Week 2: Calculus II Notes - UCLA Mathematics
Christian ParkinsonGRE Prep: Calculus II Notes1Week 2: Calculus IINotesWe concluded the Calculus I notes with Riemann integration, Fundamental Theorem ofCalculus and some helpful integration techniques. However, ofter times, you will be asked toidentify whether an integral converges or diverges even when you cannot find the value. Thisis especially true for improper integrals: those in which the integrand had a singularity or therange goes to infinity. We begin the Calculus II notes by discussing some basic convergenceresults.Proposition 1 (p-test). Let p R. The integralZ dxxp1converges if p 1 and diverges otherwise. The integralZ 1dxp0 xconverges if p 1 and diverges otherwise.Note that as a consequence, the integralZ 0dxxpnever converges. These can all be verified by a straight-forward calculation and hint at somegood intuition. The borderline case is 1/x where neither integrals converge, so as a generalrule, forZ f (x)dx1to converge, we will need f (x) 0 faster than 1/x as x and if g(x) has a singularityat x 0, then forZ 1g(x)dx0to converge, we need g(x) to blow up slower than 1/x as x 0 . Note, these “general rules”should not be seen has hard-and-fast truths, but rather heuristics which can often make iteasy to verify if an integral converges or diverges. The following two theorems give morerigorous forms of these general rules.Theorem 2 (Comparison Test). Suppose that f, g : [0, ) [0, ) are continuous andthat f (x) g(x) for all x [0, ). ThenZ Z (a) ifg(x)dx converges, then so doesf (x)dx,00
Christian ParkinsonZGRE Prep: Calculus II Notes Z g(x)dx.f (x)dx diverges, then so does(b) if200Theorem 3 (Limit Comparison Test). Suppose that f, g : [0, ) [0, ) are continuous and thatf (x)limx g(x)exists as a number in (0, ) [excluding the endpoints]. ThenZ Z f (x)dx andg(x)dx both converge or both diverge.00Note, the previous two theorems also hold for improper integrals where the functions havesingularities at the same point but the limit in the second must be taken at the singularity.This last theorem tells us that the only important thing in determining convergence ofan integral with as a limit is the asymptotic behavior of the integrand. We have a specialnotation for this. If f, g are such that f /g L as x where L is a finite non-zeronumber, then we write f g [note: in most contexts, L is required to be 1 in order to sayf g; for our purposes, it is fine to allow L to be finite and non-zero]. We typically readthis as “f is asymptotic to g.” Note that if f g, then there are constants c, C 0 suchthat for sufficiently large x, we have cg(x) f (x) Cg(x). This shows that we can easilyderive the Limit Comparison Test from the ordinary Comparison Test. A relaxed versionof this requirement leads to a new definition. If there is C 0 such that f (x) Cg(x) forall x sufficiently large, then we write f O(g) [and say “f is big-oh of g”]. Thus f g ifff O(g) and g O(f ). Specifically, f O(g) iff f /g remains bounded as x . Finally,note that we can define the same notion for x approaching some other point. For example,sin(x) x as x 0.In order to apply these rules, it is nice to know something about the growth rates ofdifferent common functions at . We discuss a few here.Proposition 4 (Growth Rates of x, ex , log(x) at ). Let α, β, γ 0. Then1 0,x log(x)α(a) limlog(x)α 0,x xβ(b) limxβ 0.x eγx(c) limWhat these tells us is that asympotically as x , any power of x is smaller than anyexponential and any power of a logarithm is smaller than any power of x. In asymptoticnotation, we write 1 log(x)α xβ eγx where f g means that f /g 0 at . Wecan compound these into new relationships xβ xβ log(x)α . Many more such relationshipscan be proved by simply by taking the required limit. Note that if f g, then f O(g);
Christian ParkinsonGRE Prep: Calculus II Notes3the converse need not be true.Example 5. Determine the convergence/divergence of the following integralsZ 1 (a)dx.x x log(x)2Z 6x 4x3 3x2 2dx(b)7x9 x 101Z 1csc(x)1/2 dx.(c)0Solution. (a) Since x, log(x) x, there are constants c1 , c2 such that for x large enough, x c1 x,log(x) c2 x. ThusZ Z dxdx C x log(x) x2M (1 c1 c2 )xwhere C, M are some constants. The former diverges since the latter does.