10.1: Power Series and Functions (2024)

Definition \(\PageIndex{1}\): Power Series

A series of the form

\[\sum_{n=0}^∞c_nx^n=c_0+c_1x+c_2x^2+\ldots \nonumber \]

is a power series centered at \(x=0.\) A series of the form

\[\sum_{n=0}^∞c_n(x−a)^n=c_0+c_1(x−a)+c_2(x−a)^2+\ldots \nonumber \]

is a power series centered at \(x=a\).

Note \(\PageIndex{1}\): Convergence of a Power Series

Consider the power series \(\displaystyle \sum_{n=0}^∞c_n(x−a)^n.\) The series satisfies exactly one of the following properties:

1. The series converges at \(x=a\) and diverges for all \(x≠a.\)
2. The series converges for all real numbers \(x\).
3. There exists a real number \(R>0\) such that the series converges if \(|x−a|<R\) and diverges if \(|x−a|>R\). At the values \(x\) where |x−a|=R, the series may converge or diverge.

Proof

Suppose that the power series is centered at \(a=0\). (For a series centered at a value of a other than zero, the result follows by letting \(y=x−a\) and considering the series

\[ \sum_{n=1}^∞c_ny^n. \nonumber \]

We must first prove the following fact:

If there exists a real number \(d≠0\) such that \(\displaystyle \sum_{n=0}^∞c_nd^n\) converges, then the series \(\displaystyle \sum_{n=0}^∞c_nx^n\) converges absolutely for all \(x\) such that \(|x|<|d|.\)

Since \(\displaystyle \sum_{n=0}^∞c_nd^n\) converges, the nth term \(c_nd^n→0\) as \(n→∞\). Therefore, there exists an integer \(N\) such that \(|c_nd^n|≤1\) for all \(n≥N.\) Writing

\[|c_nx^n|=|c_nd^n| \left|\dfrac{x}{d}\right|^n, \nonumber \]

we conclude that, for all n≥N,

\[|c_nx^n|≤\left|\dfrac{x}{d}\right|^n. \nonumber \]

The series

\[\sum_{n=N}^∞\left|\dfrac{x}{d}\right|^n \nonumber \]

is a geometric series that converges if \(|\dfrac{x}{d}|<1.\) Therefore, by the comparison test, we conclude that \(\displaystyle \sum_{n=N}^∞c_nx^n\) also converges for \(|x|<|d|\). Since we can add a finite number of terms to a convergent series, we conclude that \(\displaystyle \sum_{n=0}^∞c_nx^n\) converges for \(|x|<|d|.\)

With this result, we can now prove the theorem. Consider the series

\[\sum_{n=0}^∞a_nx^n \nonumber \]

and let \(S\) be the set of real numbers for which the series converges. Suppose that the set \(S={0}.\) Then the series falls under case i.

Suppose that the set \(S\) is the set of all real numbers. Then the series falls under case ii. Suppose that \(S≠{0}\) and \(S\) is not the set of real numbers. Then there exists a real number \(x*≠0\) such that the series does not converge. Thus, the series cannot converge for any \(x\) such that \(|x|>|x*|\). Therefore, the set \(S\) must be a bounded set, which means that it must have a smallest upper bound. (This fact follows from the Least Upper Bound Property for the real numbers, which is beyond the scope of this text and is covered in real analysis courses.) Call that smallest upper bound \(R\). Since \(S≠{0}\), the number \(R>0\). Therefore, the series converges for all \(x\) such that \(|x|<R,\) and the series falls into case iii.

□

Definition: radius of convergence

Consider the power series \(\displaystyle \sum_{n=0}^∞c_n(x−a)^n\). The set of real numbers \(x\) where the series converges is the interval of convergence. If there exists a real number \(R>0\) such that the series converges for \(|x−a|<R\) and diverges for \(|x−a|>R,\) then \(R\) is the radius of convergence. If the series converges only at \(x=a\), we say the radius of convergence is \(R=0\). If the series converges for all real numbers \(x\), we say the radius of convergence is \(R=∞\) (Figure \(\PageIndex{1}\)).

Example \(\PageIndex{1}\): Finding the Interval and Radius of Convergence

For each of the following series, find the interval and radius of convergence.

