8.5 ABSOLUTE CONVERGENCE AND THE RATIO TEST

Transcription

1 smi98485_ch08b.qd 5/7/0 :07 PM Page Chapter 8 Infinite Series 8.5 ABSOLUTE CONVERGENCE AND THE RATIO TEST You should note that, outside of the Alternating Series Test presented in section 8.4, our other tests for convergence of series (i.e., the Integral Test and the two comparison tests) appl onl to series all of whose terms are positive. So, what do we do if we re faced with a series that has both positive and negative terms, but that is not an alternating series? For instance, loo at the series sin = sin sin + 7 sin 3 + sin This has both positive and negative terms, but the terms do not alternate signs. (Calculate the first five or si terms of the series to see this for ourself.) For an such series a, we can get around this problem b checing if the series of absolute values a is convergent. When this happens, we sa that the original series a is absolutel convergent (or converges absolutel). You should note that to test the convergence of the series of absolute values a (all of whose terms are positive), we have all of our earlier tests for positive term series available to us. Eample 5. Testing for Absolute Convergence S n Determine if ( ) + is absolutel convergent Solution It is eas to show that this alternating series is convergent. (Tr it!) From the graph of the first 0 partial sums in Figure 8.34, it appears that the series converges to approimatel To determine absolute convergence, we need to determine whether or not the series of absolute values, ( ) + is convergent. We have n ( ) + = = ( ), Figure 8.34 S n = n ( ) +. which ou should recognize as a convergent geometric series ( r = < ). This sas that the original series ( ) + converges absolutel. You might be wondering what the relationship is between convergence and absolute convergence. We ll prove shortl that ever absolutel convergent series is also convergent (as in eample 5.). However, the reverse is not true; there are man series that are convergent, but not absolutel convergent. These are called conditionall convergent series. Can ou thin of an eample of such a series? If so, it s probabl the eample that follows.

2 smi98485_ch08b.qd 5/7/0 :07 PM Page 667 Section 8.5 Absolute Convergence and the Ratio Test 667 Eample 5. A Conditionall Convergent Series Determine if the alternating harmonic series ( ) + is absolutel convergent. Solution In eample 4., we showed that this series is convergent. To test this for absolute convergence, we consider the series of absolute values, ( ) + =, which is the harmonic series. We showed in section 8. (eample.7) that the harmonic series diverges. This sas that ( ) + converges, but does not converge absolutel (i.e., it converges conditionall). Theorem 5. If a converges, then a converges. This result sas that if a series converges absolutel, then it must also converge. Because of this, when we test series, we first test for absolute convergence. If the series converges absolutel, then we need not test an further to establish convergence. Proof Notice that for an real number,, we can sa that. So, for an, we have a a a. Adding a to all the terms, we get 0 a + a a. (5.) Since a is absolutel convergent, we have that a and hence, also a = a is convergent. Define b = a + a. From (5.), 0 b a and so, b the Comparison Test, b is convergent. Observe that we ma write a = (a + a a ) = (a + a ) }{{} a = b a. Since the two series on the right-hand side are convergent, it follows that a must also be convergent. b

3 smi98485_ch08b.qd 5/7/0 :07 PM Page Chapter 8 Infinite Series Eample 5.3 Testing for Absolute Convergence Determine whether sin 3 is convergent or divergent. S n Figure 8.35 S n = n sin. 3 n Solution Notice that while this is not a positive term series, it is also not an alternating series. Because of this, our onl choice (given what we now) is to test the series for absolute convergence. From the graph of the first 0 partial sums seen in Figure 8.35, it appears that the series is converging to some value around To test for absolute convergence, we consider the series of absolute values, sin 3. Notice that sin sin 3 = 3, (5.) 3 since sin, for all. Of course, is a convergent p-series (p = 3 > ). B 3 the Comparison Test and (5.), sin 3 converges, too. Consequentl, the original series sin converges absolutel (and hence, converges). 3 The Ratio Test We now introduce a ver powerful tool for testing a series for absolute convergence. This test can be applied to a wide range of series, including the etremel important case of power series that we discuss in section 8.6. As ou ll see, this test is also remarabl eas to use. Theorem 5. (Ratio Test) Given a, with a 0 for all, suppose that lim a + a = L. Then, (i) if L <, the series converges absolutel, (ii) if L > (or L = ), the series diverges and (iii) if L =, there is no conclusion. Proof (i) For L <, pic an number r with L < r <. Then, we have lim a + a = L < r. For this to occur, there must be some number N > 0, such that for N, a + a < r. (5.3)

4 smi98485_ch08b.qd 5/7/0 :07 PM Page 669 Section 8.5 Absolute Convergence and the Ratio Test 669 S n Figure 8.36 S n = n. n Multipling both sides of (5.3) b a gives us a + < r a. In particular, taing = N gives us a N+ < r a N and taing = N + gives us a N+ < r a N+ < r a N. Liewise, a N+3 < r a N+ < r 3 a N and so on. We have a N+ < r a N, for =,, 3,.... Notice that a N r = a N r is a convergent geometric series, since 0 < r <. B the Comparison Test, it follows that a N+ = a n converges, too. This n=n+ sas that a n converges absolutel. Finall, since n=n+ N a n = a n + a n, n= n= we also get that a n converges absolutel. n= n=n+ (ii) For L >, we have lim a + a = L >. This sas that there must be some number N > 0, such that for N, a + a >. (5.4) Multipling both sides of (5.4) b a, we get a + > a > 0, for all N. Notice that if this is the case, then lim a 0. B the th-term test for divergence, we now have that a diverges. Eample 5.4 Test ( ) for convergence. Solution From the graph of the first 0 partial sums of the series of absolute values,, seen in Figure 8.36, it appears that the series of absolute values converges Using the Ratio Test

