1.9

Power Series

Let \(\{a_n\}\) be sequence. We would like to define \(f(x)=\sum_{n=0}^{\infty}a_{n}\left(x-a\right)^{n}=\lim_{N\to\infty}\underbrace{\sum_{n=0}^{N}a_{n}\left(x-a\right)^{n}}_{P_N(x)}\)

It is possible if \(\lim_{N\to\infty}P_{N}\left(x\right)\) exists. \(\text{Dom}(f)=\{x:\lim_{N\to\infty}P_{N}\left(x\right)\text{ exists }\}\)

If \(x=a\), then \(P_N(x)=a_0,\forall N\)

Always \(a\in \text{Dom}(f)\). Are there any other \(x\in \text{Dom}(f)\)?

We first assume \(a=0\), then for arbitrary \(a\), everything is the same

Then we study \(\text{Dom}(f)\) of \(f(x)=\sum_{n=0}^{N}a_{n}x^{n}\)​

Definition

Given \(\{a_n\}\), let \(\text{Convergent set}(\{a_{n}\})=\left\lbrace x\in\mathbb{R}:\sum_{n=0}^{\infty}a_{n}x^{n}<\infty\right\rbrace\) and \(\text{Absolute convergent set}(\{a_{n}\})=\left\lbrace x\in\mathbb{R}:\sum_{n=0}^{\infty}a_{n}x^{n}\text{ is absolutely convergent}\right\rbrace\)

We know that \(0\in\text{ACS}({a_{n}})\subseteq\text{CS}({a_{n}})\)

Lemma

Given \(\{a_n\}\), if \(\sum_{n=0}^\infty a_n < \infty \implies \sum_{n=0}^\infty a_n x^n\) converges absolutely \(\forall x \in (-1, 1)\).

If \(1 \in \text{CS}(\{a_n\}) \implies (-1, 1) \subseteq \text{ACS}(\{a_n\})\).

Proof

If \(\sum_{n=0}^\infty a_n < \infty\)​\(\implies \lim_{n \to \infty} a_n = 0 \implies \{a_n\}\) is bounded\(\implies |a_n| < M \ \forall n\)

Now, let \(x \in (-1, 1)\)​\(\implies |x| < 1 \implies \sum_{n=0}^\infty |a_n x^n| \leq \sum_{n=0}^\infty M |x|^n = M \sum_{n=0}^\infty |x|^n\)

\(\implies M \sum_{n=0}^\infty |x|^n = \frac{M}{1 - |x|} < \infty\)​\(\implies \sum_{n=0}^\infty a_n x^n\) is absolutely convergent.

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Corollary

If \(x_0 \in \text{CS}(\{a_n\}) \implies (-|x_0|, |x_0|) \subseteq \text{ACS}(\{a_n\}) \subseteq \text{CS}(\{a_n\})\).

Proof: \(x_0 \in \text{CS} \implies \sum_{n=0}^\infty a_n x_0^n < \infty\).

By Lemma: \(\sum_{n=0}^{\infty}a_{n}x_0^{n}\left(\frac{x}{x_0}\right)^{n}\) is absolutely convergent for \(\left| \frac{x}{x_0} \right| < 1\).

\(\implies \sum_{n=0}^\infty a_n x^n\) converges absolutely for all \(x \in (-|x_0|, |x_0|)\).

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Definition

Given \(\{a_n\}\), let \(\mathcal{R} = \sup(\text{CS}(\{a_n\}))\). This \(\mathcal{R}\) is called the "convergence ratio of \(\{a_n\}\)."

The corollary implies that \(f(x) = \sum_{n=0}^\infty a_n x^n\) is well-defined \(\forall x \in (-\mathcal{R}, \mathcal{R})\).

Moreover, \(\forall x \in (-\mathcal{R}, \mathcal{R})\), \(\sum_{n=0}^\infty a_n x^n\) is absolutely convergent.

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Theorem

Let \(\{a_n\}\) be a sequence and \(\mathcal{R}\) the radius of convergence.

Then \(f : (-\mathcal{R}, \mathcal{R}) \to \mathbb{R}\), defined by \(f(x)=\sum_{n=0}^{\infty}a_{n}x^{n}=\lim_{N\to\infty}P_{N}(x),\quad P_{N}=\sum_{n=0}^{N}a_{n}x^{n},\) is well-defined and continuous.

Proof: If \(0 \leq x < \mathcal{R} \implies \exists x_0 \in \text{CS}(\{a_n\}) : 0 \leq x < x_0 < \mathcal{R}\).

Now apply the corollary to conclude that \(f\) is well-defined in \(x \in (-|x_0|, |x_0|)\)

We have \(P_N \to f\), let us prove that \(P_N \xrightarrow{\mu} f\) on \([-r, r]\) \(\forall r < \mathcal{R}\).

Given \(\epsilon > 0\), we need \(N_0\) such that \(N>N_0\implies|P_{N}(x)-f(x)|<\epsilon\ \forall x\in[-r,r]\)

Since \(0<r < \mathcal{R}\), let \(x_0\) such that \(r < x_0 < \mathcal{R}\)​\(\implies x_0 \in \text{CS}(\{a_n\}) \implies \sum_{n=0}^\infty a_n x_0^n < \infty \implies |a_n x_0^n| < M\)

Since \(\left|\frac{r}{x_0}\right| < 1\)​\(\implies \sum_{n=0}^\infty \left|\frac{r}{x_0}\right|^n = \frac{1}{1 - \left|\frac{r}{x_0}\right|} < \infty\)​\(\implies \exists N_0\) such that \(\sum_{n=N_0+1}^\infty \left|\frac{r}{x_0}\right|^n < \frac{\epsilon}{M}\)

Let us see that this \(N_0\) works

If \(N > N_0\), then \(|P_{N}(x)-f(x)|=\left|\sum_{n=0}^{N}a_{n}x^{n}-\sum_{n=0}^{\infty}a_{n}x^{n}\right|=\left|\sum_{n=N+1}^{\infty}a_{n}x^{n}\right|\leq\sum_{n=N+1}^{\infty}|a_{n}||x|^{n}=\sum_{n=N+1}^{\infty}|a_{n}||x_0^{n}|\left|\frac{x}{x_0}\right|^{n}\)

\(\leq M\sum_{n=N_0+1}^{\infty}\left|\frac{x}{x_0}\right|^{n}\leq M\sum_{n=N_0+1}^{\infty}\left|\frac{r}{x_0}\right|^{n}<M\frac{\epsilon}{M}=\epsilon\)​\(\forall x\in[-r,r]\)

This shows that \(P_{N}\xrightarrow{\mu}f\) on \([-r, r] \implies f\) is continuous on \([-r, r]\).

\(\forall r < \mathcal{R} \implies f\) is continuous on \((-\mathcal{R}, \mathcal{R})\) (Theorem).

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