APEX Power Series

Section 9.6 Power Series

So far, our study of series has examined the question of “Is the sum of these infinite terms finite?,” i.e., “Does the series converge?” We now approach series from a different perspective: as a function. Given a value of

x,

we evaluate

f (x)

by finding the sum of a particular series that depends on

x

(assuming the series converges). We start this new approach to series with a definition.

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Figure 9.6.1. Video introduction to Section 9.6

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Definition 9.6.2. Power Series.

Let

{a_{n}}

be a sequence, let

x

be a variable, and let

c

be a real number.

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The power series in $x$ is the series

$\sum_{n = 0}^{\infty} a_{n} x^{n} = a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + \dots$

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The power series in $x$ centered at $c$ is the series

$\sum_{n = 0}^{\infty} a_{n} (x - c)^{n} = a_{0} + a_{1} (x - c) + a_{2} (x - c)^{2} + a_{3} (x - c)^{3} + \dots$

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Example 9.6.3. Examples of power series.

Write out the first five terms of the following power series:

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$\sum_{n = 0}^{\infty} x^{n}$

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$\sum_{n = 1}^{\infty} (- 1)^{n + 1} \frac{(x + 1)^{n}}{n}$

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$\sum_{n = 0}^{\infty} (- 1)^{n + 1} \frac{(x - π)^{2 n}}{(2 n)!}$

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Solution 1.

One of the conventions we adopt is that $x^{0} = 1$ regardless of the value of $x .$ Therefore

$\sum_{n = 0}^{\infty} x^{n} = 1 + x + x^{2} + x^{3} + x^{4} + \dots$

This is a geometric series in $x$ with $r = x .$

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This series is centered at $c = - 1 .$ Note how this series starts with $n = 1 .$ We could rewrite this series starting at $n = 0$ with the understanding that $a_{0} = 0,$ and hence the first term is $0 .$

$\begin{aligned} \sum_{n = 1}^{\infty} (- 1)^{n + 1} \frac{(x + 1)^{n}}{n} \\ = (x + 1) - \frac{(x + 1)^{2}}{2} + \frac{(x + 1)^{3}}{3} - \frac{(x + 1)^{4}}{4} + \frac{(x + 1)^{5}}{5} \dots \end{aligned}$

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This series is centered at $c = π .$ Recall that $0! = 1 .$

$\begin{aligned} \sum_{n = 0}^{\infty} (- 1)^{n + 1} \frac{(x - π)^{2 n}}{(2 n)!} \\ = - 1 + \frac{(x - π)^{2}}{2} - \frac{(x - π)^{4}}{24} + \frac{(x - π)^{6}}{6!} - \frac{(x - π)^{8}}{8!} \dots \end{aligned}$

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Solution 2. Video solution

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We introduced power series as a type of function, where a value of

x

is given and the sum of a series is returned. Of course, not every series converges. For instance, in part 1 of Example 9.6.3, we recognized the series

\sum_{n = 0}^{\infty} x^{n}

as a geometric series in

x .

Theorem 9.2.7 states that this series converges only when

| x | < 1 .

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This raises the question: “For what values of

x

will a given power series converge?,” which leads us to a theorem and definition.

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Theorem 9.6.4. Convergence of Power Series.

Let a power series

\sum_{n = 0}^{\infty} a_{n} (x - c)^{n}

be given. Then one of the following is true:

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The series converges only at $x = c .$
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There is an $R > 0$ such that the series converges for all $x$ in $(c - R, c + R)$ and diverges for all $x < c - R$ and $x > c + R .$
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The series converges for all $x .$
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Figure 9.6.5. Video presentation of Theorem 9.6.4 and Definition 9.6.6

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The value of

R

is important when understanding a power series, hence it is given a name in the following definition. Also, note that part 2 of Theorem 9.6.4 makes a statement about the interval

(c - R, c + R),

but the not the endpoints of that interval. A series may/may not converge at these endpoints.

