Many functions that arise in applications can be represented as power series. This includes not only the familiar trigonometric, exponential, logarithm, and root functions, but also the host of “special functions” of physics and engineering such as Bessel functions and elliptic functions.

THEOREM 1 Radius of Convergence

Every power series \[F(x) = \sum_{n=0}^{\infty} a_n(x-c)^n\] has a radius of convergence \(R\), which is either a nonnegative number (\(R\ge 0\)) or infinity (\(R=\infty\)). If \(R\) is finite, \(F(x)\) converges absolutely when \(|x-c|<R\) and diverges when \(|x-c|>R\). If \(R=\infty\), then \(F(x)\) converges absolutely for all \(x\).

Proof

We assume that \(c=0\) to simplify the notation. If \(F(x)\) converges only at \(x=0\), then \(R=0\). Otherwise, \(F(x)\) converges for some nonzero value \(x=B\). We claim that \(F(x)\) must then converge absolutely for all \(\left|x\right|<|B|\). To prove this, note that because \({\displaystyle F(B) = \sum_{n=0}^\infty a_nB^n}\) converges, the general term \(a_nB^n\) tends to zero. In particular, there exists \(M>0\) such that \(|a_nB^n|<M\) for all \(n\). Therefore,

\[ \sum_{n=0}^\infty \left|a_nx^n \right| = \sum_{n=0}^\infty |a_nB^n|\,\left|\frac{x}{B}\right|^n < M\sum_{n=0}^\infty \left|\frac{x}{B}\right|^n \] If \(\left|x\right|<|B|\), then \(|x/B|<1\) and the series on the right is a convergent geometric series. By the Comparison Test, the series on the left also converges. This proves that \(F(x)\) converges absolutely if \(\left|x\right| < |B|\).

Now let \(S\) be the set of numbers \(x\) such that \(F(x)\) converges. Then \(S\) contains 0, and we have shown that if \(S\) contains a number \(B \neq 0\), then \(S\) contains the open interval \((-|B|,|B|)\). If \(S\) is bounded, then \(S\) has a least upper bound \(L > 0\) (see marginal note). In this case, there exist numbers \(B \in S\) smaller than but arbitrarily close to \(L\), and thus \(S\) contains \((-B, B)\) for all \(0 < B < L\). It follows that \(S\) contains the open interval \((-L, L)\). The set \(S\) cannot contain any number \(x\) with \(\left|x\right| > L\), but \(S\) may contain one or both of the endpoints \(x = \pm L\). So in this case, \(F(x)\) has radius of convergence \(R=L\). If \(S\) is not bounded, then \(S\) contains intervals \((-B, B)\) for \(B\) arbitrarily large. In this case, \(S\) is the entire real line R, and the radius of convergence is \(R=\infty\).

Least Upper Bound Property: If \(S\) is a set of real numbers with an upper bound \(M\) (that is, \(x \leq M\) for all \(x \in S\)), then \(S\) has a least upper bound \(L\). See Appendix B.

EXAMPLE 1 Using the Ratio Test

Where does F(x)=\(\displaystyle\sum^\infty_{n=0} \dfrac{x^n}{2^n}\) converge?

Solution

Step 1.Find the radius of convergence.

Let \(a_n = \dfrac{x^n}{2^n}\) and compute the ratio \(\rho\) of the Ratio Test: \[ \rho = \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| = \lim_{n\to\infty}\left|\frac{x^{n+1}}{2^{n+1}} \right| \cdot \left| \frac{2^n}{x^n} \right| = \lim_{n\to\infty} \frac{1}{2}\left|x\right| = \frac{1}{2}\left|x\right| \] We find that \[ \rho<1\qquad {if}\quad \frac12 \left|x\right|<1, \qquad{\text{that is, if}}\quad \left|x\right|<2 \] Thus \(F(x)\) converges if \(\left|x\right| < 2\). Similarly, \(\rho > 1\) if \(\tfrac{1}{2} \left|x\right| >1\), or \(\left|x\right| >2\). Thus \(F(x)\) converges if \(\left|x\right| >2\). Therefore, the radius of convergence is \(R=2\).

Step 2. Check the endpoints.

The Ratio Test is inconclusive for \(x=\pm 2\), so we must check these cases directly: \[ \begin{align*} F(2)&=\sum_{n=0}^{\infty} \dfrac{2^n}{2^n} = 1 + 1 + 1 + 1 +1 +1 \cdots \\ F(-2)&=\sum_{n=0}^{\infty} \dfrac{(-2)^n}{2^n} = 1 - 1 + 1 - 1 +1 -1 \cdots \end{align*} \] Both series diverge. We conclude that \(F(x)\) converges only for \(\left|x\right| < 2\) (Figure 10.24).

