AP Calculus. Notes and Homework for Chapter 9

Transcription

1 AP Calculus Notes ad Homework for Chapter 9 (Mr. Surowski) I do t feel that the textbook does a particularly good job at itroducig the material o ifiite series, power series (Maclauri ad Taylor series), ad polyomial approximatios (Maclaure ad Taylor polyomials). As a result, I have tried to put these topics ito a more sesible order ad to view these cocepts i the cotext of your ow mathematical backgrouds. Cotets The Cocept of a Numerical Series. Covergece/divergece of o-egative term series Tests for covergece of o-egative term series Coditioal ad absolute covergece; alteratig series.. 4 The Cocept of a Power Series 7 3 Polyomial Approximatios; Maclauri ad Taylor Expasios 3 3. Computatios ad tricks Error aalysis ad Taylor s Theorem

2 The Cocept of a Numerical Series Way back i Algebra II you leared that certai ifiite series ot oly made sese, you could actually compute them. The primary (if ot the oly!) examples you leared were the ifiite geometric series; oe such example might have bee = Furthemore, you leared how to compute such ifiite geometric series. I the above example sice the first term is a = 3 ad sice the ratio is r = 4, oe quickly computes the sum: = 3 4 = 3 4 = 4. Perhaps ufortuately, most ifiite series are ot geometric but rather come i a variety of forms. I ll give two below; they seem similar but really exhibit very differet behaviors: Series : Series : y y=/x y y=/x² x x 3 4 (Behavior of Series ) (Behavior of Series ) To see how the above two series differ, cosider the above figures. That o the left shows that the area represeted by the sum is

3 greater tha the area uder the curve with equatio y = /x from to. Sice this area is dx x = l x =, we see that the ifiite series must diverge (to ifiity). This diverget series is ofte called the harmoic series. Likewise, we see that the series ca be represeted by a area that is dx + x = =, which shows that this series caot diverge x to ad so coverges to some umber.. Covergece/divergece of o-egative term series Series i the above discussio illustrates a importat priciple of the real umbers. Namely, if a 0, a, a,... is a sequece of real umbers such that (i) a 0 a a..., ad (ii) there is a upper boud M for each elemet of the sequece, i.e., a M for each = 0,,,..., the the sequece coverges to some limit L (which we might ot be able to compute!): lim a = L. It turs out that this series coverges to π ; this is ot particularly easy to show. 6 3

4 (, a ) M Figure So what do sequeces have to do with ifiite series? Well, this is simple: if each term a i the ifiite series a is o-egative, the the sequece of partial sums satisfies a 0 a 0 + a a 0 + a + a k a. Furthermore, if we ca establish that for some M each partial sum S k = k a satisfies S k M the we have a limit, say, lim S k = L, i which case k we write a = L. I order to test a give ifiite series a of o-egative terms for covergece, we eed to keep i mid the followig three facts. Fact : I order for a to coverge it must happe that lim a = 0. (Thik about this: if the idividual terms of the series do t get small, there s o hope that the series ca coverge. Furthermore, this fact remais true eve whe ot all of the terms of the series are oegative.) 4

5 Fact : If Fact is true, we still eed to show that there is some umber M k such that a M for all k. Fact 3: Eve whe we have verified Facts ad, we still might ot (ad usually wo t) kow the actual limit of the ifiite series a. Warig about Fact : the requiremet that lim a = 0 is a ecessary but ot sufficiet coditio for covergece. Ideed, i the above we saw that the series = diverges but that = coverges. Exercises. Elemetary Exercises from your text: Page 48, 4.. Apply Fact above to determie those series which defiitely will ot coverge. = + si ( ) ( ) si = l ( ) = = (l ) 3. Occassioally a ifiite series ca be computed by usig a partial fractio decompositio. For example, ote that ( + ) = + ad so! = ( + ) = ( ) + = ( = ) ( + ) + 3 ( 3 ) + =. 4 5

6 Such a series is called a telescopig series because of all the iteral cacellatios. Use the above idea to compute the sums of the series #45 #5 o page 496 i your text. 4. Prove that the limit lim ( k= k l ) exists; its limit is called Euler s costat ad is deoted γ ( 0.577). To prove this just draw a picture, observig that the sequece < l < for all ad k k= that the sequece a = l is a icreasig sequece. k k=. Tests for covergece of o-egative term series I this subsectio we ll gather together a few hady tests for covergece (or divergece). They are pretty ituitive, but still require practice. The Compariso Test (i) Let a be a coverget series of positve terms ad let b be a secod series of positive terms. If for some iteger N, N implies that b a, the b also coverges. (ii) Let a be a diverget series of positve terms ad let b be a secod series of positive terms. If for some iteger N, N implies that b a, the b also diverges. or sometimes the Euler-Mascheroi costat 6