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Definition 9.6.6. Radius and Interval of Convergence.

The number $R$ given in Theorem 9.6.4 is the radius of convergence of a given series. When a series converges for only $x = c,$ we say the radius of convergence is 0, i.e., $R = 0 .$ When a series converges for all $x,$ we say the series has an infinite radius of convergence, i.e., $R = \infty .$
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The interval of convergence is the set of all values of $x$ for which the series converges.
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To find the interval of convergence, we start by using the ratio test to find the radius of convergence

R .

0 < R < \infty,

we know the series converges on

(c - R, c + R),

and it remains to check for convergence at the endpoints.

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Given

\sum_{n = 0}^{\infty} a_{n} (x - c)^{n}

we apply the ratio test to

\sum_{n = 0}^{\infty} | a_{n} (x - c)^{n} |

since the ratio test requires positive terms. We find

lim_{n \to \infty} \frac{| a_{n + 1} (x - c)^{n + 1} |}{| a_{n} (x - c)^{n} |} = L | x - c |,

where

L = lim_{n \to \infty} \frac{| a_{n + 1} |}{| a_{n} |} .

It follows that the series converges absolutely (and therefore converges) for any

x

such that

L | x - c | < 1;

that is, for

x

(c - \frac{1}{L}, c + \frac{1}{L}) .

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On the other hand, suppose for some

x

that

L | x - c | > 1 .

Then, for sufficiently large

n,

| a_{n + 1} | > | a_{n} | .

This means that the terms of

\sum_{n = 0}^{\infty} a_{n} (x - c)^{n}

are growing in absolute value, and therefore cannot converge to zero. This means that the series diverges, by Theorem 9.2.23.

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From the above observations, it follows that

R = \frac{1}{L}

must be the radius of convergence.

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Key Idea 9.6.7. Determining the Radius and Interval of Convergence.

Given the power series

\sum_{n = 0}^{\infty} a_{n} (x - c)^{n},

apply the ratio test to the series

\sum_{n = 0}^{\infty} | a_{n} (x - c)^{n} | .

The result will be

L | x - c |,

where

L = lim_{n \to \infty} \frac{| a_{n + 1} |}{| a_{n} |} .

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If $L = 0,$ then the power series converges for every $x$ by the ratio test, since $L | x - c | = 0 < 1 .$
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If $L = \infty,$ then power series converges only when $x = c .$
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If $0 < L < \infty,$ then $R = 1 / L$ is the radius of convergence: by the ratio test, the series converges when $| x - c | < R .$
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To determine the interval of convergence, plug the endpoints ( $x = c - R$ and $x = c + R$ ) into the power series, and test the resulting series for convergence. If the series converges, we include the endpoint. If it diverges, we exclude the endpoint.
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Key Idea 9.6.7 allows us to find the radius of convergence

R

of a series by applying the Ratio Test (or any applicable test) to the absolute value of the terms of the series. We practice this in the following example.

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Example 9.6.8. Determining the radius and interval of convergence.

Find the radius and interval of convergence for each of the following series:

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$\sum_{n = 0}^{\infty} \frac{x^{n}}{n!}$
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$\sum_{n = 1}^{\infty} (- 1)^{n + 1} \frac{x^{n}}{n}$
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$\sum_{n = 0}^{\infty} 2^{n} (x - 3)^{n}$
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$\sum_{n = 0}^{\infty} n! x^{n}$
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Solution 1.

