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Analytic functions

Let $\{a_{n}(x)\}_{n=1}^{\infty}$ be a sequence of functions defined on the interval $I$. Then consider the partial sums

$\displaystyle S_{1}(x) = a_{1}(x),S_{2}(x) = a_{1}(x)+a_{2}(x),S_{3}(x) = a_{1}(x)+a_{2}(x)+a_{3}(x),\ldots $

If $\lim_{n \rightarrow \infty}S_{n}(x) = S(x)$ for all $x$ in $I$, then we say $\{S_{n}(x)\}_{n=1}^{\infty}$ converges and $S(x)$ is called the sum of the series . For all $x$ in $I$, there exists $N$ such that

$\displaystyle n \geq N$   implies$\displaystyle \ \vert S_{n}(x) - S(x)\vert < \epsilon $

Then $\sum a_{n}(x)$ is called uniformly convergent.

Let $S(x) - S_{n}(x) = R(x)$. Then the uniform convergence of the series is expressed as follows: For each $x$ in the interval $I$, there exists a number $N$ such that $n \geq N implies that \vert R(x)\vert < \epsilon $.

Theorem 4..1 (Weierstrass uniformly convergent criterion)   For any $x$ in some interval $I$, there exists a sequence of real functions $\{M_{n}\}$ satisfying the followings
$(1) \ \vert a_{n}(x)\vert \leq M_{n} , \ n = 1,2,\ldots$
$(2) \ \sum M_{n} < \infty$
Then $\sum a_{n}(x)$ converges uniformly and absolutely on $I$.

Proof

$\displaystyle \vert R_{n}(x)\vert$ $\displaystyle =$ $\displaystyle \vert a_{n+1}(x) + a_{n+2}(x) + \cdots \vert \leq \vert a_{n+1}(x)\vert + \vert a_{n+2}(x)\vert + \cdots$  
  $\displaystyle \leq$ $\displaystyle M_{n+1} + M_{n+2} + \cdots$  

Since $\sum M_{n} < \infty$, for all $\epsilon > 0$, there exists $N$ such that $n \geq N$ implies

$\displaystyle M_{n+1} + M_{n+2} + \cdots < \epsilon. $

Here $N$ is indepedent of $x$, for $n > N$, $\vert R_{n}(x)\vert < \epsilon$ $\ \blacksquare$

Theorem 4..2 (Continuity of power series)   If $a_{n}(x)$ is continuous on the interval $I$ and $\sum a_{n}(x)$ converges uniformly on $I$, then $S(x) = \sum a_{n}(x)$ is continuous on $I$.

Theorem 4..3 (Term-by-term integration of series of functions)   If $a_{n}(x)$ is continuous on the interval $I$ and $\sum a_{n}(x)$ is uniformly convergent on $I$, then

$\displaystyle \int (\sum a_{n}(x))dx = \sum \int a_{n}(x)dx $

Theorem 4..4 (Term-by-term differentiation of series of functions)   If $a_{n}(x)$ is continuous on the interval $I$ and $\sum a_{n}(x)$ is uniformly convergent on $I$, then

$\displaystyle (\sum a_{n}(x))^{\prime} = \sum a_{n}^{\prime}(x) $

If $a_{n}(x) = a_{n}x^{n}$, then the series $\sum a_{n}(x) = \sum a_n x^n$ is called a power series. For example,

    $\displaystyle e^{x} = 1 + x + \frac{x^{2}}{2!} + \cdots + \frac{x^{n}}{n!}+ \cdots = \sum_{n=0}^{\infty} \frac{x^{n}}{n!}$  
    $\displaystyle \sin{x} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots = \sum_{n=0}^{\infty} \frac{(-1)^{n}x^{2n+1}}{(2n+1)!}$  
    $\displaystyle \cos{x} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots = \sum_{n=0}^{\infty} \frac{(-1)^{n}x^{2n}}{(2n)!}$  
    $\displaystyle \log{(1+x)} = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots = \sum_{n=0}^{\infty} (-1)^{n}\frac{x^{n+1}}{n+1}$  

Given $\displaystyle{\sum a_n x^n}$, by the ratio test, if

$\displaystyle \lim_{n \rightarrow \infty}\mid \frac{c_{n+1}}{c_{n}}\mid = \lim_... ... = \vert x\vert \lim_{n \rightarrow \infty}\mid \frac{a_{n+1}}{a_{n}} \mid < 1,$

then $\displaystyle{\sum a_n x^n}$ converges. Thus, if $\displaystyle{\vert x\vert < \lim_{n \rightarrow \infty}\vert\frac{a_{n}}{a_{n+1}}\vert}$, then $\displaystyle{\sum \vert a_{n}x^{n}\vert < \infty}$
If $\displaystyle{\vert x\vert > \lim_{n \rightarrow \infty}\vert\frac{a_{n}}{a_{n+1}}\vert}$, then $\displaystyle{\sum a_{n}x^{n}}$ is divergent. Then the limit

$\displaystyle \rho = \lim_{n \rightarrow \infty} \mid \frac{a_{n}}{a_{n+1}} \mid$

is called the radius of convergence

Example 4..1   Find the radius of convergence of the following power series
$(a) \ \sum \frac{x^{n}}{n^{2}} \ \ \ (b) \ \sum \frac{x^{n}}{n!}$

SOLUTION
\begin{displaymath}\begin{array}{lll} (a)& \ \rho &= \lim_{n \rightarrow \infty}... ...infty} \vert\frac{(n+1)!}{n!}\vert \\ &&= \infty . \end{array}\end{displaymath}

Theorem 4..5 (Properties of power series)   Let $rho$ be the radius of convergence of $\sum_{n=0}^{\infty}a_{n}x^{n}$. Suppose that $f(x) = \sum_{n=0}^{\infty}a_{n}x^{n}$. Then
(1) For all $r \ (0 < r < \rho)$, $\sum a_{n}x^{n}$ converges uniformly on $(-r,r)$.
(2) $f(x)$ is the class $C^{\infty}$ on $(-\rho,\rho)$.
(3) $f(x)$ is continuous on $(-\rho,\rho)$.
(4) $f(x)$ is term-by-term differentiable on $(-\rho,\rho)$ and

$\displaystyle f^{\prime}(x) = \sum_{n=1}^{\infty}na_{n}x^{n-1} $

(5) $f(x)$ is term-by-term integrable on $(-\rho,\rho)$ and

$\displaystyle \int_{0}^{x}f(t)dt = \sum_{n=0}^{\infty}\int_{0}^{x}a_{n}t^{n}dt = \sum_{n=0}^{\infty}\frac{a_{n}}{n+1}x^{n+1} $

(6) Taylor series of $f(x)$ on $(-\rho,\rho)$ is the same as the power series $\sum_{n=0}^{\infty}a_{n}x^{n}$. Thus,

$\displaystyle a_{n} = \frac{f^{(n)}(0)}{n!} $

If $f(x)$ is expressed as the power series of the positive radius $\sum_{n=0}^{\infty}a_{n}(x - a)^{n}$, then $f(x)$ is called analytic at $a$.


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