Calculus Of One Real Variable
A Tutorial
By Pheng Kim Ving
Chapter 15: Representations Of Functions By Power Series
Section 15.1: Power Series


15.1
Power Series

 

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1. Power Series

 

Functions Expressed As Geometric Series

 

Consider the geometric series 1 + r + r2 + r3 + r4 + r5 + ..., with first term 1 and common ratio r. As we've discussed earlier
in Section 14.2 Geometric Series, if –1 < r < 1 then the series converges to 1/(1 – r):

 

 

Fig. 1.1

 

–1 < x < 1,

 

 

Remark that 1/(1 – x) > 0, and thus 1 + x + x2 + x3 + x4 + x5 + ... > 0, for all x where –1 < x < 1.

 

Series That Converge To Functions

 

At any point x in (–1, 1), the sum of the series 1 + x + x2 + x3 + x4 + x5 + ... equals the value of the function 1/(1 – x). Refer
to Fig. 1.2. For example, at x = 1/2:

 

 

For this particular series of constants, each term from the second on is half the previous term. The sequence of partial sums s1,
s2, s3, ... increases and approaches 2. It starts with s1 = 1, which is half of 2, and increases at each step by half the previous
 

Fig. 1.2

 

On (–1, 1), series 1 + x + x2 + x3 + x4 + x5
+ ... converges to function 1/(1 – x).

 

term (not half the previous partial sum). So at each step it jumps up toward 2 by half the remaining distance between it and 2.
At x = 1/2 the series converges to 2. Another example is at x = –1/2:

 

 

For this particular series of constants, the absolute value of each term from the second on is half that of the previous term. The
sequence of partial sums t1, t2, t3, ... jumps down below and up above 2/3 alternatingly and approaches 2/3. It starts with s1 =
1, and decreases by half the previous term and increases by half the absolute value of the previous term alternatingly. The
distance between it and 2/3 decreases toward 0. At x = –1/2 the series converges to 2/3.

 

Since at every x in (–1, 1) the series 1 + x + x2 + x3 + x4 + x5 + ... converges to the function 1/(1 – x), we say that the series
1 + x + x2 + x3 + x4 + x5 + ... converges to the function 1/(1 – x) on the interval (–1, 1). In general, if at every point of an
interval a series converges to a function, we say that the series converges to the function on the interval.

 

Functions Represented By Power Series

 

The values of polynomials are easier to calculate than those of other functions like the exponential function ex or the sine
function sin x. The derivatives and integrals of polynomials are also easier to find than those of other types of functions. For
this particular function 1/(1 – x), these 3 items are also easy to find. However the fact that it can be expressed as an infinite
polynomial on an interval of x suggests that possibly other functions can too, which is undoubtedly useful. If a function can be
expressed as an infinite polynomial on an interval, then of course an approximate value (eg, partial sum) of the polynomial at
any point x in that interval is an approximate value of the function at x. This situation has led to the investigation of the
expressing of functions in terms of infinite polynomials. It turned out that indeed many of the elementary functions in calculus
that are infinitely often differentiable (have derivatives of all orders) can be expressed in terms of, or represented by, infinite
polynomials, called power series. The functions ex and sin x are two such functions.

 

Power Series In x

 

For the geometric series 1 + x + x2 + x3 + x4 + x5 + ..., the common ratio is x. We say that it's a geometric series in x.
Notice that x is a variable.

 

Now let's represent the function 6/(5 + x) by a series. We'll write 6/(5 + x) in the form a/(1 – r), which is the sum of the
geometric series with first term a and common ratio r for |r| < 1. We have:

 

 

The series in Eq. [1.4] is a geometric series in –x/5, since the coefficients of all the powers (–x/5)n are equal. Of course its
first term is 6/5 and its common ratio is –x/5. The series in Eq. [1.5] is also a geometric series in –x/5, but it's not a geometric
series in x, since the coefficients of the powers xn aren't all equal. It's called a power series in x, obviously because each term
is the product of a constant and a power of x. Notice that x is a variable.