7 As a result of the Compariso Test, by comparig with the coverget series we see immediately that coverges. I exactly the + = = same way, it follows that diverges. = A much more useful test is... The Limit Compariso Test (i) Let a be a coverget series of positve terms ad let secod series of positive terms. If for some R, 0 R < b be a the (ii) Let b lim = R, a b also coverges. (This is reasoable as it says that asymptotically the series b is o larger tha R times the coverget series a be a diverget series of positve terms ad let secod series of positive terms. If for some R, 0 < R a.) b be a the b lim = R, a b also diverges. (This is reasoable as it says that asymptotically the series b is at least R times the diverget series a.) Let s look at a few examples! Before goig ito these examples, recall that 7

8 = diverges ad = coverges. The series coverges. We test this agaist the coverget series + =. Ideed, = ( ) lim + ( ) =, (after some work), provig covergece. The series + diverges, as lim ( ) + showig that the terms of the series ( ) =, + are asymptotically much bigger tha the terms of the already diverget series hece The series coverges. We compare it with the coverget series + diverges. = 9/ = ad = : 8

9 lim provig covergece. The series = ( ) ( 9/ ) = lim = 0, 7/ diverges. Watch this: (l ) ( ) lim (l ) ( ) = lim (l ) x = lim x (l x) l Hôspital = lim x = lim x l Hôspital = lim x d dx d dx x (l x) x ( l x) d dx d dx x ( l x) x = lim x =. This says that, asymptotically, the series tha the diverget harmoic series ad hece itself must diverge. = = is ifiitely larger (l ) The ext test will provide us with a rich assortmet of series to test with. (So far, we ve oly bee testig agaist the coverget series ad = 9

10 the diverget series = =.) The p-series Test. Let p be a real umber. The { coverges if p > p diverges if p. The proof of the above is simple; just as we proved that by comparig it with coverged = dx x (which coverges) ad that diverged dx by comparig with x (which diverges), we see that will have p dx the same behavior as the improper itegral. But, where p, we xp have { dx = x p x p p if p > = p if p <. We already kow that diverges, so we re doe! = = The p-test works very well i cojuctio with the Limit Compariso Test. We ll look at a couple of more examples coverges. We compare it with the series 7/ 3/, = = which, by the p-test coverges: ( ) lim ( 7/ ) = lim =, 3/ 0

11 provig covergece diverges. We compare it with the series = 7/3 by the p-test diverges: ( ) = 3, which, lim provig divergece. ( 7/3 ) = lim =, 3 There is oe more very useful test, oe which works particularly well with expressios cotaiig expoetials ad/or factorials. This method is based o the fact that if r <, the the ifiite geometric series a + ar + ar a + coverges (to ). I this test we do ot eed to assume that r the series cosists oly of o-egative terms. The Ratio Test Let The (i) if R <, the (ii) if R >, the a be a ifiite series. Assume that a coverges; a diverges; a + lim = R. a (iii) if R =, the this test i icoclusive. The reasoig behid the above is simple. First of all, i case (i) we see that a is asymptotically a geometric series with ratio R < ad

12 hece coverges (but we still probably wo t kow what the series coverges to). I case (ii) the a will diverge sice asymptotically each term is R times the previous oe, which certaily implies that lim a 0, prevetig covergece. Note that i the two cases ad we have lim a + a = =, 3 which is why this case i iclusive. We tur agai to some examples. Cosider the series Therefore, = Cosider the series 3 Ideed, we have i the first case i the first case, ad that ( + ) 3. We have! = a + lim = lim a ( + ) 3 coverges.! = ( (+) 3 (+)! ( (+) 3! = lim ( + ) 3 ( + )( + ) 3 = 0.. We have 3 a + lim = lim a a + lim = lim a ( ( + ) ( ) = lim ) (+) ( ) = lim ) ) + =, + =, i the secod case, despite the fact that the first series diverges ad the secod series coverges. =

