# 数学代写|实分析作业代写Real analysis代考|Power Series Expansions

#### Doug I. Jones

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## 数学代写|实分析作业代写Real analysis代考|Power Series Expansions

In this section, we turn our attention to the study of power series and the representation of functions by means of power series. Because of their special nature, power series possess certain properties which are not valid for series of functions. We begin with the following definition.

DEFINITION 8.7.1 Let $\left{a_k\right}_{k=0}^{\infty}$ be a sequence of real numbers, and let $c \in R$. A series of the form
$$\sum_{k=0}^{\infty} a_k(x-c)^k=a_0+a_1(x-c)+a_2(x-c)^2+a_3(x-c)^3+\cdots$$
is called a power series in $(x-c)$. When $c=0$, the series is called a power series in $x$. The numbers $a_k$ are called the coefficients of the power series.
Even though the study of representation of functions by means of power series dates back to the mid seventeenth century, the rigorous study of convergence is much more recent. Certainly Newton and his successors were concerned with questions involving the convergence of a power series to its defining functions. It was Cauchy however who with his formal development of series brought mathematical rigor to the subject. As an application of his root and ratio test, Cauchy was among the first to use these tests to determine the interval of convergence of a power series. This is accomplished as follows: Consider a power series $\sum_{k=0}^{\infty} a_k(x-c)^k$. Applying the root test to this series gives
$$\varlimsup_{k \rightarrow \infty} \sqrt[k]{\left|a_k\right||x-c|^k}=|x-c| \varlimsup_{k \rightarrow \infty} \sqrt[k]{\left|a_k\right|}=|x-c| \alpha$$
where $\alpha=\varlimsup_{k \rightarrow \infty} \sqrt[k]{\left|a_k\right|}$. Thus by Theorem 7.3 .4 the series converges absolutely if $\alpha|x-c|<1$, and diverges if $\alpha|x-c|>1$. If $\alpha=0$, then $\alpha|x-c|<1$ for all $x \in \mathbb{R}$. If $0<\alpha<\infty$, then
$$\alpha|x-c|<1 \quad \text { if and only if } \quad|x-c|<\frac{1}{\alpha}$$

## 数学代写|实分析作业代写Real analysis代考|Abel’s Theorem

Suppose we are given a power series $\sum a_k(x-c)^k$ with radius of convergence $R>0$. By setting
$$f(x)=\sum_{k=0}^{\infty} a_k(x-c)^k,$$
we obtain a function that is defined for all $x,|x-c|<R$. Functions that are defined in terms of a power series (as in (14)) are usually referred to as real analytic functions. Since the series converges uniformly for all $x$ with $|x-c| \leq \rho$, for any $\rho, 0<\rho<R$, the function $f$ is continuous on $|x-c| \leq \rho$. Since this holds for all $\rho<R$, the function $f$ is continuous on $|x-c|<R$. If the series (14) also converges at an endpoint, say at $x=c+R$, then $f$ is continuous not only in $(c-R, c+R)$ but also at $x=c+R$. This follows from the following theorem of Abel. For convenience, we take $c=0$ and $R=1$.
THEOREM 8.7.5 (Abel’s Theorem) Suppose $f(x)=\sum_{k=0}^{\infty} a_k x^k$ has radius of convergence $R=1$, and that $\sum_{k=0}^{\infty} a_k$ converges. Then
$$\lim {x \rightarrow 1^{-}} f(x)=\sum{k=0}^{\infty} a_k .$$

Proof. Set $s_{-1}=0$, and for $n=0,1,2, \ldots$ let $s_n=\sum_{k=0}^n a_k$. Then by the partial summation formula $(7.2 .1)$
\begin{aligned} \sum_{k=0}^n a_k x^k & =\sum_{k=0}^{n-1} s_k\left(x^k-x^{k+1}\right)+s_n x^n \ & =(1-x) \sum_{k=0}^{n-1} s_k x^k+s_n x^n . \end{aligned}
Since the sequence $\left{s_n\right}$ converges, if we let $n \rightarrow \infty$, then for all $x,|x|<1$,
$$f(x)=(1-x) \sum_{k=0}^{\infty} s_k x^k .$$

# 实分析代写

## 数学代写|实分析作业代写Real analysis代考|Power Series Expansions

$$\sum_{k=0}^{\infty} a_k(x-c)^k=a_0+a_1(x-c)+a_2(x-c)^2+a_3$$

$\sum_{k=0}^{\infty} a_k(x-c)^k$. 对这个系列应用根测试给出
$$\varlimsup_{k \rightarrow \infty} \sqrt[k]{\left|a_k\right||x-c|^k}=|x-c| \overline{\varlimsup_{k \rightarrow \infty}} \sqrt[k]{\left|a_k\right|}=|x-c| \alpha$$

$\alpha|x-c|<1 \quad$ if and only if $\quad|x-c|<\frac{1}{\alpha}$

## 数学代写|实分析作业代写Real analysis代考|Abel’s Theorem

$$f(x)=\sum_{k=0}^{\infty} a_k(x-c)^k,$$

$(c-R, c+R)$ 但也在 $x=c+R$. 这是从以下 Abel 定理得出的。为了方便，我们取 $c=0$ 和 $R=1$. 定理 8.7.5 (阿贝尔定理) 假设 $f(x)=\sum_{k=0}^{\infty} a_k x^k$ 有 收敛半径 $R=1$ ，然后 $\sum_{k=0}^{\infty} a_k$ 收敛。然后
$$\lim x \rightarrow 1^{-} f(x)=\sum k=0^{\infty} a_k .$$

$$\sum_{k=0}^n a_k x^k=\sum_{k=0}^{n-1} s_k\left(x^k-x^{k+1}\right)+s_n x^n$$

$$f(x)=(1-x) \sum_{k=0}^{\infty} s_k x^k .$$

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