 

Power Series In x – c

 

We note that in Eq. [1.3], x is the signed distance from point x to the origin, and the origin is the centre or midpoint of the
2-unit-long interval (–1, 1), as shown in Fig. 1.3. Thus if, for example, we consider the point 4 on the x-axis, then it's the
centre of the 2-unit-long interval (4 – 1, 4 + 1) = (3, 5), as shown in Fig. 1.4. If x is in (3, 5), then x – 4 is the signed distance
from point x to centre 4 and |x – 4| < 1, then replacing x in Eq. [1.3] by x – 4 we get:

 

 

Fig. 1.3

 

–1 < x < 1.

 

Fig. 1.4

 

3 < x < 5.

 

The series in Eq. [1.7] is a power series in x – 4 or a power series centered at x = 4 or a power series about the point x = 4.
The series in Eq. [1.3] is a power series in x (or x – 0) or a power series centered at x = 0 or at the origin or a power series
about the point x = 0 or about the origin, because:

 

1 + x + x2 + x3 + ... = 1 + (x – 0) + (x – 0)2 + (x – 0)3 + ...

 

Notice that x is a variable.


Definition 1.1 – Power Series

 

A series of the form:

 

 

where the an's are constants, c is a constant, and x is a variable, is called a power series centered at x = c or about x = c or in x – c or in powers of x – c. The constants an's are called the coefficients of the power series. Of course the coefficient a0 is the constant term of the power series.

 

 

Remarks 1.1

 

1. The terms of a power series except the first term a0 are functions, more precisely power functions, hence the name “power
    series”. This is in contrast to series of constants, whose terms are all constants.

 

2. Refer to Fig. 1.5. A power series centered at c is defined at any x as an infinite sum of terms each of which is the product
    of a constant and a positive-or-0 integer power of the signed distance x – c from x to c. Now refer to Fig. 1.6. If c = 0,
    then the power is the power of the signed distance x – 0 = x from x to the origin, which is the (signed) value of x. Recall
    that a function f (x) involves the (signed) value of x.

 

Fig. 1.5

 

Power series centered at c involves signed distance x – c from x to c.

 

Fig. 1.6

 

Power series centered at 0 involves (signed) value of x.

 

3. For a power series in x – c, x is a variable. For each value of x, the series is a series of constants, which may converge or
    may diverge. For example, the series:

 

 

    which is a power series (in x). Here c = 0. For each value of x, the series is a series of constants, which converges if that
    value is in (–1, 1) and diverges otherwise.

 

4. For a power series in x – c, each value of x generates a series of constants.

 

5. The series [1.8] is also called a power series centered  at c because, as will be seen later on, it converges for all x in an
    interval centered  at c.

 

6. If we change the constants an's, we'll obtain a different power series. Given any point c, there are infinitely many power
    series centered at it.

 

Power Series As Functions

 

The terms of the power series a0 + a1(x – c) + a2(x – c)2 + a3(x – c)3 + … are power functions of x. The series may converge
for some values of x and diverge for other values of x. At any x where the series converges, its sum is a unique number. This
sum generally varies with x, and thus is a function of x. This function is called the sum function defined by the series. We'll
see later on that the set of all x's where the series converges is an interval centered at c. Let I  be that interval. The function
f (x) defined for any x in I  by:

 

f (x) = a0 + a1(x – c) + a2(x – c)2 + a3(x – c)3 + …

 

is the sum function of the series with domain I. We say that the series converges to f on I. See Fig. 1.7, where I = (m, n).
We can think of a power series as a function on the interval where it converges.

 

Fig. 1.7

 

x in I,

 

f(x) = a0 + a1(x – c) + a2(x – c)2 +
          a3(x – c)3 + ... .

 

For any point x in (m, n), the value f (x) of f at x equals the sum of the series at x, which is an infinite polynomial that's
expressed in terms of the signed distance x – c from x to c and the constants an's.

Power Series As Functions

 

 

 

Remarks 1.2

 

1. Recall that a series converges to a function on an interval if at every point of the interval, the series converges to the
    function, ie, the sum of the series equals the value of the function.

