Power Series

$$\sum_{m =0}^{\infty} a_{m}(x-x_{0})^m$$

The above infinite series is called power series.
$x$ is a variable, $a_{m}$ are called coefficient of the series, and $x_{0}$ is called the center of the series.

Sum of series

We only care about converging series.

Convergence of series

For the series given by $$\sum_{m = 0}^\infty u_{m} = u_{0} + u_{1} + u_{2} \dots + u_{n} + u_{n+1}\dots$$if the limit $$\lim_{ n \to \infty } \left| \frac{u_{n+1}}{u_{n}} \right| = L$$exists, then the ratio test asserts the convergence for $L < 1$ and divergence for $L > 1$ .

In case of the general power series $$\sum_{m=0}^\infty a_{m}x^m$$we have $$\lim_{ n \to \infty } \left| \frac{a_{n+1}x^{n+1}}{a_{n}x^n}\right| = \lim_{ n \to \infty } \left| \frac{a_{n+1}}{a_{n}}\right|\cdot |x| = L$$
Then the power series converges if $L < 1$ , ie, $| x | < R$
where $$R = \lim_{ n \to \infty } \left| \frac{a_{n}}{a_{n+1}}\right|$$is called the radius of convergence

Idea of a series solution

Consider a second order differential equation $$y'' + P(x)y' + Q(x)y = 0 \tag{1}$$
We can write $P$ and $Q$ in terms of power of $x$ (this is possible when $P$ and $Q$ are nice functions, we will see later)
Then we can write a solution as $$y = \sum_{m=0}^\infty a_{m}x^m$$with unknown coefficients, so that we get (taking differential if series is convergent) $$\begin{align}y' &= \sum_{m=1}^\infty ma_{m}x^{m-1} \ \ y'' &= \sum_{m=2}^\infty m(m-1)a_{m}x^{m-2} \end{align}$$and then by plugging it into the original differential equation we equate individual powers to 0 and determine the coefficients $a_{m}$ and hence we find a solution

Solving something with power series

$y^{'} = 2 x y$

Assume that, $$y = \sum_{m=0}^\infty a_{m}x^m = a_{0} + a_{1}x + a_{2}x^2 + a_{3}x^3\dots$$
On differentiation, $$y' = \sum_{m=1}^\infty ma_{m}x^{m-1} = a_{1} + 2a_{2}x + 3a_{3}x^2 \dots$$
Thus on substitution, we have $$a_{1} + (2a_{2} - 2a_{0})x + (3a_{3} - 2a_{1})x^2 + (4a_{4} - 2a_{2})x^3 \dots = 0$$
From this we get $$ a_{1} = 0, a_{2} = a_{0}, 3a_{2} = 2a_{1}, 4a_{4} = 2a_{2}\dots$$
Hence all odd degree coefficients become 0, and all even degree coefficients remain arbitrary, with $a_{2} = a_{0}, a_{4} = \frac{a_{0}}{2!}, a_{6} = \frac{a_{0}}{3!}$ and thus we obtain a general solution of the form $$y = a_{0}\left( 1 + x^2 + \frac{x^4}{2!} + \frac{x^6}{3!} + \dots \right) = a_{0}e^{x^2} $$

Ordinary and singular points

Considering equation $(1)$ , if $P$ and $Q$ both have a valid power series expansion in the neighborhood of a point $x_{0}$ , then they are said to be analytic at $x_{0}$ .
In this case, $x_{0}$ is called an ordinary point.
Any non-ordinary point is called singular.

Analytic Function

A function $f (x)$ is said to be real analytic on an open set $D$ in the real line if for every $x_{0} \in D$ , one can write $$f(x) = \sum_{n=0}^\infty a_{n}(x-x_{0})^n = a_{0} + a_{1}(x-x_{0}) + a_{2}(x-x_{0})^2 \dots $$in which coefficients $a_{0}, a_{1}, a_{2}$ are real and the series is convergent to $f (x)$ for $x$ in the neighborhood of $x_{0}$ .

Let's take some examples where we try to find solution around an ordinary point

Example $$y'' + \frac{2x}{1+x^2}y' - \frac{2}{1+x^2}y = 0$$

Considering $P (x) = 2 x (1 + x^{2})^{- 1}$ , we can expand as $$P(x) = 2x(1-x^2 + x^4 - x^6\dots)$$and similarly, $$Q(x) = -2(1-x^2 + x^4 - x^6\dots)$$
We find them to by analytic function at $x = 0$ , and hence that is an ordinary point

Differentiating the series, and plugging it into the differential equation, we get $$\sum_{m=0}^\infty[(m+2)(m+1)a_{m+2}x^{m+2} + (m+2)(m+1)a_{m+2}x^{m} + 2(m+1)a_{m+1}x^{m+1} - 2a_{m}x^m] = 0$$
Coefficient of $x^{0}$ is $$2a_{2} - 2a_{0} = 0 \implies a_{2} = a_{0}$$
Coefficient of $x^{1}$ gives $a_{3} = 0$
Coefficient of $x^{m}$ $$\begin{align}m(m-1)a_{m}+(m+2)(m+1)a_{m+2} + 2ma_{m} - 2a_{m} = 0 \ \ (m+2)(m+1)a_{m+2} = -(m^2 + m -2)a_{m} \ \implies a_{m+2} = - \frac{m^2 + m -2}{(m+2)(m+1)}a_{m} \ \ \implies a_{m+2} = -\frac{(m-1)}{(m+1)}a_{m}\end{align}$$
What we obtained in the end, is called a recurrence relation.
From this we notice $$a_{m} = (-1)^{\frac{m-2}{2}} \frac{1}{m-1}a_{0}$$for all even $m$ , and for all odd $m \geq 3$ , $a_{m} = 0$

Hence solution $$\begin{align} y &= \sum_{m=0}^\infty \ &= a_{0} + a_{1}x + a_{2}x^2 \dots \ &= a_{0} + a_{1}x + a_{2}x^2 + 0x^3 + \left( -\frac{1}{3} \right)x^4 + 0a_{5} \dots \ &= a_{0}\left( 1 + x^2 - \frac{x^4}{3} \dots \right) + a_{1}x \ \ &= a_{0}y_{1} + a_{1}y_{2} \end{align}$$where $y_{1} = 1 + x^{2} - \frac{x^{4}}{3}$ and $y_{2} = x$

Note

If we solved the above equation using reduction of order (considering we know one solution to be $x$ ), $$v(x) = \int \frac{1}{x^2} e^{-\int \frac{2x}{1+x^2}dx}$$
ie, $$v(x) = - \left( \frac{1}{x} + \tan^{-1}x \right) = -(1 + x\tan^{-1}x)$$
And if we use the expansion of $\tan^{- 1} x$ , $$\tan^{-1}x = x - \frac{x^3}{3} + \frac{x^5}{5} + \dots$$we can compare the result of $y_{1}$ and this, and they are the same.
The series is convergent only for $| x | < 1$

Theorem

Let $x_{0}$ be an ordinary point of the differential equation $$y'' + P(x)y' + Q(x)y = 0$$and let $α$ and $β$ be arbitrary constants.
Then there exists a unique function $y (x)$ that is analytic at $x_{0}$ and is a solution to the above DE in certain neighborhoods of $x_{0}$ and satisfies initial conditions $y (x_{0}) = α$ and $y^{'} (x_{0}) = β$

Furthermore, if the power-series expansion of $P$ and $Q$ are valid on an open interval $$|x-x_{0}| < R; R > 0$$then the power series expansion of $y (x)$ is also valid on that interval.