EXAMPLE 2

Where does \(F(x)=\sum\limits_{n=1}^\infty \frac{(-1)^n}{4^n\,n} (x-5)^n\) converge?

Solution We compute \(\rho\) with \(a_n = \frac{(-1)^n}{4^n\,n}(x-5)^n\): \[ \begin{align*} \rho = \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| &= \lim_{n\to\infty} \left| \frac{(x-5)^{n+1}}{4^{n+1}(n+1)}\,\frac{4^n n}{(x-5)^n} \right|\\ &=|x-5| \lim_{n\to\infty} \left| \frac{n}{4(n+1)} \right|\\&= \frac14 |x-5| \end{align*} \] We find that \[ \rho<1 \qquad \textrm{if}\quad \frac14|x-5|<1,\qquad\textrm{that is, if}\quad |x-5|<4 \]

Thus \(F(x)\) converges absolutely on the open interval \((1,9)\) of radius \(4\) with center \(c=5\). In other words, the radius of convergence is \(R=4\). Next, we check the endpoints: \[ \begin{align*} &x=9{:}&\quad &\sum_{n=1}^\infty \frac{(-1)^n}{4^nn} (9-5)^n = \sum_{n=1}^\infty \frac{(-1)^n}{n}&\quad&\textrm{converges (Leibniz Test)}\\ &x=1{:}&\quad &\sum_{n=1}^\infty \frac{(-1)^n}{4^nn} (-4)^n = \sum_{n=1}^\infty \frac{1}{n}&\quad&\textrm{diverges (harmonic series)} \end{align*} \] We conclude that \(F(x)\) converges for \(x\) in the half-open interval \((1,9]\) shown in Figure 10.25.

EXAMPLE 3 An Even Power Series

Where does \(\displaystyle\sum^\infty_{n=0} \frac{x^{2n}}{(2n)!}\) converge?

Solution Although this power series has only even powers of \(x\), we can still apply the Ratio Test with \(a_n = {x^{2n}}/{(2n)!}\). We have \[a_{n+1} = \frac{x^{2(n+1)}}{(2(n+1))!}=\frac{x^{2n+2}}{(2n+2)!}\] Furthermore, \((2n+2)! = (2n+ 2)(2n +1)(2n)!\), so \[ \begin{align*} \rho =\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| = \lim_{n\to\infty} \frac{x^{2n+2}}{(2n+2)!}\frac{(2n)!}{x^{2n}} = \left|x\right|^2\lim_{n\to\infty} \frac1{(2n+2)(2n+1)} =0 \end{align*} \] Thus \(\rho = 0\) for all \(x\), and \(F(x)\) converges for all \(x\). The radius of convergence is \(R=\infty\).

When a function \(f(x)\) is represented by a power series on an interval \(I\), we refer to the power series expansion of \(f(x)\) on \(I\).

EXAMPLE 4 Geometric Series

Prove that \[ \frac1{1-2x}=\sum_{n=0}^\infty 2^nx^n \qquad\textrm{for \(\left|x\right|<\frac12\)} \]

Solution Substitute \(2x\) for \(x\) in Eq. (2): \[ \begin{equation*} \frac{1}{1-2x} = \sum_{n=0}^\infty (2x)^n = \sum_{n=0}^\infty 2^nx^n\tag{3} \end{equation*} \]

Expansion (2) is valid for \(\left|x\right|<1\), so Eq. (3) is valid for \(|2x|<1\), or \(\left|x\right|<\frac12\).

EXAMPLE 5

Find a power series expansion with center \(c=0\) for \[f(x)=\frac1{2+x^2}\] and find the interval of convergence.

Solution We need to rewrite \(f(x)\) so we can use Eq. (2). We have \[ \frac1{2+x^2} = \frac12\,\left(\frac1{ 1 + \frac12x^2}\right) = \frac12 \left(\frac1{1 - \big(-\frac12x^2\big)}\right)=\frac12\left(\frac{1}{1-u}\right) \] where \(u =-\tfrac12 x^2\). Now substitute \(u = -\frac12x^2\) for \(x\) in Eq. (2) to obtain \[ \begin{align*} f(x)=\frac1{2+x^2} &= \frac12\sum_{n=0}^\infty \biggl(-\frac{x^2}2\biggr)^n\\ & = \sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{2^{n+1}} \end{align*} \] This expansion is valid if \(|{-x^2}/{2}|<1\), or \(\left|x\right|<\sqrt{2}\). The interval of convergence is \((-\sqrt{2},\sqrt{2})\).

The proof of Theorem 2 is somewhat technical and is omitted. See Exercise 66 for a proof that \(F(x)\) is continuous.

