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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 金融代写|金融数值计算代写Market Risk, Numerical Analysis for Finance代考|Power series: radius of convergence

The problem dealt with in this section is as follows. Consider a sequence of real numbers $\left(a_n\right){n \geq 0}$ and the function defined by the so-called power series: $$f(x)=\sum{n=0}^{\infty} a_n\left(x-x_0\right)^n .$$
Given $x_0 \in \mathbb{R}$, it is important to find all the values $x \in \mathbb{R}$ such that the series of functions (2.15) converges.
Example 2.33. With $x_0=0$, the power series:
$$\sum_{n=0}^{\infty} \frac{((2 n) !)^2}{16^n(n !)^4} x^n$$
converges for $|x|<1$.
Remark 2.34. It is not restrictive, by using a translation, to consider the following simplified-form power series, obtained from (2.15) with $x_0=0$ :
$$\sum_{n=0}^{\infty} a_n x^n$$
Obviously, the choice of $x$ in (2.16) determinates the convergence of the series. The following Lemma $2.35$ is of some importance.

Lemma 2.35. If (2.16) converges for $x=r_0$ then, for any $0 \leq r<\left|r_0\right|$, it is absolutely and uniformly convergent in $[-r, r]$.

Proof. It is assumed the convergence of the numerical series $\sum_{n=0}^{\infty} a_n r_0^n$, that is to say, there exists a positive constant $K$ such that $\left|a_n r_0^n\right| \leq K$. Since $\left|\frac{r}{r_0}\right|<1$, then the geometrical series $\sum_{n=0}^{\infty}\left(\frac{r}{r_0}\right)^n$ converges. Now, for any $n \geq 0$ and any $x \in[-r, r]:$
$$\left|a_n x^n\right|=\left|a_n r_0^n\right|\left|\frac{x}{r_0}\right|^n \leq K\left|\frac{x}{r_0}\right|^n \leq K\left|\frac{r}{r_0}\right|^n .$$
By Theorem 2.27, inequality (2.17) implies that (2.16) is uniformly convergent. Due to positivity, the convergence is also absolute.

From Lemma $2.35$ it follows the fundamental Theorem 2.36, due to Cauchy and Hadamard ${ }^3$, which explains the behaviour of a power series:

Theorem $2.36$ (Cauchy-Hadamard). Given the power series (2.16), then only one of the following alternatives holds:
(i) series (2.16) converges for any $x$;
(ii) series (2.16) converges only for $x=0$;
(iii) there exists a positive number $r$ such that series (2.16) converges for any $x \in]-r, r[$ and diverges for any $x \in]-\infty,-r[\cup] r,+\infty[$.

## 金融代写|金融数值计算代写Market Risk, Numerical Analysis for Finance代考|Taylor-Maclaurin series

Our starting point, here, is the Taylor $^4$ formula with Lagrange ${ }^5$ remainder term. Let $f: I \rightarrow \mathbb{R}$ be a function that admits derivatives of any order at $x_0 \in I$. The Taylor-Lagrange theorem states that, if $x \in I$, then there exists a real number $\xi$, between $x$ and $x_0$, such that:
\begin{aligned} f(x) &=P_n\left(f(x), x_0\right)+R_n\left(f(x), x_0\right) \ &=\sum_{k=0}^n \frac{f^{(k)}\left(x_0\right)}{k !}\left(x-x_0\right)^k+\frac{f^{(n+1)}(\xi)}{(n+1) !}\left(x-x_0\right)^{n+1} . \end{aligned}
Since $f$ has derivatives of any order, we may form the limit of (2.23) as $n \rightarrow \infty$; a condition is stated in Theorem $2.45$ to detect when the passage to the limit is effective.

Theorem 2.45. If $f$ has derivatives of any order in the open interval $I$, with $x_0, x \in I$, and if:
$$\lim {n \rightarrow \infty} R_n\left(f(x), x_0\right)=\lim {n \rightarrow \infty} \frac{f^{(n+1)}(\xi)}{(n+1) !}\left(x-x_0\right)^{n+1}=0,$$

then:
$$f(x)=\sum_{n=0}^{\infty} y \frac{f^{(n)}\left(x_0\right)}{n !}\left(x-x_0\right)^n .$$
Definition 2.46. A function $f(x)$, defined on an open interval $I$, is analytic at $x_0 \in I$, if its Taylor series about $x_0$ converges to $f(x)$ in some neighbourhood of $x_0$.

Remark 2.47. Assuming the existence of the derivatives of any order is not enough to infer that a function is analytic and, thus, it can be represented with a convergent power series. For instance, the function:
has derivatives of any order in $x_0=0$, but such derivates are all zero, therefore the Taylor series reduces to the zero function. This happens because the Lagrange remainder does not vanish as $n \rightarrow \infty$.

Note that the majority of the functions, that interest us, does not possess the behaviour shown in Remark 2.47. The series expansion of the most important, commonly used, functions can be inferred from Equation (2.23), i.e., from the Taylor-Lagrange theorem. And Theorem $2.45$ yields a sufficient condition to ensure that a given function is analytic.

# 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|幂级数:收敛半径

$$\sum_{n=0}^{\infty} \frac{((2 n) !)^2}{16^n(n !)^4} x^n$$

$$\sum_{n=0}^{\infty} a_n x^n$$

$$\left|a_n x^n\right|=\left|a_n r_0^n\right|\left|\frac{x}{r_0}\right|^n \leq K\left|\frac{x}{r_0}\right|^n \leq K\left|\frac{r}{r_0}\right|^n .$$
，根据定理2.27，不等式(2.17)意味着(2.16)是一致收敛的。由于正性，收敛性也是绝对的 从引理$2.35$，它遵循基本定理2.36，由于Cauchy和Hadamard ${ }^3$，它解释了幂级数的行为:

(i)级数(2.16)对任何$x$收敛;
(ii)级数(2.16)只对$x=0$收敛;
(iii)存在一个正数$r$，使得级数(2.16)对任何$x \in]-r, r[$收敛，对任何$x \in]-\infty,-r[\cup] r,+\infty[$发散

## 金融代写|金融数值计算代写市场风险，金融的数值分析代考|Taylor-Maclaurin系列

\begin{aligned} f(x) &=P_n\left(f(x), x_0\right)+R_n\left(f(x), x_0\right) \ &=\sum_{k=0}^n \frac{f^{(k)}\left(x_0\right)}{k !}\left(x-x_0\right)^k+\frac{f^{(n+1)}(\xi)}{(n+1) !}\left(x-x_0\right)^{n+1} . \end{aligned}

$$\lim {n \rightarrow \infty} R_n\left(f(x), x_0\right)=\lim {n \rightarrow \infty} \frac{f^{(n+1)}(\xi)}{(n+1) !}\left(x-x_0\right)^{n+1}=0,$$

then:
$$f(x)=\sum_{n=0}^{\infty} y \frac{f^{(n)}\left(x_0\right)}{n !}\left(x-x_0\right)^n .$$

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assignmentutor™您的专属作业导师
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