13 a + lim = lim a ( (+) 3 + ) ( 3 ) = lim ( + ) 3 = 3 <. It follows, therefore, that = also coverges. 3 Cosider the series =!. We have a + lim = lim a which certaily proves divergece. ( ) (+)! ( +! ) = lim + =, Here s a iterestig oe: (l ). Use the fact that l > for all = 9 which implies that for 9, (l ) <. Now what? Exercises. From the Text, page 5: 9 44, 5 54 (Note that some of the limits might require the use of l Hôpital s Rule.). From the Text, page 53: 7 7,. 3

14 .3 Coditioal ad absolute covergece; alteratig series I this subsectio we shall cosider series of the form a where the idividual terms a are ot ecessarily o-egative. We shall first make the followig useful defiitio. A ifiite series a is called absolutely coverget if the series the followig result. a coverges. This is importat because of Theorem. If the series a is absolutely coverget, the it is coverget. 4 We cosider a couple of simple illustratios of the above theorem. ( ) The series = will coverge by the above theorem, together with the p-test. ( ) The series does ot coverge absolutely; however, we ll see below that this series coverges. A ifiite series a which coverges but is ot absolutely cover- 4 Here s a heuristic proof of this result. We may express the series a as a differece a = b c, where the terms b are the positive terms amog a 0, a, a,... ad where the terms c are the egative terms amog a 0, a, a,.... Clearly both of these series coverge sice a is absolutely coverget ad hece a itself is coverget. 4

15 get is called coditioally coverget. There are plety of coditioally coverget series, as guarateed by the followig theorem. Theorem. [Alteratig Series Test] Let a 0 a a 0 ad satisfy lim a = 0. The the alteratig series ( ) a coverges. 5 We ll coclude this subsectios with two illustratios of the Alteratig Series Test. We kow that the harmoic series diverges; however, sice the = terms of this series decrease ad ted to zero, the Alteratig Series ( ) Test guaratees that coverges. We ll show later o that = this actually coverges to l (see page ). The series ca be show to diverge by applyig the Limit + Compariso Test with a compariso with the harmoic series (do this!). However, the terms decrease ad ted to zero ad so by the ( ) Alteratig Series Test the series + coverges. Error Boud. Give the coverget alteratig series S = a a + a 3 + ( ) a + we ca easily obtai a upper boud o the error icurred by the approximatio S a a + a 3 + ( ) a. 5 The proof of this is pretty simple. First of all, ote that the eve partial sums (a 0 a ) (a 0 a ) + (a a 3 ) (a 0 a ) + (a a 3 ) + (a 4 a 5 ), so it suffices to show that these are all bouded by some umber (see Figure, page 3). However, ote that so we re doe. a 0 ((a a ) + (a 3 a 4 ) + (a 5 a 6 ) + + (a 3 a )) a a 0, }{{} This is positive! 5

16 Ideed, sice S ( a a + a 3 + ( ) a ) = a+ a + +, ad sice a + a + + a +, we arrive at the error upper boud: S ( a a + a 3 + ( ) a ) a+. As a represetative example, cosider the coverget alteratig series P = 4 ( ) + = 4 ( ) +. If oe approximates P by usig the first 00 terms of the series, the P 99 ( ) + 0 (which is arguably uimpressive). I a exercise (see page ) you ll lear that P = π. Exercises. From the Text, page 53: 8, Use the alteratig series test to ifer the covergece of the followig series: (a) (b)

17 The Cocept of a Power Series Let us retur to the familiar geometric series, with ratio r satisfyig r <. a + ar + ar + = a r. The first thig I would like to do is simply make a mior cosmetic chage: rather tha writig r i the above sum, we shall write x: a + ax + ax + = a, x <. x I other words, if we set f(x) = a + ax + ax + = ax ad set g(x) = a x, the the followig facts emerge: (a) The domai of f is < x <, ad the domai of g is x. (b) f(x) = g(x) for all x i the iterval < x <. We say, therefore, that ax is the power series represetatio of g(x), valid o the iterval < x <. So what is a power series ayway? Well, it s just a expressio of the form a x, where a 0, a, a,... are just real costats. For ay particular value of x this ifiite sum may or may ot coverge; we ll have plety to say about issues of covergece. 7

