9.2: Properties of Power Series
- Page ID
- 175623
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\dsum}{\displaystyle\sum\limits} \)
\( \newcommand{\dint}{\displaystyle\int\limits} \)
\( \newcommand{\dlim}{\displaystyle\lim\limits} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)- Combine power series by addition or subtraction.
- Create a new power series by multiplying by a power of the variable, multiplying by a constant, or by substitution.
- Multiply two power series together.
- Differentiate and integrate power series term-by-term.
In the preceding section on power series and functions, we showed how to represent certain functions using power series. This section discusses how power series can be combined, differentiated, or integrated to create new power series. This is particularly useful for a couple of reasons. First, it allows us to find power series representations for certain elementary functions by writing those functions in terms of functions with known power series. For example, given the power series representation for \(f(x)=\frac{1}{1−x}\), we can find a power series representation for \(f^{\prime}(x)=\frac{1}{(1−x)^2}\). Second, creating a power series allows us to define new functions that cannot be written in terms of elementary functions. This capability is handy for solving differential equations for which there is no solution in terms of elementary functions.
Combining Power Series
If we have two power series with the same interval of convergence, we can add or subtract the two series to create a new one with the same interval of convergence. Similarly, we can multiply a power series by a power of \(x\) or evaluate a power series at \(x^m\) for a positive integer \(m\) to create a new power series (we did this a little bit in the previous section). Being able to do this allows us to find power series representations for certain functions by using power series representations of other functions. For example, since we know the power series representation for \(f(x)=\frac{1}{1−x}\), we can find power series representations for related functions, such as\[y=\dfrac{3x}{1−x^2} \nonumber \]and\[y=\dfrac{1}{(x−1)(x−3)}. \nonumber \]In the theorem below, we state results regarding the addition or subtraction of two power series, compositions involving power series, and multiplication of a power series by a power of the variable. For simplicity, we state the theorem for power series centered at \(x=0\). Similar results hold for power series centered at \(x=a\).
Suppose that the two power series \(\displaystyle \sum_{n=0}^\infty c_n x^n\) and \(\displaystyle \sum_{n=0}^\infty d_n x^n\) converge to the functions \(f\) and \(g\), respectively, on a common interval \(I\). Then, the following statements are true.
- The power series \(\displaystyle \sum_{n=0}^\infty (c_n x^n \pm d_n x^n)\) converges to \(f \pm g\) on \(I\).
- For any integer \(m \geq 0\) and any real number \(b\), the power series \(\displaystyle \sum_{n=0}^\infty b x^m c_n x^n\) converges to \(b x^m f(x)\) on \(I\).
- For any integer \(m \geq 0\) and any real number \(b\), the series \(\displaystyle \sum_{n=0}^\infty c_n (bx^m)^n\) converges to \(f(bx^m)\) for all \(x\) such that \(bx^m\) is in \(I\).
- Proof
-
We prove \(i\) for the case of the series \(\displaystyle \sum_{n=0}^\infty (c_n x^n + d_n x^n)\). Suppose that \(\displaystyle \sum_{n=0}^ \infty c_nx^n\) and \(\displaystyle \sum_{n=0}^ \infty d_nx^n\) converge to the functions \(f\) and \(g\), respectively, on the interval \(I\). Let \(x\) be a point in \(I\) and let \(S_N(x)\) and \(T_N(x)\) denote the \(N^{\text{th}}\) partial sums of the series \(\displaystyle \sum_{n=0}^ \infty c_n x^n\) and \(\displaystyle \sum_{n=0}^ \infty d_n x^n\), respectively. Then the sequence \(\{S_N(x)\}\) converges to \(f(x)\) and the sequence \(\{T_N(x)\}\) converges to \(g(x)\). Furthermore, the \(N^{\text{th}}\) partial sum of \(\displaystyle \sum_{n=0}^ \infty (c_nx^n+d_nx^n)\) is\[ \begin{array}{rcl}
\displaystyle \sum_{n=0}^N(c_n x^n + d_n x^n) & = & \displaystyle \sum_{n=0}^N c_n x^n + \sum_{n=0}^N d_n x^n\\[6pt]
& = & S_N(x)+T_N(x). \\[6pt]
\end{array} \nonumber \]Because\[ \begin{array}{rcl}
\displaystyle \lim_{N \to \infty }(S_N(x)+T_N(x)) & = & \displaystyle \lim_{N \to \infty }S_N(x) + \lim_{N \to \infty }T_N(x) \\[6pt]
& = & f(x)+g(x), \\[6pt]
\end{array} \nonumber \]we conclude that the series \(\displaystyle \sum_{n=0}^ \infty (c_nx^n+d_nx^n)\) converges to \(f(x)+g(x)\).
