9.4: Power Series

A power series is an infinite series whose terms involve constants \(a_n\) and powers of \(x-c\), where \(x\) is a variable and \(c\) is a constant: \(\sum\;a_n\,(x-c)^n\). In many cases \(c\) will be 0. For example, the geometric progression

\[\sum_{n=0}^{\infty} \;r^n ~=~ 1 \;+\; r \;+\; r^2 \;+\; r^3 \;+\; \cdots ~=~ \frac{1}{1-r} \nonumber \]

converges when \(\abs{r} < 1\), i.e. for \(-1<r<1\), as shown in Section 9.1. Replacing the constant \(r\) by a variable \(x\) yields the power series

\[\label{eqn:1over1minusx} \sum_{n=0}^{\infty} \;x^n ~=~ 1 \;+\; x \;+\; x^2 \;+\; x^3 \;+\; \cdots ~=~ \frac{1}{1-x} \]

that converges to \(\frac{1}{1-x}\) when \(-1<x<1\). Note that the series diverges for \(\abs{x}\ge 1\), by the n-th Term Test.

In general a power series of the form \(\sum f_n(x)\), where \(f_n(x)=a_n (x-c)^n\) is a sequence of functions, has an interval of convergence defined as the set of all \(x\) such that the series converges. The interval can be any combination of open or closed, as well as the extreme cases of a single point or all real numbers. On its interval of convergence the power series is thus a function of \(x\). The radius of convergence \(R\) of a power series is defined as half the length of the interval of convergence. In the case where the interval of convergence is all of \(\Reals\) you would say \(R=\infty\).

For example, for the above power series \(\sum_{n=0}^{\infty} f_n(x)\), where \(f_n(x) = x^n\) for \(n\ge 0\), the interval of convergence is \(-1<x<1\), so the radius of convergence is \(R=1\). Notice that

\[f(x) ~=~ \sum_{n=0}^{\infty} x^n \quad\text{for $-1<x<1$} \nonumber \]

is thus a well-defined function on the interval \((-1,1)\), where it happens to equal \(\frac{1}{1-x}\). This power series can be thought of as a polynomial of infinite degree.

To find the interval of convergence of a power series \(\sum f_n(x)\), you typically would use the Ratio Test on the absolute values of the terms (since the Ratio Test requires positive terms):

\[\label{eqn:ratiotestpower} r(x) ~=~ \lim_{n \to \infty} \;\Biggl|\frac{f_{n+1}(x)}{f_n(x)}\Biggr| \]

Note that the limit \(r(x)\) in this case is a function of \(x\). When taking the limit, though, treat \(x\) as fixed. By the Ratio Test the power series will then converge for all \(x\) such that \(r(x) < 1\), and diverge when \(r(x)>1\). When \(r(x)=1\) the test is inconclusive, so you would have to check those cases individually to see if those values of \(x\) should be added to the interval of convergence (along with the points where \(r(x) < 1\)).

Example \(\PageIndex{1}\): expseries

Add text here.

Solution

Find the interval of convergence of the power series \(~\displaystyle\sum_{n=0}^{\infty} \,\dfrac{x^n}{n\,!}~\).

Solution

For \(f_n(x) = \frac{x^n}{n\,!}\),

\[r(x) ~=~ \lim_{n \to \infty} \;\Biggl|\frac{f_{n+1}(x)}{f_n(x)}\Biggr| ~=~ \lim_{n \to \infty} \;\Biggl|\frac{x^{n+1}/(n+1)\,!}{x^n/n\,!}\Biggr| ~=~ \abs{x} \;\cdot\; \lim_{n \to \infty} \;\Biggl|\frac{1}{n+1}\Biggr| ~=~ \abs{x} \;\cdot\; 0 ~=~ 0 \nonumber \]

for any fixed \(x\). Thus, \(r(x) = 0 < 1\) for all \(x\), so the interval of convergence is all of \(\Reals\), i.e. \(-\infty < x < \infty\).

Example \(\PageIndex{1}\)

Find the interval of convergence of the power series \(~\displaystyle\sum_{n=1}^{\infty} \,\dfrac{x^n}{n}~\).