 

2. The terms of a power series are functions of x for all real x, while the series itself is a function of x only on an interval
    where the series converges. Of course this interval may be the entire real line (the set of all real numbers) itself, depending
    on the series.

 

Representations Of Functions By Power Series

 

If at a point a power series converges to the value of a function at the point then we say that the power series represents the function at that point. If at every point in an interval a power series converges to the same function then we say that the power series represents the function on that interval.

 

 

Example 1.1

 

As shown in Figs. 1.1 and 1.2, the power series 1 + x + x2 + x3 + x4 + x5 + ... represents the function 1/(1 – x) on the interval
(–1, 1).

 

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2. Centre, Radius, And Interval Of Convergence

 

Example 2.1

 

For the power series:

 

 

find where it converges absolutely, where it converges conditionally, and where it diverges.

 

Note

We'll use the letter t for the terms of the power series, since the letter a is already used for the coefficients of the series.

 

Solution
Let's use the RaT (Ratio Test). Let tn be the nth term of the given power series. Then:

 

 

EOS

 

In Example 2.1 above we utilize the ratio test and the examinations at the 2 endpoints of the interval (5, 9), which is centered at
c = 7, the same point at which the power series is centered, to determine the behavior of the series on the entire real line. The
ratio test determines the behavior of the series on the real line except at the 2 endpoints of the interval, where of course it's not
applicable because the limit is 1. At each endpoint we examine the behavior of the individual series that's obtained at that
endpoint.

 

In the examinations at the endpoints x = 5 and x = 9, we don't use the RaT for the series obtained there, since the limit at both
places is 1:

 

 

It's easy to check that the limit is also 1 for the particular series obtained at x = 9. Observe that we test for absolute
convergence because the RaT is used only for ultimately positive series. There's no conclusion on absolute convergence, so no
conclusion on convergence (if absolute convergence, then convergence; if not absolute convergence, then either convergence
or divergence). The RaT can never determine the behavior of the series at the endpoints. We won't waste time trying to employ
it at the endpoints.

 

This and other examples suggest that for a power series centered at c, exactly one of the following three alternatives holds:

 

 

This suggestion has led to the investigation for and the establishment of Theorem 2.1 below, which confirms that the suggested
set of alternatives is indeed true.

 

Where Power Series Converge Absolutely

 

For a power series centered at c, at x = c the sum of the absolute values of its terms equlas |a0|, since all the terms from the
2nd one on are 0. The power series centered at c always converges absolutely at x = c. Now refer to Figs. 2.1 and 2.2.
Theorem 2.1 below states that if it also converges at a point x1 other than c, then it converges absolutely at least on the open
interval (x1, x2) centered at c if x1 < c as in Fig. 2.1 or on the open interval (x2, x1) centered at c if x1 > c as in Fig. 2.2.

 

Fig. 2.1

 

Absolute Convergence At Least On (x1, x2) Centered At c.

 

Fig. 2.2

 

Absolute Convergence At Least On (x2, x1) Centered At c.

 

Theorem 2.1 – Where Power Series Converge Absolutely

 

A power series:

 

 

converges absolutely at least on an open interval centered at c and having x1 as one endpoint.

 

 

Proof
At x = c the sum of the absolute values of the terms of the series is a0, since all the terms from the 2nd one on are 0. This
means that the power series converges at x = c.

 

EOP

 

{2.1} Section 14.2 Theorem 3.1.

 

Fig. 2.3

 

|x – c| < |x1 – c|,
c + (c – x1) = 2c – x1.

 

Fig. 2.4

 

|x – c| < |x1 – c|,
c + (c – x1) = 2c – x1.

 

Where Power Series Diverge

 

Corollary 2.1 below asserts that if a power series diverges at a point x1, which of course must be other than c, then it diverges
for all x that is farther away from c on either side of c than x1 is. It's actually simply the contrapositive of Theorem 2.1.

 

Corollary 2.1 – Where Power Series Diverge

 

If the power series:

 

@@ 

diverges at a point x1, then x1 is different from c, and the series diverges at all x where |x – c| > |x1 – c| (x is farther away from c on either side of c than x1 is).