We examine products of power series in a later theorem.
As with all of the theorems in this section, the two series in question must share a common interval of convergence; however, these theorems hold over the intersection of the respective intervals of convergence for the given series. Thus, if one series converges on \( \left( -4, 7 \right) \) and the other converges on \( \left[ -1,10 \right) \), the theorems from this section would apply to the common interval of convergence, \( \left[ -1,7 \right) \).
We begin by showing several applications of the previous theorem and how to find the interval of convergence of a power series given the interval of convergence of a related power series.
Suppose that \(\displaystyle \sum_{n=0}^ \infty a_n x^n\) is a power series whose interval of convergence is \((−1,1)\), and suppose that \(\displaystyle \sum_{n=0}^ \infty b_n x^n\) is a power series whose interval of convergence is \((−2,2)\)
- Find the interval of convergence of the series \(\displaystyle \sum_{n=0}^ \infty (a_nx^n+b_nx^n)\).
- Find the interval of convergence of the series \(\displaystyle \sum_{n=0}^ \infty a_n3^nx^n\).
- Solutions
-
- Since the interval \((−1,1)\) is a common interval of convergence of the series \(\displaystyle \sum_{n=0}^ \infty a_nx^n\) and \(\displaystyle \sum_{n=0}^ \infty b_nx^n\), the interval of convergence of the series \(\displaystyle \sum_{n=0}^ \infty (a_nx^n+b_nx^n)\) is \((−1,1)\).
- Since \(\displaystyle \sum_{n=0}^ \infty a_nx^n\) is a power series centered at zero with radius of convergence \(1\), it converges for all \(x\) in the interval \((−1,1)\). By the previous theorem, the series\[ \sum_{n=0}^ \infty a_n3^nx^n=\sum_{n=0}^ \infty a_n(3x)^n \nonumber \]converges if \(3x\) is in the interval \((−1,1)\). Hence, the series converges for all \(x\) in the interval \(\left(−\frac{1}{3},\frac{1}{3}\right)\).
Suppose that \(\displaystyle \sum_{n=0}^ \infty a_nx^n\) has an interval of convergence of \((−1,1)\). Find the interval of convergence of \(\displaystyle \sum_{n=0}^ \infty a_n\left(\frac{x}{2}\right)^n\).
- Answer
-
Interval of convergence is \((−2,2)\).
In the next example, we show how to construct power series for functions related to \(f\). Specifically, we consider functions related to the function \(f(x)=\frac{1}{1−x}\) and we use the fact that\[\dfrac{1}{1−x}=\sum_{n=0}^ \infty x^n=1+x+x^2+x^3+\cdots \nonumber \]for \(|x| \lt 1\).
Use the power series representation for \(f(x)=\frac{1}{1−x}\) to construct a power series for each of the following functions. Find the interval of convergence of the power series.
- \(f(x)=\frac{3x}{1+x^2}\)
- \(f(x)=\frac{1}{(x−1)(x−3)}\)
- Solutions
-
- First write \(f(x)\) as\[ f(x)=3x\left(\dfrac{1}{1−(−x^2)}\right). \nonumber \]Using the power series representation for \(f(x)=\frac{1}{1−x}\) and parts ii. and iii. of the previous theorem, we find that a power series representation for \(f\) is given by\[ \sum_{n=0}^ \infty 3x(−x^2)^n=\sum_{n=0}^ \infty 3(−1)^nx^{2n+1}. \nonumber \]Since the interval of convergence of the series for \(\frac{1}{1−x}\) is \((−1,1)\), the interval of convergence for this new series is the set of real numbers \(x\) such that \(∣x^2∣ \lt 1\). Therefore, the interval of convergence is \((−1,1)\).
- To find the power series representation, use partial fractions to write \(f(x)=\frac{1}{(x-1)(x−3)}\) as the sum of two fractions. We have\[ \dfrac{1}{(x−1)(x−3)}=\dfrac{−1/2}{x−1}+\dfrac{1/2}{x−3}=\dfrac{1/2}{1−x}−\dfrac{1/2}{3−x}=\dfrac{1/2}{1−x}−\dfrac{1/6}{1−\dfrac{x}{3}}. \nonumber \]First, we obtain\[ \dfrac{1/2}{1−x}=\sum_{n=0}^ \infty \dfrac{1}{2}x^n \text{ for } |x| \lt 1. \nonumber \]Then, we have\[ \dfrac{1/6}{1−x/3}=\sum_{n=0}^ \infty \dfrac{1}{6}\left(\dfrac{x}{3}\right)^n \text{ for } |x| \lt 3. \nonumber \]Since we are combining these two power series, the interval of convergence of the difference must be the smaller of these two intervals. Using this fact, we have\[ \dfrac{1}{(x−1)(x−3)}=\sum_{n=0}^ \infty \left(\dfrac{1}{2}−\dfrac{1}{6 \cdot 3^n}\right)x^n \nonumber \]where the interval of convergence is \((−1,1)\).
