• A Group Task

  • This is a Group Task during the 2025 Spring Discrete Optimization course in Nankai.

This semester I also chose the course “C Programming Language Design”. This course requires five submissions of homework. The first three assignments were quite simple, but the last two were a bit more interesting.

Homework 4

1. 了解csv格式文件。
2. 通过excel软件，创建两个表格，内容是M*N, N*Q 浮点数矩阵，数值随意。保存为a.xlsx,b.xlsx文件。其中 M，N，Q是自然数，均不小于4。
3. 将上述excel文件另存为 a.csv,b.csv文件.
4. 编写程序，通过命令行方式，读取这两个csv数据文件，并实现矩阵乘法，将原矩阵数据和矩阵乘法结果数据输出到屏幕。
5. 程序中矩阵维数，不得是定值，需通过读取csv文件获得。
6. 源程序打包zip，运行结果截图。

我的C语言程序(用 Dev C++ 写的)：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

double** readcsv(const char* filename, int* rows, int* cols)
{
    FILE* file = fopen(filename, "r");
    if (file == NULL)
	{
        perror("Error opening file");
        return NULL;
    }

    char line[1024];
    int row = 0;
    int col = 0;

    while (fgets(line, sizeof(line), file))
	{
        row++;
        col = 0;
        char* token = strtok(line, ",");
        while (token!= NULL)
		{
            col++;
            token = strtok(NULL, ",");
        }
    }

    fseek(file, 0, SEEK_SET);

    double** matrix = (double**)malloc(row * sizeof(double*));
    for (int i = 0; i < row; i++)
	{
        matrix[i] = (double*)malloc(col * sizeof(double));
    }

    row = 0;
    while (fgets(line, sizeof(line), file))
	{
        col = 0;
        char* token = strtok(line, ",");
        while (token!= NULL)
		{
            matrix[row][col] = atof(token);
            col++;
            token = strtok(NULL, ",");
        }
        row++;
    }

    fclose(file);

    *rows = row;
    *cols = col;
    return matrix;
}

double** plusMat(double** matrix1, int rows1, int cols1, double** matrix2, int rows2, int cols2) {
    if (cols1!= rows2)
	{
        printf("Error: Incompatible matrix dimensions for multiplication.\n");
        return NULL;
    }

    double** result = (double**)malloc(rows1 * sizeof(double*));
    for (int i = 0; i < rows1; i++)
	{
        result[i] = (double*)malloc(cols2 * sizeof(double));
        for (int j = 0; j < cols2; j++)
		{
            result[i][j] = 0;
            for (int k = 0; k < cols1; k++)
			{
                result[i][j] += matrix1[i][k] * matrix2[k][j];
            }
        }
    }

    return result;
}

void freeMat(double** matrix, int rows)
{
    for (int i = 0; i < rows; i++)
	{
        free(matrix[i]);
    }
    free(matrix);
}

void printMat(double** matrix, int rows, int cols)
{
    for (int i = 0; i < rows; i++)
	{
        for (int j = 0; j < cols; j++)
		{
            printf("%.10f ", matrix[i][j]);
        }
        printf("\n");
    }
}

int main(int argc, char* argv[])
{
    int rows1, cols1;
    //char *filename1="A.csv";
    double** matrix1 = readcsv("a.csv", &rows1, &cols1);
    if (matrix1 == NULL)
	{
        return 1;
    }

    int rows2, cols2;
    //char *filename2="B.csv";
    double** matrix2 = readcsv("b.csv", &rows2, &cols2);
    if (matrix2 == NULL)
	{
        freeMat(matrix1, rows1);
        return 1;
    }

    double** result = plusMat(matrix1, rows1, cols1, matrix2, rows2, cols2);
    if (result == NULL)
	{
        freeMat(matrix1, rows1);
        freeMat(matrix2, rows2);
        return 1;
    }

    printf("Matrix A:\n");
    printMat(matrix1, rows1, cols1);

    printf("\nMatrix B:\n");
    printMat(matrix2, rows2, cols2);

    printf("\nA * B:\n");
    printMat(result, rows1, cols2);

    freeMat(matrix1, rows1);
    freeMat(matrix2, rows2);
    freeMat(result, rows1);

    return 0;
}

Homework 5

1. 了解json文件格式
2. 使用C编写程序解析test.json
3. 打印输出json内容（按照字段输出）
4. 可使用各类开源库cJSON，Morn，Melon等，不限制

