Differential Equations - Series Solutions: Power Series - i

Before looking at series solutions to a differential equation we will first need to do a cursory review of power series. A power series is a series in the form,

$$\begin{array}{}f\left(x\right)=\sum _{n=0}^{\infty }{a}_n{\left(x-x_0\right)}^n& \text{(1)}\end{array}$$

where, $x_0$ and $a_n$ are numbers. We can see from this that a power series is a function of $x$. The function notation is not always included, but sometimes it is so we put it into the definition above.

Before proceeding with our review we should probably first recall just what series really are. Recall that series are really just summations. One way to write our power series is then,

$$\begin{array}{}\begin{array}{cc}f\left(x\right)& =\sum _{n=0}^{\infty }{a}_n{\left(x-x_0\right)}^n\\ & =a_0+a_1\left(x-x_0\right)+a_2{\left(x-x_0\right)}^2+a_3{\left(x-x_0\right)}^3+\cdots \end{array}& \text{(2)}\end{array}$$

Notice as well that if we needed to for some reason we could always write the power series as,

$$\begin{array}{cc}f\left(x\right)& =\sum _{n=0}^{\infty }{a}_n{\left(x-x_0\right)}^n\\ & =a_0+a_1\left(x-x_0\right)+a_2{\left(x-x_0\right)}^2+a_3{\left(x-x_0\right)}^3+\cdots \\ & =a_0+\sum _{n=1}^{\infty }{a}_n{\left(x-x_0\right)}^n\end{array}$$

All that we’re doing here is noticing that if we ignore the first term (corresponding to $n=0$) the remainder is just a series that starts at $n=1$. When we do this we say that we’ve stripped out the $n=0$, or first, term. We don’t need to stop at the first term either. If we strip out the first three terms we’ll get,

$$\sum _{n=0}^{\infty }{a}_n{\left(x-x_0\right)}^n=a_0+a_1\left(x-x_0\right)+a_2{\left(x-x_0\right)}^2+\sum _{n=3}^{\infty }{a}_n{\left(x-x_0\right)}^n$$

There are times when we’ll want to do this so make sure that you can do it.

Now, since power series are functions of $x$ and we know that not every series will in fact exist, it then makes sense to ask if a power series will exist for all $x$. This question is answered by looking at the convergence of the power series. We say that a power series converges for $x=c$ if the series,

$$\sum _{n=0}^{\infty }{a}_n{\left(c-x_0\right)}^n$$

converges. Recall that this series will converge if the limit of partial sums,

$$\underset{N\to \infty }{lim}\sum _{n=0}^N{a}_n{\left(c-x_0\right)}^n$$

exists and is finite. In other words, a power series will converge for $x=c$ if

$$\sum _{n=0}^{\infty }{a}_n{\left(c-x_0\right)}^n$$

is a finite number.

Note that a power series will always converge if $x=x_0$. In this case the power series will become

$$\sum _{n=0}^{\infty }{a}_n{\left(x_0-x_0\right)}^n=a_0$$

With this we now know that power series are guaranteed to exist for at least one value of $x$. We have the following fact about the convergence of a power series.

Fact

Given a power series, $\text{(1)}$, there will exist a number $0\le \rho \le \infty$ so that the power series will converge for $\left|x-x_0\right|<\rho$and diverge for $\left|x-x_0\right|>\rho$. This number is called the radius of convergence.

Determining the radius of convergence for most power series is usually quite simple if we use the ratio test.

Ratio Test

Given a power series compute,
$$L=\left|x-x_0\right|\underset{n\to \infty }{lim}\left|\frac{a_{n+1}}{a_n}\right|$$then,
$$\begin{array}{cccc}& L<1& \Rightarrow & \text{the series converges}\\ & L>1& \Rightarrow & \text{the series diverges}\\ & L=1& \Rightarrow & \text{the series may converge or diverge}\end{array}$$

Let’s take a quick look at how this can be used to determine the radius of convergence.