(b) Notice that if we divide the integrand by 1/x3 (hence multiply by x3 ) and take thelimit, we seex9 4x6 3x5 2x31 4/x3 3/x4 2/x6 lim 1/7.x x 7x9 x 107 1/x8 10/x9R Thus by the Limit Comparison Test, our integral converges iff 1 dxconverges. Thex3latter converges, so the integral in (b) converges as well. (c) Since sin(x) x as x 0, we see that csc(x)1/2 1/ x. Thus the given integralconverges iffZ 1dx x0limconverges which it does.Next, we discuss applications of the integral. We have already established that it can beused to calculate area under curves but there are a few other applications that show up oftenon the math subject GRE; most prominently arc-length calculations and surfaces/volumesof revolution. We discuss the formulas for these briefly.Proposition 6 (Arc Length). Let f : [a, b] R be continuously differentiable. Thenthe arc-length of the graph (x, f (x)) between a and b is given byZ bpL 1 f 0 (x)2 dx.a
Christian ParkinsonGRE Prep: Calculus II Notes4More generally, if x, y : [a, b] R2 are continuously differentiable, then the length of thecurve parameterized by (x(t), y(t)) for t [a, b] is given byZ bpx0 (t)2 y 0 (t)2 dt.L aProposition 7 (Surface / Volume of Revolution). Suppose that f : [a, b] R iscontinuous. If we were tor rotate the graph (x, f (x)) about the x-axis this will create asurface. The surface area of this surface isZ bpf (x) 1 f 0 (x)2 dxA 2πaand the volume encapsulated by this surface isZ bf (x)2 dx.V πaThese formulas are easy to derive when you have a good picture in your mind; otherwiseyou can use them as a black box. Note that they agree with our intuition.Example 8. What is the length of the graph of f (x) 1 x2 for x [ 1, 1]?Solution. Intuitively, this graph is half of a unit circle so the length should be π. Using theformula, we seeZ 1rZ 1dxx2 L dx arcsin(1) arcsin( 1) π/2 ( π/2) π,1 21 x1 x2 1 1where the antiderivative can easily be found using the trig. substitution x sin(t).Example 9. Let f (x) x for x [0, 1] What surface area and volume result from rotatingthe graph of f about the x-axis?Solution. Intuitively, this will create a right, circular cone of “height” 1 and radius 1. Youmay recall from geometry, the volume is π3 r2 h π3 . According to our formula,ZV π01x2 dx π.3Likewise, the surface area (not including the “bottom” of the cone; i.e. the unit circle whichforms the base) will be πr r2 h2 π 2 and our formula givesZ 1 A 2πx 1 12 dx π 2.0Note: much like the arc-length formula, the surface area and volume formulas can beadjusted to account for parameterized coordinates or to account for rotation around the
Christian ParkinsonGRE Prep: Calculus II Notes5y-axis or any other line ax by c. Also, you may be asked for the volume created whenthe area between two graphs (x, f (x)) and (x, g(x)) is rotated around the x-axis. This willbe given byZbf (x)2 g(x)2 dx.πaThe reasoning should be somewhat clear from the geometry.From here we move onto sequences and series.Definition 10 (Sequence). A sequence is a map a : N R which assigns a real numberto each natural number. The value that a assigns to n N is typically denoted by an ratherthan a(n) though both may be used. In a slight abuse of notation, we then often refer to anas a sequence or otherwise denote sequences by (an ), sometimes accompanied by a range forn, e.g., (an )n 0 or (an ) n 1 .Note that sequences can begin at n 0 or n 1; both are common start points. Sequences themselves (i.e. independent of series) account for many questions on the mathsubject GRE. A common question will give a recursive formula (i.e. a formula for an interms of the preceeding entries: an 1 , an 2 , . . .) for a sequence and ask the student to identify the sequence explicitly.Example 11. Suppose that a0 0 and an an 1 (2n 1) for all n 1. Find a closedform expression for an .Solution. To identify the sequence, we just write out terms:a0 0, a1 0 1 1, a2 1 3 4, a3 4 5 9, . . . .From here it is easy to surmise that the sequence is given by an n2 . To prove this, onecould use mathematical induction; however, proofs are not necessary on the GRE so thiswould be a waste of time. We will talk more about general methods for identifying sequenceswhen we touch on Discrete Math.Of particular interest is the behavior of sequence for large n.Definition 12 (Convergence of Sequences). A sequence (an ) is said to converge toL R if for all ε 0, there is N N such that for all n N , we have an L ε. If nosuch L exists, the sequence does not have a limit and is said to diverge. If such L does exist,it is called the limit of the sequence (an ) and (an ) is said to be convergent. In this case, wewritelim an L.n Note that sometimes we drop the n part since it is implied: we may say, lim an Lor an L.