5 smi98485_ch08b.qd 5/7/0 :07 PM Page Chapter 8 Infinite Series to about. From the Ratio Test, we have lim a + a = lim = lim + = lim + = < and so, the series converges absolutel, as epected from Figure Since + = S n Figure 8.37 S n = n ( )!. e H ISTORICAL N OTES Srinivasa Ramanujan (887 90) Indian mathematician whose incredible discoveries about infinite series still mstif mathematicians. Largel self-taught, Ramanujan filled noteboos with conjectures about series, continued fractions and the Riemann zeta function. Ramanujan rarel gave a proof or even justification of his results. Nevertheless, the famous English mathematician G. H. Hard said, The must be true because, if the weren t true, no one would have had the imagination to invent them. (See Eercise 43.) n The Ratio Test is particularl useful when the general term of a series contains an eponential term, as in eample 5.4 or a factorial, as in the following eample. Eample 5.5 Using the Ratio Test Test ( )! for convergence. e Solution From the graph of the first 0 partial sums of the series seen in Figure 8.37, it appears that the series is diverging. (Loo closel at the scale on the -ais and compute a table of values for ourself.) We can confirm this suspicion with the Ratio Test. We have ( + )! lim a + a = lim e + ( + )! e = lim! e +! = lim e ( + )! e! = e lim + =. B the Ratio Test, the series diverges, as we suspected. Recall that in the statement of the Ratio Test (Theorem 5.), we said that if lim a + a =, then the Ratio Test ields no conclusion. B this, we mean that in such cases, the series ma or ma not converge and further testing is required. Eample 5.6 Use the Ratio Test for the harmonic series Solution We have A Divergent Series for Which the Ratio Test Fails lim a + a = lim +. = lim + =. Since ( + )! = ( + )! and e + = e e. In this case, the Ratio Test ields no conclusion, although we alread now that the harmonic series diverges.

6 smi98485_ch08b.qd 5/7/0 :07 PM Page 67 Section 8.5 Absolute Convergence and the Ratio Test 67 Eample 5.7 Use the Ratio Test to test the series. Solution Here, we have lim a + a = lim ( + ) = lim + + =. So again, the Ratio Test ields no conclusion, although we alread now that this is a convergent p-series. A Convergent Series for Which the Ratio Test Fails Loo carefull at eamples 5.6 and 5.7. You should recognize that the Ratio Test will be inconclusive for an p-series. Of course, we don t need the Ratio Test for these series. We now present one final test for convergence of series. Theorem 5.3 (Root Test) Given a, suppose that lim a =L. Then, (i) if L <, the series converges absolutel, (ii) if L > (or L = ), the series diverges and (iii) if L =, there is no conclusion. Notice how similar the conclusion is to the conclusion of the Ratio Test. The proof is also similar to that of the Ratio Test and we leave this as an eercise. Eample 5.8 Use the Root Test to determine the convergence or divergence of the series ( ) Using the Root Test Solution In this case, we consider lim a = lim = lim 5 = 5 <. B the Root Test, the series is absolutel convergent. B this point in our stud of series, it ma seem as if we have thrown at ou a dizzing arra of different series and tests for convergence or divergence. Just how are ou to eep all of these straight? The onl suggestion we have is that ou wor through man problems. We provide a good assortment in the eercise set that follows this section. Some

7 smi98485_ch08b.qd 5/7/0 :07 PM Page Chapter 8 Infinite Series of these require the methods of this section, while others are drawn from the preceding sections (just to eep ou thining about the big picture). For the sae of convenience, we summarize our convergence tests in the table that follows. Test When to use Conclusions Section Geometric Series a ar Converges to if r < ; r 8. diverges if r. th Term Test All series If lim 0, the series diverges. 8. Integral Test a where f () = a and a and f () d 8.3 p-series Comparison Test Limit Comparison Test Alternating Series Test f is continuous, decreasing and f () 0 both converge or both diverge. p Converges for p > ; diverges for p. 8.3 a and b, where 0 a b a and b, where If b converges, then a converges. 8.3 If a diverges, then b diverges. a and b 8.3 a a, b > 0 and lim = L > 0 both converge or both diverge. b ( ) + a where a > 0 for all If lim a = 0 and a + a for all, 8.4 then the series converges. Absolute Convergence Series with some positive and some If a converges, then 8.5 negative terms (including alternating series) a converges (absolutel). Ratio Test An series (especiall those involving For lim a + a = L, 8.5 eponentials and/or factorials) if L <, a converges absolutel if L >, a diverges, if L =, no conclusion. Root Test An series (especiall those involving For lim a =L, 8.5 eponentials) if L <, a converges absolutel if L >, a diverges, if L =, no conclusion.

8 smi98485_ch08b.qd 5/7/0 :07 PM Page 673 Section 8.5 Absolute Convergence and the Ratio Test 673 EXERCISES 8.5. Suppose that two series have identical terms ecept that in series A all terms are positive and in series B some terms are positive and some terms are negative. Eplain wh series B is more liel to converge. In light of this, eplain wh Theorem 5. is true.. In the Ratio Test, if lim a + a >, which is bigger, a + or a? Eplain wh this implies that the series a diverges. 3. In the Ratio Test, if lim a + a = L <, which is bigger, a + or a? This inequalit could also hold if L =. Compare the relative sizes of a + and a if L = 0.8 versus L =. Eplain wh L = 0.8 would be more liel to correspond to a convergent series than L =. 4. In man series of interest, the terms of the series involve powers of (e.g., ), eponentials (e.g., ) or factorials (e.g.,!). For which tpe(s) of terms is the Ratio Test liel to produce a result (i.e., a limit different than )? Briefl eplain. In eercises 5 4, determine if the series is absolutel convergent, conditionall convergent or divergent. 5. ( ) 3 6. ( ) 6!! ( ) 8. ( ) + + ( ) 3! ( ) ( ) 6. 3 ( ) ( ) 3 ( ) + + ( ) 0! ( ) ( ) 3 ( 3 4 ) ( ) + + = 0.. e 4. e sin 8. cos π ( ) = ln ( ) ( ) +! ( ) + 0 ()! ( ) ( + ) ( ) e ( e = cos 3 ) sin π ( ) ln ( ) + 3 ( ) + 4! ( ) 4 ( + )! ( 3) In the 90s, the Indian mathematician Srinivasa Ramanujan discovered the formula 8 π = (4)!(03 + 6,390). 980 (!) Approimate the series with onl the = 0 term and show that ou get 6 digits of π correct. Approimate the series using the = 0 and = terms and show that ou get 4 digits of π correct. In general, each term of this remarable series increases the accurac b 8 digits.

