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[math] \newcommand{\ex}[1]{\item } \newcommand{\sx}{\item} \newcommand{\x}{\sx} \newcommand{\sxlab}[1]{} \newcommand{\xlab}{\sxlab} \newcommand{\prov}[1] {\quad #1} \newcommand{\provx}[1] {\quad \mbox{#1}} \newcommand{\intext}[1]{\quad \mbox{#1} \quad} \newcommand{\R}{\mathrm{\bf R}} \newcommand{\Q}{\mathrm{\bf Q}} \newcommand{\Z}{\mathrm{\bf Z}} \newcommand{\C}{\mathrm{\bf C}} \newcommand{\dt}{\textbf} \newcommand{\goesto}{\rightarrow} \newcommand{\ddxof}[1]{\frac{d #1}{d x}} \newcommand{\ddx}{\frac{d}{dx}} \newcommand{\ddt}{\frac{d}{dt}} \newcommand{\dydx}{\ddxof y} \newcommand{\nxder}[3]{\frac{d^{#1}{#2}}{d{#3}^{#1}}} \newcommand{\deriv}[2]{\frac{d^{#1}{#2}}{dx^{#1}}} \newcommand{\dist}{\mathrm{distance}} \newcommand{\arccot}{\mathrm{arccot\:}} \newcommand{\arccsc}{\mathrm{arccsc\:}} \newcommand{\arcsec}{\mathrm{arcsec\:}} \newcommand{\arctanh}{\mathrm{arctanh\:}} \newcommand{\arcsinh}{\mathrm{arcsinh\:}} \newcommand{\arccosh}{\mathrm{arccosh\:}} \newcommand{\sech}{\mathrm{sech\:}} \newcommand{\csch}{\mathrm{csch\:}} \newcommand{\conj}[1]{\overline{#1}} \newcommand{\mathds}{\mathbb} [/math]

Functions Defined by Power Series.

For every power series

[[math]] \sum_{i=0}^\infty a_i(x - a)^i, [[/math]]

the \textbf{function defined by the power series} is the function [math]f[/math] which, to every real number [math]c[/math] at which the power series converges, assigns the real number [math]f(c)[/math] given by

[[math]] f(c) = \sum_{i=0}^\infty a_i(c - a)^i. [[/math]]

The domain of [math]f[/math] is obviously equal to the interval of convergence of the power series. Speaking more casually, we say simply that the function [math]f[/math] is defined by the equation

[[math]] f(x) = \sum_{i=0}^\infty a_i(x - a)^i. [[/math]]

As an example, let [math]f[/math] be the function defined by

[[math]] f(x) = \sum_{i=0}^\infty \frac{x^i}{ i!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots . [[/math]]

This power series was studied in Example 1 of Section 6 and was shown to converge for all values of [math]x[/math]. Thus the domain of the function which the series defines is the set of all real numbers. Functions defined by power series have excellent analytic properties. One of the most important is the fact that every such function is differentiable and that its derivative is the function defined by the power series obtained by differentiating the original series term by term. That is, if

[[math]] f(x) = \sum_{i=0}^\infty a_{i}(x-a)^i = a_0 + a_{1}(x-a) + a_{2}(x-a)^2 + \cdots , [[/math]]

then

[[math]] f'(x) = \sum_{i=1}^\infty i a_{i} (x-a)^{i-1} = a_{1} + 2a_{2}(x-a) + 3a_{3}(x-a)^2 + \cdots . [[/math]]

This is not a trivial result. To prove it, we begin with the following theorem:

Theorem

Suppose that [math]\rho \lt \rho'[/math], and let [math]c[/math] be an arbitrary real number such that [math]\rho \lt c \lt \rho'[/math]. Then the series [math]\sum_{i=0}^\infty a_{i}c^i[/math] diverges, whereas the series [math]\sum_{i=1}^\infty i a_{i}c^{i-1}[/math] converges absolutely. Since [math]c[/math] is positive,

[[math]] c \sum_{i=1}^\infty |ia_{i}c^{i-1}| = \sum_{i=1}^\infty |ia_{i}c^i|, [[/math]]

and it follows that the series [math]\sum_{i=1}^\infty = ia_{i}c^i[/math] is also absolutely convergent. However, it is obvious that, for every positive integer [math]i[/math],

