7/31/2019 Chapter_4 Enggng Math
1/57
MTK3001-Chapter 4
fnh 2011
Chapter 4: Infinite Series
4.1 Sequences
A sequence is simply a list of numbers. For example,
are all sequences. The numbers in a sequence are called term of the sequence.
Finite :
Infinite:
Frequently, we list the term of sequence in brace * and +, so that we would write*+
rather than just
When we define a sequence to be a list, we imply that the terms in the list are
, that there is
a first, a second, a third, a fourth, term in the sequence. This means that we can use the natural
number to the terms of the sequence. It means that we can represent any sequenceby writing for the first term, for the second term, for the third term, for the fourth term andso on; the th term, or general term, is written as . For the sequence , for example, wehave
For many of the sequences that we encounter, there is a simple rule that relates to . In the aboveexample, we see that
And, in general, that
The sequence *+ can therefore be written more compactly as *+, with the understandingthat runs through unless otherwise stated.
7/31/2019 Chapter_4 Enggng Math
2/57
MTK3001-Chapter 4
fnh 2011
When we know the rule for the general th term of a sequence, we can use it to specify the sequencein a slightly different way from before, giving both the first few terms and the th. We hardly need thisfor our sequence , since it pattern is so obvious, but we could if needed specify it in moredetail in the form
The observation that depends on the index also gives us another way of defining a sequence : as a whose domain is the set of positive integers. For example, we could write the sequence*+ as the function
Example 4.1: Write down the first five terms of the sequence *+
Example 4.2: Rewrite the following sequences, giving both the first seven terms and general th term.1. 2. 3. 4.
7/31/2019 Chapter_4 Enggng Math
3/57
MTK3001-Chapter 4
fnh 2011
The term You should note the use of the expression in the 3rdexample, especially if you haventseen it before. You will encounter this expression ( and the similar expression ) veryfrequently in this chapter on sequence and series. They provide simple but very useful ways of
. This is because so that the values of are as takes the values Similarly, the values of arerespectively.
4.1.1 Graphs of sequencesThe graph of the sequence
{} { }can be thought as the graph of the function , where do not confuse this graph with the graph of
, for
in
: they are different because
Example 4.3: Graph the following sequences
1. * +2. 3. *+4. Solution :
7/31/2019 Chapter_4 Enggng Math
4/57
MTK3001-Chapter 4
fnh 2011
With these graphs, and as a preliminary to the next section, let us consider the following
somewhat vague question: How do the above sequences behave as gets larger and larger?The following observations seem to summarize their behavior adequately.
1. The terms in the sequence
* +increase indefinitely, or without bound they have no
limiting value.2. The terms in the sequence also increase but towards a limiting value of 1.3. The terms in the sequence *+ oscillate between -1 and 1, without settling down to
any one limiting value..
4. The terms in the sequence oscillate too, but they also tend towards a limitingvalue of 1.
4.2Convergence of sequencesSuppose that we examine some sequence
*+, and that we find that the values
stabilise at
or around some particular limiting value as gets larger and larger. Rephrasing this, supposethat we find that as increases, get close to , and that the larger becomes, the closer gets to .If this happens, then we say that the sequence converges to the limit , or that the limit of as increases is , and we write either
If the values of
do not stabilize at any one value, then we say that the sequence
*
+
diverges, or that does not exist.Example 4.4: Determine whether the following sequences converge or diverge.1. 2. 3. Solution:
7/31/2019 Chapter_4 Enggng Math
5/57
MTK3001-Chapter 4
fnh 2011
4.3 Methods
We have seen above how to find limits in some fairly simple examples. However, most examples
in practice are more complicated than those, so we need to be more systematic in our
approach. In this section, we introduce some general ideas which will greatly increase the
number of limits we will able to deal with.
4.3.1 Using lHopitals ruleLHopitals rule applies to the limit of a function of a real variable as approaches a finite limitor , when the limit yields a so-called
With sequences, we are dealing with limits as the variable approaches infinity, ratherthan as a
variable
approaches a limit, but it turns out that we can still make use of
lHopitals rule.
This idea is simply this: If the limiting process yields one of the indeterminate forms as increases, replace by , apply lHopitals rule to the resulting function of, and use the answerit gives.
Example 4.5: Determine whether the sequence converges or diverges.Solution: We can see immediately that yield the indeterminate form .We attempt to use lHopitals rule, by replacing
by
everywhere, and calculating the limit of
the resulting function of as . Differentiating top and bottom separately and applying therule, we have
and we conclude that =0.4.3.2 Absolute valuesFor any sequence
*+,
|| Example 4.6: Consider . We have
||
7/31/2019 Chapter_4 Enggng Math
6/57
MTK3001-Chapter 4
fnh 2011
and hence
|| Therefore, by the result above,
also. That is, the sequence converges to 0.But note very carefully that the method is . Look again whatyou have learned in function and limit.
4.3.3 Properties Of Convergent Sequences
Let *+ and *+ be sequences, and let and , and let .Then we have the following properties:
1) 2) 3) 4) 5)
4.3.4 Squeeze lawThis law says : if the terms of a sequence *+ are squeezed between the terms of two othersequences *+ and *+ and if*+ and *+ both converge then *+also converges to . More formally:
Example 4.7: Consider . We know that
For all . Therefore,
7/31/2019 Chapter_4 Enggng Math
7/57
MTK3001-Chapter 4
fnh 2011
Also,
Therefore, by the squeeze law.
4.4 SeriesCan we add the terms *+ together to obtain a meaningful, finite value? We will now introducethe terminology and notation necessary to analyse the question.