Solution

For \(f_n(x) = \frac{x^n}{n}\),

\[r(x) ~=~ \lim_{n \to \infty} \;\Biggl|\frac{f_{n+1}(x)}{f_n(x)}\Biggr| ~=~ \lim_{n \to \infty} \;\Biggl|\frac{x^{n+1}/(n+1)}{x^n/n}\Biggr| ~=~ \lim_{n \to \infty} \;\Biggl|\frac{nx}{n+1}\Biggr| ~=~ \abs{x} \;\cdot\; \lim_{n \to \infty} \;\Biggl|\frac{n}{n+1}\Biggr| ~=~ \abs{x} \;\cdot\; 1 ~=~ \abs{x} \nonumber \]

for any fixed \(x\). Thus, the series converges when \(r(x) = \abs{x} < 1\) and diverges when \(r(x) = \abs{x} > 1\).
The cases \(r(x) = \abs{x} = 1\) need to be checked individually. For \(x=1\) the series is \(\sum_{n=1}^{\infty} \frac{1}{n}\), which diverges. For \(x=-1\) the series is \(\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}\), which converges. Thus, the interval of convergence is \(-1 \le x < 1\).

Example \(\PageIndex{1}\)

Find the interval of convergence of the power series \(~\displaystyle\sum_{n=0}^{\infty} \,n\,!\;x^n~\).

Solution

For \(f_n(x) = n\,!\;x^n\),

\[r(x) ~=~ \lim_{n \to \infty} \;\Biggl|\frac{f_{n+1}(x)}{f_n(x)}\Biggr| ~=~ \lim_{n \to \infty} \;\Biggl|\frac{(n+1)\,!\;x^{n+1}}{n\,!\;x^n}\Biggr| ~=~ \abs{x} \;\cdot\; \lim_{n \to \infty} \;\abs{n+1} ~=~ \begin{cases} ~0 & \text{if $~x=0$,}\\~\infty & \text{if $~x \ne 0$.}\end{cases} \nonumber \]

Thus, \(r(x) = \infty > 1\) for all \(x \ne 0\), so the series diverges for \(x \ne 0\). So since \(r(x) = 0 < 1\) only for \(x=0\), the interval of convergence is the single point \(x=0\).

It turns out that power series can be both differentiated and integrated term by term:⁹

Notice that the above statement says nothing about the convergence of \(f'(x)\) or \(\int f(x)\,\dx\) at the endpoints of the interval \(\abs{x-c} < R\). In each case convergence at the endpoints can be checked individually.

Example \(\PageIndex{1}\): seriesderivxn

Write the power series form of the derivative of \(~f(x) = \displaystyle\sum_{n=0}^{\infty} \,x^n~\) and find its interval of convergence. Can \(f'(x)\) be written in a non-series form?.

Solution

Differentiate \(f(x)\) term by term:

\[f'(x) ~=~ \ddx\,\left(\sum_{n=0}^{\infty} \,x^n\right) ~=~ \ddx\,(1 \,+\, x \,+\, x^2 \,+\, x^3 \,+\, \cdots) ~=~ 0 \,+\, 1 \,+\, 2x \,+\, 3x^2 \,+\, \cdots ~=~ \sum_{n=1}^{\infty} \,n x^{n-1} \nonumber \]

Since \(f(x)\) converges for \(-1<x<1\) then so does \(f'(x)\). Checking the endpoints, at \(x=1\) and \(x=-1\) the series for \(f'(x)\) are \(\sum_{n=1}^{\infty}\,n\) and \(\sum_{n=1}^{\infty} (-1)^{n-1}\,n\), respectively, both of which diverge by the n-th Term Test. Thus, the interval of convergence for \(f'(x)\) is\(-1<x<1\).

