8.2: Review of Power Series

Many applications give rise to differential equations with solutions that can’t be expressed in terms of elementary functions such as polynomials, rational functions, exponential and logarithmic functions, and trigonometric and inverse trigonometric functions. The solutions of some of the most important of these equations can be expressed in terms of power series. In this section we review relevant properties of power series. We’ll omit proofs, which can be found in any standard calculus text.

Definition 8.2.1

An infinite series of the form

\[\label{eq:7.1.1} \sum_{n=0}^\infty a_n(x-x_0)^n,\]

where \(x_0\) and \(a_0\), \(a_1,\) …, \(a_n,\) …are constants, is called a power series in \(x-x_0.\) We say that the power series Equation \ref{eq:7.1.1} converges for a given \(x\) if the limit

\[\lim_{N\to\infty} \sum_{n=0}^Na_n(x-x_0)^n \nonumber\]

exists\(;\) otherwise, we say that the power series diverges for the given \(x.\)

A power series in \(x-x_0\) must converge if \(x=x_0\), since the positive powers of \(x-x_0\) are all zero in this case. This may be the only value of \(x\) for which the power series converges. However, the next theorem shows that if the power series converges for some \(x\ne x_0\) then the set of all values of \(x\) for which it converges forms an interval.

Theorem 8.2.1

For any power series

\[\sum_{n=0}^\infty a_n(x-x_0)^n, \nonumber\]

exactly one of these statements is true\(:\)

1. The power series converges only for \(x=x_0.\)
2. The power series converges for all values of \(x.\)
3. There’s a positive number \(R\) such that the power series converges if \(|x-x_0|<R\) and diverges if \(|x-x_{0}|>R\).

In case (iii) we say that \(R\) is the radius of convergence of the power series. For convenience, we include the other two cases in this definition by defining \(R=0\) in case (i) and \(R=\infty\) in case (ii). We define the open interval of convergence of \(\sum_{n=0}^\infty a_n(x-x_0)^n\) to be

\[(x_{0}-R, x_{0}+R)\quad\text{if}\quad 0<R<\infty ,\quad\text{or}\quad (-\infty, \infty )\quad\text{if}\quad R=\infty \nonumber.\]

If \(R\) is finite, no general statement can be made concerning convergence at the endpoints \(x=x_0\pm R\) of the open interval of convergence; the series may converge at one or both points, or diverge at both.

As you'll recall from calculus, the next theorem provides a useful method for determining the radius of convergence of a power series.

Theorem 8.2.2 Ratio Test

Consider the series \(\sum_{n=0}^\infty c_n\) and suppose:

\[\lim_{n\to\infty}\left|c_{n+1}\over c_n\right|=L,\nonumber \]

I. If \(L<1\), then \(\sum_{n=0}^\infty c_n\) converges.

II. If \(L>1\) or the limit approaches \(\infty\), then \(\sum_{n=0}^\infty c_n\) diverges.

III. If \(L=1\), then the ratio test gives no information and we have to consider another test.

Example 8.2.1

Find the radius of convergence of the series:

1. \[\sum_{n=0}^\infty n!x^n \nonumber\]
2. \[\sum_{n=10}^\infty (-1)^n {x^n\over n!} \nonumber\]
3. \[\sum_{n=0}^\infty 2^nn^2 (x-1)^n.\nonumber\]

Solution a

Here \(c_n=n!x^n\), so

\[\lim_{n\to\infty}\left|c_{n+1}\over c_n\right|=\lim_{n\to\infty} \left|{(n+1)!x^{n+1}\over n!x^n}\right|=\lim_{n\to\infty}(n+1)|x|. \nonumber\]

This limit will approach \(\infty\) for all \(x\) except \(x=0\), where the limit would be \(0\), so the series only converges at the point \(x=0\).

Hence, \(R=0\).

Solution b

Here \(c_n={(-1)^nx^n\over n!}\), so

\[\lim_{n\to\infty}\left|c_{n+1}\over c_n\right|=\lim_{n\to\infty} \left|{(n)!x^{n+1}\over (n+1)!x^n}\right|=|x|\lim_{n\to\infty}{1\over n+1}=0 < 1\nonumber\]

for all x. Hence, \(R=\infty\).