1. \(\displaystyle \sum_{n=0}^∞\dfrac{x^n}{n!}\)
2. \(\displaystyle \sum_{n=0}^∞n!x^n\)
3. \(\displaystyle \sum_{n=0}^∞\dfrac{(x−2)^n}{(n+1)3^n}\)

Solution

a. To check for convergence, apply the ratio test. We have

\[ \begin{align*} ρ &=\lim_{n→∞} \left|\dfrac{\dfrac{x^{n+1}}{(n+1)!}}{\dfrac{x^n}{n!}}\right| \\[4pt]
&=\lim_{n→∞} \left|\dfrac{x^{n+1}}{(n+1)!}⋅\dfrac{n!}{x^n}\right| \\[4pt]
&=\lim_{n→∞}\left|\dfrac{x^{n+1}}{(n+1)⋅n!}⋅\dfrac{n!}{x^n}\right| \\[4pt]
&=\lim_{n→∞}\left|\dfrac{x}{n+1}\right| \\[4pt]
&=|x|\lim_{n→∞}\dfrac{1}{n+1} \\[4pt]
&=0<1\end{align*}\]

for all values of \(x\). Therefore, the series converges for all real numbers \(x\). The interval of convergence is \((−∞,∞)\) and the radius of convergence is \(R=∞.\)

b. Apply the ratio test. For \(x≠0\), we see that

\[ \begin{align*} ρ &=\lim_{n→∞}\left|\dfrac{(n+1)!x^{n+1}}{n!x^n}\right| \\[4pt]
&=\lim_{n→∞}|(n+1)x| \\[4pt]
&=|x|\lim_{n→∞}(n+1) \\[4pt]
&=∞. \end{align*}\]

Therefore, the series diverges for all \(x≠0\). Since the series is centered at \(x=0\), it must converge there, so the series converges only for \(x≠0\). The interval of convergence is the single value \(x=0\) and the radius of convergence is \(R=0\).

c. In order to apply the ratio test, consider

\[ \begin{align*} ρ &=\lim_{n→∞}\left|\dfrac{\dfrac{(x−2)^{n+1}}{(n+2)3^{n+1}}}{\dfrac{(x−2)^n}{(n+1)3^n}}\right| \\[4pt]
&=\lim_{n→∞} \left|\dfrac{(x−2)^{n+1}}{(n+2)3^{n+1}}⋅\dfrac{(n+1)3^n}{(x−2)^n}\right| \\[4pt]
&=\lim_{n→∞} \left|\dfrac{(x−2)(n+1)}{3(n+2)}\right|\\[4pt]
&=\dfrac{|x−2|}{3}.\end{align*}\]

The ratio \(ρ<1\) if \(|x−2|<3\). Since \(|x−2|<3\) implies that \(−3<x−2<3,\) the series converges absolutely if \(−1<x<5\). The ratio \(ρ>1\) if \(|x−2|>3\). Therefore, the series diverges if \(x<−1\) or \(x>5\). The ratio test is inconclusive if \(ρ=1\). The ratio \(ρ=1\) if and only if \(x=−1\) or \(x=5\). We need to test these values of \(x\) separately. For \(x=−1\), the series is given by

\[ \sum_{n=0}^∞\dfrac{(−1)^n}{n+1}=1−\dfrac{1}{2}+\dfrac{1}{3}−\dfrac{1}{4}+\ldots . \nonumber \]

Since this is the alternating harmonic series, it converges. Thus, the series converges at \(x=−1\). For \(x=5\), the series is given by

\[ \sum_{n=0}^∞\dfrac{1}{n+1}=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\ldots . \nonumber \]

This is the harmonic series, which is divergent. Therefore, the power series diverges at \(x=5\). We conclude that the interval of convergence is \([−1,5)\) and the radius of convergence is \(R=3\).

Exercise \(\PageIndex{1}\)

Find the interval and radius of convergence for the series

\[ \sum_{n=1}^∞\dfrac{x^n}{\sqrt{n}}. \nonumber \]

Hint

Apply the ratio test to check for absolute convergence.