We apply the Ratio Test to the series $\sum_{n = 1}^{\infty} | \frac{x^{n}}{n!} | :$

$\begin{aligned} lim_{n \to \infty} \frac{| x^{n + 1} / (n + 1)! |}{| x^{n} / n! |} & = lim_{n \to \infty} | \frac{x^{n + 1}}{x^{n}} \cdot \frac{n!}{(n + 1)!} | \\ = lim_{n \to \infty} | \frac{x}{n + 1} | \\ = 0 for all x . \end{aligned}$

The Ratio Test shows us that regardless of the choice of $x,$ the series converges. Therefore the radius of convergence is $R = \infty,$ and the interval of convergence is $(- \infty, \infty) .$

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We apply the Ratio Test to the series $\sum_{n = 1}^{\infty} | (- 1)^{n + 1} \frac{x^{n}}{n} | = \sum_{n = 1}^{\infty} | \frac{x^{n}}{n} | :$

$\begin{aligned} lim_{n \to \infty} \frac{| x^{n + 1} / (n + 1) |}{| x^{n} / n |} & = lim_{n \to \infty} | \frac{x^{n + 1}}{x^{n}} \cdot \frac{n}{n + 1} | \\ = lim_{n \to \infty} (\frac{n}{n + 1}) | x | \\ = | x | . \end{aligned}$

The Ratio Test states a series converges if the limit of $| a_{n + 1} / a_{n} | = L < 1 .$ We found the limit above to be $| x |;$ therefore, the power series converges when $| x | < 1,$ or when $x$ is in $(- 1, 1) .$ Thus the radius of convergence is $R = 1 .$ To determine the interval of convergence, we need to check the endpoints of $(- 1, 1) .$ When $x = - 1,$ we have the opposite of the Harmonic Series:

$\begin{aligned} \sum_{n = 1}^{\infty} (- 1)^{n + 1} \frac{(- 1)^{n}}{n} & = \sum_{n = 1}^{\infty} \frac{- 1}{n} \\ = - \infty . \end{aligned}$

The series diverges when $x = - 1 .$ When $x = 1,$ we have the series $\sum_{n = 1}^{\infty} (- 1)^{n + 1} \frac{(1)^{n}}{n},$ which is the Alternating Harmonic Series, which converges. Therefore the interval of convergence is $(- 1, 1] .$

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We apply the Ratio Test to the series $\sum_{n = 0}^{\infty} | 2^{n} (x - 3)^{n} | :$

$\begin{aligned} lim_{n \to \infty} \frac{| 2^{n + 1} (x - 3)^{n + 1} |}{| 2^{n} (x - 3)^{n} |} & = lim_{n \to \infty} | \frac{2^{n + 1}}{2^{n}} \cdot \frac{(x - 3)^{n + 1}}{(x - 3)^{n}} | \\ = lim_{n \to \infty} | 2 (x - 3) | . \end{aligned}$

According to the Ratio Test, the series converges when $| 2 (x - 3) | < 1 ⟹ | x - 3 | < 1 / 2 .$ The series is centered at 3, and $x$ must be within $1 / 2$ of 3 in order for the series to converge. Therefore the radius of convergence is $R = 1 / 2,$ and we know that the series converges absolutely for all $x$ in $(3 - 1 / 2, 3 + 1 / 2) = (2.5, 3.5) .$ We check for convergence at the endpoints to find the interval of convergence. When $x = 2.5,$ we have:

$\begin{aligned} \sum_{n = 0}^{\infty} 2^{n} (2.5 - 3)^{n} & = \sum_{n = 0}^{\infty} 2^{n} (- 1 / 2)^{n} \\ = \sum_{n = 0}^{\infty} (- 1)^{n}, \end{aligned}$

which diverges. A similar process shows that the series also diverges at $x = 3.5 .$ Therefore the interval of convergence is $(2.5, 3.5) .$

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We apply the Ratio Test to $\sum_{n = 0}^{\infty} | n! x^{n} | :$

$\begin{aligned} lim_{n \to \infty} \frac{| (n + 1)! x^{n + 1} |}{| n! x^{n} |} & = lim_{n \to \infty} | (n + 1) x | \\ = \infty for all x, except x = 0 . \end{aligned}$

The Ratio Test shows that the series diverges for all $x$ except $x = 0 .$ Therefore the radius of convergence is $R = 0 .$

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Solution 2. Video solution

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We can use a power series to define a function:

f (x) = \sum_{n = 0}^{\infty} a_{n} x^{n}

where the domain of

f

is a subset of the interval of convergence of the power series. One can apply calculus techniques to such functions; in particular, we can find derivatives and antiderivatives.