Solution c

Here \(c_n=2^nn^2(x-1)^n\), so

\[\lim_{n\to\infty}\left|c_{n+1}\over c_n\right|=\lim_{n\to\infty} \left|{2^{n+1}(n+1)^2(x-1)^{n+1}\over 2^nn^2(x-1)^n}\right|=2|x-1|\lim_{n\to\infty}{(n+1)^2\over n^2}=2|x-1|\nonumber\]

which converges when \(2|x-1|<1\) which is true when \(|x-1|<{1\over 2}\).

Hence, \(R=1/2\).

Taylor Series

Definition 8.2.2

If a function \(f\) has derivatives of all orders at a point \(x=x_0\), then the Taylor series of \(f\) about \(x_0\) is defined by

\[\sum_{n=0}^\infty {f^{(n)}(x_0)\over n!}(x-x_0)^n. \nonumber \]

In the special case where \(x_0=0\), this series is also called the Maclaurin series of \(f\):

\[\sum_{n=0}^\infty {f^{(n)}(0)\over n!}x^n. \nonumber \]

Taylor series for most of the common elementary functions converge to the functions on their open intervals of convergence. For example, you should be familiar with the following Maclaurin series:

\[\label{eq:7.1.2} e^{x} = \sum_{n=0}^{\infty} \frac{x^{n}}{n!}, \quad -\infty<x<\infty \]

\[\label{eq:7.1.3} \sin x = \sum_{n=0}^{\infty} (-1)^{n} \frac{x^{2n+1}}{(2n+1)!}, \quad -\infty<x<\infty \]

\[\label{eq:7.1.4} \cos x = \sum_{n=0}^{\infty} (-1)^{n} \frac{x^{2n}}{(2n)!} \quad -\infty<x<\infty \]

\[\label{eq:7.1.5} \frac{1}{1-x} = \sum_{n=0}^{\infty} x^{n} \quad -1<x<1\]

Differentiation of Power Series

A power series with a positive radius of convergence defines a function

\[f(x)=\sum_{n=0}^\infty a_n(x-x_0)^n \nonumber\]

on its open interval of convergence. We say that the series represents \(f\) on the open interval of convergence. A function \(f\) represented by a power series may be a familiar elementary function as in Equations \ref{eq:7.1.2} - \ref{eq:7.1.5}; however, it often happens that \(f\) isn’t a familiar function, so the series actually defines \(f\).

The next theorem (again, from calculus) shows that a function represented by a power series has derivatives of all orders on the open interval of convergence of the power series, and provides power series representations of the derivatives.

Theorem 8.2.3

A power series

\[f(x)=\sum_{n=0}^\infty a_n(x-x_0)^n \nonumber\]

with positive radius of convergence \(R\) has derivatives of all orders in its open interval of convergence, and successive derivatives can be obtained by repeatedly differentiating term by term\(;\) that is,

\[\begin{align} f'(x)&={\sum_{n=1}^\infty na_n(x-x_0)^{n-1}}\label{eq:7.1.6},\\ f''(x)&={\sum_{n=2}^\infty n(n-1)a_n(x-x_0)^{n-2}},\label{eq:7.1.7}\\ &\vdots&\nonumber\\ f^{(k)}(x)&={\sum_{n=k}^\infty n(n-1)\cdots(n-k+1)a_n(x-x_0)^{n-k}}\label{eq:7.1.8}.\end{align}\nonumber \]

Moreover, all of these series have the same radius of convergence \(R.\)

Example 8.2.2

Let \(f(x)=\sin x\). From Equation \ref{eq:7.1.3},

\[f(x)=\sum_{n=0}^\infty(-1)^n {x^{2n+1}\over(2n+1)!}. \nonumber\]

From Equation \ref{eq:7.1.6},

\[f'(x)=\sum_{n=0}^\infty(-1)^n{d\over dx}\left[x^{2n+1}\over(2n+1)!\right]= \sum_{n=0}^\infty(-1)^n {x^{2n}\over(2n)!}, \nonumber\]

which is the series Equation \ref{eq:7.1.4} for \(\cos x\).

Uniqueness of Power Series

The next theorem shows that if \(f\) is defined by a power series in \(x-x_0\) with a positive radius of convergence, then the power series is the Taylor series of \(f\) about \(x_0\).