任务中的 test.json

我采用的是 cJSON 库来读取 json 文件。

其中 cJSON 库的下载来自 Github 中的资源，在进行编写我们的程序之前(我是用 VisualStudio 来编写的)，需要将下载的 cJSON 库中的 cJSON.c 和 cJSON.h 添加到我们建立的项目中的源代码里，同时在编写程序过程中我也遇到了 cJSON 库无法调用的情形，这需要更改该项目的“包含目录”(在 Visual Studio 中更改)。

首先用 fopen 函数打开文件 test.json ，读取文件内容到一个字符串 file 中，然后使用 cJSON_Parse 函数解析这个字符串得到一个 cJSON 对象。接着，程序调用 print_json 函数来打印这个 cJSON 对象的内容。 print_json 函数递归地遍历cJSON对象的所有子节点，并打印它们的类型和值。最后，程序释放 cJSON 对象所占用的内存并退出。

值得注意的是，如果只是按照上述程序读入最终输出的 json 文件中的内容中的英文和数字均输出正常，但是 json 文件中的中文内容输出结果为乱码。我一直不知道原因在哪，最终的解决做法是将 json 文件中的文件编码在 Visual Studio 中改成 UTF-8(带签名) 这样的格式才行，感觉可能是由于编码格式不同所导致的最终中文显示的错乱。

我的C语言程序(用 Visual Studio 写的)：

#include <iostream>

extern "C" {
#include <string.h>
#include <stdio.h>
#include "cJSON.h"
}

void print_json(cJSON* json) {
    if (json == NULL) return;

    if (json->type == cJSON_Object) {
        cJSON* child = json->child;
        while (child) {
            printf("%s: ", child->string);
            print_json(child);
            child = child->next;
        }
    }
    else if (json->type == cJSON_Array) {
        int i;
        for (i = 0; i < cJSON_GetArraySize(json); i++) {
            printf("[%d]: ", i);
            print_json(cJSON_GetArrayItem(json, i));
        }
    }
    else if (json->type == cJSON_String) {
        printf("%s", json->valuestring);
    }
    else if (json->type == cJSON_Number) {
        printf("%d", json->valueint);
    }
    else {
        printf("Unknown type");
    }
    printf("\n");
}

int main() {
    FILE* file = fopen("test.json", "r");
    if (file == NULL) {
        printf("Failed to open file\n");
        return 1;
    }

    fseek(file, 0, SEEK_END);
    long file_size = ftell(file);
    fseek(file, 0, SEEK_SET);

    char* json_str = (char*)malloc(file_size + 1);
    fread(json_str, 1, file_size, file);
    fclose(file);

    json_str[file_size] = '\0';

    cJSON* json = cJSON_Parse(json_str);
    free(json_str);

    if (json == NULL) {
        printf("Failed to parse JSON\n");
        return 1;
    }

    print_json(json);

    cJSON_Delete(json);
    return 0;
}

Remark

If you have any questions or need further assistance, feel free to ask! Here is my email: nkusherr1 at gmail.com

During the process of writing my complex variable paper, since the main content I was dealing with was about the residue theorem, it was inevitable to use contour integrals. At first, I just thought it would be fine to verbally describe what the contour looked like. However, considering that the final exam for Complex Variable II was a paper, I thought it would be better to be more careful. Therefore, I taught myself to draw with LaTeX.