 

 

Proof

EOP

 

Centre, Radius, And Interval Of Convergence

 

Example 2.1, the comment following it, Theorem 2.1, and Corollary 2.1 show that the set of values of x where a power series
centered at c converges is an interval centered at c, that the series diverges everywhere else outside of that interval, and that
that interval may be either just the isolated point c or finite with positive length or infinite.

 

Definition 2.1 – Centre, Radius, And Interval Of Convergence

 

The power series:

 

 

The point c is said to be the centre of convergence of the series. The number R in case 2 is said to be the radius of convergence of the series. For case 1, we say that the radius of convergence is 0. For case 3, we say that the radius of convergence is infinite. The interval I is said to be the interval of convergence of the series.

 

 

Remarks 2.1

 

1. In case 1, the series converges only at c and diverges everywhere else. In case 2, it converges in I and diverges
     everywhere else outside of I. The behavior of the series at the endpoints c – R and c + R depends on the individual series.
     Some series converge at both endpoints and so I = [c – R, c + R], certain series diverge at both endpoints and so I =
     (c – R, c + R), other series converge at one endpoint and diverge at the other and so either I = [c – R, c + R) or I = (c –
     R, c + R]. Of course an endpoint is included in I only if the series converges at that endpoint. In case 3, the series
     converges everywhere on R (converges for all x.)

 

2. The interval I is the interval of convergence of a power series in x – c if the series converges for all x in I and diverges for
     all x outside of I. Of course I is centered at c.

 

3. The radius of convergence is always R whether the interval of convergence is (c – R, c + R) or [c – R, c + R) or (c – R, c +
    R] or [c – R, c + R].

 

Power Series As Functions

 

We want to emphasize that a power series restricted to its interval of convergence is a function.

 

Power Series As Functions

 

 

 

Determining The Centre, Radius, And Interval Of Convergence

 

 

 

Determining The Centre, Radius, And Interval Of Convergence

 

Consider the power series:

 

 

Note that the an's are the coefficients, not terms, of the series.

 

 

Remarks 2.2

 

1. To determine the radius of convergence, we can employ either one of the equations in Eq. [2.1] or the RaT directly as done
    in Example 2.1. In any case, we can find the limit first, then obtain the value of R from it as its reciprocal.

 

2. Knowing the radius of convergence R doesn't imply knowing the interval of convergence, as then the interval of convergence
    may be either (c – R, c + R) or [c – R, c + R) or (c – R, c + R] or [c – R, c + R].

 

3. Observe that the radius and thus the interval of convergence depend only on the coefficients an's and not on both the an’s
    and the powers of x – c. We can think of the reason as being like this: the an's are constant and thus fixed and x is variable,
    and hence the an's must “decide” where x has to be to make the series converge, and that's possible because x can change.

 

Example 2.2

 

Determine the centre, radius, and interval of convergence of the series:

 

 

Note

We already know all the 3 attributes of this geometric series. This example is simply an easy start of our practice of our newly
acquired skills.

 

Solution
The centre of convergence of the given series is c = 0. Let an = 1. Then:

 

EOS

 

In Example 2.2, if x = 0 then the sum of the series is 0. If 0 < x < 1, then as n gets larger and larger, xn gets smaller and
smaller towards 0 from the right of 0 fast enough that the series converges. If – 1< x < 0, then as n gets larger and larger, xn
gets smaller and smaller towards 0 from the right and the left of 0 alternately fast enough that the series converges. In other
cases, the limit of xn isn't 0 and thus the series diverges.

 

Example 2.3

 

Find the centre, radius, and interval of convergence of the following power series:

 

 

Solution
a) The centre of convergence of the series is c = 0. Let an = n!. Then:

 

EOS

 

As Example 2.3 shows, when the radius of convergence is 0 or infinite, we don't have to worry about the endpoints to
determine the interval of convergence.

 

In part a, in the case of – 1 < x < 0 or 0 < x < 1, no matter how small x is in size, the product n!xn either doesn't approach 0
or doesn't approach 0 fast enough. This is because xn doesn't approach 0 faster enough than n! approaches infinity. Other
cases are obvious.