Theorem 8.2.4

If the power series

\[f(x)=\sum_{n=0}^\infty a_n(x-x_0)^n \nonumber\]

has a positive radius of convergence, then

\[\label{eq:7.1.9} a_n={f^{(n)}(x_0)\over n!};\]

that is, \(\sum_{n=0}^\infty a_n(x-x_0)^n\) is the Taylor series of \(f\) about \(x_0\).

This result can be obtained by setting \(x=x_0\) in Equation \ref{eq:7.1.8}, which yields

\[f^{(k)}(x_0)=k(k-1)\cdots1\cdot a_k=k!a_k. \nonumber\]

This implies that

\[a_k={f^{(k)}(x_0)\over k!}.\nonumber\]

Except for notation, this is the same as Equation \ref{eq:7.1.9}.

Shifting the Summation Index in a Power Series

For any integer \(k\), the power series

\[\sum _ { n = n _ { 0 } } ^ { \infty } b _ { n } \left( x - x _ { 0 } \right) ^ { n - k } \nonumber \]

can be rewritten as

\[\sum _ { n = n _ { 0 } - k } ^ { \infty } b _ { n + k } \left( x - x _ { 0 } \right) ^ { n } \nonumber \]

that is, replacing \(n\) by \(n + k\) in the general term and subtracting k from the lower limit of summation leaves the series unchanged.

Example 8.2.3

Rewrite the following power series from Equation \ref{eq:7.1.7} and Equation \ref{eq:7.1.8} so that the general term in each is a constant multiple of \((x-x_0)^n\):

\[(a) \sum_{n=2}^\infty n(n-1)a_n(x-x_0)^{n-2}\quad (b) \sum_{n=k}^\infty n(n-1)\cdots(n-k+1)a_n(x-x_0)^{n-k}. \nonumber \]

Solution a

Replacing \(n\) by \(n+2\) in the general term and subtracting \(2\) from the lower limit of summation yields

\[\sum_{n=2}^\infty n(n-1)a_n(x-x_0)^{n-2}= \sum_{n=0}^\infty (n+2)(n+1)a_{n+2}(x-x_0)^n. \nonumber \]

Solution b

Replacing \(n\) by \(n+k\) in the general term and subtracting \(k\) from the lower limit of summation yields

\[\sum_{n=k}^\infty n(n-1)\cdots(n-k+1)a_n(x-x_0)^{n-k}= \sum_{n=0}^\infty (n+k)(n+k-1)\cdots(n+1)a_{n+k}(x-x_0)^n. \nonumber \]

Example 8.2.4

Given that

\[f(x)=\sum_{n=0}^\infty a_nx^n, \nonumber\]

write the function \(xf''\) as a power series in which the general term is a constant multiple of \(x^n\).

Solution

From Theorem 8.2.4 with \(x_0=0\),

\[f''(x)=\sum_{n=2}^\infty n(n-1)a_nx^{n-2}.\nonumber\]

Therefore

\[xf''(x)=\sum_{n=2}^\infty n(n-1)a_nx^{n-1}.\nonumber\]

Replacing \(n\) by \(n+1\) in the general term and subtracting \(1\) from the lower limit of summation yields

\[xf''(x)=\sum_{n=1}^\infty (n+1)na_{n+1}x^n.\nonumber\]

Linear Combinations of Power Series

If a power series is multiplied by a constant, then the constant can be placed inside the summation; that is,

\[c\sum_{n=0}^\infty a_n(x-x_0)^n=\sum_{n=0}^\infty ca_n(x-x_0)^n.\nonumber\]

Two power series

\[f(x)=\sum_{n=0}^\infty a_n(x-x_0)^n \quad\mbox{ and }\quad g(x)=\sum_{n=0}^\infty b_n(x-x_0)^n\nonumber\]

with positive radii of convergence can be added term by term at points common to their open intervals of convergence; thus, if the first series converges for \(|x-x_0|<R_{1}\) and the second converges for \(|x-x_{0}|<R_{2}\), then

\[f(x)+g(x)=\sum_{n=0}^\infty(a_n+b_n)(x-x_0)^n\nonumber\]

for \(|x-x_0|<R\), where \(R\) is the smaller of \(R_{1}\) and \(R_{2}\). More generally, linear combinations of power series can be formed term by term; for example,

\[c_1f(x)+c_2g(x)=\sum_{n=0}^\infty(c_1a_n+c_2b_n)(x-x_0)^n.\nonumber\]

Example 8.2.5

Find the Maclaurin series for \(\cosh x\) as a linear combination of the Maclaurin series for \(e^x\) and \(e^{-x}\).