Use the series for \(f(x)=\frac{1}{1−x}\) on \(|x| \lt 1\) to construct a series for \(\frac{1}{(1−x)(x−2)}\). Determine the interval of convergence.
- Answer
-
\(\displaystyle \sum_{n=0}^ \infty \left(−1+\frac{1}{2^{n+1}}\right)x^n\). The interval of convergence is \((−1,1)\).
In Example \(\PageIndex{2}\), we showed how to find power series for specific functions. In Example \(\PageIndex{3}\), we show how to do the opposite: given a power series, determine which function it represents.
Consider the power series \(\displaystyle \sum_{n=0}^ \infty 2^nx^n\). Find the function \(f\) represented by this series. Determine the interval of convergence of the series.
- Solution
-
Writing the given series as\[ \sum_{n=0}^ \infty 2^nx^n=\sum_{n=0}^ \infty (2x)^n, \nonumber \]we can recognize this series as the power series for\[ f(x)=\dfrac{1}{1−2x}. \nonumber \]Since this is a geometric series, the series converges if and only if \(|2x|<1\). Therefore, the interval of convergence is \(\left(−\frac{1}{2},\frac{1}{2}\right)\).
Find the function represented by the power series \(\displaystyle \sum_{n=0}^ \infty \frac{1}{3^n}x^n\). Determine its interval of convergence.
- Answer
-
\(f(x)=\frac{3}{3−x}\). The interval of convergence is \((−3,3)\).
When winning the lottery, an individual can sometimes receive winnings in one lump sum or smaller payments over fixed intervals. For example, you could receive 20 million dollars today or 1.5 million each year for the next 20 years. Which is the better deal?
Luckily, our new knowledge will allow us to use series to compare values of payments over time with a lump sum payment today. We will compute how much future payments are worth in today's dollars, assuming we can invest winnings and earn interest. The value of future payments in today's dollars is known as the present value of those payments.
Suppose you win the lottery and are given the following three options:
- Receive 20 million dollars today;
- Receive 1.5 million dollars per year over the next 20 years; or
- Receive 1 million dollars per year indefinitely (being passed on to your heirs).
Which is the best deal, assuming the annual interest rate is 5%? We answer this by working through the following sequence of questions.
- How much is the 1.5 million dollars received annually over 20 years worth in terms of today's dollars, assuming an annual interest rate of 5%?
- Use the answer to part a. to find a general formula for the present value of payments of \(C\) dollars received each year over the next \(n\) years, assuming an average annual interest rate \(r\).
- Find a formula for the present value if annual payments of \(C\) dollars continue indefinitely, assuming an average annual interest rate \(r\).
- Use the answer to part c. to determine the present value of 1 million dollars paid annually indefinitely.
- Use your answers to parts a. and d. to determine which of the three options is best.