9 smi98485_ch08b.qd 5/7/0 :07 PM Page Chapter 8 Infinite Series 44. Prove that Ramanujan s series in eercise 43 converges. 45. To show that! that ( + lim 46. Determine whether diverges. converges, use the Ratio Test and the fact ) ( = lim + ) = e.! 3 5 ( ) converges or 47. One reason that it is important to distinguish absolute from conditional convergence of a series is the rearrangement of series, to be eplored in this eercise. Show that the series ( ) is absolutel convergent and find its sum S. Find the sum S + of the positive terms of the series. Find the sum S of the negative terms of the series. Verif that S = S + + S. This ma seem obvious, since for the finite sums ou are most familiar with, the order of addition never matters. However, ou cannot separate the positive and negative terms for conditionall convergent series. For eample, show that ( ) converges (conditionall) but that the + series of positive terms and the series of negative terms both diverge. Eplain in words wh this will alwas happen for conditionall convergent series. Thus, the order of terms matters for conditionall convergent series. B eploring further, we can uncover a trul remarable fact: for conditionall convergent series, ou can reorder the terms so that the partial sums converge to an real number. To illustrate this, suppose we want to reorder the series ( ) so that the partial sums + π ( converge to. Start b pulling out positive terms ) such that the partial sum is within 0. of π. 3 5 Net, tae the first negative term ( ) and positive terms such that the partial sum is within 0.05 of π. Then tae the net negative term ( ) 4 and positive terms such that the partial sum is within 0.0 of π. Argue that ou could continue in this fashion to reorder the terms so that the partial sums converge to π. (Especiall eplain wh ou will never run out of positive terms.) Then eplain wh ou cannot do the same with the absolutel convergent series ( ). 8.6 POWER SERIES We now want to epand our discussion of series to the case where the terms of the series are functions of the variable. Pa close attention to what we are about to introduce, for this is the culmination of all our hard wor in the preceding five sections. The primar reason for studing series is that we can use them to represent functions. This opens up all inds of possibilities for us, from approimating the values of transcendental functions to calculating derivatives and integrals of such functions, to studing differential equations. As well, defining functions as convergent series produces an eplosion of new functions available to us. In fact, man functions of great significance in applications (for instance, Bessel functions) are defined as a series. We tae the first few steps in this section. As a start, consider the series ( ) = + ( ) + ( ) + ( ) 3 +. Notice that for each fied, this is a geometric series with r = ( ). Recall that this sas that the series will converge whenever r = < and will diverge whenever r =. Further, for each with < (i.e., < < 3), the series converges to a r = ( ) = 3.

10 smi98485_ch08b.qd 5/7/0 :07 PM Page 675 Section 8.6 Power Series f() P () P () 3 P 3 () Figure 8.38 = and the first three partial 3 sums of ( ). That is, for each in the interval (, 3), we have ( ) = 3. For all other values of, the series diverges. In Figure 8.38, we show a graph of f () = 3, along with the first three partial sums P n, of this series, where P n () = n ( ) = + ( ) + ( ) + +( ) n, on the interval [, 3]. Notice that as n gets larger, P n () appears to get closer to f (), for an given in the interval (, 3). Further, as n gets larger, P n () tends to sta close to f () for a larger range of -values. Mae certain that ou understand what we ve observed here: we have taen a series and noticed that it is equivalent to (i.e., it converges to) a nown function on a certain interval. You might as wh anone would care if ou could do that. Certainl, f () = is a perfectl good function and anthing ou d want to do with it will most 3 definitel be easier using the algebraic epression than using the equivalent series 3 representation, ( ). However, imagine what benefits ou might find if ou could tae a given function (sa, one that ou don t now a whole lot about) and find an equivalent series representation. This is precisel what we are going to do in section 8.7. For instance, we will be able to show that for all, e =! = + +! + 3 3! (6.) 4! So, who cares? Well, suppose ou wanted to calculate e How would ou do that? Of course, ou d use a calculator. But, haven t ou ever wondered how our calculator does it? The problem is that e is not an algebraic function. That is, we can t compute its values b using algebraic operations (i.e., addition, subtraction, multiplication, division and nth roots). Over the net few sections, we begin to eplore this question. For the moment, let us sa this: if we have the series representation (6.) for e, then for an given, we can compute an approimation to e, simpl b summing the first few terms of the equivalent power series. This is eas to do, since the partial sums of the series are simpl polnomials. In general, an series of the form Power series b ( c) = b 0 + b ( c) + b ( c) + b 3 ( c) 3 + is called a power series in powers of ( c). We refer to the constants b, = 0,,,... as the coefficients of the series. The first question is: for what values of does the series converge? Saing this another wa, the power series b ( c) defines a function of. Its domain is the set of all for which the series converges. The primar tool for investigating the convergence or divergence of a power series is the Ratio Test. Notice again that the partial sums of a power series are all polnomials (the simplest functions around).

11 smi98485_ch08b.qd 5/7/0 :07 PM Page Chapter 8 Infinite Series Eample 6. Determine the values of for which the power series 3 + converges. Solution Using the Ratio Test, we have lim a + a = lim ( + ) ( + ) = lim = lim = 3 <, for < 3 or 3 < < 3. So, the series converges absolutel for 3 < < 3 and diverges for > 3 (i.e., for > 3 or < 3). Since the Ratio Test gives no conclusion for the endpoints =±3, we must test these separatel. For = 3, we have the series 3 + = = Since lim 3 = 0, the series diverges b the th-term test for divergence. The series diverges when = 3, for the same reason. Thus, the power series converges for all in the interval ( 3, 3) and diverges for all outside this interval. Determining Where a Power Series Converges Observe that eample 6. has something in common with the introductor eample. In both cases, the series have the form b ( c) and there is an interval of the form (c r, c + r) on which the series converges and outside of which the series diverges. (In the case of eample 6., notice that c = 0.) This interval on which a power series converges is called the interval of convergence. The constant r is called the radius of convergence (i.e., r is half the length of the interval of convergence). It turns out that there is such an interval for ever power series. We have the following result. 3. Since + = and 3 + = Theorem 6. Given an power series, b ( c), there are eactl three possibilities: (i) The series converges for all (, ) and the radius of convergence is r = ; (ii) The series converges onl for = c (and diverges for all other values of ) and the radius of convergence is r = 0; or (iii) The series converges for (c r, c + r) and diverges for < c r and for > c + r, for some number r with 0 < r <. The proof of the theorem can be found in Appendi A.