18 Our primary tool i determiig the covergece properies of a power series a x will be the Ratio Test. Recall that the series a x will coverge if which meas that a + x + > lim a x a +, = x lim a a x is absolutely coverget for all x satisfyig x < lim a a +. a The quatity R = lim is sometimes called the radius of covergece of the power series a x. Agai, as log as R < x < R, we a + are guarateed that a x is absolutely coverget ad hece coverget. A couple of simple examples will help to shed some light. The power series R = a lim = a + ( ) x + lim has radius of covergece ( + ( + +3 ) ) = lim ( + 3) ( + )( + ) =. This meas that the above power series has radius of covergece ad so the series is absolutely coverget for < x <. 8

19 The power series R = x has radius of covergece ( ) a lim = a + lim ( + + ) = lim + =, so i this case the radius of covergece is, which guaratees that the power series coverges for all x satisfyig < x <. ( ) x Fially, cosider the power series. I this case the radius! of covergece is similarly computed: R = a lim = a + lim ( (!) (+)! ) = lim + =. This ifiite radius of covergece meas that the power series actually coverges for all real umbers x. Here, we cosider (x + ), has radius of covergece R = ( + ) + lim =. ( ) x This meas that the series will coverge where x + <, i.e., where 4 < x < 0.! Fially, we cosider the power series. The radius of cover- gece is x R = lim + ( + ) =. 9

20 But this is a power series i x ad so will coverge if x <. This gives covergece o the iterval < x <. I the above examples we computed itervals withi which we are guarateed covergece of the power series. Next, ote that for values of x outside the radius of covergece we caot have covergece, for the the limit of the ratios will be greater tha, prevetig the idividual terms from approachig 0. This raises the questio of covergece at the edpoits. We shall ivestigate this i the examples already cosidered above. ( ) x We have see that the power series has radius of covergece R = meaig that this series will coverge i the iterval + < x <. What if x =? What if x =? Well, we ca take these up separately, usig the methods of the previous two subsectios. If x =, we have the series ( ) ( ) + = +, which diverges (use the Limit Compariso Test agaist the harmoic series). If x =, the the series becomes ( ) +, which coverges by the Alteratig Series Test. We therefore kow the full story ad ca state the iterval of covergece: ( ) x + has iterval of covergece < x. 0

21 We have see that the power series x has radius of covergece r =. Agai, we should ask about the behavior of the series whe x = ±. If x = we have the series ( ) = ( ) which diverges, whereas whe x =, we have Therefore, = which also diverges. x has iterval of covergece < x <. ( ) x The power series has ifiite radius of covergece so! there are o edpoits to check. Before closig this sectio, we should metio that ot all power series are of the form a x ; they may appear i a traslated format, say, oe like a (x a), where a is a costat. For example, cosider the series o page 9 above; let us determie the iterval of covergece. We have already see that this series coverges o 4 < x < 0. If x = 4, the this series is the alteratig harmoic series, which we kow to coverge. O the other had, if x = 0, the the series becomes the diverget harmoic series. Summarizig, the iterval of covergece is 4 x < 0.

22 Here we ll give a argumet that the alteratig harmoic series coverges to l (see page 5.) 6 We start with the geometric series: x + x x 3 + = + x. Itegrate both sides ad get a power series for l + x : x x + x3 3 = dx = l + x ; + x of course you eed to verify that the costat of itegratio is 0. It s very easy to verify that the iterval of covergece of the power series o the left-had side above is < x. Therefore, substitutig for x gives + 3 = l. Exercises. From the Text, page 5: 7, 5 8 (ote: 5 8 are just geometric series).. I this exercise you ll obtai a ifiite series represetatio for π; see page 6. (a) Write dow the ifiite geometric power series i x which coverges to + x. (b) Itegrate both sides. (c) What is ta ()? 3. Compute the iterval of covergece of the power series + x + 3x 4 + 4x 6 + = x ( ). Ca you fid a fuctio to which this power series coverges o its radius of covergece? 6 The umber l.59 is almost as ubiquitous as π or e. Perhaps surprisigly, this umber is idelibly imprited ito the cosciousess of commuicatio egieers, as this represets the theoretical miimum E b /N 0 (i decibels, ad is closely related to the sigal-to-oise ratio of a trasmissio) above which error-free iformatio ca be set. =