Figure \(\PageIndex{1}\): (credit: modification of work by Robert Huffstutter, Flickr)
- Solutions
-
- Consider the payment of 1.5 million dollars made at the end of the first year. If you receive that payment today instead of one year from now, you could invest that money and earn 5% interest. Therefore, the present value of that money \(P_1\) satisfies \(P_1(1+0.05)=1.5\) million dollars. We conclude that\[P_1=\dfrac{1.5}{1.05}=1.429 \text{ million dollars.} \nonumber \]Similarly, consider the payment of 1.5 million dollars made at the end of the second year. If you receive that payment today, you could invest that money for two years, earning 5% interest compounded annually. Therefore, the present value of that money \(P_2\) satisfies \(P_2(1+0.05)^2=1.5\) million dollars. We conclude that\[P_2=1.5(1.05)^2 = 1.361 \text{ million dollars.} \nonumber \]The value of the future payments today is the sum of the present values \(P_1\), \(P_2\), \(\ldots\), \(P_{20}\) of each of those annual payments. The present value \(P_k\) satisfies\[P_k=\dfrac{1.5}{(1.05)^k}. \nonumber \]Therefore,\[P=\dfrac{1.5}{1.05}+\dfrac{1.5}{(1.05)^2}+\ldots+\dfrac{1.5}{(1.05)^{20}} = 18.693 \text{ million dollars.} \nonumber \]
- Using the result from part a., we see that the present value \(P\) of \(C\) dollars paid annually for \(n\) years, assuming an annual interest rate \(r\), is given by\[P=\dfrac{C}{1+r}+\dfrac{C}{(1+r)^2}+\cdots+\dfrac{C}{(1+r)^n} \text{ dollars.} \nonumber \]
- Using the result from part b. we see that the present value of an annuity that continues indefinitely is given by the infinite series\[P=\sum_{n=0}^ \infty \dfrac{C}{(1+r)^{n+1}}.\nonumber \]We can view the present value as a power series in \(r\), which converges as long as \(\Bigg|\frac{1}{1+r}\Bigg| \lt 1\). Since \(r \gt 0\), this series converges. Rewriting the series as\[P=\dfrac{C}{(1+r)}\sum_{n=0}^ \infty \left(\dfrac{1}{1+r}\right)^n,\nonumber \]we recognize this series as the power series for\[f(r)=\dfrac{1}{1−\left(\frac{1}{1+r}\right)}=\dfrac{1}{\left(\frac{r}{1+r}\right)}=\dfrac{1+r}{r}. \nonumber \]We conclude that the present value of this annuity is\[P=\dfrac{C}{1+r} \cdot \dfrac{1+r}{r}=\dfrac{C}{r}. \nonumber \]
- From the result to part c., we conclude that the present value \(P\) of \(C=1\) million dollars paid out every year indefinitely, assuming an annual interest rate \(r=0.05\), is given by\[P=\dfrac{1}{0.05}=20 \nonumber \]million dollars.
- From part a., we see that receiving 1.5 million dollars over 20 years is worth 18.693 million dollars in today's dollars. From part d., we see that indefinitely receiving $1 million annually is worth $20 million in today's dollars. Therefore, either receiving a lump sum payment of $20 million today or receiving $1 million indefinitely have the same present value.
Multiplication of Power Series
We can also create new power series by multiplying power series. Multiplying two power series provides another way of finding power series representations for functions. The way we multiply them is similar to how we multiply polynomials. For example, suppose we want to multiply\[\sum_{n=0}^ \infty c_nx^n = c_0+c_1x+c_2x^2+\cdots \nonumber \]and\[\sum_{n=0}^ \infty d_nx^n=d_0+d_1 x+d_2x^2+\cdots. \nonumber \]It appears that the product should satisfy\[\left(\sum_{n=0}^ \infty c_nx^n\right)\left(\sum_{n=−0}^ \infty d_nx^n\right)=(c_0+c_1x+c_2x^2+\cdots) \cdot (d_0+d_1x+d_2x^2+\cdots)=c_0d^0+(c_1d^0+c_0d^1)x+(c_2d^0+c_1d^1+c_0d^2)x^2+\cdots. \nonumber \]In the following theorem, we state the main result regarding multiplying power series, showing that if \(\displaystyle \sum_{n=0}^ \infty c_nx^n\) and \(\displaystyle \sum_{n=0}^ \infty d_nx^n\) converge on a common interval \(I\), then we can multiply the series in this way, and the resulting series also converges on the interval \(I\).
Suppose that the power series \(\displaystyle \sum_{n=0}^ \infty c_n x^n\) and \(\displaystyle \sum_{n=0}^ \infty d_n x^n\) converge to \(f\) and \(g\), respectively, on a common interval \(I\). Let\[e_n=c_0d_n+c_1d_{n−1}+c_2d_{n−2}+\cdots+c_{n−1}d_1+c_nd_0=\sum_{k=0}^nc_kd_{n−k}. \nonumber \]Then\[\left(\sum_{n=0}^ \infty c_nx^n\right)\left(\sum_{n=0}^ \infty d_nx^n\right)=\sum_{n=0}^ \infty e_nx^n \nonumber \]and\[\sum_{n=0}^ \infty e_nx^n \text{ converges to } f(x) \cdot g(x) \text{ on } I. \nonumber \]The series \(\displaystyle \sum_{n=0}^ \infty e_nx^n\) is known as the Cauchy product of the series \(\displaystyle \sum_{n=0}^ \infty c_nx^n\) and \(\displaystyle \sum_{n=0}^ \infty d_nx^n\).1
We omit the proof of this theorem as it is beyond the level of this text and is typically covered in a more advanced course. We now provide an example of this theorem by finding the power series representation for\[f(x)=\dfrac{1}{(1−x)(1−x^2)} \nonumber \]using the power series representations for\[y=\dfrac{1}{1−x} \text{ and } y=\dfrac{1}{1−x^2}. \nonumber \]
Multiply the power series representation\[\dfrac{1}{1−x}=\sum_{n=0}^ \infty x^n=1+x+x^2+x^3+\cdots \nonumber \]for \(|x| \lt 1\) with the power series representation\[\dfrac{1}{1−x^2}=\sum_{n=0}^ \infty \left( x^2 \right) ^n=1+x^2+x^4+x^6+\cdots \nonumber \]for \(|x| \lt 1\) to construct a power series for \(f(x)=\frac{1}{(1−x)(1−x^2)}\) on the interval \((−1,1)\).