An infinite series is an expression of the form
The numbers are called the terms of the series. Since all our series will have aninfinite number of terms, we will omit the word infinite from now on, and just refer to series.
Consider the sum
This is a sum, of terms, and we know that such a sum makes sense, and can becalculated numerically if we know the specific values of . But what about theinfinite sum
-we dont know what this means, or whether it means anything, or how to calculate it if it does
mean something: .So our basic problem is: Can we make sense of the idea? We will see that the ans wer is:sometimes yes, and sometimes no.
4.4.1 Partial sums
Given the infinite series , we write down the sequence of finite sums
7/31/2019 Chapter_4 Enggng Math
8/57
MTK3001-Chapter 4
fnh 2011
The numbers
Are called the partial sums of the finite series, and is the th partial sum. The sequence
* +Of these numbers is called the sequence of partial sums.
As increases, the partial sum Includes more and more terms of the series. Thus if tends toward a limit as , that is,if
Then it is reasonable to view this limit as the sum of the terms in the series. This gives usour definition of an infinite sum.
Example 4.8: Determine whether the series
( ) ( ) ( ) ( )
( ) Converges or diverges. If it converges, find the sum.
Solution:
We need to find, if possible, a closed form expression for , and then to see if exists. We have
( ) ( ) ( ) ( )
7/31/2019 Chapter_4 Enggng Math
9/57
MTK3001-Chapter 4
fnh 2011
( ) ( ) ( ) (This is an example of a telescoping sum-you will have seen such sums before-in which a large
summation of terms, with careful grouping, collapses to give a simple answer.)
To determine if the series converges or diverges, we calculate (which also will give usthe sum, if it exists). We have
( ) And therefore converges, and has a sum of 1/3.So far we have determined whether or not a series converges by carefully examining
, the
sequence of partial sums. However, in practice, it is relatively rare that the th partial sum canbe written in a sufficiently simple form for us to be able to do this. For most series, convergenceir divergence has to be determined indirectly, using convergence tests, some of which we will
discuss later. These tests do not generally give us an exact value for the sum, but once
convergence is established, the sum of a series can generally be approximated to any desired
degree of accuracy by summing sufficiently many terms, or by other means.
4.5 Special SeriesIt is important to understand thoroughly the convergence properties of certain commonly
occurring classes of series. In this section, we will examine two very important such classes, the
geometric series and the -series. As the chapter proceeds, we will see that these two classesprovide the basis for analyzing the behavior of many other more complex series.4.5.1 Geometric seriesA geometric series is one of the form
Where and are some fixed real numbers. For example
Is a geometric series with and and ( )
( ) ( )
( )
Is a geometric series with and
7/31/2019 Chapter_4 Enggng Math
10/57
MTK3001-Chapter 4
fnh 2011
The fundamental facts about the convergence of geometric series are the following:
The general geometric series
converges if|| and diverges if| | If the series converges, then the sum is
Example 4.9:
Find whether the following two geometric series converge or diverge, and find their sums if
they converge:
1. 2.
Solution:
7/31/2019 Chapter_4 Enggng Math
11/57
MTK3001-Chapter 4
fnh 2011
Example 4.10:
Show that the following are geometric series, and discuss their convergence
1.
2. Solution:
7/31/2019 Chapter_4 Enggng Math
12/57
MTK3001-Chapter 4
fnh 2011
4.5.2 -seriesA series of the form
For any fixed real number , is called a -series.If we take the special case we get
which is known as the harmonic series.
The fundamental fact about convergence of-series is the following:A -series converges if and diverges if Thus, the harmonic series diverges because , while the series converges, because Example 4.11:
Write down , and decide whether the following
-series converge or diverge.
1. 2. 3. 4. Solution:
7/31/2019 Chapter_4 Enggng Math
13/57
MTK3001-Chapter 4
fnh 2011
4.6Properties of Convergent SeriesLet and be convergent series, and let . Then1. converges, and 2.
converges, and
We could also add the following rule to the list (it actually follows from (1)):
If diverges and , the also diverges.Example 4.12:
Determine whether the following series converge or diverge. If the series converge, find their
sum, if possible.
1.
2. Solution:
7/31/2019 Chapter_4 Enggng Math
14/57
MTK3001-Chapter 4
fnh 2011
4.7The th- term testThe th- Term test is a simple test for the divergenceof a series. The good aspect of the testssimplicity is that is usually easy to apply; the bad aspect is that there are many cases where it
gives us no information at all! The test says the following:
We normally use the th term test in the following way: asked to analyse the convergence of aspecific series , we look first not at the series , but at the sequence *+. Weattempt to find the limit of the sequence ( this is usually an easier problem than thatof handling the series). If we are able to show either that does not exist, or that does exist but has a non-zero value, then we conclude immediately from the thTerm Test that the series
diverges- and this solves the entire problem for us.
What if we find that the answer is that . That is,the th-Term test has not helped at all with the original problem!We then need to apply other methods (which will probably be harde to apply, but will be more
likely to yield definite information). We will turn to study of such methods shortly.
4.8Tests for ConvergenceOur discussion so far has not gone deep enough to allow us to analyse the convergence of any
but a few quite special classes of series (though these classes are important, and must be
understood well for success in applying other methods). To th-Term Test, these tests will insome cases yield inconclusive results, but they are still very useful.We will break up our set of convergence tests into two groups:
Tests for series of terms, such as
Test for with terms of either sign. An important special case here is atest for that is, series whose terms are alternately positive andnegative, such as
7/31/2019 Chapter_4 Enggng Math
15/57
MTK3001-Chapter 4
fnh 2011
4.9Series of Positive TermsIn this section, we will introduce three tests which apply to series of positive terms.