 

In part b, in the case of x < – 1 or x > 1, no matter how large x is in size, the quotient xn/n! approaches 0 fast enough. This is
because the size of xn approaches infinity slower enough than n! approaches infinity. Other cases are obvious.

 

Example 2.4

 

Determine the centre, radius, and interval of convergence of the series:

 

 

Note

The given series isn't in the form of a power series. So we'll first re-write it in the form of a power series.

 

Solution

 

which converges. Consequently the interval of convergence is I = [– 5, 2].
EOS

 

 

It's the opposite situation for the case of x < – 5 or x > 2.

 

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3. Power Series For A Given Interval Of Convergence

 

Example 3.1

 

Find a power series that has (1, 5) as its interval of convergence. Note: many answers are possible.

 

Note
The required series is of the form:

 

 

Solution

The centre of convergence is c = (1 + 5)/2 = 3. So the required power series is of the form:

 

 

We have:

 

EOS

 

Remarks 3.1
 
In Example 3.1:
 

1. We have to examine the series obtained at the 2 endpoints to make sure that indeed the series found has the given interval
    as its interval of convergence. In this example with the open interval (1, 5), the 2 series must both diverge.

 

 

    which diverges; at x = 5 the series is:

 

   

 

    which diverges. So for any given non-0 constant k, an = k/2n will do. There are indeed many possible answers.

 

Example 3.2

 

Find a power series that has [1, 5) as its interval of convergence. Note: many answers are possible.

 

Note

This time the power series has to converge at x = 1. It still has to diverge at x = 5.

 

Solution
The centre of convergence is c = (1 + 5)/2 = 3. So the required power series is of the form:

 

 

We have:

 

EOS

 

Remarks 3.2
 
In Example 3.2:

 

 

2. Similarly to part 3 of Remarks 3.1 On Example 3.1, it's easy to check that for any given non-0 constant k, an = k/(2nn) will
    do. There are indeed many possible answers.

 

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4. Determining Power Series Representing Given Functions

 

In the following two examples, we'll use the formula for the geometric series: a/(1 – r) = a + ar + ar2 + ar3 + ..., –1 < r < 1,
to determine a power series that represents a given function.

Example 4.1

 

Determine a power series that represents the function 1/(1 – x) on an interval centered at x = 10 and determine that interval.
Write the series in the sigma notation.

 

Note

We know that 1/(1 – x) is represented by 1 + x + x2 + x3 + ... on (–1, 1). Now we try to see if it's represented by another
power series on an interval centered at x = 10. If it is, that series must be diffrerent from 1 + x + x2 + x3 + ..., which
represents it only on (–1, 1) and thus can't equal it at, for example, x = 10.

 

The centre of convergence of the series will be x = 10. So it'll be of the form a0 + a1(x – 10) + a2(x – 10)2 + a3(x – 10)3 + ... .
Thus we'll algebraically manipulate 1/(1 – x) to introduce the term x – 10 in its denominator.

 

Solution

EOS

 

In Example 4.1, the series found is a power series in x – 10 and a geometric series in – (x – 10)/9. This example shows that
indeed the function 1/(1 – x) is represented by another series on the interval (1, 19) centered at (1 + 19)/2 = 10. For all x in
(1, 19), the function 1/(1 – x) equals the series – 1/9 + (1/92)(x – 10) – (1/93)(x – 10)2 + (1/94)(x – 10)3 – ... . For example, at
x = 10 the function equals –1/9 and the series equals –1/9 also.

 

Note that the denominator in the original function is 1 – x, while the terms of the power series contain powers of x – 10, not of
1 – x. Remark that we don’t write – (x – 10)/9 as (10 – x)/9, because we should have the powers of x – 10, not of 10 – x, in
the series, in accordance with the definition of power series.

 

Example 4.2

 

Find a power series that represents the function 1/(1 – x) on an interval centered at x = –100 and determine that interval.
Express the series in the sigma notation.