Latex绘图主要是用tikz包

在开头需使用

\usepage{tikz}

Example 1

在计算欧拉积分时，我们需要考虑如下围道

围道示意图如下：

具体的 Latex 代码如下：

\begin{center}
	\begin{tikzpicture}[>={Stealth[length=0.3cm]},
		decoration={markings,
		mark= at position .1 with {\arrow{<}},
		mark= at position .45 with {\arrow{<}},
		mark= at position .83 with {\arrow{<}},
		}
	]
		\def\angle{40}
		\def\bigradius{5}
		\def\littleradius{1}
		\filldraw[postaction = {decorate}, thick ,fill=gray!40]
			(\angle:\littleradius) node[right]{$\epsilon$}
			-- (\angle:\bigradius) node[right]{$R$}
			arc(\angle:0:\bigradius) node[below]{$R$}
			-- (\littleradius,0) node[below]{$\epsilon$}
			arc (0:\angle:\littleradius) -- cycle;

		\draw[dashed,thick] (0,0) node[xshift=15pt, yshift=6pt]{$\theta$} -- (\angle:\littleradius);
		\node at(3,-0.3){$\Gamma_1$};
		\node at(5,2){$\Gamma_2$};
		\node at(2.5,2.5){$\Gamma_3$};
		\node at(1.5,0.5){$\Gamma_4$};
		\draw[-Latex] (-0.1*\bigradius,0) -- (1.2*\bigradius,0) node[below]{$x$} ;
		\draw[-Latex] (0,-0.1*\bigradius) -- (0,4) node[right]{$iy$};
	\end{tikzpicture}
\end{center}

Example 2

在计算高斯积分时，我们可以考虑一种不同于课本方式的围道计算，如下图所示：

具体的 Latex 代码如下：

\begin{center}
	\begin{tikzpicture}
		\def\gap{0.2}
		\def\bigradius{3}
		\def\littleradius{0.5}
		% Axes

		\draw[->](-1,-3) -- (5,3) -- (3,3) -- (-3,-3) -- (-1,-3) --cycle;
		\fill[gray!40](-1,-3) -- (5,3) -- (3,3) -- (-3,-3) -- (-1,-3) --cycle;


		\draw [thick,->] (-4, 0) -- (6,0);
		\draw [thick,->] (0, -4) -- (0,4);

		\node at (0.5,4){$iy$};
		\node at (6.2,0.3){$x$};
		\node at (1.3,-1.3){$\Gamma_1$};
		\node at (4,3.3){$\Gamma_2$};
		\node at (1.1,1.6){$\Gamma_3$};
		\node at (-2,-3.3){$\Gamma_4$};
		\draw(1,0) node[shape=circle,draw,inner sep=1pt]{};
		\node at (1.2,0.3){$\frac{1}{2}$};
		\draw(3,0) node[shape=circle,draw,inner sep=1pt]{};
		\node at (3.2,-0.3){$R$};
		\draw(5,0) node[shape=circle,draw,inner sep=1pt]{};
		\node at (5.2,-0.3){$R+1$};
		\draw(0,3) node[shape=circle,draw,inner sep=1pt]{};
		\node at (0.3,3.3){$iR$};
		\draw[dashed, thick] (0,3)--(3,3)--(3,0);
		\draw[dashed, thick] (5,0)--(5,3);
		%\draw[fill=gray,domain=-3:3]plot({\x-2},\x);