12 smi98485_ch08b.qd 5/7/0 :07 PM Page 677 Section 8.6 Power Series 677 Eample 6. Determine the interval and radius of convergence for the power series 0! ( ). Solution lim a + a From the Ratio Test, we have = lim 0 + ( ) +! ( + )! 0 ( )! Since ( ) = 0 lim + = ( ) ( ) ( + )! and ( + )! = ( + )! = 0 lim + = 0 <, for all. This sas that the series converges absolutel for all. Thus, the interval of convergence for this series is (, ) and the radius of convergence is r =. The interval of convergence for a power series can be a closed interval, an open interval or a half-open interval, as in the following eample. Eample 6.3 A Half-Open Interval of Convergence Determine the interval and radius of convergence for the power series 4. Solution From the Ratio Test, we have lim a + a = lim + 4 ( + )4 + = 4 lim + = 4 <. So, we are guaranteed absolute convergence for < 4 and divergence for > 4. It remains onl to test the endpoints of the interval: =±4. For = 4, we have 4 = 4 4 = which ou will recognize as the harmonic series, which diverges. For = 4, we have 4 = ( 4) ( ) =, 4 which is the alternating harmonic series, which we now converges (see eample 4.). So, in this case, the interval of convergence is the half-open interval [ 4, 4) and the radius of convergence is r = 4. Notice that (as stated in Theorem 6.) ever power series, a ( c) converges at least for = c since for = c, we have the trivial case a ( c) = a (c c) = a 0 + a 0 = a = a 0. Finding the Interval and Radius of Convergence,

13 smi98485_ch08b.qd 5/7/0 :07 PM Page Chapter 8 Infinite Series Eample 6.4 Determine the radius of convergence for the power series!( 5). Solution From the Ratio Test, we have lim a + a = lim ( + )!( 5) +!( 5) ( + )! 5 = lim! = lim [( + ) 5 ] { 0, if = 5 =, if 5. Thus, this power series converges onl for = 5 and so, its radius of convergence is r = 0. Suppose that the power series b ( c) has radius of convergence r > 0. Then the series converges absolutel for all in the interval (c r, c + r) and might converge at one or both of the endpoints, = c r and = c + r. Notice that since the series converges for each (c r, c + r), it defines a function f on the interval (c r, c + r), f () = b ( c) = b 0 + b ( c) + b ( c) + b 3 ( c) 3 +. A Power Series That Converges at Onl One Point It turns out that such a function is continuous and differentiable, although the proof is beond the level of this course. In fact, we differentiate eactl the wa ou might epect, Differentiating a power series Integrating a power series f () = d d f () = d d [b 0 + b ( c) + b ( c) + b 3 ( c) 3 + ] = b + b ( c) + 3b 3 ( c) + = b ( c), where the radius of convergence of the resulting series is also r. Since we find the derivative b differentiating each term in the series, we call this term-b-term differentiation. Liewise, we can integrate a convergent power series term-b-term, f ()d = b ( c) d = b ( c) d ( c) + = b + K, + where the radius of convergence of the resulting series is again r and where K is a constant of integration. The proof of these two results can be found in a tet on advanced calculus. It s important to recognize that these two results are not obvious. The are not simpl an application of the rule that a derivative or integral of a sum is simpl the sum of the derivatives or integrals, respectivel, since a series is not a sum, but rather, a limit of a sum. (What s the difference, anwa?) Further, these results are true for power series, but are not true for series in general.

14 smi98485_ch08b.qd 5/7/0 :07 PM Page 679 Section 8.6 Power Series 679 Eample 6.5 Find the interval of convergence of the series sin( 3 ) derivatives does not converge for an. Solution Notice that sin( 3 ), for all, and show that the series of since sin( 3 ). Further, is a convergent p-series (p = > ) and so, it follows from the Comparison Test that sin( 3 ) converges absolutel, for all. On the other hand, the series of derivatives (found b differentiating the series term b term) is d d [ sin( 3 ] ) = 3 cos( 3 ) = [ cos( 3 )], which diverges for all, b the th-term test for divergence, since the terms do not tend to zero as, for an. Keep in mind that sin( 3 ) is not a power series. (Wh not?) The result of eample 6.5 (a convergent series whose series of derivatives diverges) cannot occur with an power series with radius of convergence r > 0. In the following eample, we find that once we have a convergent power series representation for a given function, we can use this to obtain power series representations for an number of other functions, b differentiating and integrating the series term b term. Eample 6.6 A Convergent Series Whose Series of Derivatives Diverges Use the power series ( ) to find power series representations of + and tan. Differentiating and Integrating a Power Series ( + ), Solution Notice that ( ) = ( ) is a geometric series with ratio r =. This series converges, then, whenever r = = <, to a r = ( ) = +. That is, for < <, + = ( ). (6.) Differentiating both sides of (6.), we get ( + ) = ( ), for < <.