[[math]] |a_{i}c^i| \leq i |a_{i}c^i| = |ia_{i}c^i|. [[/math]]

The Comparison Test therefore implies that the series [math]\sum_{i=1}^\infty |a_{i}c^i|[/math] converges, and this fact implies the convergence of [math]\sum_{i=0}^\infty a_{i}c^i[/math], which is a contradiction. Hence the original assumption is false, and we conclude that

[[math]] \begin{equation} \rho' \leq \rho. \label{eq9.7.1} \end{equation} [[/math]]

Next, suppose that [math]\rho' \lt \rho[/math]. We shall derive a contradiction from this assumption also, which, together with the inequality (1), proves that [math]\rho' = \rho[/math]. Let [math]b[/math] and [math]c[/math] be any two real numbers such that [math]\rho' \lt b \lt c \lt \rho[/math]. It follows from the definition of [math]\rho'[/math] that the series [math]\sum_{i=1}^\infty ia_{i}b^{i-1}[/math] diverges. Similarly, from the definition of [math]\rho[/math], we know that the series [math]\sum_{i=0}^\infty a_{i}c^i[/math] converges, and therefore [math]\lim_{i \rightarrow \infty} a_ic^i = 0[/math]. Because [math]c[/math] is positive, it follows that there exists a positive integer [math]N[/math] such that, for every integer [math]i \geq N[/math],

[[math]] |a_{i}c^i| \lt c. [[/math]]

But [math]|a_ic^i| = |a_i|c^i[/math], and so the preceding inequality becomes [math]|a_i|c^i \lt c[/math], or, equivalently,

[[math]] |a_i| \lt \frac{1}{c^{i-1}}. [[/math]]

Hence, since [math]b[/math] is also positive, we obtain

[[math]] |ia_{i}b^{i-1}| = ib^{i-1} |a_i| \lt i \frac{b^{i-1}}{c^{i-1}} = i\Bigl(\frac{b}{c}\Bigr)^{i-1} , [[/math]]

for every integer [math]i \geq N[/math]. Let us set [math]\frac{b}{c} = r[/math]. Then [math]0 \lt r \lt 1[/math], and we have shown that

[[math]] |ia_{i}b^{i-1}| \lt ir^{i-1}, \;\;\;\mbox{for every integer}\; i \geq N . [[/math]]

However, it is shown in Example 3, page 508, that the series [math]\sum_{i=1}^\infty ir^{i-1}[/math] converges if [math]|r| \lt 1[/math]. Hence the preceding inequality and the Comparison Text imply that the series [math]\sum_{i=1}^\infty |ia_{i}b^{i-1}|[/math] converges, and this contradicts the above conclusion that [math]\sum_{i=1}^\infty ia_{i}b^{i-1}[/math] diverges. This completes the proof that [math]\rho' = \rho[/math], and, as we have remarked, also proves (7.1).

Show Proof

{{{4}}}

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Note that Theorem (7.1) does not state that a power series [math]\sum_{i=0}^\infty a_{i}(x-a)^i[/math] and its derived series have the same intertval of convergence, but only that they have the same radius of convergence. For example, in Example 1(b), page 514, the interval of convergence of the power series [math]\sum_{k=1}^\infty (-1)^{k-1} \frac{x^k}{k}[/math] is shown to be the half-open interval [math](-1,1][/math]. However, the derived series is

[[math]] \begin{eqnarray*} \sum_{k=1}^\infty (-1)^{k-1} k \frac{x^{k-1}}{k} &=& \sum_{k=1}^\infty ( -1)^{k-1} x^{k-1} \\ &=& 1 - x + x^2 - x^3 + \cdots , \end{eqnarray*} [[/math]]

which does not converge for [math]x = 1[/math]. It is a geometric series having the open interval of convergence [math](-1, 1)[/math]. Let [math]\sum_{i=0}^\infty a_{i}(x-a)^i[/math] be a power series, and let [math]f[/math] and [math]g[/math] be the two functions defined respectively by [math]f(x) = \sum_{i=0}^\infty a_{i}(x-a)^i[/math] and by [math]g(x) = \sum_{i=1}^\infty ia_{i}(x -a)^{i-1}[/math]. We have proved that there is an interval, which, with the possible exception of its endpoints, is the common domain of [math]f[/math] and [math]g[/math]. However, we have not yet proved that the function [math]g[/math] is the derivative of the function [math]f[/math]. This fact is the content of the following theorem.