	\end{tikzpicture}
\end{center}

Example 3

在计算高斯积分时，我们发现可以对高斯积分进行推广，进而得到 $e^{-a\sqrt{\pi}}$ 的拉普拉斯逆变换，在计算中，我们需要考虑如下图所示的积分围道：

具体的 Latex 代码如下：

\begin{center}
	\begin{tikzpicture}[>={Stealth[length=0.3cm]},
        decoration={markings,
        mark= at position .15 with {\arrow{>}},
        mark= at position .3 with {\arrow{>}},
        %mark= at position .42 with {\arrow{>}},
        mark= at position .49 with {\arrow{>}},
        mark= at position .63 with {\arrow{>}},
        mark= at position 0.9 with {\arrow{>}},
        }]
        \def\radius{3.4}
        \def\bigradius{3.4}
        \def\littleradius{0.5}
        \def\angle{60}
        \def\gap{0.3}
    \draw[dashed,thick]
        let
            \n0 = {cos(\angle)},
        in
            (\radius*\n0,-\radius*1.2) node[right]{$c - i\infty$}-- (\radius*\n0,\radius*1.2)node[right]{$c+i\infty$};
    % contour
    \filldraw[postaction = {decorate}, thick ,fill=gray!40]
    let
    \n1 = {asin(\gap/2/\bigradius)},
    \n2 = {asin(\gap/2/\littleradius)}
    in
    (0.5*\bigradius,\bigradius)node[right]{$B$}--(0,\bigradius) arc(90:180-\n1:\bigradius) --(-\n2:-\littleradius) arc (-\n2:-(360-\n2):-\littleradius) --(\n1:-\bigradius) arc(180+\n1:270:\bigradius)--(0.5*\bigradius,-\bigradius)node[right]{$A$}--(0.5*\bigradius,\bigradius);

    \draw (0,0) -- (0.5*\bigradius,\bigradius) node[xshift=-20pt,yshift=-20pt]{$R$};
    \draw (0,0) -- (2*\angle:\littleradius) node[xshift=0pt,yshift=-8pt]{$\delta$};
    \node[xshift=-10pt,yshift=10pt] at (-\bigradius,0){$C_R$};
    \node[xshift=-10pt,yshift=10pt] at (-\littleradius,0){$C_{\delta}$};

	\draw[->](1.5,3.6)--(0.2,3.6);
	\node at (0.8,3.9){$L_1$};
	\draw[->](0.2,-3.6)--(1.5,-3.6);
	\node at (0.8,-3.9){$L_2$};
	\node at(-2,2.1){$\Gamma_1$};
	\node at(-2,-2.1){$\Gamma_2$};
	\node at(-2,0.5){$\lambda_1$};
	\node at(-2,-0.5){$\lambda_2$};
	\node at(2,0.5){$L$};
    % pole
    %\fill (0,0)  circle (3pt) ;
    %\draw[-Latex] (0.6,-0.8) node[right]{pole} --(0.15,-0.15);

    % axis
    \draw[-Latex] (-1.8*\radius,0) -- (1.3*\radius,0) node[below]{$x$} ;
    \draw[-Latex] (0,-1.2*\radius) -- (0,1.2*\radius) node[right]{$iy$};
	\end{tikzpicture}
\end{center}

Example 3

在计算狄利克雷积分时，我们需要考虑一种围道计算，如下图所示：

具体的 Latex 代码如下：

\begin{center}
	  \begin{tikzpicture}[>={Stealth[length=0.3cm,bend]},
				decoration={markings,
				mark= at position .1 with {\arrow{>}},
				mark= at position .32 with {\arrow{>}},
				mark= at position .55 with {\arrow{>}},
				% mark= at position 0.86 with {\arrow{>}},
				}
			]
				\def\gap{0.3}
				\def\bigradius{3}
				\def\littleradius{0.5}
			\filldraw[postaction = {decorate}, thick ,fill=gray!40]
			let
				\n1 = {asin(\gap/2/\bigradius)},
				\n2 = {asin(\gap/2/\littleradius)}
			in
				(0+\n1:\bigradius) node[above right]{$R$}  arc (0+\n1:360-\n1:\bigradius) node[below left]{$C_{R}$} -- (0-\n2:\littleradius) arc (360-\n2:0+\n2:\littleradius) node[above right]{$\delta$} -- cycle;
				\draw[thick,->] (300:\littleradius) arc (300:130:\littleradius) node[above]{$C_{\delta}$};
			%\fill (-1.2,0)  circle (3pt) ;
			%\draw[->] (-1.2,-1) node[right]{pole} --(-1.2,-0.15);  % <--
			\draw[-Latex] (-1.5*\bigradius,0) -- (1.5*\bigradius,0) node[below]{$x$} ;
			\draw[-Latex] (0,-1.2*\bigradius) -- (0,1.2*\bigradius) node[right]{$iy$};
	  \end{tikzpicture}
\end{center}

Remark

If you have any questions or need further assistance, feel free to ask! Here is my email: nkusherr1 at gmail.com

This is the essay of the Class “Complex Analysis II”.

The topic of the paper I have written is about three classic integral propositions and their generalizations.