Christian ParkinsonGRE Prep: Calculus II Notes6The rules for limits are sequences are the same as those for limits of functions. We repeatthem here.Proposition 13 (Rules for limits of sequences).convergent sequences. ThenSuppose (an ) and (bn ) are two(a) lim (an bn ) lim an lim bn .n n (b) lim (an bn ) n lim ann n lim bn .n limn anprovided that lim bn 6 0.n n limn bn (d) lim [αan ] α lim an for any constant α R.(c) limn anbn n (e) If h : R R is continuous, then lim h(an ) h n lim an . That is, limits can sliden inside continuous functions. [Note: this is actually a perfectly good definition for continuous function on R. By the sequential criterion theorem, a function h is continuousiff limn h(an ) h (limn an ) for all convergent sequences (an ).]We also have analogous theorems like the squeeze theorem which we omit for brevity.Each sequence also defines a series. A series is a formal infinite sum. Infinite sums are ofgreat interest since, for example, they can be used to approximate functions which cannotbe explicity calculated in a finite number of steps; e.g. ex or cos(x).thpartial sumDefinition 14 (Partial Sums & Series). Given a sequence (an ) n 1 , the Nof (an ) is given byNXsN an .n 1The series (or infinite sum) of (an ) is given (formally at least) bys lim sNN Xn 1an limN NXan .n 1If the partial sums sn have a limit, then the series is said to converge; otherwise it is said todiverge.In some cases, it is easy enough to look at the partial sums and explicitly take the limit.Example 15. Calculate the infinite sums of an 12nand bn 1.n(n 1)
Christian ParkinsonGRE Prep: Calculus II Notes7Solution. For an , we see37152N 11s1 , s2 , s3 , s4 , . . . , sN .248162NThus sN 1 and so we say X1 1.2nn 1For bn , we note that by partial fractions,bn 11 .n n 1Thus writing out the partial sum, we see 11 11 111sN 1 ··· .22 33 4NN 1We notice that the sum “telescopes”; that is, all intermediate terms are canceled and we areleft with1 1.sN 1 N 1Thus we also have X1 1.n(n 1)n 1For examples of series that diverge, consider an n, whence sN N (N2 1) oran ( 1)n 1 where s1 1, s2 0, s3 1, s4 0, . . . and the partial sums continue tooscillate and thus do not converge.While it is fairly easy to tell if a sequence converges or diverges, it can be difficult to tellwhether a series converges or diverges. For example, it may be unclear initially whether XX1 X ( 1)nsin(n),ornlog(n)n 2n 1n 1converge or diverge since we can’t find a nice closed form for the partial sums. To this end,we have built up several tests for convergence/divergence. The first test tells you that if theseries is to converge, we at least need the summand to approach zero.Theorem 16 (Divergence Test). XThenan diverges.Suppose that (an ) is a sequence such that an 6 0.n 1PP nTheorem 21 tells us, for example, that orn 1 n 1n 1 sin(n) will diverge. However,the converse is not true: having a summand which goes to zero does not guarantee convergence for the series.
Christian ParkinsonGRE Prep: Calculus II Notes8Proposition 17 (Harmonic Series). The Harmonic series given by X1nn 1diverges. [A nice document with roughly twenty proofs of this fact can be found here.]Indeed, this is part of a larger fact.Proposition 18 (p-series). Let p R. The series X1npn 1converges is p 1 and diverges for p 1.You may recognize this as a direct analog to one of the above propositions for convergence of integrals (as we will soon see, this is no coincidence) and wonder whether the otherintegral convergence tests have analogs for sequence. The answer is yes.Theorem 19 (Comparison Test for Series). Suppose that (an ) and (bn ) are sequenceswith non-negative terms such that an bn for all n N. Then(a) if Xbn converges then so doesn 1(b) if X Xan .n 1an diverges then so doesn 1 Xbn .n 1Theorem 20 (Limit Comparison Test for Series). Suppose that (an ) and (bn ) arenon-negative sequences such that bn 0 for all n sufficiently large. Iflimn then Xn 1an and Xan L (0, )bnbn either both converge or both diverge.n 1One feature that is more important for series than for integrals is sign changes.reP The( 1)n sults above apply for non-negative sequences but they do not address series like n 1nn[such series with summandsoftheform( 1)awhereaarenon-negativetermsarecallednnP 1 diverges (by the p-series test), with the alternating sign,alternating series]. While n 1 nthere is cancellation occuring and perhaps there is enough cancellation that the alternatingseries converges. This is indeed the case; we state this as a proposition after a few otherrelated definition and results.