15 smi98485_ch08b.qd 5/7/0 :07 PM Page Chapter 8 Infinite Series Multipling both sides b gives us a new power series representation: ( + ) = ( ) +, valid for < <. Notice that we can also obtain a new power series from (6.) b substitution. For instance, if we replace with, we get + = ( ) ( ) = ( ), (6.3) valid for < < (which is equivalent to having < or < < ). Integrating both sides of (6.3) gives us + d = ( ) ( ) + d = + c. (6.4) + You should recognize the integral on the left-hand side of (6.4) as tan. That is, tan ( ) + = + c, for < <. (6.5) + Taing = 0 gives us tan ( ) = + c = c, + so that c = tan 0 = 0. Equation (6.5) now gives us a power series representation for tan, namel: tan ( ) + = = , for < <. Notice that woring as in eample 6.6, we can produce power series representations of an number of functions. In section 8.7, we present a sstematic method for producing power series representations for a wide range of functions. EXERCISES 8.6. Power series have the form a ( c). Eplain wh the farther is from c, the larger the terms of the series are and the less liel the series is to converge. Describe how this general trend relates to the radius of convergence.. Appling the Ratio Test to a ( c) requires ou to evaluate lim a + ( c) a. For = c, this limit equals 0 and the series converges. As increases or decreases, c increases. If the series has a finite radius of convergence r > 0, what is the value of the limit when c =r? Eplain how the limit changes when c < r and c > r and how this determines the convergence or divergence of the series. 3. As shown in eample 6., 0! ( ) converges for all. If = 00, the value of ( ) = 000 gets ver large ver fast, as increases. Eplain wh, for the series to converge, the value of! must get large faster than 000. To illustrate how fast the factorial grows, compute 50!, 00! and 00! (if our calculator can). 4. In a power series representation of + about c = 0, eplain wh the radius of convergence cannot be greater than (thin about the domain of + ).

16 smi98485_ch08b.qd 5/7/0 :07 PM Page 68 Section 8.6 Power Series 68 In eercises 5 4, find a power series representation of f () about c = 0 (refer to eample 6.6). Also, determine the radius and interval of convergence and graph f () together with the partial sums 3 a and 6 a. 5. f () = 6. f () = 3 7. f () = f () = 9. f () = 3 0. f () = 3 +. f () = f () = 4 +. f () = f () = 3 6 In eercises 5 0, determine the interval of convergence and the function to which the given power series converges. 5. ( + ) 6. ( 3) ( ) 8. ( ) ( ) 0. (3 + ) ( 3 4 In eercises 38, determine the radius and interval of convergence ! ( ) ( ) 3 ( ) 6.!( + ) 8. ( ) 30. ( 3) 3. 3! ) ( ) + ( + ) 4!( ) ( ) ( + ) ! ()! 34. ( + ) (!) ()! ( + )! ( ) In eercises 39 46, find a power series representation and radius of convergence b integrating or differentiating one of the series from eercises f () = 3 tan 40. f () = ln( ) 4. f () = 3 4. f () = ( ) ( ) 43. f () = ln( + ) 44. f () = ln( + 4) 45. f () = 46. f () = ( + 4) (4 + ) In eercises 47 50, find the interval of convergence of the (nonpower) series and the corresponding series of derivatives cos( 3 ) 48. e 50. cos(/) e 5. For an constants a and b > 0, determine the interval and radius of convergence of ( a). b 5. Prove that if a has radius of convergence r, with 0 < r <, then a has radius of convergence r. 53. If a has radius of convergence r, with 0 < r <, determine the radius of convergence of a ( c) for an constant c. 54. If a has radius of convergence r, with 0 < r <, determine the radius of convergence of ) a ( b constant b 0. for an 55. Show that f () = + ( ) = + has the power series representation f () =

17 smi98485_ch08b.qd 5/7/0 :07 PM Page Chapter 8 Infinite Series Find the radius of convergence. Set = and discuss the 000 interesting decimal representation of,00, , Use long division to show that = Even great mathematicians can mae mistaes. Leonhard Euler started with the equation + = 0, rewrote it as / + = 0, found power series representations for each function and concluded that =0. Substitute = to show that the conclusion is false, then find the mistae in Euler s derivation. 58. If our CAS or calculator has a command named Talor, use it to verif our answers to eercises Note that the radius of convergence in each of eercises 5 9 is. Given that the functions in eercises 5, 6, 8 and 9 are undefined at =, eplain wh the radius of convergence can t be larger than. The restricted radius in eercise 7 can be understood using comple numbers. Show that + = 0 for =±i, where i =. In general, a comple number a + bi is associated with the point (a, b). Find the distance between the comple numbers 0 and i b finding the distance between the associated points (0, 0) and (0, ). Discuss how this compares to the radius of convergence. Then use the ideas in this eercise to quicl conjecture the radius of convergence of power series with center c = 0 for the functions f () = 4 + 4, f () = and f () = For each series f (), compare the intervals of convergence of f () and f () d, where the antiderivative is taen term b term. (a) f () = ( ) ; (b) f () = ; (c) f () =. As stated in the tet, the radius of convergence remains the same after integration (or differentiation). Based on the eamples in this eercise, does integration mae it more or less liel that the series will converge at the endpoints? Conversel, will differentiation mae it more or less liel that the series will converge at the endpoints? 8.7 TAYLOR SERIES You ma still be wondering about the reason wh we have developed series. Each time we have developed a new concept, we have wored hard to build a case for wh we want to do what we re doing. For eample, in developing the derivative, we set out to find the slope of a tangent line and to find instantaneous velocit, onl to find that the were essentiall the same thing. When we developed the definite integral, we did so in the course of tring to find area under the curve. But, we have not et completel revealed wh we re pursuing series, even though we ve been developing them for more than five sections now. Well, the punchline is close at hand. In this section, we develop a compelling reason for considering series. The are not merel another mathematical curiosit, but rather, are an essential means for eploring and computing with transcendental functions (e.g., sin, cos, ln, e, etc.). Suppose that the power series b ( c) has radius of convergence r > 0. As we ve observed, this means that the series converges absolutel to some function f on the interval (c r, c + r). We have f () = b ( c) = b 0 + b ( c) + b ( c) + b 3 ( c) 3 + b 4 ( c) 4 +, for each (c r, c + r). Differentiating term b term, we get that f () = b ( c) = b + b ( c) + 3b 3 ( c) + 4b 4 ( c) 3 +,