Abstract

The Residue Theorem ingeniously combines Laurent series and integral calculation through Cauchy’s Integral Theorem, providing a brand-new approach for us to calculate complex generalized integrals and even improper integrals, significantly reducing the difficulty of solving complex integrals solely through mathematical analysis methods. This article will study three classic integrals - the Euler integral, the Gaussian integral, and the Dirichlet integral, as well as their extended forms, from the perspective of the Residue Theorem, based on my learning of the Residue Theorem this semester and the propositions I have accumulated by reading related literature.

Recalling

我们先来回顾一下留数定理：

关于留数的定义：

设 $z_0$ 是 $f(z_0)$ 的孤立奇点，于是 $f(z)$ 在 $ V(z_0,R)-{z_0} $ 中有Laurent展开

此时

称为$f$在$z_0$处的留数，记为$Res(f,z_0)$.

关于留数定理的内容：

设$\Gamma$为一条正向简单闭路径，内部为$D$，${z_k}{1\leq k\leq n}$是$D$中有限个点，今若$f$在 $\bar{D}-\left{z_k\right}{1\leq k\leq n}$ 上解析，则

关于在极点处留数的计算方法

设$a$是$f$的$n$阶极点，$n\geq1$.并设在$a$附近我们有$f(z)=\frac{g(z)}{(z-a)^n}$，其中$g(z)$在$a$解析且$g(a)\neq0$.则

接着我们来研究以下几个经典的积分命题

Euler Integral

欧拉积分这个例子来自南开数分教材第$19$章$B$组第$11$题.

对于$\forall \lambda>0,x>0,\alpha\in(-\frac{\pi}{2},\frac{\pi}{2})$，有 \(\left\{ \begin{aligned} \int_{0}^{+\infty}t^{x-1}e^{-\lambda t\cos\alpha}\cos(\lambda t\sin\alpha)dt=\frac{\Gamma(x)}{\lambda^x}\cos\alpha x\\ \int_{0}^{+\infty}t^{x-1}e^{-\lambda t\cos\alpha}\sin(\lambda t\sin\alpha)dt=\frac{\Gamma(x)}{\lambda^x}\sin\alpha x \end{aligned} \right.\)

Gauss Integral

高斯积分(概率积分)

我们也可以对高斯积分进行进一步的推广：

对于$\forall a,b\in\mathbb{C}$且$Re(a)>0$

在研究高斯积分的过程中，我们也可以发现菲涅尔积分公式：

同时我们也可以发现一些Laplace逆变换.

Dirichlet Integral

狄利克雷积分这个例子来自南开复变教材第$4$章习题$33-(v),(vi)$.

更进一步，我们有其推广形式：

对$\forall m\in\mathbb{N}^{*}$，

不过说实话这个狄利克雷积分挺搞的，复变函数期末考考，数学分析III期末考也考，结果我都算错了，太难绷了.

Epilogue

留数定理堪称复变函数领域的一座巍峨丰碑。它犹如一条精妙的纽带，将复变函数在孤立奇点处看似微观局部的留数特性，与宏观层面的闭曲线积分紧密相连，展现出一种高屋建瓴的理论架构。在理论深度上，它是柯西积分定理等经典理论的卓越升华，极大地拓展了复变函数积分理论的边界，为深入探究函数在奇点附近的行为以及复杂区域上的积分开辟了崭新通途。其影响力更是跨越复变函数的范畴，在调和分析、数论等数学分支中若隐若现地编织起联系的网络，促进了数学学科内部的深度交融。

正如高斯所言：“数学中的一些美丽定理具有这样的特性：它们极易从事实中归纳出来，但证明却隐藏的极深。” 留数定理便是这样一个美丽且深刻的定理，它建立了函数在孤立奇点处的留数与闭曲线积分之间的联系，这种联系看似简洁明了，但背后的证明和理论基础却蕴含着深刻的数学思想。

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For more information about this essay, please refer to the following sources(Chinese version):

Click Here to get my essay.