- Solution
-
We need to multiply\[(1+x+x^2+x^3+\cdots)(1+x^2+x^4+x^6+\cdots).\nonumber \]We note that this product will have a constant term and terms for \( x \), \( x^2 \), \( x^3 \), and so on. Hence, we see that the product is given by\[\begin{array}{rcl}
(1+x+x^2+x^3+\cdots)(1+x^2+x^4+x^6+\cdots) & = & 1+x+(1+1)x^2+(1+1)x^3+(1+1+1)x^4+(1+1+1)x^5+\cdots \\[6pt]
& = & 1+x+2x^2+2x^3+3x^4+3x^5+\cdots. \\[6pt]
\end{array} \nonumber \]Since the series for \(y=\frac{1}{1−x}\) and \(y=\frac{1}{1−x^2}\) both converge on the interval \((−1,1)\), the series for the product also converges on the interval \((−1,1)\).
As you progress in Mathematics, there will be a few moments (specifically, in Differential Equations) when you must find the product of two power series. It's a common convention to list the expansion's first three or four nonzero terms.
Multiply the series \(\frac{1}{1−x}= \displaystyle \sum_{n=0}^ \infty x^n\) by itself to construct a series for \(\frac{1}{(1−x)(1−x)}\).
- Answer
-
\(1+2x+3x^2+4x^3+\cdots\)
Differentiating and Integrating Power Series
Consider a power series \(\displaystyle \sum_{n=0}^ \infty c_n x^n = c_0+c_1x+c_2x^2+\cdots\) that converges on some interval \(I\), and let \(f\) be the function defined by this series. Here we address two questions about \(f\):
- Is \(f\) differentiable, and if so, how do we determine the derivative \(f^{\prime}\)?
- How do we evaluate the indefinite integral \( \int f(x)\,dx\)?
We know we can evaluate the derivative by differentiating each term separately for a polynomial with a finite number of terms. Similarly, we can evaluate the indefinite integral by integrating each term separately. We show we can do the same for convergent power series. That is, if\[f(x)=c_nx^n=c_0+c_1x+c_2x^2+\cdots \nonumber \]converges on some interval \( I \), then\[f^{\prime}(x)=c_1+2c_2x+3c_3x^2+\cdots \nonumber \]and\[ \int f(x)\,dx=C+c_0x+c_1\dfrac{x^2}{2}+c_2\dfrac{x^3}{3}+\cdots. \nonumber \]Evaluating the derivative and indefinite integral in this way is called term-by-term differentiation of a power series and term-by-term integration of a power series, respectively.
The ability to differentiate and integrate power series term-by-term also allows us to use known power series representations to find power series representations for other functions. For example, given the power series for \(f(x)=\frac{1}{1−x}\), we can differentiate term-by-term to find the power series for \(f^{\prime}(x)=\frac{1}{(1−x)^2}\). Similarly, using the power series for \(g(x)=\frac{1}{1+x}\), we can integrate term-by-term to find the power series for \(G(x)=\ln(1+x)\), an antiderivative of \( g \). We show how to do this in Example \(\PageIndex{6}\) and Example \(\PageIndex{7}\). First, we state the following theorem, which provides the main result regarding power series differentiation and integration.