Solution

By definition,

\[\cosh x={1\over2}e^x+{1\over2}e^{-x}. \nonumber\]

Since

\[e^x=\sum_{n=0}^\infty {x^n\over n!}\quad\mbox{ and }\quad e^{-x}=\sum_{n=0}^\infty (-1)^n {x^n\over n!}, \nonumber\]

it follows that

\[\label{eq:7.1.12} \cosh x=\sum_{n=0}^\infty {1\over2}[1+(-1)^n]{x^n\over n!}.\]

Since

\[{1\over2}[1+(-1)^n]=\left\{\begin{array}{cl}1&\mbox{ if } n=2m,\mbox{ an even integer},\\ 0&\mbox{ if }n=2m+1,\mbox{ an odd integer}, \end{array}\right. \nonumber\]

we can rewrite Equation \ref{eq:7.1.12} more simply as

\[\cosh x=\sum_{m=0}^\infty{x^{2m}\over(2m)!}. \nonumber\]

This result is valid on \((-\infty,\infty)\), since this is the open interval of convergence of the Maclaurin series for both \(e^x\) and \(e^{-x}\).

Example 8.2.6

Suppose

\[y=\sum_{n=0}^\infty a_n x^n \nonumber\]

on an open interval \(I\) that contains the origin.

1. Express \[(2-x)y''+2y \nonumber\] as a power series in \(x\) on \(I\).
2. Use the result of (a) to find necessary and sufficient conditions on the coefficients \(\{a_n\}\) for \(y\) to be a solution of the homogeneous equation

   \[\label{eq:7.1.13} (2-x)y''+2y=0\]

   on \(I\).

Solution a

From Equation \ref{eq:7.1.7} with \(x_0=0\),

\[y''=\sum_{n=2}^\infty n(n-1)a_nx^{n-2}. \nonumber\]

Therefore

\[(2-x)y''+2y= 2y''-xy''+2y={\sum_{n=2}^\infty 2n(n-1)a_nx^{n-2} -\sum_{n=2}^\infty n(n-1)a_nx^{n-1} +\sum_{n=0}^\infty 2a_n x^n}.\nonumber\]

Since series are just polynomials, adding them is just like adding polynomials; that is, we add like terms. To add series we want to "run out" terms until each series starts at the same power of \(x\).

So, we get

\[{\sum_{n=2}^\infty 2n(n-1)a_nx^{n-2} -\sum_{n=2}^\infty n(n-1)a_nx^{n-1} +\sum_{n=0}^\infty 2a_n x^n}\nonumber\]

\[=4a_2+{\sum_{n=3}^\infty 2n(n-1)a_nx^{n-2}-\sum_{n=2}^\infty n(n-1)a_nx^{n-1}+2a_0+\sum_{n=1}^\infty 2a_n x^n}\nonumber\]

To combine the three series we need to shift indices so all series start at the same value of \(n\); it doesn't matter where we chose to start the series. In this case let's choose a start of \(n=2\). So, we will need to reindex the first and third series; thus,

\[{\sum_{n=3}^\infty 2n(n-1)a_n x^{n-2}}=\sum_{n=2}^\infty 2(n+1)na_{n+1}x^{n-1}\nonumber\]

and

\[{\sum_{n=1}^\infty 2a_n x^n}=\sum_{n=2}^\infty 2a_{n-1}x^{n-1}\nonumber\]

Now all three series will start with \(n=2\); thus,

\[\label{eq:7.1.17} (2-x)y''+2y=(4a_2+2a_0)+\sum_{n=2}^\infty [2(n+1)na_{n+1}-n(n-1)a_n+2a_{n-1}]x^{n-1}.\]

Solution b

From Equation \ref{eq:7.1.17} we see that \(y\) satisfies Equation \ref{eq:7.1.13} on \(I\) if

\[4a_2+2a_0=0\nonumber\] and \[\label{eq:7.1.18} 2(n+1)na_{n+1}-n(n-1)a_n+2a_{n-1}=0,\quad n=2,3,4, \dots.\nonumber\]

Example 8.2.7

Suppose

\[y=\sum_{n=0}^\infty a_n (x-1)^n \nonumber\]

on an open interval \(I\) that contains \(x_0=1\). Express the function

\[\label{eq:7.1.19} (1+x)y''+2(x-1)^2y'+3y\]

as a power series in \(x-1\) on \(I\).