THEOREM 2 Term-by-Term Differentiation and Integration

Assume that \[ F(x) = \sum_{n=0}^{\infty} a_n(x-c)^n \] has radius of convergence \(R>0\). Then \(F(x)\) is differentiable on \((c-R, c+R)\) [or for all \(x\) if \(R=\infty\)]. Furthermore, we can integrate and differentiate term by term. For \(x \in (c-R, c+R)\), \[ \begin{align*} F'(x) &= \sum_{n=1}^\infty\, na_n(x-c)^{n-1}\\ \int F(x)\, dx &= A+\sum_{n=0}^\infty \frac{a_n}{n+1}(x-c)^{n+1}\qquad (\textrm{\(A\) any constant}) \end{align*} \] These series have the same radius of convergence \(R\).

EXAMPLE 6 Differentiating a Power Series

Prove that for \(-1 <x < 1\), \[ \frac1{(1-x)^2} = 1+2x+3x^2+4x^3+5x^4+\cdots \]

Solution The geometric series has radius of convergence \(R=1\): \[\frac{1}{1-x} =1+x+x^2+x^3+x^4 +\cdots \] By Theorem 2, we can differentiate term by term for \(\left|x\right|<1\) to obtain \[ \begin{align*} \frac{d}{dx} \,\Big(\frac{1}{1-x}\Big) &= \frac{d}{dx} (1+x+x^2+x^3+x^4 +\cdots )\notag \\ \frac1{(1-x)^2} &= 1+2x+3x^2+4x^3+5x^4+\cdots \end{align*} \]

EXAMPLE 7 Power Series for Arctangent

Prove that for \(-1 <x < 1\), \[ \begin{align*} \tan^{-1}x =\sum_{n=0}^\infty \frac{(-1)^{n}x^{2n+1}}{2n+1} = x-\frac{x^3}3+\frac{x^5}5-\frac{x^7}7+\cdots\tag{4} \end{align*} \]

Solution Recall that \(\tan^{-1}x\) is an antiderivative of \((1+x^2)^{-1}\). We obtain a power series expansion of this antiderivative by substituting \(-x^2\) for \(x\) in the geometric series of Eq.(2): \[ \frac{1}{1+x^2}= 1-x^2+x^4-x^6+\cdots \]

This expansion is valid for \(|x^2|<1\)–that is, for \(\left|x\right|<1\). By Theorem 2, we can integrate series term by term. The resulting expansion is also valid for \(\left|x\right|<1\): \[ \begin{align*} \tan^{-1}x &= \int\frac{dx}{1+x^2} = \int (1-x^2+x^4-x^6+\cdots )\, dx\\ & = A+x - \frac{x^3}3+\frac{x^5}5-\frac{x^7}7+\cdots \end{align*} \]

Setting \(x=0\), we obtain \(A = \tan^{-1}0 =0\). Thus Eq. (4) is valid for \(-1 < x < 1\).

GRAPHICAL INSIGHT

Let’s examine the expansion of the previous example graphically. The partial sums of the power series for \(f(x) = \tan^{-1}x\) are \[ S_N(x) = x - \frac{x^3}3+\frac{x^5}5-\frac{x^7}7 + \cdots + (-1)^N \frac{x^{2N+1}}{2N+1} \]

For large \(N\) we can expect \(S_N(x)\) to provide a good approximation to \(f(x) = \tan^{-1}x\) on the interval \((-1,1)\), where the power series expansion is valid. Figure 10.26 confirms this expectation: The graphs of \(S_{50}(x)\) and \(S_{51}(x)\) are nearly indistinguishable from the graph of \(\tan^{-1}x\) on \((-1,1)\). Thus we may use the partial sums to approximate the arctangent. For example, \(\tan^{-1}~(0.3)\) is approximated by \[ S_4(0.3) = 0.3-\frac{(0.3)^3}3+\frac{(0.3)^5}5 - \frac{(0.3)^7}7+ \frac{(0.3)^9}9 \approx 0.2914569 \]

Since the power series is an alternating series, the error is less than the first omitted term: \[ |{\tan^{-1}(0.3)-S_4(0.3)}| < \frac{(0.3)^{11}}{11} \approx 1.61 \times 10^{-7} \] The situation changes drastically in the region \(\left|x\right|>1\), where the power series diverges and the partial sums \(S_N(x)\) deviate sharply from \(\tan^{-1}x\).

The solution in Example 8 is called the “Bessel function of order 1.” The Bessel function of order \(n\) is a solution of \[ x^2y''+xy'+(x^2-n^2)y=0 \] These functions have applications in many areas of physics and engineering.