18 smi98485_ch08b.qd 5/7/0 :07 PM Page 683 Section 8.7 Talor Series 683 again, for each (c r, c + r). Liewise, we get f () = b ( )( c) = b + 3 b 3 ( c) + 4 3b 4 ( c) + and f () = b ( )( )( c) 3 = 3 b b 4 ( c) + and so on (all valid for c r < < c + r ). Notice that if we substitute = c in each of the above derivatives, all the terms of the series drop out, ecept one. We get f (c) = b 0, f (c) = b, f (c) = b, Talor Series Epansion of f () about = c f (c) = 3! b 3 and so on. Observe that in general, we have f () (c) =! b. (7.) Solving (7.) for b, we have b = f () (c), for = 0,,,...! To summarize, we found that if b ( c) is a convergent power series with radius of convergence r > 0, then the series converges to some function f that we can write as f () = b ( c) f () (c) = ( c), for (c r, c + r).! Now, thin about this problem from another angle. Instead of starting with a series, suppose that ou start with an infinitel differentiable function, f (i.e., f can be differentiated infinitel often). Then, we can construct the series f () (c) ( c),! called a Talor series epansion for f. (See the historical note on Broo Talor in section 7..) There are two important questions we need to answer. Does a series constructed in this wa converge? If so, what is its radius of convergence? If the series converges, it converges to a function. What is that function (for instance, is it f )? We can answer the first of these questions on a case-b-case basis, usuall b appling the Ratio Test. The second question will require further insight. Eample 7. Constructing a Talor Series Epansion Construct the Talor series epansion for f () = e, about = 0 (i.e., tae c = 0). Solution Here, we have the etremel simple case where f () = e, f () = e and so on, f () () = e, for = 0,,,...

19 smi98485_ch08b.qd 5/7/0 :07 PM Page Chapter 8 Infinite Series Remar 5.3 The special case of a Talor series epansion about = 0 is often called a Maclaurin series. (See the historical note about Colin Maclaurin in section 8.3.) That is, the series f () (0) is the Maclaurin series! epansion for f. This gives us the Talor series f () (0) ( 0) =! From the Ratio Test, we have lim a + a = lim + ( + )! e 0! =! = lim!.! ( + )! = lim = (0) = 0 <, for all. + So, the Talor series! converges for all real numbers. At this point, though, we do not now the function to which the series converges. (Could it be e?) Before we present an further eamples of Talor series, let s see if we can determine the function to which a given Talor series converges. First, notice that the partial sums of a Talor series (lie an power series) are simpl polnomials. We define P n () = n f () (c) ( c)! = f (c) + f (c)( c) + f (c)! ( c) + + f (n) (c) ( c) n. n! Observe that P n () is a polnomial of degree n, as f () (c) is a constant for each. We! refer to P n as the Talor polnomial of degree n for f epanded about = c e P () 4 Figure 8.39a = e and = P (). Eample 7. For f () = e, find the Talor polnomial of degree n epanded about = 0. Solution As in eample 7., we have that f () () = e, for all. So, we have the nth degree Talor polnomial is P n () = = Constructing and Graphing Talor Polnomials n n f () (0) ( 0) =! n e 0!! = + +! + 3 n + + 3! n!. Since we established in eample 7. that the Talor series for f () = e about = 0 converges for all, this sas that the sequence of partial sums (i.e., the sequence of Talor polnomials) converges for all. In an effort to determine the function to which the Talor polnomials are converging, we have plotted P (), P (), P 3 () and P 4 (), together with the graph of f () = e in Figures 8.39a d, respectivel. Notice that as n gets larger, the graphs of P n () appear (at least on the interval displaed) to be approaching the graph of f () = e. Since we now that the Talor series converges and the graphical evidence suggests that the partial sums of the series are approaching f () = e, it is reasonable to conjecture that the series converges to e.

20 smi98485_ch08b.qd 5/7/0 :07 PM Page 685 Section 8.7 Talor Series e 8 e 8 e P 4 () P () 4 4 P 3 () Figure 8.39b = e and = P (). Figure 8.39c = e and = P 3 (). Figure 8.39d = e and = P 4 (). This is, in fact, eactl what is happening, as we can prove using the following two theorems. Theorem 7. (Talor s Theorem) Suppose that f has (n + ) derivatives on the interval (c r, c + r), for some r > 0. Then, for (c r, c + r), f () P n () and the error in using P n () to approimate f () is R n () = f () P n () = f (n+) (z) (n + )! ( c)n+, (7.) for some number z between and c. The error term R n () in (7.) is often called the remainder term. Note that this term loos ver much lie the first neglected term of the Talor series, ecept that f (n+) is evaluated at some (unnown) number z between and c, instead of at c. This remainder term serves two purposes: it enables us to obtain an estimate of the error in using a Talor polnomial to approimate a given function and as we ll see in the net theorem, it gives us the means to prove that a Talor series for a given function f converges to f. The proof of Talor s Theorem is somewhat technical and so we leave it for the end of the section. Note: If we could show that lim R n() = 0, for all in (c r, c + r), n then we would have that 0 = lim R n() = lim [ f () P n()] = f () lim P n() n n n or lim P n() = f (), for all (c r, c + r). n

21 smi98485_ch08b.qd 5/7/0 :07 PM Page Chapter 8 Infinite Series Remar 7. Observe that for n = 0, Talor s Theorem simplifies to a ver familiar result. We have R 0 () = f () P 0 () = f (z) (0 + )! ( c)0+. Since P 0 () = f (c), we have simpl f () f (c) = f (z)( c). Dividing b ( c), gives us f () f (c) = f (z), c which is the conclusion of the Mean Value Theorem. In this wa, observe that Talor s Theorem is a generalization of the Mean Value Theorem. That is, the sequence of partial sums of the Talor series (i.e., the sequence of Talor polnomials) converges to f () for each (c r, c + r). We summarize this in Theorem 7.. Theorem 7. Suppose that f has derivatives of all orders in the interval (c r, c + r), for some r > 0 and lim R n() = 0, for all in (c r, c + r). Then, the Talor series for f n epanded about = c converges to f (), that is, for all in (c r, c + r). We now return to the Talor series epansion of f () = e about = 0, constructed in eample 7. and investigated further in eample 7. and prove that it converges to e, as we had suspected. Eample 7.3 f () = f () (c) ( c),! Proving That a Talor Series Converges to the Desired Function Show that the Talor series for f () = e epanded about = 0 converges to e. Solution We alread found the indicated Talor series,! in eample 7.. Here, we have f () () = e, for all = 0,,,... This gives us the remainder term R n () = f (n+) (z) (n + )! ( e z 0)n+ = (n + )! n+, (7.3) where z is somewhere between and 0 (and depends also on the value of n). We first find a bound on the size of e z. Notice that if > 0, then 0 < z < and so, e z < e. If 0, then z 0, so that e z e 0 =. We define M to be the larger of these two bounds on e z. That is, we let M = ma{e, }. Then, for an and an n, we have e z M. Together with (7.3), this gives us the error estimate e z R n () = (n + )! n+ M n+ (n + )!. (7.4) To prove that the Talor series converges to e, we want to use (7.4) to show that lim R n() = 0, for all. However, for an given, how can we compute n lim n n+? While we cannot do so directl, we can use the following indirect (n + )! approach. We consider the series n+ as follows. Appling the Ratio Test, (n + )! n=0