Solution

Since we want a power series in \(x-1\), we rewrite the coefficient of \(y''\) in Equation \ref{eq:7.1.19} as \(1+x=2+(x-1)\), so Equation \ref{eq:7.1.19} becomes

\[2y''+(x-1)y''+2(x-1)^2y'+3y. \nonumber\]

From Equation \ref{eq:7.1.6} and Equation \ref{eq:7.1.7} with \(x_0=1\),

\[y'=\sum_{n=1}^\infty na_n(x-1)^{n-1}\quad\mbox{ and }\quad y ''=\sum_{n=2}^\infty n(n-1)a_n(x-1)^{n-2}. \nonumber\]

Therefore

\[\begin{aligned} 2y '' &= \sum_{n=2}^\infty 2n(n-1)a_n(x-1)^{n-2},\\ (x-1)y '' &= \sum_{n=2}^\infty n(n-1)a_n(x-1)^{n-1},\\ 2(x-1)^2y' &= \sum_{n=1}^\infty2na_n(x-1)^{n+1},\\ 3y &= \sum_{n=0}^\infty 3a_n (x-1)^n.\end{aligned} \nonumber \]

Running out terms so all series start at the same power of \(x-1\) gives us

\[\begin{aligned} 2y '' &= \sum_{n=2}^\infty 2n(n-1)a_n(x-1)^{n-2}=4a_2+12a_3(x-1)+\sum_{n=4}^\infty 2n(n-1)a_n(x-1)^{n-2},\\ (x-1)y '' &= \sum_{n=2}^\infty n(n-1)a_n(x-1)^{n-1}=2a_2(x-1)+\sum_{n=3}^\infty n(n-1)a_n(x-1)^{n-1},\\ 2(x-1)^2y' &= \sum_{n=1}^\infty2na_n(x-1)^{n+1},\\ 3y &= \sum_{n=0}^\infty 3a_n (x-1)^n=3a_0+3a_1(x-1)+\sum_{n=2}^\infty 3a_n (x-1)^n.\end{aligned} \nonumber \]

We now need to shift indices so all four series start at the same value of \(n\); let's choose to start at \(n=1\), so we need to reindex the first, second, and fourth series as the third one already starts at \(n=1\). This yields

\[\begin{aligned} 2y '' &= \sum_{n=2}^\infty 2n(n-1)a_n(x-1)^{n-2}=4a_2+12a_3(x-1)+\sum_{n=1}^\infty 2(n+3)(n+2)a_{n+3}(x-1)^{n+1},\\ (x-1)y '' &= \sum_{n=2}^\infty n(n-1)a_n(x-1)^{n-1}=2a_2(x-1)+\sum_{n=1}^\infty (n+2)(n+1)a_{n+2}(x-1)^{n+1},\\ 2(x-1)^2y' &= \sum_{n=1}^\infty2na_n(x-1)^{n+1},\\ 3y &= \sum_{n=0}^\infty 3a_n (x-1)^n=3a_0+3a_1(x-1)+\sum_{n=1}^\infty 3a_{n+1} (x-1)^{n+1}.\end{aligned} \nonumber \]

We can now add these together, giving us

\[4a_2+12a_3(x-1)+\sum_{n=1}^\infty 2(n+3)(n+2)a_{n+3}(x-1)^{n+1}+2a_2(x-1)+\sum_{n=1}^\infty (n+2)(n+1)a_{n+2}(x-1)^{n+1}+\sum_{n=1}^\infty2na_n(x-1)^{n+1}+3a_0+3a_1(x-1)+\sum_{n=1}^\infty 3a_{n+1} (x-1)^{n+1}\nonumber\]

and combining like terms gives us

\[(1+x)y''+2(x-1)^2y'+3y=2y''+(x-1)y''+2(x-1)^2y'+3y=(4a_2+3a_0)+(12a_3+2a_2+3a_1)(x-1)+\sum_{n=1}^\infty [2(n+3)(n+2)a_{n+3}+(n+2)(n+1)a_{n+2}+2na_n+3a_{n+1}](x-1)^{n+1}\nonumber\]