EXAMPLE 8

Find a power series solution to the initial value problem \[ \begin{equation*} x^2y''+xy'+(x^2-1)y=0,\qquad y'(0) = 1 \tag{5} \end{equation*} \]

Solution Assume that Eq.(5) has a power series solution \(F(x)= \sum\limits_{n=0}^\infty a_nx^n\). Then \[ \begin{align*} y'=F'(x) &= \sum_{n=0}^\infty na_nx^{n-1} = a_1 + 2a_2 x + 3a_3 x^2 + \cdots\\ y''=F''(x) &= \sum_{n=0}^\infty n(n-1)a_nx^{n-2} = 2a_2 + 6a_3 x + 12a_4 x^2 + \cdots \end{align*} \]

Now substitute the series for \(y\), \(y'\), and \(y''\) into the differential equation (5) to determine the recursion relation satisfied by the coefficients \(a_n\): \[ \begin{eqnarray*} &&x^2y''+xy'+(x^2-1)y \\ &&\qquad = x^2\sum_{n=0}^\infty n(n-1)a_nx^{n-2}+x\sum_{n=0}^\infty na_nx^{n-1} +(x^2-1)\sum_{n=0}^\infty a_nx^n\\ &&\qquad= \sum_{n=0}^\infty n(n-1)a_nx^n+ \sum_{n=0}^\infty na_nx^n-\sum_{n=0}^\infty a_nx^n+ \sum_{n=0}^\infty a_nx^{n+2}\\ &&\qquad= \sum_{n=0}^\infty (n^2-1)a_nx^n + \sum_{n=2}^\infty a_{n-2}x^{n}=0\tag{6} \end{eqnarray*} \]

The differential equation is satisfied if \[ \sum_{n=0}^\infty (n^2-1)a_nx^n = -\sum_{n=2}^\infty a_{n-2}x^n \]

The first few terms on each side of this equation are \[ -a_0+0\cdot x + 3a_2x^2+8a_3x^3 + 15a_4x^4 + \cdots = 0+0\cdot x -a_0x^2-a_1x^3-a_2x^4-\cdots \]

Matching up the coefficients of \(x^n\), we find that \[ \begin{align*} -a_0 = 0,\qquad 3a_2 = -a_0,\qquad 8a_3 = -a_1,\qquad 15a_4 = -a_2\tag{7} \end{align*} \]

In general, \((n^2-1)a_n = -a_{n-2}\), and this yields the recursion relation \[ \boxed{\bbox[#FAF8ED,5pt]{ a_n = -\frac{a_{n-2}}{n^2-1}\qquad \textrm{for \(n\ge 2\)}}}\tag {8} \]

Note that \(a_0=0\) by Eq. (7). The recursion relation forces all of the even coefficients \(a_2\), \(a_4\), \(a_6, \dots\) to be zero: \[ \textrm{\(a_2=\frac{a_0}{2^2-1}\) so \(a_2=0\), and then \(a_4=\frac{a_2}{4^2-1}=0\) so \(a_4=0\), etc.} \]

As for the odd coefficients, \(a_1\) may be chosen arbitrarily. Because \(F'(0) = a_1\), we set \(a_1=1\) to obtain a solution \(y = F(x)\) satisfying \(F'(0)=1\). Now apply Eq. (8): \[ \begin{alignat*}{2} n&=3{:}\qquad& a_3 &= -\frac{a_1}{3^2-1}= -\frac{1}{3^2-1}\\ n&=5{:} \qquad&a_5 &= -\frac{a_3}{5^2-1} = \frac{1}{(5^2-1)(3^2-1)}\\ n&=7{:} \qquad&a_7 &= -\frac{a_5}{7^2-1} = -\frac{1}{(7^2-1)(3^2-1)(5^2-1)} \end{alignat*} \]

This shows the general pattern of coefficients. To express the coefficients in a compact form, let \(n=2k+1\). Then the denominator in the recursion relation (8) can be written \[ n^2-1=(2k+1)^2-1=4k^2+4k=4k(k+1) \] and \[ a_{2k+1} = - \frac{a_{2k-1}}{4k(k+1)} \]

Applying this recursion relation \(k\) times, we obtain the closed formula \[ a_{2k+1} = (-1)^k \left( \frac{1}{4k(k+1)}\right) \left(\frac{1}{4(k-1)k}\right)\cdots \left(\frac{1}{4(1)(2)}\right) = \frac{(-1)^k}{4^k\,k!\,(k+1)!} \]

Thus we obtain a power series representation of our solution: \[ F(x) = \sum_{k=0}^\infty\,\frac{(-1)^k}{4^kk!(k+1)!}x^{2k+1} \]

A straightforward application of the Ratio Test shows that \(F(x)\) has an infinite radius of convergence. Therefore, \(F(x)\) is a solution of the initial value problem for all \(x\).