Remark

If you have any questions or need further assistance, feel free to ask! Here is my email: nkusherr1 at gmail.com

This is the third class of Complex Analysis, hosted by Shengxuan Tang. In this class, we will focus on the following topics:

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For more information about the 3rd Class of Complex Analysis, please refer to the following sources(Chinese version):

Complex Analysis (begin from page 17 to page ?)

If you have any questions or need further assistance, feel free to ask! Here is my email: nkusherr1 at gmail.com

This is an analysis problem from 2022~2023 boling class mathematical-analysis-III mid-term examination paper. This post will show you the problem and its connecting problem from Mathematical-analysis-III of Nankai University Exercise 14(B)-13.

Problem from Exercise Paper

Problem

设 \(\alpha\gt0\)，讨论级数

的收敛性(若收敛，精确到条件收敛和绝对收敛)

Problem from Textbook

For the more information about the following problem, you can refer to the book Mathematical-analysis-III of Nankai University Exercise 14(B)-13.

Problem

设$\alpha\in(0,1)$，$p>max\lbrace\alpha,1-\alpha\rbrace$，则

收敛.

Solution for PDF

Click Here and get the solution about these two similar series problems.

Remark

If you have any questions or need further assistance, feel free to ask! Here is my email: nkusherr1 at gmail.com

This semester, I participated in the Functional Analysis Seminar set up by Mr. Liu Rui. The main content was to discuss some cutting-edge topics related to Functional Analysis.

And the fist part of this seminar is to study a paper called Expressivity of Spiking Neural Networks held by one of his graduate student.

the Paper

Abstract

The synergy between spiking neural networks and neuromorphic hardware holds promise for the development of energy-efficient AI applications. Inspired by this potential, we revisit the foundational aspects to study the capabilities of spiking neural networks where information is encoded in the firing time of neurons. Under the Spike Response Model as a mathematical model of a spiking neuron with a linear response function, we compare the expressive power of artificial and spiking neural networks, where we initially show that they realize piecewise linear mappings. In contrast to ReLU networks, we prove that spiking neural networks can realize both continuous and discontinuous functions. Moreover, we provide complexity bounds on the size of spiking neural networks to emulate multi-layer (ReLU) neural networks. Restricting to the continuous setting, we also establish complexity bounds in the reverse direction for one-layer spiking neural networks.

Keywords: Expressivity, Approximation Theory, Spiking Neural Networks, Deep (ReLU) Neural Networks, Temporal Coding, Linear Regions

Click Here to read the original paper.

the Notes

As you know, I’ve always been in the habit of taking notes.

But the size of my note is too large. So I I uploaded my notes to my Baidu Netdisk.

CLick Here to download my note.

Remark

About this Seminar, I have something to share.

At first, there were more than ten participants in this course, but I don’t know if it was because the content was too abstract or for some other reason. After three classes, only the graduate students and I were left (LOL).

But Neural Network Learning is really interesting

By the way, if you have any questions or need further assistance, feel free to ask! Here is my email: nkusherr1 at gmail.com

These are two Integral Problems in our textbook. It’s a little difficult but very interesting. Let’s see.

Problems

Exercise 1

For the more information about the following problem, you can refer to the book Mathematical-analysis-III of Nankai University Exercise 15(B)-12(4).

研究下列广义积分的敛散性：

Exercise 2

For the more information about the following problem, you can refer to the book Mathematical-analysis-III of Nankai University Exercise 15(B)-15.

设(\mathbf{A}=(a_{ij}))是(n)阶实对称正定矩阵，$b_1,…,b_n$和$c$均是实数.求

其中$\exp(u)$表示$e^u$.

Solution for PDF

Click Here and get the solution about these two Integral Problems.

Remark

If you have any questions or need further assistance, feel free to ask! Here is my email: nkusherr1 at gmail.com

This is the Midterm Exam of Mathematical Analysis III(Boling Class) at Nankai University in the autumn semester of 2024-2025.

On the whole, the difficulty level of this test is between that of last year and the year before.