4.9.1 Guessing the answerOur first two tests will require us to make an initial intelligent at whether our seriesconverges or diverges. Once we have made this initial guess, we apply the tests to try to provethat we are right. Its useful, therefore, for us to discuss first how we can go about guessing
efficiently, and with a good chance of correctness.
Is a guess acceptable?
You, might think that guesswork is incompatible with mathematical argument, so is a guess
acceptable? The answer is: On its own, no! But the point is that once we have made our guess,
we go on to apply a test to the guess. If the test succeeds, then the fact that weguessed no longer matters. If the test fails, then we made a bad guess, or we didnt use the
right test, and we need to try again!
How, then, do we guess successfully? In the long run, the only recipe for success is to see lots of
worked examples, and to try working many yourself. But below are three informal guidelines to
get started. Bear in mind that most such rules are not infallible: you must expect that some of
your guesses will turn out to be bad ones, although the hope is that your good guesses will
increase with practice.
Rule 1. If the term following the sigma-sign is multiplied by a constant factor, the constant can
be moved out to the left of the sigma-sign, and the series analysed without it.
For example,
The rule says that
Then, since diverges (its the harmonic series), we guess that also diverges.
In fact, this first rule gives us more than a mere
at the answer-it gives us aresult
to be correct-because we saw in our discussion of the properties of convergentseries in Section 4.6 that converges if and only if converges
Rule 2. Terms which are small in value can be disregarded when they are added to ones which
are large in value.
7/31/2019 Chapter_4 Enggng Math
16/57
MTK3001-Chapter 4
fnh 2011
For examples,
We would guess that
Since 1 is small compared with when is of any significant size. Then, since converges, and converges (it is geometric series with ), we guess that will probably converge as well (though we still to this to be certain). Simirlarly, we would guess that
since 30 is small compared with
when
is large. Then, since
______________we guess that also ____________.
Again, we guess that
Then,
Rule 3. If polynomial terms in
appear, then all but the highest power of
in each one can
usually be ignored. (Note that this is more or less a special case of Rule 2, but it occurs so often
that its worth stating it on its own.)
Here are some examples to illustrate.
We would guess that
Then, since
______________ we guess that also
____________.
Likewise, we guess that
7/31/2019 Chapter_4 Enggng Math
17/57
MTK3001-Chapter 4
fnh 2011
Then, this series is ______________
What about
4.9.2 The Comparison Ratio testLet be a series of positive terms, be a known convergent series ofpositive terms, a, and let be a known divergent series of positive terms.1) If then converges, provided that is finite.2) If or if then diverges.How do we use the Comparison Ratio test?
i) Guess at the behavior of the given series --that is , guess whether converges ordiverges.
ii) Apply the Comparison ratio test to attempt to prove the guess is correct.
But how do we obtain the known series that is required to apply the test a convergent series
if we guessed convergence, an a divergent series
if we guessed divergence?
The answer is that the process of guessing in Step (i) should already have suggested a suitableseries or : if our guess took the form seems to behave like the known series then the series is almost certainly the series or that we need forStep (ii).
7/31/2019 Chapter_4 Enggng Math
18/57
MTK3001-Chapter 4
fnh 2011
Example: We return to an earlier example, where we guessed that
which is a known divergent series. As the comment above suggest, the harmonic series should serve as the required known divergent series in the Comparison ratio test.
Thus, we let and , and apply (2) of the test: we have
[ ] Therefore, diverges, by the Comparison Ratio Test.Note that, as we said earlier, the fact that we made a guess along the way :the successful application of the test stands on its own.Example: We also guessed earlier that
Which is ______________________
Thus,
7/31/2019 Chapter_4 Enggng Math
19/57
MTK3001-Chapter 4
fnh 2011
Exercise: Guess the behavior of the following series, and then use the Comparison ratio Test to
attempt to determine whether they converge or diverge.
1.
2. 3. ||
7/31/2019 Chapter_4 Enggng Math
20/57
MTK3001-Chapter 4
fnh 2011
4.9.3 The Comparison testLet be a series of positive terms, be a known convergent series ofpositive terms, a, and let be a known divergent series of positive terms.1) If for each and converges, then converges.2) If for each and diverges, then diverges.
To apply the test, we still need to guess the behavior of the series as earlier, and ourguess should provide the series or required to apply the test.
It is generally good to use this test the Comparison Ratio Test failed. The greatest difficulty once we have made our guess is in showing that the require
inequality holds. An approach that is often helpful is to try to prove that (which gives ) or (which gives ).
Example: We guessed earlier that the series
||
Which is a convergent -series. We will apply the Comparison test to confirm this guess.Now || for all , so that
||
Now, taking
|| In the comparison test, we find that since ____________, so does_________________.
Exercise: Use the Comparison Test to determine whether the following series converge or
diverge.
1. 2.
7/31/2019 Chapter_4 Enggng Math
21/57
MTK3001-Chapter 4
fnh 2011
4.9.4 DAlemberts Ratio testOur third testis quite different in character from the earlier tests: it doesnt require us to make a
guess about the behavior of our series, and it doesnt require the use of any known convergent
or divergent series. Instead, it compares the terms of the given series with each other.
Let be a series of positive terms, and let assuming that the limit exists. (if the limit does not exist, then we cannot apply the test). The
test is as follows:
1) If the series converges2) If the series diverges3) If
, the ratio test fails, and the series may converge or it may diverge.
Example: Consider here we have
And so,
[ ]
[ ]
Therefore, this series is ____________________, by dAlemberts ratio test.
A couple of observations about the test are worth noting.
This test is often useful when
involves
(as in the example above) or
th powers.
This test will fail whenever has the form Exercise: use dAlemberts ratio Test, if possible, to determine whether the following series
converge or diverge.
