Calculus Of One Real Variable – By Pheng Kim Ving
Chapter 15: Representations Of Functions By Power Series – Section 15.2: Derivatives And Integrals Of
Power Series


15.2
Derivatives And Integrals Of Power Series

 

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1. Derivatives Of Power Series

 

Centre, Radius, And Interval Of Convergence Of Derivative Power Series

 

Recall that if two functions are equal on an interval, their derivatives must also be equal on that interval, except at one or both
endpoints of the interval when one or both derivatives don't exist there. Consider the function 1/(1 – x) and its power-series representation:

 

 

So its radius of convergence is 1/1 = 1, the same as that of the original power series. Let's find its interval of convergence. At x
= 0 – 1 = –1 the derivative series is:

 

 

Now consider the function ln (1 + x), of course for x > –1. As will be seen in Example 2.1 later on in this section, it can be
represented by a power series, as follows:

 

 

which diverges, At x = 0 + 1 = 1 the derivative series is:

 

 

The interval of convergence of the derivative series is (–1, 1), different from that of the original series, which is (–1, 1]. The
derivative series diverges at an endpoint where the original series converges. The derivative series loses an endpoint of the
interval of convergence of the original series. Now we see why for the derivative series obtained after differentiation we put
“at most on” before the interval of convergence of the original series.

 

In general:

 

Properties Of Derivative Power Series

 

A power series with non-0 (and thus positive) radius of convergence is differentiated term by term. The derivative series also
is a power series. If the original series represents a function, the derivative series represents the derivative of that function.

 

The centre and radius of convergence of the derivative series are the same as those of the original series, and its interval of
convergence is that of the original series possibly losing one or both endpoints. If the original series converges at an
endpoint, the derivative series may converge or may diverge there. If the original series diverges at an endpoint, the
derivative series diverges there too.

 

So to determine the interval of convergence of the derivative series, we have to check any endpoint where the original series
converges.

 

 

Determining A Power Series Representing A Given Function

 

Example 1.1

 

Determine a power series that represents the function 2/(1 – x)3 on an interval centered at x = 0. Write the series in the sigma
notation and determine the interval where the representation is valid.

 

Note
We can use algebraic manipulation on 1/(1 –
x), whose power-series representation is well known, as done in
Section 15.1 Example 5.1 and Example 5.2. However we note that 2/(1 –
x)3 can be obtained by differentiating 1/(1 – x) twice
and that the power-series representation of 1/(1 –
x) is known. So we'll try differentiations on 1/(1 – x) and its series. Perhaps,
in this case, it'll be simpler than algebraic manipulation.

 

Solution

EOS

 

Remarks 1.1

 

On Example 1.1.

 

1. Indeed, in this case, differentiation must be simpler than algebraic manipulation.

 

2. Because each original series diverges at both endpoints of its interval of convergence, for the derivative series we don't need
    to put “at most on” before the interval of convergence of the original series, as the derivative series must diverge there also.
    We don't worry about determining the behavior of the derivative series at any endpoint. If the original series converges at
    one or both endpoints, we must check the derivative series obtained there, as it may converge or may diverge there. For
    example:

 

   

 

    The original series converges at the endpoint x = – b. The derivative series may converge or may diverge there. So for the
    derivative series we put “at most on” before the interval of convergence of the original series. We would have to check the
    behavior of the derivative series obtained at x = – b to determine the interval of convergence of the derivative power series.

 

3. To keep the pattern of the coefficients clear, we don't multiply out their factors. This is very useful when we have to write
    the series in the sigma notation.

 

4. In the sigma notation, the running index and the exponent on x are the same. In general, in the sigma notation we prefer
    to have the exponent on x or (x – c) be the same as the running index.