 

Note
As (x – (– 100)) = x + 100, the required series will be of the form a0 + a1(x + 100) + a2(x + 100)2 + a3(x + 100)3 + ... .

 

Solution

EOS

 

In Example 4.2, the centre of the interval (–201, 1) is ((–201) + 1)/2 = –100. Examples 4.1 and 4.2 suggest that for any point c
in the domain of the function 1/(1 – x) on the real line, there's a power series that represents the function on an interval
centered at x = c. Of course the point x = 1 is excluded because it's not in the domain of the function (if we try to obtain x – 1
in the denominator we'll have 1 – (x – 1) – 1 = – (x – 1) = 1 – x; the denominator – (x – 1) can't be used; the denominator
1 – x is nothing new).

 

Example 4.3

 

Represent the function f(x) = x2/(3 + x) by a power series. Use the sigma notation for the answer and state the interval on
which the representation is valid.

 

Note

When the centre of convergence isn't specified by the problem statement, it's understood to be x = 0. We'll utilize the
power-series representation of 1/(1 – x) and algebraic manipulation. For this purpose, we note that x2/(3 + x) = x2(1/(3 + x)).
So we first determine a representation of 1/(3 + x), and then multiply by x2 to get the representation of x2/(3 + x).

 

Solution

 

EOS

 

In Example 4.3, for the coefficients, eg –1/32, we keep the denominator in the power form, eg –1/32, we don't write them in
the multiplied-out form, eg –1/9. This is to keep their pattern clear, which is very useful when we write the series in the sigma
notation.

 

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5. Algebraic Operations On Power Series

 

Addition, Subtraction, And Constant Multiplication Of Power Series

 

Addition and subtraction of two power series that have the same centre of convergence, done at the same point x that belongs
to the intervals of convergence of both series, and multiplication of a power series by a constant can be performed just like
series of constants; see Section 14.2 Definition 4.1. For example:

 

 

The nth term of the sum is the sum of the nth terms of the original series.

 

Definition 5.1 – Addition, Subtraction, And Constant Multiplication Of Power Series

 

 

 

Theorem 5.1 – Properties Of Sums, Differences, And Constant Multiples Of Power Series

 

 

 

Proof
1) This property is clearly a straightforward consequence of Definition 5.1 above.

 

2) We'll prove that the interval of convergence of the sum is the smaller of those of the two original series, which implies that
    its radius of convergence is the smaller of their radiuses of convergence . The proof for the other two operations is similar.

  

EOP

 

{3.1} Section 14.2 Theorem 4.1.

Multiplication And Division Of Power Series

 

Consider the addition of numbers. For example, (m + n) + ( p + q) = (m + p) + (n + q). The nth term of the sum of two finite
series is the sum of the nth terms of the two original series. This has extended to infinite series. Now consider the
multiplication of numbers. For example, (m + n)(p + q) = m( p + q) + n( p + q). The nth term of the product of two finite
series isn't the product of the nth terms of the original series. The definition of multiplication of infinite series of course is
naturally based on that of finite series.

 

Let's multiply two (infinite) series having the same centre of convergence c:

 

(a0 + a1(x – c) + a2(x – c)2 + …)(b0 + b1(x – c) + b2(x – c)2 + …)
     = a0b0 + a0b1(x – c) + a0b2(x – c)2 + … + a1b0(x – c) + a1b1(x – c)2 + a1b2(x – c)3 + … + a2b0(x – c)2 + a2b1(x – c)3
        + a2b2(x – c)4 + …
     = a0b0 + (a0b1 + a1b0)(x – c) + (a0b2 + a1b1 + a2b0)(x – c)2 + …

 

 

Definition 5.2 – Cauchy Product Of Power Series

 

 

 

Note on the formula giving pn: pn is the nth coefficient of the product, ai is the ith coefficient of the first original series, bn–i is
the (n – i)th coefficient of the second original series. For pn, the subscript i of a increases from 0 to n and the subscript n – i of
b decreases from n to 0. In any term in the formula, the sum of the subscripts of a and b add up to n.

 

The radius of convergence of a Cauchy product equals at least the smaller of those of the original series. This assertion is
discussed in a textbook on more advanced calculus or on mathematical analysis.