Since \(f(x)=\frac{1}{1-x}\) for \(-1<x<1\) then \(f'(x)=\frac{1}{(1-x)^2}\) for \(-1<x<1\). Thus,

\[\sum_{n=1}^{\infty} \,n x^{n-1} ~=~ 1 \,+\, 2x \,+\, 3x^2 \,+\, \cdots ~=~ \frac{1}{(1-x)^2} \quad\text{for $~-1<x<1$.} \nonumber \]

Many physical applications—especially those involving oscillations and mechanical vibrations—involve solving differential equations of the form

\[\label{eqn:besseldiffeq} \frac{d^2y}{\dx^2} \;+\; \frac{1}{x}\,\dydx \;+\; y ~=~ 0 ~~, \]

known as Bessel’s equation. This equation has a solution \(J_0(x)\), known as Bessel’s function of order zero,¹⁰ defined in terms of a power series:

\[\label{eqn:besselzero} J_0(x) ~=~ \bigsum{n = 0}{\infty}~ (-1)^n\,\frac{x^{2n}}{(n\,!)^2\,\cdot\,2^{2n}} ~=~ 1 \;-\; \frac{x^2}{2^2} \;+\; \frac{x^4}{2^2 \,\cdot\, 4^2} \;-\; \frac{x^6}{2^2 \,\cdot\, 4^2 \,\cdot\, 6^2} \;+\; \cdots \]

The Ratio Test shows that \(J_0(x)\) converges for all \(x\), since for any fixed \(x\),

\[r(x) ~=~ \lim_{n \to \infty} \;\left|\dfrac{(-1)^{n+1}\,\dfrac{x^{2n+2}}{((n+1)\,!)^2\,\cdot\,2^{2n+2}}} {(-1)^n\,\dfrac{x^{2n}}{(n\,!)^2\,\cdot\,2^{2n}}}\right| ~=~ x^2 \;\cdot\; \lim_{n \to \infty} \;\Biggl|\frac{1}{4\,(n+1)^2}\Biggr| ~=~ 0 ~<~ 1 ~. \nonumber \]

The general Bessel equation of order \(\bm{m}\),

\[\label{eqn:besseldiffeqgen} \frac{d^2y}{\dx^2} \;+\; \frac{1}{x}\,\dydx \;+\; \left(1 - \frac{m^2}{x^2}\right)\,y ~=~ 0 \]

for \(m=0,1,2,\ldots\), has a solution \(J_m(x)\), called Bessel’s function of order \(\bm{m}\):

\[\label{eqn:besselm}\index{Bessel functions!order $m$} J_m(x) ~=~ \bigsum{n = 0}{\infty}~ (-1)^n\,\frac{1}{n\,! \,\cdot\, (n+m)\,!}\, \left(\frac{x}{2}\right)^{2n+m} \]

For example, the Bessel function \(J_1(x)\) of order 1 is

\[J_1(x) ~=~ \bigsum{n = 0}{\infty}~ (-1)^n\,\frac{1}{n\,! \,\cdot\, (n+1)\,!}\, \left(\frac{x}{2}\right)^{2n+1} ~=~ \frac{x}{2} \;-\; \frac{x^3}{2^2 \,\cdot\, 4} \;+\; \frac{x^5}{2^2 \,\cdot\, 4^2 \,\cdot\, 6} \;-\; \frac{x^7}{2^2 \,\cdot\, 4^2 \,\cdot\, 6^2 \,\cdot\, 8} \;+\; \cdots \nonumber \]

Term by term differentiation shows that \(J_0'(x) = -J_1(x)\):

\[\begin{aligned} \ddx\,(J_0(x)) ~&=~ \ddx\,\left(1 \;-\; \frac{x^2}{2^2} \;+\; \frac{x^4}{2^2 \,\cdot\, 4^2} \;-\; \frac{x^6}{2^2 \,\cdot\, 4^2 \,\cdot\, 6^2} \;+\; \frac{x^8}{2^2 \,\cdot\, 4^2 \,\cdot\, 6^2 \,\cdot\, 8^2} \;-\; \cdots\right)\\ &=~ -\frac{x}{2} \;+\; \frac{x^3}{2^2 \,\cdot\, 4} \;-\; \frac{x^5}{2^2 \,\cdot\, 4^2 \,\cdot\, 6} \;+\; \frac{x^7}{2^2 \,\cdot\, 4^2 \,\cdot\, 6^2 \,\cdot\, 8} \;-\; \cdots ~=~ -J_1(x)\end{aligned} \nonumber \]

Graphs of \(J_0(x)\) and \(J_1(x)\) are shown in Figure [fig:bessel] below. As you can see, \(J_0(x)\) and \(J_1(x)\) behave as sort of “poor man’s” cosine and sine functions, respectively.