Problems and Solutions

Exercise 1

研究下列级数的收敛性

Solution 1-(1)-1

已知$\displaystyle\sum_{n=1}^{+\infty}\frac{\sin n}{\sqrt{n}}$ 收敛，考虑：

又

而$\displaystyle\sum_{n=1}^{+\infty}\frac{1}{n^2\sqrt{n}}$收敛，故 \(\displaystyle\sum_{n=1}^{+\infty}\left|\frac{\cos\theta_n}{n^2\sqrt{n}}\right|\) 收敛，故$\displaystyle\sum_{n=1}^{+\infty}\frac{\cos\theta_n}{n^2\sqrt{n}}$收敛，即有$\displaystyle\sum_{n=1}^{+\infty}\frac{\sin(n+\frac{1}{n^2})}{\sqrt{n}}$收敛.

Solution 1-(1)-2

\(\displaystyle\sum_{n=1}^{+\infty}\frac{\sin(n+\frac{1}{n^2})}{\sqrt{n}}=\sum_{n=1}^{+\infty}\frac{\sin n\cos\frac{1}{n^2}}{\sqrt{n}}+\sum_{n=1}^{+\infty}\frac{\cos n\sin\frac{1}{n^2}}{\sqrt{n}}\)

由于$\displaystyle\sum_{n=1}^{+\infty}\frac{\sin n}{\sqrt{n}}$收敛，$\displaystyle\cos\frac{1}{n^2}$单调有界，故由Abel判别法可知$\displaystyle\sum_{n=1}^{+\infty}\frac{\sin n\cos\frac{1}{n^2}}{\sqrt{n}}$收敛.

由于 \(\displaystyle\left|\frac{\cos n\sin\frac{1}{n^2}}{\sqrt{n}}\right|\leq\left|\frac{sin\frac{1}{n^2}}{\sqrt{n}}\right|\leq\frac{1}{n^2\sqrt{n}}\)且$\displaystyle\sum_{n=1}^{+\infty}\frac{1}{n^2\sqrt{n}}$收敛，故$\displaystyle\sum_{n=1}^{+\infty}\frac{\cos n\sin\frac{1}{n^2}}{\sqrt{n}}$收敛.

又由于 \(\displaystyle\sum_{n=1}^{+\infty}\frac{\sin(n+\frac{1}{n^2})}{\sqrt{n}}=\sum_{n=1}^{+\infty}\frac{\sin n\cos\frac{1}{n^2}}{\sqrt{n}}+\sum_{n=1}^{+\infty}\frac{\cos n\sin\frac{1}{n^2}}{\sqrt{n}}\)

故$\displaystyle\sum_{n=1}^{+\infty}\frac{\sin(n+\frac{1}{n^2})}{\sqrt{n}}$收敛.

Solution 1-(2)

\(\begin{aligned} &(1+\frac{1}{n+1})^{2n}-(1+\frac{2}{n+a})^n=e^{2n\ln\left(1+\frac{1}{n+1}\right)}-e^{\ln\left(1+\frac{2}{n+a}\right)}\\ =&e^{2n\left(\frac{1}{n+1}-\frac{1}{2(n+1)^2}+o(\frac{1}{n^2})\right)}-e^{n\left(\frac{2}{n+a}-\frac{4}{2(n+a)^2}+o(\frac{1}{n^2})\right)}\\ =&e^2{\left(e^{\frac{2n}{n+1}-2-\frac{n}{(n+1)^2}+o(\frac{1}{n})}-e^{\frac{2n}{n+a}-2-\frac{2n}{(n+a)^2}+o(\frac{1}{n})}\right)}\\ =&e^2\left(1-\frac{2}{n+1}-\frac{n}{(n+1)^2}+o(\frac{1}{n^2})-1+\frac{2a}{n+a}+\frac{2n}{(n+a)^2}+o(\frac{1}{n^2})\right)\\ =&e^2\left(\frac{2a(n+1)-2(n+a)}{(n+1)(n+a)}+\frac{2}{n}-\frac{1}{n}+o(\frac{1}{n})\right)=e^2\left(\frac{2a-2+1}{n}+o(\frac{1}{n})\right)\sim e^2\frac{2a-1}{n} \end{aligned}\) 由上式可知当且仅当$2a-1=0\Rightarrow a=\frac{1}{2}$时收敛，其余情况均发散.