 

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2. Integrals Of Power Series

 

Centre, Radius, And Interval Of Convergence Of Integral Power Series

 

Recall that if two functions are equal on an interval, their integrals must also be equal on that interval up to a constant (differ
by a constant). Consider the function 1/(1 – x) and its power-series representation:

 

 

 

In this case the integral series converges at an endpoint where the original series diverges. The integral series gains an
endpoint. Now we see why for the integral series obtained after integration we put “at least on” before the interval of
convergence of the original series.

 

 

thus:

 

 

In general:

 

Properties Of Integral Power Series

 

A power series with non-0 (and thus positive) radius of convergence is integrated term by term. The integral series also is a
power series. If the original series represents a function, the integral series represents the integral of that function.

 

The centre and radius of convergence of the integral series are the same as those of the original series, and its interval of
convergence is that of the original series possibly gaining one or both endpoints. If the original series converges at an
endpoint, the integral series converges there too. If the original series diverges at an endpoint, the integral series may
converge or may diverge there.

 

So to determine the interval of convergence of the integral series, we have to check any endpoint where the original series
diverges.

 

 

Since the derivative power series can lose one or both endpoints of the interval of convergence of the original power series and
can't gain any, it must be that the integral power series can gain one or both endpoints and can't lose any, because the original
is the derivative of the integral.

 

Fig. 2.1 provides a way to easily remember the properties of derivative and integral power series, without confusion as to
which can only lose endpoints and which can only gain endpoints of the interval of convergence of the original power series.
Think of the derivative as “smaller” than the original and thus it can only lose and can't gain endpoints, and think of the integral
as “larger” than the original and thus it can only gain and can't lose endpounts.

 

Fig. 2.1

 

Derivative Can Only Lose And
Integral Can Only Gain Endpoints.

 

Determining A Power Series Representing A Given Function

 

Example 2.1

 

1. Determine a power-series representation of the function ln (1 + x) on an interval centered at x = 0. Write the series in the
    sigma notation and determine the interval where the representation is valid.
2. Find the sum of the alternating harmonic series.

 

Note
We know that ln (1 + x) is an integral of 1/(1 + x) = 1/(1 – (– x)) and we know the power-series representation of 1/(1 – x).
So we'll try to integrate 1/(1 – x) and its power-series representation.

 

Solution

 

    The centre of convergence of the power series is 0. Let an = (–1)n+1/n. Then:

 

 

EOS

 

Remark 2.1

 

In Example 2.1, We can multiply both sides of:

 

 

Recall that multiplying a series by a non-0 constant doesn't change the behavior of the series. So the interval of convergence of
of a non-0 constant multiple of a power series is the same as that of the original series. Note that at any point x in their interval
of convergence, the non-0 constant multiple may have a different value (sum) than that of the original. The range of the non-0
constant multiple may be different from that of the original. The interval of convergence is the domain (of the sum function), not
the range.

 

Example 2.2

 

1. Determine a power-series representation of the function arctan x on an interval centered at x = 0. Write the series in the
    sigma notation and determine the interval where the representation is valid.
2. Find the sum of the converging alternating series:

 

 

Note
We know that arctan x is the integral of 1/(1 + x2). We know the power-series representation of 1/(1 – x) for |x| < 1. We see
that 1/(1 + x2) can be obtained by replacing x in 1/(1 – x) by –x2, and that |–x2| < 1 iff |x| < 1. So we’ll try integration.

Solution

 

    The centre of convergence of this power series is 0. Let an = (-1)n/(2n + 1). Then:

 

 


EOS

 

Remark 2.2

 

Examples 1.1, 2.1, and 2.2 show that to determine a power-series representation of a given function, we can try differentiation
or integration of a known series-representation of a function. Whether to try differentiation or integration depends on the given
function.

 

Some Power-Series Representations And Sums Of Some Series

 

We gather here the power-series representations and the sums of series obtained in Examples 2.1 and 2.2.