Bessel functions \(J_0(x)\) and \(J_1(x)\)

[sec9dot4]

For Exercises 1-8 find the interval of convergence of the given power series.

4

\(\bigsum{n = 1}{\infty}~ \dfrac{n x^n}{(n + 1)^2}\)

\(\bigsum{n = 1}{\infty}~ \dfrac{n x^n}{2^n}\)

\(\bigsum{n = 1}{\infty}~ n^2 \,(x - 2)^n\vphantom{\dfrac{(x + 4)^n}{2^n}}\)

\(\bigsum{n = 0}{\infty}~ \dfrac{(x + 4)^n}{2^n}\)

4

\(\bigsum{n = 1}{\infty}~ \dfrac{(x + 1)^n}{n^n}\)

\(\bigsum{n = 1}{\infty}~ n^n \,x^n\vphantom{\dfrac{(x + 1)^n}{n^n}}\)

\(\bigsum{n = 0}{\infty}~ (-1)^n\,x^n\vphantom{\dfrac{(x + 1)^n}{n^n}}\)

\(\bigsum{n = 1}{\infty}~ \dfrac{n x^n}{n + 1}\)

Note that power series of the form \(\sum_{n=0}^{\infty} a_n x^n\) have an issue at \(x=0\) when \(n=0\): \(0^0\) is an indeterminate form—it can equal anything (or nothing). What value has it implicitly been assigned so far? What would be the technically correct way to write the series \(\sum_{n=0}^{\infty} a_n x^n\) so that this issue goes away?

Show that

\[\sum_{n=1}^{\infty}\,n\,x^2 ~=~ \frac{x}{(1-x)^2} \quad\text{for $~-1<x<1$.} \nonumber \]

Write the following infinite series as a rational number:

\[\frac{1}{10} \;+\; \frac{2}{100} \;+\; \frac{3}{1000} \;+\; \frac{4}{10000} \;+\; \cdots \nonumber \]

[[1.]]

Differentiating term by term, verify that the Bessel function \(J_0(x)\) satisfies Bessel’s equation (see equation ([eqn:besseldiffeq])).

Show that for all \(m \ge 1\) the Bessel functions \(J_m(x)\) converge for all \(x\).

For all \(m \ge 1\) verify that the Bessel functions \(J_m(x)\) satisfy the general Bessel equation of order \(m\) (see equation([eqn:besseldiffeqgen])).

[exer:besselderiv] For the Bessel functions \(J_0(x)\) and \(J_1(x)\) show that:

1. \(\Ddx\,(x\,J_1(x)) ~=~ x\,J_0(x)\)
2. \(\int J_0(x)\,J_1(x)~\dx ~=~ -\frac{1}{2}\,J_0^2(x)\)
3. \(\int x\,J_0(x)\,J_1(x)~\dx ~=~ -\frac{1}{2}\,x\,J_0^2(x) ~+~ \frac{1}{2}\,\int J_0^2(x)~\dx\)
4. For all integers \(n\ge 2\),

   \[\int x^n\,J_0(x)~\dx ~=~ x^n\,J_1(x) ~+~ (n-1)\,x^{n-1}\,J_0(x) ~-~ (n-1)^2\,\int x^{n-2}\,J_0(x)~\dx ~. \nonumber \]

   (Hint: Use part (a) and integration by parts twice.)

For all integers \(m\ge 2\) show that the Bessel functions \(J_m(x)\) satisfy the relations:

1. \(m\,J_m(x) ~+~ x\,J_m'(x) ~=~ x\,J_{m-1}(x)\)
2. \(J_{m-1}(x) ~-~ J_{m+1}(x) ~=~ 2J_{m}'(x)\)

Use long division to obtain the first three terms of \(\frac{1}{x J_0^2(x)}\), then integrate term by term to show that

\[J_0(x)\,\int \frac{\dx}{x J_0^2(x)} ~=~ J_0(x)\,\cdot\,\ln\,x ~+~ \frac{x^2}{4} ~-~ \frac{3x^4}{128} ~+~ \cdots ~. \nonumber \]

This function is a Bessel function of the second kind and is another solution of Bessel’s equation.