1.
7/31/2019 Chapter_4 Enggng Math
22/57
MTK3001-Chapter 4
fnh 2011
2. 3. Solution:
7/31/2019 Chapter_4 Enggng Math
23/57
MTK3001-Chapter 4
fnh 2011
Summary
Assignment
7/31/2019 Chapter_4 Enggng Math
24/57
MTK3001-Chapter 4
fnh 2011
4.10 General SeriesSo far we have only discussed convergence tests for series of positive terms. We will now
discuss general series, series containing terms of either sign. By far the most important class of
general series is the class ofalternating series, in which the terms strictly alternate in sign.
Examples of alternating series are:
Alternating series arise very often in practice, and we will shortly introduce a convergence test
specifically for such series. But first, we will see what can be said about general series.
4.10.1 Absolute ConvergenceA general series is said to be absolutely convergent if the series || converges.Putting this informally: if we have a series of terms of either sign, we consider the newseries || got by making all the terms positive, then we say that the original series isabsolutely convergent.
For example, the series
Is absolutely convergent, because the new series obtained by making all the terms positive
that is,
----converges because _______________
Exercise : Determine whether these series are absolutely convergence
1. 2. 3.
7/31/2019 Chapter_4 Enggng Math
25/57
MTK3001-Chapter 4
fnh 2011
Solution:
7/31/2019 Chapter_4 Enggng Math
26/57
MTK3001-Chapter 4
fnh 2011
4.10.2 Using absolute convergenceWe began this section by discussing the convergence of a series , where the terms were of either sign, and we instead introduced a different series, namely || , into thediscussion, and analysed its convergence. Why? How does this help us to obtain information
about the series that we were originally interested in?The answer lies in the following important fact:If is absolutely convergent, then is also convergent (in the ordinary sense).
That isand this is how we apply the fact in practiceif we examine the new series || ,with positive terms, and can show that it converges, then it follows that the original series , with terms of either sign, also converges. Thus, of the three worked examples above,we can conclude immediately that in the first and second cases the original series
-----_____________
What about cases like the third example, where we found that || __________? Theanswer is that we have obtained no new information about the original series . That is,the original series may converge or may divergeand we will have to use some othekind of argument to tell which. (The Alternating Series Test, which we will introduce shortly, is
the remaining candidate test in this situation.)
4.10.3 The Ratio Test for absolute convergenceThe following version of dAlemberts Ratio Test is useful for investigating absolute
convergence.
Let be a general series, and let
if the limit exists.
If , then converges absolutely, and therefore converges. If or , then diverges. If , then the test gives no information about .
7/31/2019 Chapter_4 Enggng Math
27/57
MTK3001-Chapter 4
fnh 2011
As with the ratio test for series of positive terms, this test is often good to use when contains factor , or th powers, but can be expected to fail when has theform
Example : Anlyse the convergence of the following series.
1. 2. Solution:
1. The term
contains
, so we try the Ratio Test: we have
|| || And so
Therefore, converges absolutely, and, in particular, converges in theordinary sense.
2. Try it yourself
7/31/2019 Chapter_4 Enggng Math
28/57
MTK3001-Chapter 4
fnh 2011
Exercise: Are the following series absolutely convergent or not?
1. 2. 4.10.4 The Alternating Series TestRefer back to
examined the new series
||
And found that it diverged, and we concluded that this gave us no help in determining
whether the original series converges.The alternating series test frequently enables us to resolve the question of convergence
in such cases. It is as follows.
If
satisfies each of the following three conditions, then
converges:
1. is an alternating series,2. and3. || ||4.10.5 Practical tipsChecking condition (1) is normally straightforwardit is normally obvious on inspection
whether or not a given series is alternating.
Condition (2) presents a problem that we have already studied: that of finding the limit
of a sequence, so in principle, we know how to deal with it.
Checking condition (3)that is, determining whether || || for all is usuallythe hardest of the three conditions to handle, and requires a few comments. Depending
on the form of the term at least the following two methods can be tried:
7/31/2019 Chapter_4 Enggng Math
29/57
MTK3001-Chapter 4
fnh 2011
Proving that || || ; Converting || into a function of the real number , by replacing
each occurrence of by , and showing that , at least for sufficiently large. (We are not concerned about negative values of , forexample.)
All three conditions are needed
It is tempting to think that it is unnecessary check to both (2) and (3) when
applying the Alternating Series Test, because the conditions see, rather similar. But
all three conditions are necessary: if any one fails, then the test fails.
Note that if the series does not satisfy condition (2), that is, if
does not
exist, or if , then the series diverges by the th-Term Test.Example: Apply the Alternating Series Test to
Solution:
1. The series clearly alternates.2. (you should check this if it is not clear).3. We finally to shoe that || ||.
Method 1:
|| ||
For . Thereore, || || , which implies that || ||.Hence, by the Alternating Series Test, converges.
7/31/2019 Chapter_4 Enggng Math
30/57
MTK3001-Chapter 4
fnh 2011
Method 2: Replacing each occurance of in || by , we obtain the function . Then, we find
for . Therefore, is decreasing for , and so || ||, for . Therefore, by the Alternating Series Test, converges.Exercise:
Apply the Alternating Series Test to the following series.
1. 2. 3. Solution:
7/31/2019 Chapter_4 Enggng Math
31/57
MTK3001-Chapter 4
fnh 2011
7/31/2019 Chapter_4 Enggng Math
32/57
MTK3001-Chapter 4
fnh 2011
4.10.6 Conditional convergenceLet be an alternating series. We say that is conditionally convergent if converges and | | diverges.Consider again , . We show that
diverges and that converges.