 

Two Power Series And Sum Of Each Of Two Series

 

 

 

 

Series Developments Of Two Numbers

 

 

 

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3. Determining A Function Represented By A Given Series

 

Consider the power series 1 + x + x2 + x3 + ... If we didn't know the formula for its sum, we would have to settle with an
approximation of the sum, eg by a partial sum. Now it's good that the series equals the function 1/(1 – x) on the interval (–1,
1). Its exact sum at any point x in (–1, 1) is the value of 1/(1 – x) at that point x, which can easily be calculated. So for a
given power series, it may also be useful if there's a function equal to it on some interval.

 

The process here is the reverse to those in Examples 1.1, 2.1, and 2.2.

 

Example 3.1

 

Consider the power series:

 

1 – 3x + 9x2 – 27x3 + 81x4 – 243x5 + ...

 

a. Write it in the sigma notation.
b. Determine its interval of convergence.

c. Determine a function represented by it. Specify the interval where the representation is valid.

 

Solution

 

    which diverges. Thus the required interval of convergence is (–1/3. 1/3).

 

 

EOS

 

In Example 3.1 part c, we don't use differentiation or integration, we simply use algebraic manipulation.

 

Example 3.2

 

Consider the power series:

 

 

a. Express it in the sigma notation.
b. Determine its interval of convergence.
c. Find a function represented by it. Specify the interval where the representation applies.

 

Solution

 

 

 

EOS

 

In Example 3.2 part c, we factor the common factor x, we write a series as the derivative of its integral, for as many series as
needed till we reach a series that's recognized to represent some function. Then we replace that series by the function. Then
we differentiate, as many functions as there are series written as the derivatives of their integrals. And we obtain the function
that's represented by the given series.

 

When we replace the series 1 + x + x2 + x3 + x4 + x5 + ... by the function 1/(1 – x), the condition on x is “–1 < x < 1”. So the
condition on the derivative of 1/(1 – x) is also “–1 < x < 1”, for the derivative can't gain endpoints. This is in accordance with
the finding in part b that the interval of convergence of the given series is (–1, 1), the series that's to represent the function.

 

Example 3.3

 

Consider the series:

 

 

a. Write it in the sigma notation.
b. Find its interval of convergence.

c. Determine a function that it represents. State the interval where the representation is valid.

 

Solution

 

    which diverges. Thus the interval of convergence is [–1, 1).

 

 

    So the function is continuous there, and as the series converges and is right-continuous there, the sum of the series and the
    value of the function there are equal. At x = 1 the series diverges. Thus:

 

EOS

 

In Example 3.3 part c, we multiply the given series by x2, we write the resulting series as the integral of its derivative, we get a
series that's recognized to represent some function. Then we replace that series by the function. Next we integrate the function.
We obtain a function represented by the resulting series. And we divide that function by x2 to obtain the function that's
represented by the given series.

 

When we replace the series t + t2 + t3 + t4 + t5 + t6 + ... by the function 1/(1 – t) – 1, the condition on t is “–1 < t < 1”. So the
condition on the integral of 1/(1 – t) – 1 is “at least on –1 < t < 1”, for the integral may gain one or both endpoints where the
given series converges. Since the interval of convergence of the given series as found in part b is [–1, 1), we check the
endpoint –1 where it converges and we don't check the endpoint 1 where it diverges, to determine the interval where the
representation of the function by it is valid.

 

The condition on x is the same as that on t, because t is just a dummy variable used to represent x to avoid confusion with the
upper limit of integration x.

 

Determining The Sum Of A Power Series

 

A power series involves a variable, say x. So its sum involves x too. That is, its sum is a function of x. Its sum is a function
represented by it. Thus to determine the sum of a power series is to determine a function that's represented by the power
series. For example, in Example 3.2, the line “Determine the function represented by the power series:” can be re-phrased as
“Determine the sum of the power series:”. The solution would be the same.