Theorem

THEOREM. If the radius of convergence [math]\rho[/math] of the power series [math]\sum_{i=0}^\infty a_i (x-a)^i[/math] is not zero, then the function [math]f[/math] defined by [math]f(x) = \sum_{i=0}^\infty a_i(x-a)^i[/math] is differentiable at ecery [math]x[/math] such that [math]|x-a| \lt \rho[/math] and

[[math]] f'(x) = \sum_{i=1}^\infty ia_i(x-a)^{i-1}. [[/math]]

Show Proof

{{{4}}}

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Example

(a) Show that

[[math]] e^x = \sum_{i=0}^\infty \frac{x^i}{i!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots . [[/math]]

for every real number [math]x[/math], and (b) show that

[[math]] \ln(1 + x) = \sum_{k=1}^\infty (-1)^{k-1} \frac{x^k}{k} = x - \frac{x^2}{2!} + \frac{x^3}{3!} - \cdots . [[/math]]

for every real number [math]x[/math] such that [math]|x| \lt 1[/math]. For (a), let [math]f[/math] be the function defined by [math]f(x) = \sum_{i=0}^\infty \frac{x^i}{i!}[/math]. We have already shown that the domain of [math]f[/math] is the set of all real numbers; i.e., the radius of convergence is infinite. It follows from the preceding theorem that

[[math]] f'(x) = \sum_{i=1}^\infty i \frac{x^{i-1}}{i!} , \;\;\;\mbox{for every real number}\; x. [[/math]]

Since [math]\frac{i}{i!} = \frac{1}{(i- 1)!}[/math], we obtain

[[math]] \sum_{i=1}^\infty \frac{ix^{i-1}}{i!} = \sum_{i=1}^\infty \frac{x^{i-1}}{(i-1)!} = \sum_{k=0}^\infty \frac{x^k}{k!}, [[/math]]

where the last equation is obtained by replacing [math]i - 1[/math] by [math]k[/math]. Thus we have proved that

[[math]] f'(x) = f(x)\;\;\;\mbox{for every real number}\; x. [[/math]]

The function [math]f[/math] therefore satisfies the differential equation [math]\frac{dy}{dx} = y[/math], whose general solution is [math]y = ce^x[/math]. Hence [math]f(x) = ce^x[/math] for some constant [math]c[/math]. But it is obvious from the series which defines [math]f[/math] that [math]f(0) = 1[/math]. It follows that [math]c = 1[/math], and (a) is proved. A similar technique is used for (b). Let [math]f(x) = \sum_{k=1}^\infty (-1)^{k-1} \frac{x^k}{k}[/math]. The domain of [math]f[/math] is the half-open interval [math](-1, 1][/math], and the radius of convergence is 1. Hence

[[math]] \begin{eqnarray*} f'(x) &=& \sum_{k=1}^\infty (-1)^{k-1} \frac{k-1}{k} = \sum_{k=1}^\infty (-1 )^{k-1} x^{k-1}\\ &=& 1 - x + x^2 - x^3 + \cdots , \end{eqnarray*} [[/math]]

for [math]|x| \lt 1[/math]. The latter is a geometric series with sum equal to [math]\frac{1}{1 + x}[/math]. Hence we have shown that

[[math]] f'(x) = \frac{dx}{1+x}, \;\;\;\mbox{for every $x$ such that}\; |x| \lt 1. [[/math]]

Integration yields

[[math]] f(x) = \int \frac{dx}{1+x} = \ln |1 +x| + c. [[/math]]

Since [math]|x| \lt 1[/math], we have [math]|1 + x| = (1 + x)[/math]. From the series which defines [math]f[/math] we see that [math]f(0) = 0[/math]. Hence

[[math]] 0 = f(0) = \ln(1 + 0) + c = 0 + c = c. [[/math]]

lt follows that [math]f(x) = \ln(1 + x)[/math] for every real number [math]x[/math] such that [math]|x| \lt 1[/math], and (b) is therefore established.