Therefore, , is conditionally convergent.Exercise : Determine whether the following series converge absolutely, converge
conditionally or diverge.
1. 2. 3.
7/31/2019 Chapter_4 Enggng Math
33/57
MTK3001-Chapter 4
fnh 2011
4.11 Power SeriesWeve spent quite a bit of time talking about series now and with only a couple of exceptions
weve spent most of that time talking about how to determine if a series will converge or not.
Its now time to start looking at some specific kinds of series and well eventually reach the point
where we can talk about a couple of application of series.
In this section we are going to start talking about power series. A power series about a, or just
power series, is any series that can be written in the form,
where a and cn are numbers. The cns are often called the coefficients of the series. The first
thing to notice about a power series is that it is a function ofx. That is different from any other
kind of series that weve looked at to this point. In all the prior sections weve only allowednumbers in the series and now we are allowing variables to be in the series as well. This will not
change how things work however. Everything that we know about series still holds.
In the discussion of power series convergence is still a major question that well be dealing
with. The difference is that the convergence of the series will now depend upon the value ofx
that we put into the series. A power series may converge for some values ofxand not for other
values ofx.
Before we get too far into power series there is some terminology that we need to get out of
the way.
First, as we will see in our examples, we will be able to show that there is a number R so that
the power series will converge for, | | and will diverge for | | . This numberis called the radius of convergence for the series. Note that the series may or may not converge
if| | . What happens at these points will not change the radius of convergence.Secondly, the interval of all xs, including the end points if need be, for which the power series
converges is called the interval of convergence of the series.
These two concepts are fairly closely tied together. If we know that the radius of convergence
of a power series is R then we have the following.
7/31/2019 Chapter_4 Enggng Math
34/57
MTK3001-Chapter 4
fnh 2011
The interval of convergence must then contain the interval since we knowthat the power series will converge for these values. We also know that the interval of
convergence cant contain xs in the ranges and since we know thepower series diverges for these value ofx. Therefore, to completely identify the interval of
convergence all that we have to do is determine if the power series will converge for
or . If the power series converges for one or both of these values then well need toinclude those in the interval of convergence.Before getting into some examples lets take a quick look at the convergence of a power series
for the case of . In this case the power series becomes,
and so the power series converges. Note that we had to strip out the first term since it was theonly non-zero term in the series.
It is important to note that no matter what else is happening in the power series we are
guaranteed to get convergence for . The series may not converge for any other value ofx,but it will always converge for .Lets work some examples. Well put quite a bit of detail into the first example and then not
put quite as much detail in the remaining examples.
Example 1 Determine the radius of convergence and interval of convergence for the following power
series.
Solution
Okay, we know that this power series will converge for
, but thats it at this point. To determine the
remainder of thexsfor which well get convergence we can use any of the tests that weve discussed tothis point. After application of the test that we choose to work with we will arrive at condition(s) onx
that we can use to determine which values ofx for which the power series will converge and which
values ofxfor which the power series will diverge. From this we can get the radius of convergence and
most of the interval of convergence (with the possible exception of the endpoints.
7/31/2019 Chapter_4 Enggng Math
35/57
MTK3001-Chapter 4
fnh 2011
With all that said, the best tests to use here are almost always the ratio or root test. Most of the power
series that well be looking at are set up for one or the other. In this case well use the ratio test.
Before going any farther with the limit lets notice that sincexis not dependent on the limit and so it
can be factored out of the limit. Notice as well that in doing this we'll need to keep the absolute value
bars on it since we need to make sure everything stays positive and x could well be a value that will
make things negative. The limit is then,
| |
| |
So, the ratio test tells us that if the series will converge, if the series will diverge, and if we dont know what will happen. So, we have, | | | |
| | | | Well deal with the case in a bit. Notice that we now have the radius of convergence for this
power series. These are exactly the conditions required for the radius of convergence. The radius of
convergence for this power series is .Now, lets get the interval of convergence. Well get most (if not all) of the interval by solving the first
inequality from above.
So, most of the interval of validity is given by . All we need to do is determine if the powerseries will converge or diverge at the endpoints of this interval. Note that these values ofx will
correspond to the value ofxthat will give .
7/31/2019 Chapter_4 Enggng Math
36/57
MTK3001-Chapter 4
fnh 2011
The way to determine convergence at these points is to simply plug them into the original power series
and see if the series converges or diverges using any test necessary.
In this case the series is,
This series is divergent by the Divergence Test since . In this case the series is,
This series is also divergent by the Divergence Test since doesnt exist.So, in this case the power series will not converge for either endpoint. The interval of convergence is
then, Exercise:
1. Determine the radius of convergence and interval of convergence for the following powerseries.
2. Determine the radius of convergence and interval of convergence for the following power
series.
7/31/2019 Chapter_4 Enggng Math
37/57
MTK3001-Chapter 4
fnh 2011
4.11.1 Power Series and function
We opened the last section by saying that we were going to start thinking about applications of series
and then promptly spent the section talking about convergence again. Its now time to actually start
with the applications of series.
With this section we will start talking about how to represent functions with power series. The natural
question of why we might want to do this will be answered in a couple of sections once we actually
learn how to do this.
Lets start off with one that we already know how to do, although when we first ran across this series
we didnt think of it as a power series nor did we acknowledge that it represented a function.