23 3 Polyomial Approximatios; Maclauri ad Taylor Expasios So far we have see istaces of power series ad, by the use of various tricks (especially differetiatio ad itegratio) we see how to get ew power series ad eve determie the represetig fuctio. For example, from the geometric power series + x + x + x 3 + =, valid for < x <, x we see that a differetiatio ad itegratio give ew power series covergig to kow fuctios o their itervals of covergece: + x + 3x + 4x 3 + = d ( ) = dx x ad, x + x + x3 3 + = dx x ( x), valid for < x <, = l x, valid for x <. Ideed, it was essetially from the above equatio that we get the somewhat surprisig result that = l. Playig the same sort of variatio o the geometric series: x + x 4 x 6 + = + x, gives, by itegratio, the result x x3 3 + x5 5 x7 7 + = ta x. By replacig x by x a we ca get similar represetatios i terms of power series i x a. A simple example might simply be to write 3

24 (x ) (x ) + (x )3 3 (Do you see how to get this?) = dx x = l x, valid for 0 < x. Examples like the above are edless, but all ivolve our startig with some series whose sum is kow. What we would like to do below is to reverse this process, startig with a kow differetiable fuctio f ad try to sythesize from this a power series i x a, where the behavior of f at x = a is kow. Way back i our study of the liearizatio of a fuctio we saw that it was occassioally coveiet ad useful to approximate a fuctio by oe of its taget lies. More precisely, if f is a differetiable fuctio, ad if a is a value i its domai, the we have the approximatio f (a) f(x) f(a), which results i x a f(x) f(a) + f (a)(x a) for x ear a. A graph of this situatio should help remid the studet of how good (or bad) such a approximatio might be: 4 y y=f(x) 3 (a,f(a)) x y=f(x)+f'(a)(x-a) -4 4

25 Notice that as log as x does ot move too far away from the poit a, the the above approximatio is pretty good. Let s take a secod look at the above. Note that i approximatig a fuctio f by a liear fuctio L ear the poit a, the (i) The graph of L will pass through the poit (a, f(a)), i.e., L(a) = f(a), ad (ii) The slope of the lie y = L(x) will be the same as the derivative of f at x = a, i.e., L (a) = f (a). That is the say, the best liear fuctio L to use i approximatig f ear a is oe whose 0-th ad first derivatives at x = a are the same as for f: L(a) = f(a) ad L (a) = f (a). So what if istead of usig a straight lie to approximate f we were to use a quadratic fuctio Q? What, the, would be the atural requiremets? Well, i aalogy with what was said above we would require f ad Q to have the same first three derivatives (0-th, first, ad secod): Q(a) = f(a), Q (a) = f (a), ad Q (a) = f (a). Such a quadratic fuctio is actually very easy to build: the result would be that Q(x) = f(a) + f (a)(x a) + f (a) (x a).! (The reader should pause here to verify that the above quadratic fuctio really does have the same first three derivatives as f at x = a.) Notice the im- This secod-order approximatio is depicted here. provemet over the liear approximatio. 5

26 4 y y=f(x) 3 (a,f(a)) x - - y=f(x)+f'(a)(x-a)+f''(a)(x-)²/ -3-4 I geeral, we may approximate a fuctio with a polyomial P (x) of degree by isistig that this polyomial have all of its first + derivatives at x = a equal those of f: P (a) = f(a), P (a) = f (a), P (a) = f (a),, P () (a) = f () (a), where, i geeral, f (k) (x) deotes the k-th derivative of f at x. It is easy to see that the followig gives a recipe for P (x): P (x) = f(a)+f (a)(x a)+ f (a)! (x a)+ f (a) (x a) 3 + +f () (a)(x a) 3! = k=0 f (k) (a) (x a) k.! We expect, the, to have a pretty good approximatio 6

27 f(x) k=0 f (k) (a) (x a) k.! f (k) (a) The polyomial P (x) = (x a) k is called the Taylor polyomial of degree for f at x = a. If a = 0, the above polyomial! k=0 becomes f(x) k=0 f (k) (a) x k! ad is usually called the Maclauri polyomial of degree for f. What if, istead of stoppig at a degree polyomial, we cotiued the process idefiitely to obtai a power series? This is possible ad we obtai k=0 f (k) (a) (x a) k the Taylor series for f at x = a, ad! k=0 f (k) (a) x k the Maclauri series for f.! Warig. It is very temptig to assume that the Taylor series for a fuctio f will actually coverge to f(x) o its iterval of covergece, that is, f(x) = k=0 f (k) (a) (x a) k.! For most of the fuctios we ve cosidered here, this is true, but the geeral result ca fail. 7 As a result, we shall adopt the otatio 7 As a example, cosider the fuctio f defied by settig { e /x if x 0 f(x) = 0 if x = 0. Oe ca show that all derivatives of f vaish at x = 0 ad so caot equal its Maclauri series i ay 7