 

The multuplication of power series certainly is more complicated than the operations discussed earlier. The division of power
series is even more complicated than the multiplication. It's investigated in a more advanced calculus course or in a
mathematical analysis course.

 

Determining Power Series Representing Given Functions

 

Example 5.1 – Using Direct Multiplication

 

Determine a power-series representation of the function 1/(1 – 3x)2 on an interval centered at 0 by using direct multiplication of
power series. Use the sigma notation for the series and state the interval of the validity of the representation.

 

Solution

EOS

 

To find the product, we can also use the formula for the Cauchy product, as shown in Example 5.2 below.

 

Example 5.2 – Using The Cauchy Product

 

Determine a power-series representation of the function 1/(1 – 3x)2 on an interval centered at 0 by using the Cauchy product
of power series. Use the sigma notation for the answer and state the interval of the validity of the representation.

 

Solution

EOS

 

In solving problems, if you're not required to use a specific method, you can use either the direct multiplication or the formula
for the Cauchy product.

 

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6. Tasks Performed

 

To help us understand this section more clearly, we list and distinguish the tasks performed, as follows.

 

1. In Part 1, we define what a power series is.

 

2. In Part 2, given a power series, we determine its interval of convergence.

 

3. In Part 3, conversely, given an interval, we determine a power series that has the interval as its interval of convergence.

 

4. In Part 4, 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.

 

5. In Part 5, 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.

 

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Problems & Solutions

 

1. Determine the centre, radius, and interval of convergence of the following power series:

 

  

 

Solution

 

The centre of convergence of the given power series is c = 1. Let an = 1/(n2 + 1). We have:

 

 

which also converges. Thus the interval of convergence is I = [0, 2].

 

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2. Find the centre, radius, and interval of convergence of this power series:

 

  

 

Solution

 

The centre of convergence is c = –1. Let an = 1/(5n+2n). Then:

 

 

which, like the harmonic series, diverges. Thus the interval of convergence is I = [– 6, 4).

 

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3. Determine the centre, radius, and interval of convergence of:

 

  

 

Solution

 

 

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4. Find the centre, radius, and interval of convergence of the following power series:

 

  

 

Solution

 

The centre of convergence of the given power series is c = –3. Let an = n!. Then:

 

 

So the radius of convergence is R = 0 and thus the interval of convergence is I = {–3}.

 

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5. Determine the centre, radius, and interval of convergence of this series:

 

  

 

Solution

 

 

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6. Find a power series that has (1, 5] as its interval of convergence. Note: many answers are possible.

 

Solution

The centre of convergence is c = (1 + 5)/2 = 3. So the required power series is of the form:

 

 

We have:

 

 

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7. Find a power series that has [1, 5] as its interval of convergence. Note: many answers are possible.

 

Solution

The centre of convergence is c = (1 + 5)/2 = 3. So the required power series is of the form:

 

 

We have:

 

 

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8. Determine a power series that represents the function 4/(3 – x) on an interval centered at x = 0 and determine that interval.
    Write the series in the sigma notation.

 

Solution

 

 

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9. Find a power series that represents the function 4/(3 – x) on an interval centered at x = 5 and determine that interval. Write
    the series in the sigma notation.

 

Solution

 

 

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10. Represent the function x3/(x – 3)2 by a power series by using algebraic manipulation and algebraic operations on series.
      Use the sigma notation for the answer and state the interval where the representation is valid.

 

Solution

 

 

 

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11.
a. A set of real numbers is the set of convergence of a series if the series converges for all x in the set and diverges for all x
    outside of it. For each of the following series, use the RaT on the absolute values to find the set of convergence of the series.

 

 

    converges at x = x1 then it converges absolutely for all x where |x – c| > |x1 – c|.

 

c. Prove that if the above series diverges at x = x1 then it diverges for all x where |x – c| < |x1 – c|.

 

Solution

 

a.

i. Let tn = 3n/xn. Then:

 

 

ii. Let tn = n!/xn. Then:

 

 

 

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