22 smi98485_ch08b.qd 5/7/0 :07 PM Page 687 Section 8.7 Talor Series 687 we have lim a n+ n a = lim n n n+ (n + )! = lim (n + )! n+ n n + = 0, for all. This then sas that the series n+ converges absolutel for all. B n=0 (n + )! the th-term test for divergence, since this last series converges, its general term must tend to 0 as n. That is, lim n n+ (n + )! = 0 and so, from (7.4), lim n R n() = 0, for all. From Theorem 7., we now have that the Talor series converges to e for all. That is, e =! = + +! + 3 3! + 4 4! +. (7.5) The tric, if there is one, in finding a Talor series epansion is in accuratel calculating enough derivatives for ou to recognize the general form of the nth derivative. So, tae our time and BE CAREFUL! Once this is done, ou need onl show that R n () 0, as n, for all to ensure that the series converges to the function ou are epanding. One of the reasons for calculating Talor series is that we can use their partial sums to compute approimate values of a function. M M! Eample 7.4 Use the Talor series for e in (7.5) to obtain an approimation to the number e. Solution We have e = e =! = We list some partial sums of this series in the accompaning table. From this we get the ver accurate approimation e Using a Talor Series to Obtain an Approimation of e!. Eample 7.5 A Talor Series Epansion of sin Find the Talor series for f () = sin, epanded about = π and prove that the series converges to sin for all. Solution In this case, the Talor series is ) f ()( π! ( π ).

23 smi98485_ch08b.qd 5/7/0 :07 PM Page Chapter 8 Infinite Series First, we compute some derivatives and their value at = π. We have f () = sin f () = cos f () = sin f () = cos f (4) () = sin f ( π ) =, f ( π ) = 0, f ( π ) =, f ( π ) = 0, f (4)( π ) = and so on. Recognizing that ever other term is zero and ever other term is ±, we see that the Talor series is f ()( π! ) ( π ) = ( π ) + ( π ) 4 ( π ) 6 + 4! 6! = ( ) ( π ). ()! In order to test the series for convergence, we consider the remainder term f (n+) ( (z) R n () = π ) n+ (n + )!, (7.6) for some z between and π. From our derivative calculations, note that f (n+) (z) = { ±cos z, if n is even ±sin z, if n is odd. From this, it follows that f (n+) (z), for ever n. (Notice that this is true whether n is even or odd.) From (7.6), we now have R n () = f (n+) (z) (n + )! π n+ (n + )! as n, for ever, as in eample 7.3. This sas that sin = ( ) ( π ()! π n+ 0, ) = ( π ) + ( π ) 4, 4! for all. In Figures 8.40a d, we show graphs of f () = sin together with the Talor polnomials P (), P 4 (), P 6 () and P 8 () (the first few partial sums of the series). Notice that the higher the degree of the Talor polnomial is, the larger the interval is over which the polnomial provides a close approimation to f () = sin.

24 smi98485_ch08b.qd 5/7/0 :07 PM Page 689 Section 8.7 Talor Series 689 P 4 () P () sin sin Figure 8.40a = sin and = P (). Figure 8.40b = sin and = P 4 (). P 8 () sin sin P 6 () Figure 8.40c = sin and = P 6 (). Figure 8.40d = sin and = P 8 (). In the following eample, we illustrate how to use Talor s Theorem to estimate the error in using a Talor polnomial to approimate the value of a function. Eample 7.6 Estimating the Error in a Talor Polnomial Approimation Use a Talor polnomial to approimate the value of ln(.) and estimate the error in this approimation. Solution First, note that since ln is nown eactl and is close to. (wh would this matter?), we epand f () = ln in a Talor series about =. We compute an adequate number of derivatives so that the pattern becomes clear. We have f () = ln f() = 0 f () = f () = f () = f () = f () = 3 f () = f (4) () = 3 4 f (4) () = 3! f (5) () = 4! 5. f (5) () = 4!. f () () = ( ) + ( )! f () () = ( ) + ( )!

25 smi98485_ch08b.qd 5/7/0 :07 PM Page Chapter 8 Infinite Series We get the Talor series f () () ( )! = ( ) ( ) + 3! ( )3 + ( )! + +( ) ( ) +! = ( ) + ( ). We could use the remainder term to show that the series converges to f () = ln, for 0 < <, but this is not the original question here. (This is left as an eercise.) As an illustration, we construct the Talor polnomial, P 4 (), P 4 () = 4 ( ) + ( ) from above = ( ) ( ) + 3 ( )3 4 ( )4. 4 ln 3 P 4 () We show a graph of = ln and = P 4 () in Figure 8.4. Taing =. gives us the approimation ln(.) P 4 (.) = 0. (0.) + 3 (0.)3 4 (0.) We can use the remainder term to estimate the error in this approimation. We have Error = ln(.) P 4 (.) = R 4 (.) = f (4+) (z) (. )4+ 4! z 5 (4 + )! = (0.) 5, 5! Figure 8.4 = ln and = P 4 (). where z is between and.. This gives us the following bound on the error: Error = (0.)5 5z 5 < (0.)5 5( 5 ) = , since < z <. implies that z < =. Another wa to thin of this is that our approimation of ln(.) is off b no more than ± A more significant problem related to eample 7.6 is to determine how man terms of the Talor series are needed in order to guarantee a given accurac. We use the remainder term to accomplish this in the following eample. Eample 7.7 Finding the Number of Terms Needed for a Given Accurac Find the number of terms in the Talor series for f () = ln epanded about = that will guarantee an accurac of at least 0 0 in the approimation of ln(.) and ln(.5).