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Theorem 9.6.9. Derivatives and Indefinite Integrals of Power Series Functions.

Let

f (x) = \sum_{n = 0}^{\infty} a_{n} (x - c)^{n}

be a function defined by a power series, with radius of convergence

R .

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$f (x)$ is continuous and differentiable on $(c - R, c + R) .$
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$f^{'} (x) = \sum_{n = 1}^{\infty} a_{n} \cdot n \cdot (x - c)^{n - 1},$ with radius of convergence $R .$
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$\int f (x) d x = C + \sum_{n = 0}^{\infty} a_{n} \frac{(x - c)^{n + 1}}{n + 1},$ with radius of convergence $R .$
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Figure 9.6.10. Video presentation of Theorem 9.6.9

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A few notes about Theorem 9.6.9:

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The theorem states that differentiation and integration do not change the radius of convergence. It does not state anything about the interval of convergence. They are not always the same.
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Notice how the summation for $f^{'} (x)$ starts with $n = 1 .$ This is because the constant term $a_{0}$ of $f (x)$ becomes $0$ through differentiation.
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Differentiation and integration are simply calculated term-by-term using the Power Rules.
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Example 9.6.11. Derivatives and indefinite integrals of power series.

Let

f (x) = \sum_{n = 0}^{\infty} x^{n} .

Find

f^{'} (x)

and

F (x) = \int f (x) d x,

along with their respective intervals of convergence.

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Solution 1.

We find the derivative and indefinite integral of

f (x),

following Theorem 9.6.9.

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$\begin{aligned} f^{'} (x) & = \sum_{n = 1}^{\infty} n x^{n - 1} = 1 + 2 x + 3 x^{2} + 4 x^{3} + \dots \\ = \sum_{n = 0}^{\infty} (n + 1) x^{n} . \end{aligned}$

In Example 9.6.3, we recognized that $\sum_{n = 0}^{\infty} x^{n}$ is a geometric series in $x .$ We know that such a geometric series converges when $| x | < 1;$ that is, the interval of convergence is $(- 1, 1) .$ To determine the interval of convergence of $f^{'} (x),$ we consider the endpoints of $(- 1, 1) :$

$f^{'} (- 1) = 1 - 2 + 3 - 4 + \dots, which diverges.$

$f^{'} (1) = 1 + 2 + 3 + 4 + \dots, which diverges.$

Therefore, the interval of convergence of $f^{'} (x)$ is $(- 1, 1) .$

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$F (x) = \int f (x) d x = C + \sum_{n = 0}^{\infty} \frac{x^{n + 1}}{n + 1} = C + x + \frac{x^{2}}{2} + \frac{x^{3}}{3} + \dots$ To find the interval of convergence of $F (x),$ we again consider the endpoints of $(- 1, 1) :$

$F (- 1) = C - 1 + 1 / 2 - 1 / 3 + 1 / 4 + \dots$

The value of $C$ is irrelevant; notice that the rest of the series is an Alternating Series that whose terms converge to 0. By the Alternating Series Test, this series converges. (In fact, we can recognize that the terms of the series after $C$ are the opposite of the Alternating Harmonic Series. We can thus say that $F (- 1) = C - \ln (2) .$ )

$F (1) = C + 1 + 1 / 2 + 1 / 3 + 1 / 4 + \dots$

Notice that this summation is $C +$ the Harmonic Series, which diverges. Since $F$ converges for $x = - 1$ and diverges for $x = 1,$ the interval of convergence of $F (x)$ is $[- 1, 1) .$

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Solution 2. Video solution

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The previous example showed how to take the derivative and indefinite integral of a power series without motivation for why we care about such operations. We may care for the sheer mathematical enjoyment “that we can”, which is motivation enough for many. However, we would be remiss to not recognize that we can learn a great deal from taking derivatives and indefinite integrals.

