# 金融代写|金融数值计算代写Market Risk, Numerical Analysis for Finance代考|ORIE5650

#### Doug I. Jones

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## 金融代写|金融数值计算代写Market Risk, Numerical Analysis for Finance代考|Abel theorem and series summation

We present here an important theorem, due to $\mathrm{Abel}^8$, which explains the behaviour of a given power series, with positive radius of convergence, at the boundary of the interval of convergence. In the previous Examples $2.39$ and $2.40$, we observed different behaviours at the boundary of the convergence interval: they can be explained by Abel Theorem 2.57, for the proof of which we refer to [9].

Theorem 2.57 (Abel). Denote by $f(x)$ the sum of the power series (2.16), in which we assume that the radius of convergence is $r>0$. Assume further that the numerical series $\sum_{n=0}^{\infty} r^n a_n$ converges. Then:
$$\lim {x \rightarrow r^{-}} f(x)=\sum{n=0}^{\infty} a_n r^n$$ Proof. The generality of the proof is not affected by the choice $r=1$, as different radii can be achieved with a straightforward change of variable. Let:
$$s_n=\sum_{m=0}^{n-1} a_m$$
then:
$$s=\lim {n \rightarrow \infty} s_n=\sum{n-0}^{\infty} a_n .$$
Now, observe that $a_0=s_1$ and $a_n=s_{n+1}-s_n$ for any $n \in \mathbb{N}$. If $|x|<1$, then 1 is the radius of convergence of the power series:
$$\sum_{n=0}^{\infty} s_{n+1} x^n .$$
To show it, notice that:
$$\lim {n \rightarrow \infty}\left|\frac{s{n+2}}{s_{n+1}}\right|=\lim {n \rightarrow \infty}\left|\frac{a{n+1}+s_{n+1}}{s_{n+1}}\right|=1 .$$
When $|x|<1$, series (2.57) can be multiplied by $1-x$, yielding:
\begin{aligned} (1-x) \sum_{n=0}^{\infty} s_{n+1} x^n &=\sum_{n=0}^{\infty} s_{n+1} x^n-\sum_{n=0}^{\infty} s_{n+1} x^{n+1} \ &=\sum_{n=0}^{\infty} s_{n+1} x^n-\sum_{n=1}^{\infty} s_n x^n \ &=s_1+\sum_{n=1}^{\infty}\left(s_{n+1}-s_n\right) x^n \ &=a_0+\sum_{n=1}^{\infty} a_n x^n=f(x) \end{aligned}

## 金融代写|金融数值计算代写Market Risk, Numerical Analysis for Finance代考|Basel problem

One of the most celebrated problems in Classical Analysis is the Basel Problem, which consists in determining the exact value of the infinite series:
$$\sum_{n=1}^{\infty} \frac{1}{n^2} .$$
Mengoli $^{10}$ originally posed, in 1644 , this problem that takes its name from Basel, birthplace of Euler ${ }^{11}$ who first provided the correct solution $\frac{\pi^2}{6}$ in [19]. There exist several solutions of the Basel problem; here we present the solution of Choe [11], based on the power series expansion of $f(x)=\arcsin x$, shown in Formula (2.48), as well as on the Abel Theorem $2.57$ and on the following integral Formula (2.67), which can be proved by induction on $m \in \mathbb{N}$ :
$$\int_0^{\frac{\pi}{2}} \sin ^{2 m+1} t \mathrm{~d} t=\frac{(2 m) ! !}{(2 m+1) ! !} .$$
The first step towards solving the Basel problem is to observe that, in the sum (2.66), the attention can be confined to odd indexes only. Namely, if $E$ denotes the sum of the series (2.66), then $E$ can be computed by considering, separately, the sums on even and odd indexes:
$$\sum_{n=1}^{\infty} \frac{1}{(2 n)^2}+\sum_{n=0}^{\infty} \frac{1}{(2 n+1)^2}=E .$$
On the other hand:
$$\sum_{n=1}^{\infty} \frac{1}{(2 n)^2}=\sum_{n=1}^{\infty} \frac{1}{4 n^2}=\frac{E}{4},$$
yielding:
$$\sum_{n=0}^{\infty} \frac{1}{(2 n+1)^2}=\frac{3}{4} E .$$
Now, observe that $E=\frac{\pi^2}{6} \Longleftrightarrow \frac{3}{4} E=\frac{\pi^2}{8}$. In other words, the Basel problem is equivalent to show that:
$$\sum_{n=0}^{\infty} \frac{1}{(2 n+1)^2}=\frac{\pi^2}{8},$$
whose proof can be found in [11].

# 金融数值计算代写

## 金融代写|金融数值计算代写市场风险，金融数值分析代考|阿贝尔定理和级数求和

$$\lim {x \rightarrow r^{-}} f(x)=\sum{n=0}^{\infty} a_n r^n$$证明。证明的通用性不受选择$r=1$的影响，因为可以通过直接改变变量来获得不同的半径。让:
$$s_n=\sum_{m=0}^{n-1} a_m$$

$$s=\lim {n \rightarrow \infty} s_n=\sum{n-0}^{\infty} a_n .$$

$$\sum_{n=0}^{\infty} s_{n+1} x^n .$$

$$\lim {n \rightarrow \infty}\left|\frac{s{n+2}}{s_{n+1}}\right|=\lim {n \rightarrow \infty}\left|\frac{a{n+1}+s_{n+1}}{s_{n+1}}\right|=1 .$$

\begin{aligned} (1-x) \sum_{n=0}^{\infty} s_{n+1} x^n &=\sum_{n=0}^{\infty} s_{n+1} x^n-\sum_{n=0}^{\infty} s_{n+1} x^{n+1} \ &=\sum_{n=0}^{\infty} s_{n+1} x^n-\sum_{n=1}^{\infty} s_n x^n \ &=s_1+\sum_{n=1}^{\infty}\left(s_{n+1}-s_n\right) x^n \ &=a_0+\sum_{n=1}^{\infty} a_n x^n=f(x) \end{aligned}

## 金融代写|金融数值计算代写市场风险，金融数值分析代考|巴塞尔问题

$$\sum_{n=1}^{\infty} \frac{1}{n^2} .$$

$$\int_0^{\frac{\pi}{2}} \sin ^{2 m+1} t \mathrm{~d} t=\frac{(2 m) ! !}{(2 m+1) ! !} .$$解决巴塞尔问题的第一步是注意到，在和(2.66)中，注意力只能局限于奇数指标。即，如果 $E$ 表示级数(2.66)的和，则 $E$ 可以分别考虑偶数和奇数索引上的和来计算:
$$\sum_{n=1}^{\infty} \frac{1}{(2 n)^2}+\sum_{n=0}^{\infty} \frac{1}{(2 n+1)^2}=E .$$
:
$$\sum_{n=1}^{\infty} \frac{1}{(2 n)^2}=\sum_{n=1}^{\infty} \frac{1}{4 n^2}=\frac{E}{4},$$

$$\sum_{n=0}^{\infty} \frac{1}{(2 n+1)^2}=\frac{3}{4} E .$$现在，注意看 $E=\frac{\pi^2}{6} \Longleftrightarrow \frac{3}{4} E=\frac{\pi^2}{8}$。换句话说，巴塞尔问题相当于表明:
$$\sum_{n=0}^{\infty} \frac{1}{(2 n+1)^2}=\frac{\pi^2}{8},$$
，其证明可在[11]中找到。

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MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

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