26 smi98485_ch08b.qd 5/7/0 :07 PM Page 69 Section 8.7 Talor Series 69 Solution From our calculations in eample 7.6 and (7.), we have that for some number z between and., R n (.) = f (n+) (z) (. )n+ (n + )! = n! z n (n + )! (0.)n+ = (0.)n+ (0.)n+ < (n + )zn+ n +, since < z <. implies that z < =. Further, since we want the error to be less than 0 0, we require that R n (.) < (0.)n+ n + < 0 0. P 9 () 3 You can solve this inequalit for n b trial and error to find that n = 9 will guarantee the required accurac. Notice that larger values of n will also guarantee this accurac, since (0.) n+ n + is a decreasing function of n. We then have the approimation ln(.) P 9 (.) = 9 ( ) + (. ) = (0.) (0.) + 3 (0.)3 4 (0.)4 + 5 (0.)5 4 ln 6 (0.)6 + 7 (0.)7 8 (0.)8 + 9 (0.) , Figure 8.4 = ln and = P 9 (). which from our error estimate we now is correct to within 0 0. We show a graph of = ln and = P 9 () in Figure 8.4. In comparing Figure 8.4 with Figure 8.4, observe that while P 9 () provides an improved approimation to P 4 () over the interval of convergence (0, ), it does not provide a better approimation outside of this interval. Similarl, notice that for some number z between and.5, R n (.5) = f (n+) (z) (n + )! (.5 )n+ = n! z n (n + )! (0.5)n+ = (0.5)n+ (0.5)n+ < (n + )zn+ n +, since < z <.5 implies that z < =. So, here we require that R n (.5) < (0.5)n+ n + < 0 0. Solving this b trial and error shows that n = 8 will guarantee the required accurac. Observe that this sas that man more terms are needed to approimate f (.5) than for f (.), to obtain the same accurac. This further illustrates the general principle that the farther awa is from the point about which we epand, the slower the convergence of the Talor series will be.

27 smi98485_ch08b.qd 5/7/0 :07 PM Page Chapter 8 Infinite Series For our convenience, we have compiled a list of common Talor series in the following table. e = Interval of Talor Series Convergence Where to find! = ! 3 + 4! 4 + (, ) eamples 7. and 7.3 sin = cos = sin = ln = ( ) ( + )! + = 3! 3 + 5! 5 7! 7 + (, ) eercise 9 ( ) ()! = + 4! 4 6! 6 + (, ) eercise 7 ( ) ()! ( π ) = ( π ) + ( π ) 4 (, ) eample 7.5 4! ( ) + ( ) = ( ) ( ) + 3 ( )3 (0, ] eamples 7.6, 7.7 tan = ( ) + + = (, ) eample 6.6 Notice that once ou have found a Talor series epansion for a given function, ou can find an number of other Talor series simpl b maing a substitution. Eample 7.8 Find Talor series in powers of for e, e and e. Solution Rather than compute the Talor series for these functions from scratch, recall that we had established in eample 7.3 that e t =! t = + t +! t + 3! t 3 + 4! t 4 +, (7.7) for all t (, ). We use the variable t here instead of, so that we can more easil mae substitutions. Taing t = in (7.7), we get the new Talor series: e =! () =! = + +! + 3 3! 3 +. Similarl, letting t = in (7.7), we get the Talor series e =! ( ) =! = + +! 4 + 3! 6 +. Finall, taing t = in (7.7), we get e ( ) =! ( ) = = +!! 3 3! 3 +. Notice that all of these last three series converge for all (, ). (Wh is that?) Finding New Talor Series from Old Ones

28 smi98485_ch08b.qd 5/7/0 :07 PM Page 693 Section 8.7 Talor Series 693 Proof of Talor s Theorem Recall that we had observed that the Mean Value Theorem was a special case of Talor s Theorem. As it turns out, the proof of Talor s Theorem parallels that of the Mean Value Theorem. Both mae use of Rolle s Theorem: If g is continuous on the interval [a, b], differentiable on (a, b) and g(a) = g(b), then there is a number c (a, b) for which g (c) = 0. As with the proof of the Mean Value Theorem, for a fied (c r, c + r), we define the function g(t) = f () f (t) f (t)( t)! f (t)( t) 3! f (t)( t) 3 n! f (n) (t)( t) n ( t)n+ R n () ( c), n+ where R n () is the remainder term, R n () = f () P n (). If we tae t =, notice that g() = f () f () = 0 and if we tae t = c, we get g(c) = f () f (c) f (c)( c)! f (c)( c) 3! f (c)( c) 3 n! f (n) (c)( c) n ( c)n+ R n () ( c) n+ = f () P n () R n () = R n () R n () = 0. B Rolle s Theorem, there must be some number z between and c for which g (z) = 0. Differentiating our epression for g(t) (with respect to t!), we get (beware of all the product rules!) g (t) = 0 f (t) f (t)( ) f (t)( t) f (t)()( t)( ) f (t)( t) n! f (n) (t)(n)( t) n ( ) n! f (n+) (t)( t) n R n () (n + )( t)n ( ) ( c) n+ = n! f (n+) (t)( t) n (n + )( t)n + R n (), ( c) n+ after most of the terms cancel. So, taing t = z, we have that 0 = g (z) = n! f (n+) (z)( z) n (n + )( z)n + R n (). ( c) n+ Solving this for the remainder term, R n (), we get (n + )( z)n R n () = ( c) n+ n! f (n+) (z)( z) n and finall, R n () = n! f (n+) (z)( z) n ( c) n+ as we had claimed. = f (n+) (z) ( c)n+ (n + ) n! = f (n+) (z) (n + )! ( c)n+, (n + )( z) n