Suppose that the power series \(\displaystyle \sum_{n=0}^ \infty c_n (x−a)^n\) converges on the interval \((a−R,a+R)\) for some \(R \gt 0\). Let \( f \) be the function defined by the series\[f(x)=\sum_{n=0}^ \infty c_n(x−a)^n = c_0+c_1(x−a)+c_2(x−a)^2+c_3(x−a)^3+\cdots \nonumber \]for \(|x−a| \lt R\). Then \( f \) is differentiable on the interval \((a−R,a+R)\) and we can find \(f^{\prime}\) by differentiating the series term-by-term:\[f^{\prime}(x) = \sum_{n=1}^ \infty n c_n(x−a)^{n−1} = c_1+2c_2(x−a)+3c_3(x−a)^2+\cdots \nonumber \]for \(|x−a| \lt R\). Also, to find \( \int f(x)\, dx\), we can integrate the series term-by-term. The resulting series converges on \((a−R,a+R)\), and we have\[ \int f(x)\,dx=C+\sum_{n=0}^ \infty c_n\dfrac{(x−a)^{n+1}}{n+1}=C+c_0(x−a)+c_1\dfrac{(x−a)^2}{2}+c_2\dfrac{(x−a)^3}{3}+\cdots \nonumber \]for \(|x−a| \lt R\).
This result's proof is omitted and beyond the text's scope. Note that although this theorem guarantees the same radius of convergence when a power series is differentiated or integrated term-by-term, it says nothing about what happens at the endpoints. The differentiated and integrated power series may behave differently at the endpoints than the original series. We see this behavior in the next examples.
- Use the power series representation\[f(x)=\dfrac{1}{1−x}=\sum_{n=0}^ \infty x^n=1+x+x^2+x^3+\cdots \nonumber \]for \(|x| \lt 1\) to find a power series representation for\[g(x)=\dfrac{1}{(1−x)^2} \nonumber \]on the interval \((−1,1)\). Determine whether the resulting series converges at the endpoints.
- Use the result of part a. to evaluate the sum of the series \(\displaystyle \sum_{n=0}^ \infty \frac{n+1}{4^n}\).
- Solutions
-
- Since \(g(x)=\frac{1}{(1−x)^2}\) is the derivative of \(f(x)=\frac{1}{1−x}\), we can find a power series representation for \(g\) by differentiating the power series for \(f\) term-by-term. The result is\[\begin{array}{rcl}
g(x) & = & \dfrac{1}{(1−x)^2} \\[6pt]
& = & \dfrac{d}{\,dx}(\dfrac{1}{1−x}) \\[6pt]
& = & \displaystyle \sum_{n=0}^ \infty \dfrac{d}{\,dx}(x^n) \\[6pt]
& = & \dfrac{d}{\,dx}(1+x+x^2+x^3+\cdots) \\[6pt]
& = & 0+1+2x+3x^2+4x^3+\cdots \\[6pt]
& = & \displaystyle \sum_{n=0}^ \infty (n+1)x^n \\[6pt]
\end{array} \nonumber \]for \(|x| \lt 1\). Theorem \(\PageIndex{1}\) does not guarantee anything about the behavior of this series at the endpoints. Testing the endpoints by using the Divergence Test, we find that the series diverges at both endpoints \(x= \pm 1\). - From part a. we know that\[\sum_{n=0}^ \infty (n+1)x^n=\dfrac{1}{(1−x)^2}. \nonumber \]Therefore,\[\begin{array}{rcl}
\displaystyle \sum_{n=0}^ \infty \dfrac{n+1}{4^n} & = & \sum_{n=0}^ \infty (n+1) \left(\dfrac{1}{4} \right)^n \\[6pt]
& = & \dfrac{1}{\left(1−\frac{1}{4} \right)^2}\\[6pt]
& = & \dfrac{1}{\left(\frac{3}{4} \right)^2}\\[6pt]
& = & \dfrac{16}{9} \\[6pt]
\end{array} \nonumber \]
- Since \(g(x)=\frac{1}{(1−x)^2}\) is the derivative of \(f(x)=\frac{1}{1−x}\), we can find a power series representation for \(g\) by differentiating the power series for \(f\) term-by-term. The result is\[\begin{array}{rcl}
Differentiate the series \(\frac{1}{(1−x)^2}= \displaystyle \sum_{n=0}^ \infty (n+1)x^n\) term-by-term to find a power series representation for \(\frac{2}{(1−x)^3}\) on the interval \((−1,1)\).
- Answer
-
\(\displaystyle \sum_{n=0}^ \infty (n+2)(n+1)x^n\)
For each of the following functions \( f \), find a power series representation for \( f \) by integrating the power series for \(f^{\prime}\) and find its interval of convergence.