Example 1 Determine the radius of convergence for the following power series.$$\sum _{n=0}^{\infty }\frac{{\left(-3\right)}^n}{n{7}^{n+1}}{\left(x-5\right)}^n$$

So, in this case we have,
$$a_n=\frac{{\left(-3\right)}^n}{n{7}^{n+1}}{a}_{n+1}=\frac{{\left(-3\right)}^{n+1}}{\left(n+1\right)7^{n+2}}$$Remember that to compute $a_{n+1}$ all we do is replace all the $n$’s in $a_n$ with $n+1$. Using the ratio test then gives,
$$\begin{array}{cc}L& =\left|x-5\right|\underset{n\to \infty }{lim}\left|\frac{a_{n+1}}{a_n}\right|\\ & =\left|x-5\right|\underset{n\to \infty }{lim}\left|\frac{{\left(-3\right)}^{n+1}}{\left(n+1\right)7^{n+2}}\frac{n{7}^{n+1}}{{\left(-3\right)}^n}\right|\\ & =\left|x-5\right|\underset{n\to \infty }{lim}\left|\frac{-3}{\left(n+1\right)7}\frac{n}{1}\right|\\ & =\frac{3}{7}\left|x-5\right|\end{array}$$Now we know that the series will converge if,
$$\frac{3}{7}\left|x-5\right|<1\Rightarrow \left|x-5\right|<\frac{7}{3}$$and the series will diverge if,
$$\frac{3}{7}\left|x-5\right|>1\Rightarrow \left|x-5\right|>\frac{7}{3}$$In other words, the radius of the convergence for this series is,
$$\rho =\frac{7}{3}$$

As this last example has shown, the radius of convergence is found almost immediately upon using the ratio test.

So, why are we worried about the convergence of power series? Well in order for a series solution to a differential equation to exist at a particular $x$ it will need to be convergent at that $x$. If it’s not convergent at a given $x$ then the series solution won’t exist at that $x$. So, the convergence of power series is fairly important.

Next, we need to do a quick review of some of the basics of manipulating series. We’ll start with addition and subtraction.

There really isn’t a whole lot to addition and subtraction. All that we need to worry about is that the two series start at the same place and both have the same exponent of the $x-x_0$. If they do then we can perform addition and/or subtraction as follows,

$$\sum _{n=n_0}^{\infty }{a}_n{\left(x-x_0\right)}^n\pm \sum _{n=n_0}^{\infty }{b}_n{\left(x-x_0\right)}^n=\sum _{n=n_0}^{\infty }\left(a_n\pm b_n\right){\left(x-x_0\right)}^n$$

In other words, all we do is add or subtract the coefficients and we get the new series.

One of the rules that we’re going to have when we get around to finding series solutions to differential equations is that the only $x$ that we want in a series is the $x$ that sits in ${\left(x-x_0\right)}^n$. This means that we will need to be able to deal with series of the form,

$${\left(x-x_0\right)}^c\sum _{n=0}^{\infty }{a}_n{\left(x-x_0\right)}^n$$

where $c$ is some constant. These are actually quite easy to deal with.

$$\begin{array}{cc}{\left(x-x_0\right)}^c\sum _{n=0}^{\infty }{a}_n{\left(x-x_0\right)}^n& ={\left(x-x_0\right)}^c\left(a_0+a_1\left(x-x_0\right)+a_2{\left(x-x_0\right)}^2+\cdots \right)\\ & =a_0{\left(x-x_0\right)}^c+a_1{\left(x-x_0\right)}^{1+c}+a_2{\left(x-x_0\right)}^{2+c}+\cdots \\ & \sum _{n=0}^{\infty }{a}_n{\left(x-x_0\right)}^{n+c}\end{array}$$

So, all we need to do is to multiply the term in front into the series and add exponents. Also note that in order to do this both the coefficient in front of the series and the term inside the series must be in the form $x-x_0$. If they are not the same we can’t do this, we will eventually see how to deal with terms that aren’t in this form.