Exercise 2

判断下列积分的收敛性

Solution 2-1

首先我们有广义积分的绝对收敛和收敛是等价的，故我们只需研究下列积分的收敛性即可：

由于

故

反证：我们假设原积分收敛，则有：

进而有

而

故矛盾，即$\displaystyle\iint_{x^2+y^2\geq1}\frac{\cos(x^2)}{x^2+y^2}dxdy$发散.

Solution 2-2

令 \(r=\displaystyle\frac{\sqrt{t}}{\left|\cos\theta\right|}\) 则\(\displaystyle dr=\frac{1}{2\sqrt{t}}\frac{1}{\left|\cos\theta\right|}\)，代入$(\clubsuit)$式，我们有：

故$\displaystyle\iint_{x^2+y^2\geq1}\frac{\cos(x^2)}{x^2+y^2}dxdy$发散.

Exercise 3

研究下列积分的收敛性

Solution 3

可能的奇点：$0,+\infty$

在$x=0$处，$\displaystyle\frac{x^q}{1+x^p}\cos x\sim x^q(x\rightarrow0^{+})\Rightarrow q\gt-1$收敛.

在$x\rightarrow+\infty$处，

显然由$Cauchy$判别法易知$q-p\geq0$发散.

当$q-p\lt-1$时，

即此时有$\displaystyle\int_{1}^{+\infty}\frac{x^q}{1+x^p}\cos xdx$绝对收敛.

当$-1\leq q-p\lt0$时，

而$\displaystyle\int_{1}^{+\infty}\frac{1}{4}x^{q-p}dx$发散，$\displaystyle\int_{1}^{+\infty}\frac{1}{4}x^{q-p}\cos 2xdx$由Dilichlet判别法易知收敛，故

故此时有$\displaystyle\int_{1}^{+\infty}\frac{x^q}{1+x^p}\cos xdx$条件收敛.

综上：

Exercise 4

设$p\geq0$，数列$a_n$满足$a_1=1,a_{n+1}=n^{-p}\arctan a_n$，判断并证明级数

的收敛性

Solution 4

①$p\gt0$时，

由达朗贝尔判别法知收敛.

②$p=0$时，

取$r=-2$有$\lim_{n\rightarrow\infty}\frac{a_n^{-2}}{n}=\frac{2}{3}\Rightarrow a_n\sim\sqrt{\frac{2}{3n}}\Rightarrow\sum_{n=1}^{+\infty}a_n$发散.

Exercise 5

判断下列积分的收敛性

Solution 5

显然这是一个非绝对收敛的积分.

其中$n\leq\theta_n\leq n+1$，故由$Libiniz$判别法知收敛，即$\displaystyle\int_{0}^{+\infty}(-1)^{\left[x^3\right]}dx$条件收敛.

Exercise 6

设$G$为$\mathbb{R}^2$上的有界闭区域，$\partial G$由有线条分段光滑的简单闭曲线构成，假设$u\in C^2(G)$，且$u$在边界上恒为$0$，证明对$\forall\lambda\gt0$，

Solution 6

由于$u$在$\partial G$上恒为$0$，故由Green公式有

由Cauchy-Schwart积分不等式有：

故综上：

Solution for PDF

Click Here and get the solution of this exam.

Remark

If you have any questions or need further assistance, feel free to ask! Here is my email: nkusherr1 at gmail.com

This is the second class of Complex Analysis, hosted by Shengxuan Tang. In this class, we will focus on the following topics:

Essential Singularity

• Casorati Weierstrass Theorem
• Meromorphic Function
• Holomorphism
• Calculation of Residue

Argument Principle

Rouche Theorem

• Fundamental Theorem of Algebra

Open Mapping Theorem

Download

For more information about the 2nd Class of Complex Analysis, please refer to the following sources(Chinese version):

Complex Analysis (begin from page 11 to page 16)

If you have any questions or need further assistance, feel free to ask! Here is my email: nkusherr1 at gmail.com