- \(f(x)=\ln(1+x)\)
- \(f(x)=\tan^{−1}x\)
- Solutions
-
- For \(f(x)=\ln(1+x)\),the derivative is \(f^{\prime}(x)=\frac{1}{1+x}\). We know that\[\dfrac{1}{1+x}=\dfrac{1}{1−(−x)}=\sum_{n=0}^ \infty (−x)^n=1−x+x^2−x^3+\cdots\nonumber \]for \(|x| \lt 1\). To find a power series for \(f(x)=\ln(1+x)\), we integrate the series term-by-term.\[ \int f^{\prime}(x)\,dx= \int (1−x+x^2−x^3+\cdots)\,dx=C+x−\dfrac{x^2}{2}+\dfrac{x^3}{3}−\dfrac{x^4}{4}+\cdots\nonumber \]Since \(f(x)=\ln(1+x)\) is an antiderivative of \(\frac{1}{1+x}\), it remains to solve for the constant \(C\). Since \(\ln(1+0)=0\), we have \(C=0\). Therefore, a power series representation for \(f(x)=\ln(1+x)\) is\[\ln(1+x)=x−\dfrac{x^2}{2}+\dfrac{x^3}{3}−\dfrac{x^4}{4}+\cdots=\sum_{n=1}^ \infty (−1)^{n+1}\dfrac{x^n}{n} \text{ for } |x| \lt 1.\nonumber \]Theorem \(\PageIndex{1}\) does not guarantee anything about the behavior of this power series at the endpoints. However, checking the endpoints, we find that at \(x=1\), the series is the Alternating Harmonic Series, which converges. Also, at \(x=−1\), the series is the Harmonic Series, which diverges. It is important to note that, even though this series converges at \(x=1\), we cannot guarantee that the series converges to \(\ln(2)\). In fact, the series converges \(\ln(2)\), but showing this fact requires more advanced techniques.2 The interval of convergence is \((−1,1]\).
- The derivative of \(f(x)=\tan^{−1}x\) is \(f^{\prime}(x)=\frac{1}{1+x^2}\). We know that\[ \displaystyle \dfrac{1}{1+x^2}=\dfrac{1}{1−(−x^2)}=\sum_{n=0}^ \infty (−x^2)^n=1−x^2+x^4−x^6+\cdots \nonumber \]for \(|x| \lt 1\). To find a power series for \(f(x)=\tan^{−1}x\), we integrate this series term-by-term.\[ \int f^{\prime}(x)\,dx= \int (1−x^2+x^4−x^6+\cdots)\,dx=C+x−\dfrac{x^3}{3}+\dfrac{x^5}{5}−\dfrac{x^7}{7}+\cdots. \nonumber \]Since \(\tan^{−1}(0)=0\), we have \(C=0\). Therefore, a power series representation for \(f(x)=\tan^{−1}x\) is\[ \tan^{−1}x=x−\dfrac{x^3}{3}+\dfrac{x^5}{5}−\dfrac{x^7}{7}+\cdots=\sum_{n=0}^ \infty (−1)^n\dfrac{x^{2n+1}}{2n+1}\nonumber \]for \(|x| \lt 1\). Again, Theorem \(\PageIndex{1}\) does not guarantee anything about the convergence of this series at the endpoints. However, checking the endpoints and using the Alternating Series Test, we find that the series converges at \(x=1\) and \(x=−1\). As discussed in the footnote to part a., using Abel's Theorem, it can be shown that the series converges to \(\tan^{−1}(1)\) and \(\tan^{−1}(−1)\) at \(x=1\) and \(x=−1\), respectively. Thus, the interval of convergence is \([−1,1]\).
Integrate the power series \(\displaystyle \ln(1+x)=\sum_{n=1}^ \infty (−1)^{n+1}\frac{x^n}{n}\) term-by-term to evaluate \(\displaystyle \int \ln(1+x)\,dx\).
- Answer
-
\(\displaystyle \sum_{n=2}^ \infty \frac{(−1)^nx^n}{n(n−1)}\)
Up to this point, we have shown several techniques for finding power series representations for functions. However, how do we know that these power series are unique? That is, given a function \(f\) and a power series for \(f\) at \(a\), is it possible that there is a different power series for \(f\) at \(a\) that we could have found if we had used a different technique? The answer to this question is no. This fact should not seem surprising if we consider power series as polynomials with infinite terms. Intuitively, if\[c_0 + c_1x + c_2x^2 + \cdots = d_0 + d_1x + d_2x^2 + \cdots \nonumber \]for all values \(x\) in some open interval \( I \) about zero, then the coefficients \(c_n\) should equal \(d_n\) for \(n \geq 0\). We now state this result formally.
Let \(\displaystyle \sum_{n=0}^ \infty c_n(x−a)^n\) and \(\displaystyle \sum_{n=0}^ \infty d_n(x−a)^n\) be two convergent power series such that\[\sum_{n=0}^ \infty c_n(x−a)^n=\sum_{n=0}^ \infty d_n(x−a)^n \nonumber \]for all \( x \) in an open interval containing \(a\). Then \(c_n=d_n\) for all \(n \geq 0\).