Next, we need to talk about differentiation of a power series. By looking at $\text{(2)}$ it should be fairly easy to see how we will differentiate a power series. Since a series is just a giant summation all we need to do is differentiate the individual terms. The derivative of a power series will be,

$$\begin{array}{cc}{f}^{\prime }\left(x\right)& =a_1+2a_2\left(x-x_0\right)+3a_3{\left(x-x_0\right)}^2+\cdots \\ & =\sum _{n=1}^{\infty }n{a}_n{\left(x-x_0\right)}^{n-1}\\ & =\sum _{n=0}^{\infty }n{a}_n{\left(x-x_0\right)}^{n-1}\end{array}$$

So, all we need to do is just differentiate the term inside the series and we’re done. Notice as well that there are in fact two forms of the derivative. Since the $n=0$ term of the derivative is zero it won’t change the value of the series and so we can include it or not as we need to. In our work we will usually want the derivative to start at $n=1$, however there will be the occasional problem were it would be more convenient to start it at $n=0$.

Following how we found the first derivative it should make sense that the second derivative is,

$$\begin{array}{cc}{f}^{\prime \prime }\left(x\right)& =\sum _{n=2}^{\infty }n\left(n-1\right)a_n{\left(x-x_0\right)}^{n-2}\\ & =\sum _{n=1}^{\infty }n\left(n-1\right)a_n{\left(x-x_0\right)}^{n-2}\\ & =\sum _{n=0}^{\infty }n\left(n-1\right)a_n{\left(x-x_0\right)}^{n-2}\end{array}$$

In this case since the $n=0$ and $n=1$ terms are both zero we can start at any of three possible starting points as determined by the problem that we’re working.

Next, we need to talk about index shifts. As we will see eventually we are going to want our power series written in terms of ${\left(x-x_0\right)}^n$ and they often won’t, initially at least, be in that form. To get them into the form we need we will need to perform an index shift.

Index shifts themselves really aren’t concerned with the exponent on the $x$ term, they instead are concerned with where the series starts as the following example shows.

Example 2 Write the following as a series that starts at $n=0$ instead of $n=3$.$$\sum _{n=3}^{\infty }{n}^2{a}_{n-1}{\left(x+4\right)}^{n+2}$$

An index shift is a fairly simple manipulation to perform. First, we will notice that if we define $i=n-3$ then when $n=3$we will have $i=0$. So, what we’ll do is rewrite the series in terms of $i$ instead of $n$. We can do this by noting that $n=i+3$. So, everywhere we see an $n$ in the actual series term we will replace it with an $i+3$. Doing this gives,
$$\begin{array}{cc}\sum _{n=3}^{\infty }{n}^2{a}_{n-1}{\left(x+4\right)}^{n+2}& =\sum _{i=0}^{\infty }{\left(i+3\right)}^2{a}_{i+3-1}{\left(x+4\right)}^{i+3+2}\\ & =\sum _{i=0}^{\infty }{\left(i+3\right)}^2{a}_{i+2}{\left(x+4\right)}^{i+5}\end{array}$$The upper limit won’t change in this process since infinity minus three is still infinity.
The final step is to realize that the letter we use for the index doesn’t matter and so we can just switch back to $n$’s.
$$\sum _{n=3}^{\infty }{n}^2{a}_{n-1}{\left(x+4\right)}^{n+2}=\sum _{n=0}^{\infty }{\left(n+3\right)}^2{a}_{n+2}{\left(x+4\right)}^{n+5}$$

Now, we usually don’t go through this process to do an index shift. All we do is notice that we dropped the starting point in the series by 3 and everywhere else we saw an $n$ in the series we increased it by 3. In other words, all the $n$’s in the series move in the opposite direction that we moved the starting point.