28 f(x) k=0 f (k) (a) (x a) k! f (k) (a) to mea that f(x) is represeted by the power series (x a) k ; i! k=0 Subsectio 3. we ll worry about whether ca be replaced with =. First, however, we shall delve ito some computatios. 3. Computatios ad tricks I this subsectio we ll give some computatios of some Taylor ad Maclauri series, ad provide some iterestig shortcuts alog the way. At least half of the AP (BC) questios regardig ifiite series closely resemble those of this sectio! We ll itemize a few examples. Let f(x) = si x ad fid its Maclauri series expasio. This is simple as the derivatives (at x = 0) are easy to compute f (0) (x) = si 0 = 0, f (0) = cos 0 =, f (0) = si 0 = 0, f (0) = cos 0 =, f (4) (0) = si 0 = 0, ad the patter repeats all over agai. This immediately gives the Maclauri series for si x: si x x x3 3! + x5 5! x7 7! + = ( ) x + ( + )!. (A hady trick) If we wish to compute the Maclauri series for cos x, we could certaily follow the same procedure as for the si x i the iterval about x = 0. See the Miiproject o page 35. 8

29 above example. However, sice cos x = d dx si x, we ca likewise obtai the Maclauri series for the cos x by differetiatig that for the si x, this yields the series cos x d dx ( ) x + ( + )! = (A hady trick) Sice l( + x) = we may start with the geometric series x + x = dx + x ( ) x ()!. ( ) x = + x, ad the itegrate each term to obtai the Maclauri series for l( + x), valid if x < : l( + x) = x x + x3 3 = ( ) x dx = x + +. (Note that there is o costat occurig i the above itegratios sice whe x = 0, l( + x) = l = 0.) Sice d dx ex = e x = at x = 0, we immediately have the Maclauri series expasio for e x : e x + x + x! + x3 3! + = x!. (A hady trick) I the above series we may substitute x for x ad get the Maclauri series for e x : e x x + x4! x6 3! + = ( ) x!. 9

30 Or, if we wat the Maclauri series for somethig more complicated, like x 3 e x3, we could just multiply the above series by x 3, which would be much easier tha computig all of the derivatives of x 3 e x3 at x = 0: x 3 e x x 3 x 5 + x7! x9 3! + = ( ) x+3!. Note how easy this is compared to havig to calculate d dx (e x ) (where the chai ad product rules very much complicate thigs!). (A hady trick) Let s fid the Maclauri series expasio of the fuctio f(x) = si x. I this case, we certaily would t wat to compute x successive derivatives of the quotiet si x. However, rememberig x the Maclauri series expasio of si x ad the dividig by x will accomplish the same thig much more easily; the resultig series is si x x x 3! + x4 5! = ( ) x ( + )!. The Taylor series expasio of cos x about x = π. We have, where f(x) = cos x, that f ( ( π ) = cos π ) ( ( = 0, f π ) = si π ) ( =, f π ) = cos ( ) ( ( π = 0, f π ) = si π ) =, after which poit the cycle repeats. Therefore, the Taylor series is cos x ( x π ( ) ) x π 3 + 3! ( ) x π 5 + = 5! ( x π ( ) ( )! = ) Of course, the same thig ca be obtaied by otig that si ( x ) π = cos x, replacig x by π i the Maclauri series for si x, ad multiplyig the whole thig by. computig the Maclauri series for cos x ad the replacig x by x π.. 30

31 (A hady trick we ve essetially already see this oe) We kow that + x + x + x 3 + = x = x valid for x <. we ca get further valid sums by differetiatig the above: + x + 3x + 4x 3 + = ( + )x = d dx ( x ) = ( x), valid for x <. From the above Maclauri series, we ca easily extract the derivatives of f(x) = ( x) at x = 0. Sice the coefficiet of xk i the Maclauri series for f is f (k) (0), we coclude that f (k) (0) = k +, from which it k! k! follows immediately that f (k) (0) = (k+)!. (You ca ascertai directly, if you wish.) Further valid sums ca be obtaied by differetiatig ad/or itegratig. Exercises. From the Text, page 57: (Be sure to use hady tricks wherever possible). From the Text, page 57: From the Text, page 57: Usig the Maclauri series for ta x which was give o page 3, give a geeral formula for dk dx k ( ta x ) at x = 0. Is this easy to determie directly? 3

