Introduction

In the previous chapter, we had a discussion on the methods of solving

$$y^{\prime\prime} + a y^{\prime} + b y = f(x);$$

where $a, b$ were real numbers and $f$ was a real valued continuous function. We also looked at Euler Equations which can be reduced to the above form. The natural question is:
what if $a$ and $b$ are functions of $x$ ?

In this chapter, we have a partial answer to the above question. In general, there are no methods of finding a solution of an equation of the form

$$y^{\prime\prime} + q(x) y^\prime + r(x) y = f(x), \;\; x \in I$$

where $q(x)$ and $r(x)$ are real valued continuous functions defined on an interval $I \subset {\mathbb{R}}.$ In such a situation, we look for a class of functions $q(x)$ and $r(x)$ for which we may be able to solve. One such class of functions is called the set of analytic functions.

DEFINITION 9.1.1 (Power Series)   Let $x_0 \in {\mathbb{R}}$ and $a_0, a_1, \ldots, a_n, \ldots \in {\mathbb{R}}$ be fixed. An expression of the type

$$\sum_{n=0}^{\infty} a_n (x - x_0)^n$$ (9.1.1)

is called a power series in $x$ around $x_0.$ The point $x_0$ is called the centre, and $a_n$ 's are called the coefficients.

In short, $a_0, a_1, \ldots, a_n, \ldots$ are called the coefficient of the power series and $x_0$ is called the centre. Note here that $a_n \in {\mathbb{R}}$ is the coefficient of $(x - x_0)^n$ and that the power series converges for $x = x_0.$ So, the set

$$S = \{ x \in {\mathbb{R}}: \; \sum\limits_{n=0}^{\infty} a_n (x - x_0)^n {\mbox{ converges}} \}$$

is a non-empty. It turns out that the set $S$ is an interval in ${\mathbb{R}}.$ We are thus led to the following definition.

EXAMPLE 9.1.2  

1. Consider the power series
   $$x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots.$$
   In this case, $x_0 = 0$ is the centre, $\; a_0 = 0$ and $\; a_{2n} = 0$ for $n \geq 1.$ Also, $a_{2n + 1} = \displaystyle\frac{(-1)^n}{(2n+1)!}, \;\; n=1,2,\ldots.$ Recall that the Taylor series expansion around $x_0 = 0$ of $\sin x$ is same as the above power series.
2. Any polynomial
   $$a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n$$
   is a power series with $x_0 = 0$ as the centre, and the coefficients $a_{m} = 0$ for $m \geq n+1.$

DEFINITION 9.1.3 (Radius of Convergence)   A real number $R \geq 0$ is called the radius of convergence of the power series (9.1.1), if $\sum\limits_{n\ge 0} a_n (x - x_0)^n$ converges absolutely for all $x$ satisfying $\vert x- x_0\vert < R$ and diverges for all $x$ satisfying $\vert x - x_0\vert > R.$

From what has been said earlier, it is clear that the set of points $x$ where the power series (9.1.1) is convergent is the interval $(-R + x_0, \; x_0 + R ),$ whenever $R$ is the radius of convergence. If $R = 0,$ the power series is convergent only at $x = x_0.$

Let $R > 0$ be the radius of convergence of the power series (9.1.1). Let $I = (-R + x_0, \; x_0 + R ).$ In the interval $I,$ the power series (9.1.1) converges. Hence, it defines a real valued function and we denote it by $f(x),$ i.e.,

$$f(x) = \sum_{n=1}^\infty a_n (x-x_0)^n, \; x \in I.$$

Such a function is well defined as long as $x \in I.$ $\;\;f$ is called the function defined by the power series (9.1.1) on $I.$ Sometimes, we also use the terminology that (9.1.1) induces a function $f$ on $I.$

It is a natural question to ask how to find the radius of convergence of a power series (9.1.1). We state one such result below but we do not intend to give a proof.

THEOREM 9.1.4  

1. Let $\sum\limits_{n=1}^\infty a_n (x - x_0)^n$ be a power series with centre $x_0.$ Then there exists a real number $R \geq 0$ such that
   $$\sum_{n=1}^\infty a_n (x - x_0)^n \; {\mbox{ converges for all }} \; x \in (-R + x_0, x_0 + R).$$
   In this case, the power series $\sum\limits_{n=1}^\infty a_n (x - x_0)^n$ converges absolutely and uniformly on
   $$\vert x - x_0 \vert \leq r \; {\mbox{ for all }} \; r < R$$
   and diverges for all $x$ with
   $$\vert x - x_0 \vert > R.$$

2. Suppose $R$ is the radius of convergence of the power series (9.1.1). Suppose $\lim\limits_{n \longrightarrow \infty} \sqrt[n]{\vert a_n\vert}$ exists and equals $\ell.$
   1. If $\ell \neq 0,$ then $R = \displaystyle\frac{1}{\ell}.$
   2. If $\ell = 0,$ then the power series (9.1.1) converges for all $x \in {\mathbb{R}}.$
   Note that $\lim\limits_{n \longrightarrow \infty} \sqrt[n]{\vert a_n\vert}$ exists if $$\lim_{n \longrightarrow \infty} \left\vert \frac{a_{n+1}}{a_n} \right\vert$$ exists and
   $$\lim_{n \longrightarrow \infty} \sqrt[n]{\vert a_n\vert} = \lim_{n \longrightarrow \infty} \left\vert \frac{a_{n+1}}{a_n} \right\vert.$$

Remark 9.1.5   If the reader is familiar with the concept of $\limsup$ of a sequence, then we have a modification of the above theorem.

