Infi-2 16

Power series #definition

\begin{matrix} \sum_{n = 0}^{\infty} a_{n} (x - a)^{n} \\ Where a_{n} does not depend on x \\ is called a power series in the neighborhood of a \end{matrix}

\begin{matrix} Convergence domain is a set X such that: \\ \forall I \subseteq X : \sum_{n = 0}^{\infty} f_{n} (x) converges pointwise on I \end{matrix}

\begin{matrix} Convergence domain of \sum_{n = 0}^{\infty} a_{n} (x - a)^{n} can be one of three options: \\ 1. R, for example \sum_{n = 0}^{\infty} \frac{x^{n}}{n!} \\ 2. {a}, for example \sum_{n = 0}^{\infty} n! x^{n} \\ 3. Interval, that is symmetric around a \end{matrix}

\begin{matrix} Let R \in R : \sum_{n = 0}^{\infty} a_{n} (x - a)^{n} converges on (a - R, a + R) \\ and diverges if x \in (- \infty, a - R) \cup (a + R, \infty) \\ R is then called a convergence radius of power series \\ And there are no "holes" in the convergence domain \\ Note: endpoints of convergence might be included or excluded \\ Proof: \\ Let b \in (a - R, a + R) \\ Let | c - a | < | b - a | \\ Let \sum_{n = 0}^{\infty} a_{n} (b - a)^{n} converges \\ \sum_{n = 0}^{\infty} | a_{n} (c - a)^{n} | = \sum_{n = 0}^{\infty} | a_{n} (c - a)^{n} \frac{(b - a)^{n}}{(b - a)^{n}} | = \\ = \sum_{n = 0}^{\infty} | a_{n} (b - a)^{n} | \cdot {| \frac{c - a}{b - a} |}^{n} \\ a_{n} (b - a)^{n} \underset{n \to \infty}{\to} 0 \\ ⟹ | a_{n} (b - a)^{n} | \underset{n \to \infty}{\to} 0 ⟹ \forall n > N : | a_{n} (b - a)^{n} | < 1 \\ \forall n > N : | a_{n} (b - a)^{n} | \cdot {| \frac{c - a}{b - a} |}^{n} < {| \frac{c - a}{b - a} |}^{n} \\ | \frac{c - a}{b - a} | < 1 ⟹ \sum_{n = 0}^{\infty} {| \frac{c - a}{b - a} |}^{n} converges \\ ⟹ \sum_{n = 0}^{\infty} | a_{n} (b - a)^{n} | \cdot {| \frac{c - a}{b - a} |}^{n} converges \\ ⟹ \sum_{n = 0}^{\infty} | a_{n} (c - a)^{n} | converges ⟹ \end{matrix}

\begin{matrix} Let \sum_{n = 0}^{\infty} a_{n} (x - a)^{n} \\ Let L = lim_{n \to \infty} | \frac{a_{n + 1} (x - a)^{n + 1}}{a_{n} (x - a)^{n}} | = \\ = lim_{n \to \infty} | \frac{a_{n + 1}}{a_{n}} | \cdot | x - a | = | x - a | \cdot lim_{n \to \infty} | \frac{a_{n + 1}}{a_{n}} | \\ Let t = lim_{n \to \infty} | \frac{a_{n + 1}}{a_{n}} | \\ t \cdot | x - a | < 1 ⟹ Series converges \\ t \cdot | x - a | > 1 ⟹ Series diverges \\ ⟹ Series converges ⟺ | x - a | < \frac{1}{t} \\ ⟹ R = \frac{1}{t} = \frac{1}{lim_{n \to \infty} | \frac{a_{n + 1}}{a_{n}} |} \\ Note: \\ We can also write R = \frac{1}{lim_{n \to \infty} \sqrt[n]{| a_{n} |}} \end{matrix}

\begin{matrix} lim_{n \to \infty} | \frac{a_{n + 1}}{a_{n}} | = 0 ⟹ R = \infty \\ lim_{n \to \infty} | \frac{a_{n + 1}}{a_{n}} | = \infty ⟹ R = 0 \\ ∄ lim_{n \to \infty} | \frac{a_{n + 1}}{a_{n}} | ⟹ ? \end{matrix}

\begin{matrix} \sum_{n = 0}^{\infty} \frac{(- 1)^{n}}{(2 n)!} x^{2 n} \\ a_{n} = {\begin{matrix} 0 & n = 2 k - 1 \\ \frac{(- 1)^{n}}{(2 n)!} & n = 2 k \end{matrix} \\ ? ? ? \\ Let t = x^{2} \\ \sum_{n = 0}^{\infty} \frac{(- 1)^{n}}{(2 n)!} t^{n} \\ R = lim_{n \to \infty} \frac{1}{2 n + 1} = 0 \end{matrix}