Recall that the geometric series is
||
Dont forget as well that if|| the series diverges.Now, if we take and this becomes,
||
Turning this around we can see that we can represent the function
with the power series
|| This provision is important. We can clearly plug any number other than into the function,
however, we will only get a convergent power series if || . This means the equality in(1)will onlyhold if|| . For any other value ofxthe equality wont hold. Note as well that we can also use thisto acknowledge that the radius of convergence of this power series is and the interval ofconvergence is || .
http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeriesandFunctions.aspx#ZEqnNum731065http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeriesandFunctions.aspx#ZEqnNum731065http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeriesandFunctions.aspx#ZEqnNum731065http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeriesandFunctions.aspx#ZEqnNum7310657/31/2019 Chapter_4 Enggng Math
38/57
MTK3001-Chapter 4
fnh 2011
This idea of convergence is important here. We will be representing many functions as power series
and it will be important to recognize that the representations will often only be valid for a range ofxs
and that there may be values ofxthat we can plug into the function that we cant plug into the power
series representation.
In this section we are going to concentrate on representing functions with power series where the
functions can be related back to(2).
In this way we will hopefully become familiar with some of the kinds of manipulations that we will
sometimes need to do when working with power series.
So, lets jump into a couple of examples.
Example 1 Find a power series representation for the following function and determine its interval of
convergence.
Solution
What we need to do here is to relate this function back to(2). This is actually easier than it might look.
Recall that thexin(2)is simply a variable and can represent anything. So, a quick rewrite of gives,
and so the in holds the same place as thexin(2). Therefore, all we need to do is replace thexin(3)and weve got a power series representation for .
provided || Notice that we replaced both thexin the power series and in the interval of convergence.
All we need to do now is a little simplification.
provided || ||
http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeriesandFunctions.aspx#ZEqnNum756110http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeriesandFunctions.aspx#ZEqnNum756110http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeriesandFunctions.aspx#ZEqnNum756110http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeriesandFunctions.aspx#ZEqnNum756110http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeriesandFunctions.aspx#ZEqnNum756110http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeriesandFunctions.aspx#ZEqnNum756110http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeriesandFunctions.aspx#ZEqnNum756110http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeriesandFunctions.aspx#ZEqnNum756110http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeriesandFunctions.aspx#ZEqnNum756110http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeriesandFunctions.aspx#ZEqnNum756110http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeriesandFunctions.aspx#ZEqnNum756110http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeriesandFunctions.aspx#ZEqnNum756110http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeriesandFunctions.aspx#ZEqnNum752839http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeriesandFunctions.aspx#ZEqnNum752839http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeriesandFunctions.aspx#ZEqnNum752839http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeriesandFunctions.aspx#ZEqnNum752839http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeriesandFunctions.aspx#ZEqnNum756110http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeriesandFunctions.aspx#ZEqnNum756110http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeriesandFunctions.aspx#ZEqnNum756110http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeriesandFunctions.aspx#ZEqnNum7561107/31/2019 Chapter_4 Enggng Math
39/57
MTK3001-Chapter 4
fnh 2011
So, in this case the interval of convergence is the same as the original power series. This usually wont
happen. More often than not the new interval of convergence will be different from the original interval
of convergence.
Example 2 Find a power series representation for the following function and determine its interval of
convergence.
Solution
This function is similar to the previous function. The difference is the numerator and at first glance that
looks to be an important difference. Since(2)doesnt have an x in the numerator it appears that we
cant relate this function back to that.
However, now that weve worked the first example this one is actually very simple since we can use the
result of the answer from that example. To see how to do this lets first rewrite the function a little.
Now, from the first example weve already got a power series for the second term so lets use that to
write the function as,
provided || Notice that the presence ofxs outside of the series will NOT affect its convergence and so the interval
of convergence remains the same.
The last step is to bring the coefficient into the series and well be done. When we do this make sure
and combine thexs as well. We typically only want a singlexin a power series.
provided || As we saw in the previous example we can often use previous results to help us out. This is an
important idea to remember as it can often greatly simplify our work.
Example 3 Find a power series representation for the following function and determine its interval of
convergence.
http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeriesandFunctions.aspx#ZEqnNum756110http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeriesandFunctions.aspx#ZEqnNum756110http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeriesandFunctions.aspx#ZEqnNum756110http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeriesandFunctions.aspx#ZEqnNum7561107/31/2019 Chapter_4 Enggng Math
40/57
MTK3001-Chapter 4
fnh 2011
Solution
So, again, weve got an x in the numerator. So, as with the last example lets factor that out and see
what weve got left.
If we had a power series representation for
we could get a power series representation for .So, lets find one. Well first notice that in order to use (4) well need the number in the denominator to
be a one. Thats easy enough to get.
Now all we need to do to get a power series representation is to replace the xin(3)with
. Doing thisgives,
Now lets do a little simplification on the series.
The interval of convergence for this series is,
|| || Okay, this was the work for the power series representation for lets now find a power seriesrepresentation for the original function. All we need to do for this is to multiply the power series
representation for byxand well have it.
http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeriesandFunctions.aspx#ZEqnNum752839http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeriesandFunctions.aspx#ZEqnNum752839http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeriesandFunctions.aspx#ZEqnNum752839http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeriesandFunctions.aspx#ZEqnNum7528397/31/2019 Chapter_4 Enggng Math
41/57
MTK3001-Chapter 4
fnh 2011
The interval of convergence doesnt change and so it will be || .So, hopefully we now have an idea on how to find the power series representation for some functions.
Admittedly all of the functions could be related back to(2)but its a start.
We now need to look at some further manipulation of power series that we will need to do on
occasion. We need to discuss differentiation and integration of power series.
Lets start with differentiation of the power series,
Now, we know that if we differentiate a finite sum of terms all we need to do is differentiate each of
the terms and then add them back up. With infinite sums there are some subtleties involved that we
need to be careful with, but are somewhat beyond the scope of this course.