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Recall that

f (x) = \sum_{n = 0}^{\infty} x^{n}

in Example 9.6.11 is a geometric series. According to Theorem 9.2.7, this series converges to

1 / (1 - x)

when

| x | < 1 .

Thus we can say

f (x) = \sum_{n = 0}^{\infty} x^{n} = \frac{1}{1 - x}, on (- 1, 1) .

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Integrating the power series, (as done in Example 9.6.11,) we find

\begin{matrix} (9.6.1) & F (x) = C_{1} + \sum_{n = 0}^{\infty} \frac{x^{n + 1}}{n + 1}, \end{matrix}

while integrating the function

f (x) = 1 / (1 - x)

gives

\begin{matrix} (9.6.2) & F (x) = - \ln | 1 - x | + C_{2} . \end{matrix}

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Equating Equations (9.6.1) and (9.6.2), we have

F (x) = C_{1} + \sum_{n = 0}^{\infty} \frac{x^{n + 1}}{n + 1} = - \ln | 1 - x | + C_{2} .

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Letting

x = 0,

we have

F (0) = C_{1} = C_{2} .

This implies that we can drop the constants and conclude

\sum_{n = 0}^{\infty} \frac{x^{n + 1}}{n + 1} = - \ln | 1 - x | .

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We established in Example 9.6.11 that the series on the left converges at

x = - 1;

substituting

x = - 1

on both sides of the above equality gives

- 1 + \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + \dots = - \ln (2) .

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On the left we have the opposite of the Alternating Harmonic Series; on the right, we have

- \ln (2) .

We conclude that

1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots = \ln (2) .

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Important: We stated in Key Idea 9.2.20 (in Section 9.2) that the Alternating Harmonic Series converges to

\ln (2),

and referred to this fact again in Example 9.5.6 of Section 9.5. However, we never gave an argument for why this was the case. The work above finally shows how we conclude that the Alternating Harmonic Series converges to

\ln (2) .

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We use this type of analysis in the next example.

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Example 9.6.12. Analyzing power series functions.

Let

f (x) = \sum_{n = 0}^{\infty} \frac{x^{n}}{n!} .

Find

f^{'} (x)

and

\int f (x) d x,

and use these to analyze the behavior of

f (x) .

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Solution 1.

We start by making two notes: first, in Example 9.6.8, we found the interval of convergence of this power series is

(- \infty, \infty) .

Second, we will find it useful later to have a few terms of the series written out:

\begin{matrix} (9.6.3) & \sum_{n = 0}^{\infty} \frac{x^{n}}{n!} = 1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{6} + \frac{x^{4}}{24} + \dots \end{matrix}

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We now find the derivative:

\begin{aligned} f^{'} (x) & = \sum_{n = 1}^{\infty} n \frac{x^{n - 1}}{n!} \\ = \sum_{n = 1}^{\infty} \frac{x^{n - 1}}{(n - 1)!} = 1 + x + \frac{x^{2}}{2!} + \dots . \end{aligned}

Since the series starts at

n = 1

and each term refers to

(n - 1),

we can re-index the series starting with

n = 0 :

\begin{aligned} = \sum_{n = 0}^{\infty} \frac{x^{n}}{n!} \\ = f (x) . \end{aligned}

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We found the derivative of

f (x)

f (x) .

The only functions for which this is true are of the form

y = c e^{x}

for some constant

c .

f (0) = 1

(see Equation (9.6.3)),

c

must be 1. Therefore we conclude that

f (x) = \sum_{n = 0}^{\infty} \frac{x^{n}}{n!} = e^{x}

for all

x .