- Proof
-
Let\[\begin{array}{rcl}
f(x) & = & c_0+c_1(x−a)+c_2(x−a)^2+c_3(x−a)^3+\cdots \\[6pt]
& = & d_0+d_1(x−a)+d_2(x−a)^2+d_3(x−a)^3+\cdots. \\[6pt]
\end{array} \nonumber \]Then \(f(a)=c_0=d_0\). By Theorem \( \PageIndex{3} \), we can differentiate both series term-by-term. Therefore,\[\begin{array}{rcl}
f^{\prime}(x) & = & c_1+2c_2(x−a)+3c_3(x−a)^2+\cdots \\[6pt]
& = & d_1+2d_2(x−a)+3d_3(x−a)^2+\cdots, \\[6pt]
\end{array} \nonumber \]and thus, \(f^{\prime}(a)=c_1=d_1\). Similarly,\[\begin{array}{rcl}
f^{\prime\prime}(x) & = & 2c_2+3 \cdot 2c_3(x−a)+\cdots \\[6pt]
& = & 2d_2+3 \cdot 2d_3(x−a)+\cdots \\[6pt]
\end{array} \nonumber \]implies that \(f^{\prime\prime}(a)=2c_2=2d_2\), and therefore, \(c_2=d_2\). More generally, for any integer \(n \geq 0\), \(f^{(n)} (a)=n!c_n=n!d_n\), and consequently, \(c_n=d_n\) for all \(n \geq 0\).
In this section, we have shown how to find power series representations for certain functions using various algebraic operations, differentiation, or integration. However, we are still limited in the functions for which we can find power series representations. Next, by introducing the Taylor series, we show how to find power series representations for many more functions.
Footnotes
1 This theorem is included for completeness; however, your instructor might "approximate" this theorem (which is perfectly fine) by stating that the product of two convergent power series is best computed by finding the first three or four lowest powers of the product. That is,\[\begin{array}{rcl}
\left(\sum_{n=0}^ \infty c_nx^n\right)\left(\sum_{n=0}^ \infty d_nx^n\right) & = & \left( c_0 + c_1 x + c_2 x^2 + \cdots \right)\left( d_0 + d_1 x + d_2 x^2 + \cdots \right) \\[6pt]
& = & c_0 d_0 + \left( c_0 d_1 + c_1 d_0 \right) x + \left( c_0 d_2 + c_1 d_1 + c_2 d_0 \right) x^2 + \cdots \\[6pt]
\end{array} \nonumber \]
2 Abel's Theorem, covered in more advanced texts, deals with this more technical point.
Key Concepts
- Given two power series \(\displaystyle \sum_{n=0}^ \infty c_nx^n\) and \(\displaystyle \sum_{n=0}^ \infty d_nx^n\) that converge to functions \(f\) and \(g\) on a common interval \(I\), the sum and difference of the two series converge to \(f \pm g\), respectively, on \(I\). In addition, for any real number \(b\) and integer \(m \geq 0\), the series \(\displaystyle \sum_{n=0}^ \infty bx^mc_nx^n\) converges to \(bx^mf(x)\) and the series \(\displaystyle \sum_{n=0}^ \infty c_n(bx^m)^n\) converges to \(f(bx^m)\) whenever \(bx^m\) is in the interval \(I\).
- Given two power series that converge on an interval \((−R, R)\), the Cauchy product of the two power series converges on the interval \((−R, R)\).
- Given a power series that converges to a function \(f\) on an interval \((−R,R)\), the series can be differentiated term-by-term and the resulting series converges to \(f^{\prime}\) on \((−R,R)\). The series can also be integrated term-by-term and the resulting series converges to \( \int f(x)\,dx\) on \((−R,R)\).
Glossary
- term-by-term differentiation of a power series
- a technique for evaluating the derivative of a power series \(\displaystyle \sum_{n=0}^ \infty c_n(x−a)^n\) by considering the derivative of each term separately to create the new power series \(\displaystyle \sum_{n=1}^ \infty nc_n(x−a)^{n−1}\)
- term-by-term integration of a power series
- a technique for integrating a power series \(\displaystyle \sum_{n=0}^ \infty c_n(x−a)^n\) by integrating each term separately to create the new power series \(\displaystyle C+\sum_{n=0}^ \infty c_n\frac{(x−a)^{n+1}}{n+1}\)