Answer

The interval of convergence is \([−1,1).\) The radius of convergence is \(R=1.\)

Example \(\PageIndex{2}\): Graphing a Function and Partial Sums of its Power Series

Sketch a graph of \(f(x)=\dfrac{1}{1−x}\) and the graphs of the corresponding partial sums \( \displaystyle S_N(x)=\sum_{n=0}^Nx^n\) for \(N=2,4,6\) on the interval \((−1,1)\). Comment on the approximation \(S_N\) as \(N\) increases.

Solution

From the graph in Figure you see that as \(N\) increases, \(S_N\) becomes a better approximation for \(f(x)=\dfrac{1}{1−x}\) for \(x\) in the interval \((−1,1)\).

Exercise \(\PageIndex{2}\)

Sketch a graph of \(f(x)=\dfrac{1}{1−x^2}\) and the corresponding partial sums \(\displaystyle S_N(x)=\sum_{n=0}^Nx^{2n}\) for \(N=2,4,6\) on the interval \((−1,1).\)

Hint

\(S_N(x)=1+x^2+\ldots +x^{2N}=\dfrac{1−x^{2(N+1)}}{1−x^2}\)

Answer

Example \(\PageIndex{3}\): Representing a Function with a Power Series

Use a power series to represent each of the following functions \(f\). Find the interval of convergence.

1. \(f(x)=\dfrac{1}{1+x^3}\)
2. \(f(x)=\dfrac{x^2}{4−x^2}\)

Solution

a. You should recognize this function \(f\) as the sum of a geometric series, because

\[ \dfrac{1}{1+x^3}=\dfrac{1}{1−(−x^3)}. \nonumber \]

Using the fact that, for \(|r|<1,\dfrac{a}{1−r}\) is the sum of the geometric series

\[ \sum_{n=0}^∞ar^n=a+ar+ar^2+\ldots , \nonumber \]

we see that, for \(|−x^3|<1,\)

\[ \begin{align*} \dfrac{1}{1+x^3} =\dfrac{1}{1−(−x^3)} \\[4pt] =\sum_{n=0}^∞(−x^3)^n \\[4pt] =1−x^3+x^6−x^9+\ldots . \end{align*}\]

Since this series converges if and only if \(|−x^3|<1\), the interval of convergence is \((−1,1)\), and we have

\[ \dfrac{1}{1+x^3}=1−x^3+x^6−x^9+\ldots for|x|<1.\nonumber \]

b. This function is not in the exact form of a sum of a geometric series. However, with a little algebraic manipulation, we can relate f to a geometric series. By factoring 4 out of the two terms in the denominator, we obtain

\[ \begin{align*} \dfrac{x^2}{4−x^2} =\dfrac{x^2}{4(\dfrac{1−x^2}{4})}\\[4pt] =\dfrac{x^2}{4(1−(\dfrac{x}{2})^2)}.\end{align*}\]

Therefore, we have

\[ \begin{align*} \dfrac{x^2}{4−x^2} &=\dfrac{x^2}{4(1−(\dfrac{x}{2})^2)} \\[4pt]
&= \dfrac{\dfrac{x^2}{4}}{1−(\dfrac{x}{2})^2} \\[4pt]
&= \sum_{n=0}^∞\dfrac{x^2}{4}(\dfrac{x}{2})^{2n}. \end{align*}\]

The series converges as long as \(|(\dfrac{x}{2})^2|<1\) (note that when \(|(\dfrac{x}{2})^2|=1\) the series does not converge). Solving this inequality, we conclude that the interval of convergence is \((−2,2)\) and

\[ \begin{align*} \dfrac{x^2}{4−x^2} &=\sum_{n=0}^∞\dfrac{x^{2n+2}}{4^{n+1}}\\[4pt]
&=\dfrac{x^2}{4}+\dfrac{x^4}{4^2}+\dfrac{x^6}{4^3}+ \ldots \end{align*}\]

for \(|x|<2.\)

Exercise \(\PageIndex{3}\)

Represent the function \(f(x)=\dfrac{x^3}{2−x}\) using a power series and find the interval of convergence.

Hint

Rewrite f in the form \(f(x)=\dfrac{g(x)}{1−h(x)}\) for some functions \(g\) and \(h\).

Answer

\(\displaystyle \sum_{n=0}^∞\dfrac{x^{n+3}}{2^{n+1}}\) with interval of convergence \((−2,2)\)