Nicely enough for us however, it is knwn that if the power series representation of has a radius ofconvergence of then the term by term differentiation of the power series will also have a radiusof convergence of R and (more importantly) will in fact be the power series representation ofprovided we stay within the radius of convergence.
Again, we should make the point that if we arent dealing with a power series then we may or may not
be able to differentiate each term of the series to get the derivative of the series.
So, what all this means for is that,
Note the initial value of this series. It has been changed from to . This is anacknowledgement of the fact that the derivative of the first term is zero and hence isnt in the
http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeriesandFunctions.aspx#ZEqnNum756110http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeriesandFunctions.aspx#ZEqnNum756110http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeriesandFunctions.aspx#ZEqnNum756110http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeriesandFunctions.aspx#ZEqnNum7561107/31/2019 Chapter_4 Enggng Math
42/57
MTK3001-Chapter 4
fnh 2011
derivative. Notice however, that since the n=0 term of the above series is also zero, we could start the
series at if it was required for a particular problem. In general however, this wont be done in thisclass.
We can now find formulas for higher order derivatives as well now.
Once again, notice that the initial value ofn changes with each differentiation in order to acknowledge
that a term from the original series differentiated to zero.
Lets now briefly talk about integration. Just as with the differentiation, when weve got an infinite
series we need to be careful about just integration term by term. Much like with derivatives it turns out
that as long as were working with power series we can just integrate the terms of the series to get the
integral of the series itself. In other words,
Notice that we pick up a constant of integration, C, that is outside the series here.
Lets summarize the differentiation and integration ideas before moving on to an example or two.
Fact
If has a radius of convergence of then,
both of these also have a radius of convergence ofR.
7/31/2019 Chapter_4 Enggng Math
43/57
MTK3001-Chapter 4
fnh 2011
Now, lets see how we can use these facts to generate some more power series representations of
functions.
Example 4 Find a power series representation for the following function and determine its interval of
convergence.
Solution
To do this problem lets notice that
( )Then since weve got a power series representation for
all that well need to do is differentiate that power series to get a power series representation for.
(
)
Then since the original power series had a radius of convergence of the derivative, and henceg(x), will also have a radius of convergence of .Example 5 Find a power series representation for the following function and determine its interval ofconvergence.
Solution
In this case we need to notice that
7/31/2019 Chapter_4 Enggng Math
44/57
MTK3001-Chapter 4
fnh 2011
and the recall that we have a power series representation for
Remember we found a representation for this in Example 3. So,
We can find the constant of integration, C, by plugging in a value ofx. A good choice is since that
will make the series easy to evaluate.
So, the final answer is,
Note that it is okay to have the constant sitting outside of the series like this. In fact, there is no way to
bring it into the series so dont get excited about it.
7/31/2019 Chapter_4 Enggng Math
45/57
MTK3001-Chapter 4
fnh 2011
4.12 Taylor Series
In the previous section we started looking at writing down a power series representation of a function.
The problem with the approach in that section is that everything came down to needing to be able to
relate the function in some way to and while there are many functions out there that can be related to this function there are many more
that simply cant be related to this.
So, without taking anything away from the process we looked at in the previous section, we need to do
is come up with a more general method for writing a power series representation for a function.
So, for the time being, lets make two assumptions. First, lets assume that the function does infact have a power series representation about ,
Next, we will need to assume that the function, , has derivatives of every order and that we can infact find them all.
Now that weve assumed that a power series representation exists we need to determine what the
coefficients, cn, are. This is easier than it might at first appear to be. Lets first just evaluate everything
at . This gives,
So, all the terms except the first are zero and we now know what c0 is. Unfortunately, there isnt any
other value ofx that we can plug into the function that will allow us to quickly find any of the other
coefficients. However, if we take the derivative of the function (and its power series) then plug in we get,
and we now know c1.
Lets continue with this idea and find the second derivative.
7/31/2019 Chapter_4 Enggng Math
46/57
MTK3001-Chapter 4
fnh 2011
So, it looks like,
Using the third derivative gives,
Using the fourth derivative gives,
Hopefully by this time youve seen the pattern here. It looks like, in general, weve got the following
formula for the coefficients.
This even works for if you recall that and define .So, provided a power series representation for the function about exists the Taylor Series
for about is,Taylor Series
If we use , so we talking about the Taylor Series about , we call the series a Maclaurin Seriesfor or,
7/31/2019 Chapter_4 Enggng Math
47/57
MTK3001-Chapter 4
fnh 2011
Maclaurin Series
Before working any examples of Taylor Series we first need to address the assumption that a Taylor
Series will in fact exist for a given function. Lets start out with some notation and definitions that well
need.
To determine a condition that must be true in order for a Taylor series to exist for a function lets first
define the nth
degree Taylor polynomial of
as,
Note that this really is a polynomial of degree at most n! If we were to write out the sum without the
summation notation this would clearly be an nth degree polynomial. Well see a nice application of
Taylor polynomials in the next section.
Notice as well that for the full Taylor Series,
the n
thdegree Taylor polynomial is just the partial sum for the series.
Next, the remainder is defined to be,
So, the remainder is really just the error between the function
and the n
thdegree Taylor
polynomial for a given n.
With this definition note that we can then write the function as,
We now have the following Theorem.
7/31/2019 Chapter_4 Enggng Math
48/57
MTK3001-Chapter 4
fnh 2011
Theorem
Suppose that . Then if,
for | | , then,
on | | .In general showing that is a somewhat difficult process and so we will be assumingthat this can be done for some Rin all of the examples that well be looking at.Now lets look at some examples.
Example 1 Find the Taylor Series for about .Solution
This is actually one of the easier Taylor Series that well be asked to compute. To find the Taylor Series
for a function we will need to determine a general formula for . This is one of the few functionswhere this is easy to do right from the start.