In case, $\sqrt[n]{\vert a_n\vert}$ does not tend to a limit as $n \longrightarrow \infty,$ then the above theorem holds if we replace $\lim\limits_{n \longrightarrow \infty} \sqrt[n]{\vert a_n\vert}$ by $\limsup\limits_{n \longrightarrow \infty} \sqrt[n]{\vert a_n\vert}.$

EXAMPLE 9.1.6  

1. Consider the power series $\sum\limits_{n=0}^\infty (x+1)^n.$ Here $x_0 = -1$ is the centre and $a_n = 1$ for all $n \geq 0.$ So, $\sqrt[n]{\vert a_n\vert} = \sqrt[n]{1} = 1.$ Hence, by Theorem 9.1.4, the radius of convergence$R = 1.$
2. Consider the power series $$\sum\limits_{n \geq 0} \frac{(-1)^n (x+1)^{2n+1}}{(2n + 1)!}.$$ In this case, the centre is
   $$x_0 = -1, \; a_{n} = 0 \; {\mbox{ for }} n \; {\mbox{ even and }} \; a_{2n+1} = \frac{(-1)^n}{(2n + 1)!}.$$
   So,
   $$\lim_{n\longrightarrow \infty} \sqrt[2n+1]{\vert a_{2n+1}\vert} ... ...ox{ and }} \; \lim_{n\longrightarrow \infty} \sqrt[2n]{\vert a_{2n}\vert} = 0.$$
   Thus, $\lim\limits_{n\longrightarrow \infty} \sqrt[n]{\vert a_{n}\vert}$ exists and equals $0.$ Therefore, the power series converges for all $x \in {\mathbb{R}}.$ Note that the series converges to $\sin(x + 1).$
3. Consider the power series $\sum\limits_{n=1}^\infty x^{2n}.$ In this case, we have
   $$a_{2n} = 1 \; {\mbox{ and }} a_{2n+1} = 0 \; {\mbox{ for }} n=0,1,2,\ldots.$$
   So,
   $$\lim_{n\longrightarrow \infty} \sqrt[2n+1]{\vert a_{2n+1}\vert} ... ...ox{ and }} \; \lim_{n\longrightarrow \infty} \sqrt[2n]{\vert a_{2n}\vert} = 1.$$
   Thus, $\lim\limits_{n\longrightarrow \infty} \sqrt[n]{\vert a_{n}\vert}$ does not exist.

   We let $u = x^2.$ Then the power series $\sum\limits_{n=1}^\infty x^{2n}$ reduces to $\sum\limits_{n=1}^\infty u^{n}.$ But then from Example 9.1.6.1, we learned that $\sum\limits_{n=1}^\infty u^{n}$ converges for all $u$ with $\vert u\vert < 1.$ Therefore, the original power series converges whenever $\vert x^2 \vert < 1$ or equivalently whenever $\vert x\vert < 1.$ So, the radius of convergence is $R = 1.$ Note that

   $$\frac{1}{1-x^2} = \sum\limits_{n=1}^\infty x^{2n} \; \; {\mbox{ for }} \;\; \vert x\vert < 1.$$

4. Consider the power series $$\sum\limits_{n \geq 0} n^n x^n.$$ In this case, $\sqrt[n]{\vert a_n\vert} = \sqrt[n]{n^n} = n.$ doesn't have any finite limit as $n \longrightarrow \infty.$ Hence, the power series converges only for $x = 0.$
5. The power series $$\sum\limits_{n \geq 0} \frac{ x^{n}}{n!}$$ has coefficients $a_n = \displaystyle \frac{1}{n!}$ and it is easily seen that $\lim\limits_{n \longrightarrow \infty} \left\vert \displaystyle\frac{1}{n!} \right\vert^{\frac{1}{n}} = 0$ and the power series converges for all $x \in {\mathbb{R}}.$ Recall that it represents $e^x.$

DEFINITION 9.1.7   Let $f: I \longrightarrow {\mathbb{R}}$ be a function and $x_0 \in I.$ $f$ is called analytic around $x_0$ if there exists a $\delta > 0$ such that

$$f(x) = \sum_{n \geq 0} a_n (x - x_0)^n \;\; {\mbox{ for every $x$ with }} \vert x - x_0 \vert < \delta.$$

That is, $f$ has a power series representation in a neighbourhood of $x_0.$