Example 1 illustrates an important point. The domain of the function [math]f[/math] defined by [math]f(x) = \sum_{k=1}^\infty (-1)^{k-1} \frac{x^k}{k} [/math] is the half-open interval [math](-1, 1][/math]. On the other hand, the domain of the function [math]\ln(1 + x)[/math] is the unbounded interval [math](-1, \infty)[/math]. It is essential to realize that the equation

[[math]] \ln (1 + x) = \sum_{k=1}^\infty (-1)^{k-1} \frac{x^k}{k} [[/math]]

has been shown to hold only for values of $x$ which are interior points of the interval of convergence of the power series. It certainly does not hold at points outside the interval of convergence where the series diverges. As far as the endpoints of the interval are concerned, it can be proved that a function defined by a power series is continuous at every point of its interval of convergence. Hence the above equation does in fact hold for [math]x = 1[/math], and we therefore obtain the following formula for the sum of the alternating harmonic series:

[[math]] \ln 2 = \sum_{k=1}^\infty (-1)^{k-1} \frac{1}{k} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots . [[/math]]

Let [math]\sum_{i=0}^\infty a_{i}(x - a)^i[/math] be a power series with a nonzero radius of convergence [math]\rho[/math], and let [math]f[/math] be the function defined by the power series

[[math]] f(x) = \sum_{i=0}^\infty a_{i}(x - a)^i, [[/math]]

for every [math]x[/math] in the interval of convergence. By iterated applications of Theorem (7.2), i.e., first to the series, then to the derived series, then to the derived series of the derived series, etc., we may conclude that [math]f[/math] has derivatives of arbitrarily high order within the radius of convergence. The formula for the nth derivative is easily seen to be

Theorem

[[math]] f^{(n)}(x) = \sum_{i=n}^\infty i(i - 1) \cdots (i - n + 1)a_i(x - a)^{i-n} , [[/math]]

for every $x$ such that $

Show Proof

x - a

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Is it possible for a function [math]f[/math] to be defined by two different power series about the same number [math]a[/math]? The answer is no, provided the domain of [math]f[/math] is not just the single number [math]a[/math]. The reason, as the following corollary of Theorem (7.3) shows, is that the coefficients of any power series about [math]a[/math] which defines [math]f[/math] are uniquely determined by the function [math]f[/math].

Theorem

lf [math]f(x) = \sum_{i=0}^\infty a_i(x-a)^i[/math] and if the radius of convergence of the power series is not zero, then, for every integer [math]n \geq 0[/math],

[[math]] a_n = \frac{1}{n!} f^{(n)}(a). [[/math]]

[By the zeroth derivative [math]f^{(0)}[/math] we mean simply [math]f[/math] itself. Hence, for [math]n = 0[/math], the conclusion is the obviously true equation [math]a_0 = f(a)[/math].]

Show Proof

The radius of convergence [math]\rho[/math] is not zero, and so the formula in (7.3) holds. Since [math]i(i-1) \cdots (i - n + 1) = \frac{i!}{(i-n)!}[/math], we obtain

[[math]] \begin{eqnarray*} f^{(n)}(x) &=& \sum_{i=n}^\infty \frac{i!}{(i - n)!} a_i (x - a)^{i-n} \\ &=& n! a_n + (n + 1)! a_{n+1}(x - a) + \frac{(n+2)!}{2!} a_{n+2}(x - a)^2 + \cdots , \end{eqnarray*} [[/math]]

for every [math]x[/math] such that [math]|x-a| \lt \rho[/math]. Setting [math]x = a[/math], we obtain

[[math]] f^{(n)}(a)= n!a_n, [[/math]]

from which the desired conclusion follows.

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\end{exercise}

General references

Doyle, Peter G. (2008). "Crowell and Slesnick's Calculus with Analytic Geometry" (PDF). Retrieved Oct 29, 2024.