 

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4. Differentiability, Integrability, And Continuity Of Power Series

 

In the discussions above we've implicitly assumed that power series with non-0 (and thus positive) radiuses of convergence
are differentiable and integrable on (–R, R), where R > 0 is the radius of convergence. Rigorously speaking, these properties
have to be proved, since power series are infinite polynomials (of infinite degree) while polynomials that we dealt with
previously in the contexts of differentiablity and integrability are finite (of finite degrees). Appropriately these properties are
investigated and proved as a theorem in a more advanced calculus course or a mathematical analysis course. Here for
reference to the theorem in later sections we state it without proof.

 

We've also assumed that power series with non-0 (and thus positive) radiuses of convergence are continuous on their intervals
of convergence. If the interval of convergence of a series is (–R , R) for some R > 0, then the series is continuous at every
point in (–R, R). If the interval of convergence of a series is [–R , R), then the series is continuous at every point in (–R, R) and
right-continuous at x = –R. Etc. The continuity property is usually stated and proved as a theorem accompanying the theorem
on the differentiability and integrability properties.

 

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5. Tasks Performed

 

As pointed out in Section 15.1 Part 6, we performed in that section, among other things, the following tasks:

 

1. Given a function, for any point in its domain we determine a power series that represents it on an interval centered at the
    point, by using algebraic manipulation.

 

2. Given a function, for any point in its domain we determine a power series that represents it on an interval centered at the
    point, by using algebraic operations.

 

Now in this section we perform the following tasks:

 

3. In Part 1, given a function, for any point in its domain we determine a power series that represents it on an interval centered
    at the point, by using differentiation.

 

4. In Part 2, given a function, for any point in its domain we determine a power series that represents it on an interval centered
    at the point, by using integration.

 

5. In Part 3, conversely, given a power series, we determine the function represented by the series on an interval derived from
    the interval of convergence of the series, by using algebraic manipilation or differentiation or integration.

 

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Problems & Solutions

 

1. Determine a power series centered at x = 0 that represents the function 1/(1 + 3x). Write the series in the sigma notation
    and specify the interval where the representation is valid.

 

Solution

 

 

Note

 

With the power-series representation 1/(1 – x) = 1 + x + x2 + x3 + x4 + x5 + ..., we use algebraic manipulation. Clearly using
differentiation would lead to nowhere. Now let's see what will happen if we use integration:

 

 

replace x by –3x, –1 < –3x < 1 iff 1/3 > x > –1/3 or –1/3 < x < 1/3:

 

 

differentiate both sides:

 

 

Well, in this case, we have to also use differentiation in addition to integration, and it's way more lengthy and more complicated
than using algebraic manipulation.

 

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2. Find a power series in powers of x – 3 that represents the function 1/x2. Write the series in the sigma notation and specify
    the interval where the representation is valid.

 

Note

 

Recall that a power series in powers of x – 3 is one that's centered at x = 3.

 

Solution

 

 

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3. Establish a power series in powers of x that represents the function ln (3 – x). Put the series in the sigma notation and
    specify the interval where the representation is valid.

 

Solution

 

 

At x = –3 the series is:

 

 

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4. Determine a power-series-in-powers-of-x representation of the function (2 – x)/(1 + x). Use the sigma notation for the
    series and find the interval where the representation applies.

 

Solution

 

 

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5. Consider the power series:

 

    1 – 5x + 25x2 – 125x3 + 625x4 –3,125x5 + ...

 

a. Write it in the sigma notation.
b. Determine its interval of convergence.

c. Find a function represented by it. Specify the interval where the representation is valid.

 

Solution

 

 

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6. Consider the series:

 

    2 + 4x2 + 6x4 + 8x6 + 10x8 + 12x10 + ...

 

a. Express it in the sigma notation.
b. Find its interval of convergence.

c. Determine its sum. State the interval where the expression for the sum is valid.

 

Solution

 

 

    which diverges. So the interval of convergence of the series is (–1, 1).

 

 

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7. Consider the series:

 

 

a. Write it in the sigma notation.
b. Determine its interval of convergence.

c. Find a function represented by it. Specify the interval where the representation applies.

 

Solution

 

    which diverges. Thus the interval of convergence is [–1, 1).

 

 

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