32 5. Without explicitig computig derivatives, compute the Maclari series expasio for the ratioal fuctio f(x) =. (Hit: do a x x + x + log divisio.) 6. O page 9 we foud that if x <, the l( + x) = x x + x3 3. Sice the series o the right-had side above coverges whe x = (alteratig series test), we coclude that Show that l = = l. The above shows that if the terms i a coditioally coverget series are permuted, the the resultig sum ca be differet! 3. Error aalysis ad Taylor s Theorem I this fial subsectio we wish to address two importat questios: Questio A: If P (x) is the Maclauri (or Taylor) polyomial of degree for the fuctio f(x), how good is the approximatio f(x) P (x)? More precisely, how large ca the error f(x) P (x) be? Questio B: Whe ca we say that the Maclauri or Taylor series of f(x) actually coverges to f(x)? The aswers to these questios are highly related. The aswer to both of these questios is actually cotaied i Taylor s Theorem with Remaider. Before statig this theorem, I wat to idicate 3

33 that this theorem is really just a geeralizatio of the Mea Value Theorem, which I ll state below as a remider. Mea Value Theorem. Let f be a differetiable fuctio o some ope iterval I. If a ad x are both i I, the there exists a real umber c betwee a ad x such that f(x) f(a) = f (c). x a Put differetly, there exists a umber c betwee a ad x such that f(x) = f(a) + f (c)(x a). Havig bee remided of the Mea Value Theorem, perhaps ow Taylor s Theorem with Remaider wo t seem so strage. Here it is. Taylor s Theorem with Remaider. Let f be a ifiitely-differetiable fuctio o a ope iterval I. If a ad x are i I, ad if is a o-egative iteger, the there exists a real umber c betwee a ad x such that f(x) = f(a) + f (a)(x a) + f (c) (x a) +! + f () (a) (x a) + f (+) (c) (x a)+.! ( + )! }{{} this is the remaider We ll coclude this subsectio with some examples. As a warm-up, let s prove that the Maclauri series for cos x actually coverges to f(x) = cos x for all x. We have, by Taylor s Theorem with Remaider, that cos x = x! + x4 4! ± x (!) + f (+) (c) ( + )! x+, 33

34 for some real umber c betwee 0 ad x. Sice all derivatives of cos x are ± si x or ± cos x, we see that f (+) (c). This meas that for fixed x, if we let, the the remaider provig that f (+) (c) ( + )! x+ 0, cos x = x! + x4 4! = ( ) x ()!. Oe easily computes that the first two terms of the Maclauri series expasio of + x is + x. Let s give a upper boud o the error ( + x + x ) whe x < 0.0. By Taylor s Theorem with Remaider, we kow that the absolute value of the error is give by f (c) x, where c is betwee 0 ad x, ad where f(x) = + x. Sice f (c) = 4( + c) 3/, ad sice c is betwee 0 ad x, we see that + c.99 ad so f (c) = 4( + c).54. 3/ 4.993/ This meas that the error i the above approximatio is o more tha f (c) x.54 (0.0) < Aother way of viewig this result is that, accurate to four decimal places, + x = + x wheever x < 0.0. Exercises 34

35 . From the Text, page 486: 0, 9 3, 33.. Miiproject: Cosider the fuctio defied by settig f(x) = { e /x if x 0, if x =. (a) Compute d dx e /x, d dx e /x, d 3 dx 3 e /x. (b) Notice that each of the above derivatives are of the form e /x P (/x), where P (x) is a polyomial. (c) Show that if P (x) is a polyomial, the d dx e /x P (/x) = e /x Q(/x), where Q(x) is also a polyomial. (d) Deduce that every derivative of e /x has the form d dx e /x where P (x) is a polyomial. (e) Show that for ay polyomial P (x), = e /x P (/x), lim P (/x) = 0. x 0 e /x (f) Coclude that the Maclauri series expasio for f(x) = e /x is idetically equal to. Therefore we see that for x 0, the e /x is ever equal to its Maclauri series expasio! 35