4.2: Properties of Power Series

Combining Power Series

Theorem: Combining Power Series

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.

1. The power series \(\displaystyle \sum_{n=0}^\infty (c_n x^n \pm d_n x^n)\) converges to \(f \pm g\) on \(I\).
2. 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 \(I\).
3. 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 (in the text)

Caution: Common Interval of Convergence Required

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, \( \left[ -1,7 \right) \).

Lecture Example \(\PageIndex{1}\)

Suppose that \(\displaystyle \sum_{n=0}^ \infty a_n x^n\) is a power series whose interval of convergence is \((−4,4)\), and suppose that \(\displaystyle \sum_{n=0}^ \infty b_n x^n\) is a power series whose interval of convergence is \((−6,6)\)

1. Find the interval of convergence of the series \(\displaystyle \sum_{n=0}^ \infty (a_nx^n+b_nx^n)\).
2. Find the interval of convergence of the series \(\displaystyle \sum_{n=0}^ \infty a_n2^nx^n\).

Lecture Example \(\PageIndex{2}\)

Find a power series for each of the following functions. Find the interval of convergence of the power series.

1. \(f(x)=\frac{1 + x}{1-x}\)
2. \(f(x)=\frac{6}{x^2 - 2x −8}\)

Multiplication of Power Series

Theorem: Product of Power Series

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 \(\displaystyle \sum_{n=0}^ \infty d_nx^n\).

Lecture Example \(\PageIndex{3}\)

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 \((−1,1)\)

Note: The Expansion of Products of Power Series

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.

Differentiating and Integrating Power Series

Theorem: Term-by-Term Differentiation and Integration for Power Series

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 \(|x−a| \lt R\).

Lecture Example \(\PageIndex{4}\)

Find a power series for \( f^{\prime}(x) \) given\[ f(x) = \dfrac{x^6}{1 - x}. \nonumber \]

Advice

When trying to find a power series representation for a function that is not a rational function of the form \( \frac{f(x)}{a +bx} \), it's best to ask, "What would I integrate or differentiate to get this?"

Lecture Example \(\PageIndex{5}\)

Find a power series for the following function.\[ f(x) = \dfrac{-8x^7}{(1 + x^8)^2} \nonumber \]

Hint

This is the derivative of \( \frac{1}{1+x^8} \)

Lecture Example \(\PageIndex{6}\)

Use a power series to approximate the value of the definite integral to six decimal places.\[ \int_0^{0.3} \dfrac{x^2}{1 + x^4} \, dx \nonumber \]

Hint

\( \displaystyle \sum_{n = 0}^{\infty} \frac{(-1)^{n} x^{4n + 3}}{4n + 3} \) is an alternating series, so the error is bounded by \( b_{n + 1} \). You are correct if \( n = 2\) and the approximation is \( 0.00896891\ldots \).

Theorem

\[ \ln(1 + x) = \sum_{n = 1}^{\infty} \dfrac{(-1)^{n - 1} x^n}{n}. \quad (-1,1] \nonumber \]

Proof

Hint: \( \ln(1 + x) = \int \frac{1}{1 + x} \, dx  \)

Lecture Example \(\PageIndex{7}\)

Find a power series representation for the function and determine the radius of convergence.\[ f(x) = \dfrac{\ln(1 - x^9)}{x} \nonumber \]

Theorem

\[ \tan^{-1}(x) = \sum_{n = 0}^{\infty} (-1)^n \dfrac{x^{2n + 1}}{2n + 1}, \quad |x| \leq 1 \nonumber \]

Proof

Lecture Example \(\PageIndex{8}\)

Find a power series representation for the function and determine the radius of convergence.\[ f(x) = x^2 \tan^{-1}(x^3) \nonumber \]

Online Lecture Example \(\PageIndex{9}\)

Find a closed-form for the sum of the series\[ \sum_{n = 2}^{\infty} n(n - 1)x^{n - 2}. \nonumber \]

Theorem: Uniqueness of Power Series

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 \(n \geq 0\).