To get a formula for all we need to do is recognize that,
and so,
Therefore, the Taylor series for about is
7/31/2019 Chapter_4 Enggng Math
49/57
MTK3001-Chapter 4
fnh 2011
Example 2 Find the Taylor Series for about .Solution
There are two ways to do this problem. Both are fairly simple, however one of them requires
significantly less work. Well work both solutions since the longer one has some nice ideas that well see
in other examples.
Solution 1
As with the first example well need to get a formula for . However, unlike the first one wevegot a little more work to do. Lets first take some derivatives and evaluate them atx=0.
After a couple of computations we were able to get general formulas for both and. Weoften wont be able to get a general formula for so dont get too excited about getting thatformula. Also, as we will see it wont always be easy to get a general formula for .
So, in this case weve got general formulas so all we need to do is plug these into the Taylor Series
formula and be done with the problem.
Solution 2
The previous solution wasnt too bad and we often have to do things in that manner. However, in this
7/31/2019 Chapter_4 Enggng Math
50/57
MTK3001-Chapter 4
fnh 2011
case there is a much shorter solution method. In the previous section we used series that weve already
found to help us find a new series. Lets do the same thing with this one. We already know a Taylor
Series for about and in this case the only difference is weve got a in the exponentinstead of just an .So, all we need to do is replace thexin the Taylor Series that we found in the first example with
This is a much shorter method of arriving at the same answer so dont forget about using previously
computed series where possible (and allowed of course).
Example 3 Find the Taylor Series for about .Solution
For this example we will take advantage of the fact that we already have a Taylor Series for about . In this example, unlike the previous example, doing this directly would be significantly longerand more difficult.
To this point weve only looked at Taylor Series about
(also known as Maclaurin Series) so lets
take a look at a Taylor Series that isnt about . Also, well pick on the exponential function onemore time since it makes some of the work easier. This will be the final Taylor Series for exponentials inthis section.
Example 4 Find the Taylor Series for about .
7/31/2019 Chapter_4 Enggng Math
51/57
MTK3001-Chapter 4
fnh 2011
Solution
Finding a general formula for is fairly simple.
The Taylor Series is then,
Okay, we now need to work some examples that dont involve the exponential function since these will
tend to require a little more work.
Example 5 Find the Taylor Series for about .Solution
First well need to take some derivatives of the function and evaluate them at .
In this example, unlike the previous ones, there is not an easy formula for either then general derivative
or the evaluation of the derivative. However, there is a clear pattern to the evaluations. So, lets plug
what weve got into the Taylor series and see what we get,
7/31/2019 Chapter_4 Enggng Math
52/57
MTK3001-Chapter 4
fnh 2011
So, we only pick up terms with even powers on the xs. This doesnt really help us to get a general
formula for the Taylor Series. However, lets drop the zeroes and renumber the terms as follows to
see what we can get.
By renumbering the terms as we did we can actually come up with a general formula for the Taylor
Series and here it is,
This idea of renumbering the series terms as we did in the previous example isnt used all that often,
but occasionally is very useful. There is one more series where we need to do it so lets take a look at
that so we can get one more example down of renumbering series terms.
7/31/2019 Chapter_4 Enggng Math
53/57
MTK3001-Chapter 4
fnh 2011
We really need to work another example or two in which isnt about .Example 7 Find the Taylor Series for about .
Example 6 Find the Taylor Series for about .Solution
As with the last example well start off in the same manner.
So, we get a similar pattern for this one. Lets plug the numbers into the Taylor Series.
In this case we only get terms that have an odd exponent on xand as with the last problem once we
ignore the zero terms there is a clear pattern and formula. So renumbering the terms as we did in the
previous example we get the following Taylor Series.
7/31/2019 Chapter_4 Enggng Math
54/57
MTK3001-Chapter 4
fnh 2011
Solution
Here are the first few derivatives and the evaluations.
Note that while we got a general formula here it doesnt work for . This will happen on occasionso dont worry about it when it does.In order to plug this into the Taylor Series formula well need to strip out the term first.
Notice that we simplified the factorials in this case. You should always simplify them if there are more
7/31/2019 Chapter_4 Enggng Math
55/57
MTK3001-Chapter 4
fnh 2011
than one and its possible to simplify them.
Also, do not get excited about the term sitting in front of the series. Sometimes we need to do that
when we cant get a general formula that will hold for all values ofn.
Example 8 Find the Taylor Series for about .Solution
Again, here are the derivatives and evaluations.
Notice that all the negatives signs will cancel out in the evaluation. Also, this formula will work for all n,
unlike the previous example.
Here is the Taylor Series for this function.
7/31/2019 Chapter_4 Enggng Math
56/57
MTK3001-Chapter 4
fnh 2011
Now, lets work one of the easier examples in this section. The problem for most students is that it may
not appear to be that easy (or maybe it will appear to be too easy) at first glance.
Example 9 Find the Taylor Series for about .Solution
Here are the derivatives for this problem.
This Taylor series will terminate after . This will always happen when we are finding the TaylorSeries of a polynomial. Here is the Taylor Series for this one.
When finding the Taylor Series of a polynomial we dont do any simplification of the right hand side. We
leave it like it is. In fact, if we were to multiply everything out we just get back to the original
polynomial!
7/31/2019 Chapter_4 Enggng Math
57/57
MTK3001-Chapter 4
While its not apparent that writing the Taylor Series for a polynomial is useful there are times where
this needs to be done. The problem is that they are beyond the scope of this course and so arent
covered here. For example, there is one application to series in the field of Differential Equations where
this needs to be done on occasion.