### 数学代写|数学分析代写Mathematical Analysis代考|MATH2050

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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
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• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|数学分析代写Mathematical Analysis代考|Power Series

Let $a_k, k=0,1,2, \ldots$, be a sequence of real numbers. The series of functions
$$a_0+a_1 x+a_2 x^2+\ldots+a_k x^k+\ldots$$
is called power series with coefficients $a_0, a_1, a_2, \ldots, a_k, \ldots$
A power series satisfies one of the following:
(i) the series converges only at $x=0$;
(ii) the series converges at any $x \in \mathbb{R}$;
(iii) there exists a real number $\varrho>0$ such that the series converges for $|x|<\varrho$ and does not converge for $|x|>\varrho$.

In particular, the convergence set of the power series (1.24), i.e. the set of points $x \in \mathbb{R}$ at which (1.24) converges, is an interval centred at the origin, namely: just ${0}$ in case (i), the whole $\mathbb{R}$ in case (ii), and an interval between $-\varrho, \varrho$ in case (iii). To prove these claims let us begin with the following result.

Theorem 1 If the power series (1.24) converges at some $\xi \neq 0$, it converges totally on any closed, bounded interval contained in $(-|\xi|,|\xi|)$.
Proof The convergence of the numerical series
$$a_0+a_1 \xi+a_2 \xi^2+\ldots+a_k \xi^k+\ldots$$
implies that the sequence $a_k \xi^k$ is infinitesimal as $k \rightarrow+\infty$, and hence bounded. Put equivalently, there exists $M>0$ such that
$$\left|a_k \xi^k\right| \leq M, \quad \forall k \in \mathbb{N} .$$

## 数学代写|数学分析代写Mathematical Analysis代考|Taylor Series

Let $f(x)$ be a real function defined on an interval $(a, b)$ in $\mathbb{R}$ and let $x_0 \in(a, b)$ be a point. We seek to establish whether there exists a power series centred at $x_0$ that converges on $(a, b)$ to $f$, which is usually phrased by saying that $f$ can be expanded in power series around $x_0$ on the interval $(a, b)$.
The first result in this direction goes as follows.
Theorem 1 If the power series
$$\sum_{k=0}^{\infty} a_k\left(x-x_0\right)^k$$
has convergence radius $\varrho>0$, its sum $f(x)$ is differentiable infinitely many times for $\left|x-x_0\right|<Q$. and for any $m \in \mathbb{N}$ the mth derivative equals
$$f^{(m)}(x)=\sum_{k=m}^{\infty} k(k-1) \cdots(k-m+1) a_k\left(x-x_0\right)^{k-m}$$
Furthermore, $f$ admits a series expansion of the form
$$f(x)=\sum_{k=0}^{\infty} \frac{f^{(k)}\left(x_0\right)}{k !}\left(x-x_0\right)^k$$
Proof Formula (1.36) arises by repeatedly applying Theorem 6 of the previous section (in particular, the theorem on differentiating power series). Putting $x=x_0$ in (1.36), all terms after the first vanish, and $f^{(m)}\left(x_0\right)=m ! a_m$ for every $m \in \mathbb{N}$. Substituting $a_k=f^{(k)}\left(x_0\right) / k$ ! in (1.35) gives (1.37).

By Theorem 1 we know that if $f$ can be expanded in power series around $x_0$ on $(a, b)$, then on some neighbourhood of $x_0$ inside $(a, b)$, of the form $\left|x-x_0\right|<\varrho$, we necessarily have that
(i) $f$ is differentiable infinitely many times when $\left|x-x_0\right|<\varrho$;

# 数学分析代考

## 数学代写|数学分析代写Mathematical Analysis代考|Power Series

$$a_0+a_1 x+a_2 x^2+\ldots+a_k x^k+\ldots$$

(i) 该级数仅收敛于 $x=0$;
(ii) 该系列收敛于任何 $x \in \mathbb{R}$;
(iii) 存在实数 $\varrho>0$ 使得该系列收敛于 $|x|<\varrho$ 并且不收敛于 $|x|>\varrho$.

$$a_0+a_1 \xi+a_2 \xi^2+\ldots+a_k \xi^k+\ldots$$

$$\left|a_k \xi^k\right| \leq M, \quad \forall k \in \mathbb{N}$$

## 数学代写|数学分析代写Mathematical Analysis代考|Taylor Series

$$\sum_{k=0}^{\infty} a_k\left(x-x_0\right)^k$$

$$f^{(m)}(x)=\sum_{k=m}^{\infty} k(k-1) \cdots(k-m+1) a_k\left(x-x_0\right)^{k-m}$$

$$f(x)=\sum_{k=0}^{\infty} \frac{f^{(k)}\left(x_0\right)}{k !}\left(x-x_0\right)^k$$

(i) $f$ 可微分无数次当 $\left|x-x_0\right|<\varrho$;

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## MATLAB代写

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