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We can also find

\int f (x) d x :

\begin{aligned} \int f (x) d x & = C + \sum_{n = 0}^{\infty} \frac{x^{n + 1}}{n! (n + 1)} \\ = C + \sum_{n = 0}^{\infty} \frac{x^{n + 1}}{(n + 1)!} \end{aligned}

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We write out a few terms of this last series:

C + \sum_{n = 0}^{\infty} \frac{x^{n + 1}}{(n + 1)!} = C + x + \frac{x^{2}}{2} + \frac{x^{3}}{6} + \frac{x^{4}}{24} + \dots

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The integral of

f (x)

differs from

f (x)

only by a constant, again indicating that

f (x) = e^{x} .

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Solution 2. Video solution

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Example 9.6.12 and the work following Example 9.6.11 established relationships between a power series function and “regular” functions that we have dealt with in the past. In general, given a power series function, it is difficult (if not impossible) to express the function in terms of elementary functions. We chose examples where things worked out nicely.

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In this section’s last example, we show how to solve a simple differential equation with a power series.

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Example 9.6.13. Solving a differential equation with a power series.

Give the first 4 terms of the power series solution to

y^{'} = 2 y,

where

y (0) = 1 .

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Solution.

The differential equation

y^{'} = 2 y

describes a function

y = f (x)

where the derivative of

y

is twice

y

and

y (0) = 1 .

This is a rather simple differential equation; with a bit of thought one should realize that if

y = C e^{2 x},

then

y^{'} = 2 C e^{2 x},

and hence

y^{'} = 2 y .

By letting

C = 1

we satisfy the initial condition of

y (0) = 1 .

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Let’s ignore the fact that we already know the solution and find a power series function that satisfies the equation. The solution we seek will have the form

f (x) = \sum_{n = 0}^{\infty} a_{n} x^{n} = a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + \dots

for unknown coefficients

a_{n} .

We can find

f^{'} (x)

using Theorem 9.6.9:

f^{'} (x) = \sum_{n = 1}^{\infty} a_{n} \cdot n \cdot x^{n - 1} = a_{1} + 2 a_{2} x + 3 a_{3} x^{2} + 4 a_{4} x^{3} \dots .

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Since

f^{'} (x) = 2 f (x),

we have

\begin{aligned} a_{1} + 2 a_{2} x + 3 a_{3} x^{2} + 4 a_{4} x^{3} \dots & = 2 (a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + \dots) \\ = 2 a_{0} + 2 a_{1} x + 2 a_{2} x^{2} + 2 a_{3} x^{3} + \dots \end{aligned}

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The coefficients of like powers of

x

must be equal, so we find that

a_{1} = 2 a_{0}, 2 a_{2} = 2 a_{1}, 3 a_{3} = 2 a_{2}, 4 a_{4} = 2 a_{3}, etc.

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The initial condition

y (0) = f (0) = 1

indicates that

a_{0} = 1;

with this, we can find the values of the other coefficients:

\begin{aligned} a_{0} = 1 and a_{1} = 2 a_{0} & \Rightarrow a_{1} = 2; \\ a_{1} = 2 and 2 a_{2} = 2 a_{1} & \Rightarrow a_{2} = 4 / 2 = 2; \\ a_{2} = 2 and 3 a_{3} = 2 a_{2} & \Rightarrow a_{3} = 8 / (2 \cdot 3) = 4 / 3; \\ a_{3} = 4 / 3 and 4 a_{4} = 2 a_{3} & \Rightarrow a_{4} = 16 / (2 \cdot 3 \cdot 4) = 2 / 3 . \end{aligned}

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Thus the first 5 terms of the power series solution to the differential equation

y^{'} = 2 y

f (x) = 1 + 2 x + 2 x^{2} + \frac{4}{3} x^{3} + \frac{2}{3} x^{4} + \dots

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In Section 9.8, as we study Taylor Series, we will learn how to recognize this series as describing

y = e^{2 x} .

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Our last example illustrates that it can be difficult to recognize an elementary function by its power series expansion. It is far easier to start with a known function, expressed in terms of elementary functions, and represent it as a power series function. One may wonder why we would bother doing so, as the latter function probably seems more complicated. In the next two sections, we show both how to do this and why such a process can be beneficial.

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