Christian ParkinsonGRE Prep: Calculus II NotesDefinition 21 (AbsoluteConvergence).P lutely if the series n 1 an converges.The seriesP n 19an is said to converge abso-PP Theorem 22 (Absolute Convergence Theorem). If n 1 an converges, thenn 1 anconverges. That is, absolute convergence implies convergence. (This is a reflection of thefact that R [with the usual norm] is a Banach space.)Since the converse is not true (lack of absolute convergence does not imply lack of conP( 1)n . For this we need another theorem.vergence), this still says nothing about n 1nTheorem 23 (Alternating SeriesSuppose that (an ) is a non-negative, decreasingPTest). nsequence such that an 0. Then n 1 ( 1) an converges.The Alternating Series test is a special case of a more general theorem. By a theorem ofDirichlet, if (an ) is a non-negativedecreasing to zero, and (bn ) is any series whosePsequence partial sums are bounded, then n 1 an bn converges; the alternating series test is the casethat bn ( 1)n . Note: it is a good exercise to prove by example that the assumptionthat (an ) is decreasing is necessary; without this, the alternating series could diverge even ifan 0.Above we remarked that the rules for infinite sums very closely resemble those for improper integrals. Here is the reason why:Theorem 24 (Integral Test). Suppose that f : [0, ) R is a continuous decreasingfunction such that limx f (x) 0. ThenZ Xf (n) converges.f (x)dx converges if and only if0n 0That is, to check convergence of the sum, we simply need to check convergence of theintegral. Morally: “sums are integrals and vice versa.”Finally, there are two more tests for convergence which are very useful.Theorem 25 (Ratio Test). Suppose that (an ) is a sequence such that an 6 0 for allsufficiently large n and assume thatlimn an 1 L [0, ).anThen(a) if L 1, then Xan converges,n 1(b) if L 1, then Xn 1an diverges.(i.e., assume the limit exists)
Christian ParkinsonGRE Prep: Calculus II Notes10Theorem 26 (Root Test). Suppose that (an ) is a sequence and assume thatplim n an L [0, ). (i.e., assume the limit exists)n Then(a) if L 1, then Xan converges,n 1(b) if L 1, then Xan diverges.n 1Note that neither the ratio test or the root test can address the case that the given limitL 1; in this case, these tests are inconclusive and another test must be used. Also, it is notstrictly necessary for the limit to exist. We could replace L with the limit superior (whichwill always exists as a number in [0, ]) in either case and the theorems continue to hold.We introduce one more convergence test which is not always covered in the Calculus IIcurriculum but it can be very useful.Theorem 27 (Cauchy Condensation Test). Let (an ) be a non-negative sequence suchthat an 0. Then Xan converges if and only if X2n a2n converges.n 1n 1This test can be useful in series involving logarithms since log(2n ) n log(2).Example 28. Determine the convergence/divergence of(a) XXXX1n2 4( 1)n nπ,(d),(b),(c).3 n3/2nnn log(n log(n))arctan(n)n 1n 1n 1n 0Solution. There are several ways to check convergence or divergence. We suggest one wayfor each sum. Series (a) diverges by limit comparison with the harmonic series. Series (b)converges by the alternating series test. Series (c) converges by the ratio or root test. Series(d) converges by the root test.Note that series (c) above is a special series. The underlying sequence is a geometricprogression; i.e., a sequence of the form cr0 , cr1 , cr2 , cr3 , . . . (above we have c 1, r 1/π).Such series are so important, they are not only given their own name, they are evaluatedexplicitly.Proposition 29 (Geometric Series). Let c R \ {0}. Then cX,if r 1,n1 rcr divergent if r 1.n 0
Christian ParkinsonGRE Prep: Calculus II Notes11This is easily proven by showing (using induction) thatNXcrn n 1c(1 rN 1 )1 rand then taking the limit (or course, r 1 needs to be dealt with separately, but this is noproblem).This last proposition is a nice segue into the final topic of Calculus II: series of functionsand specifically, the Taylor series. Note that in the above proposition, the function f :( 1, 1) R defined by1, x ( 1, 1)f (x) 1 xcould just as well be defined byf (x) Xxn ,x ( 1, 1).n 0A natural question is then: what other functions has such representations as “infinite polynomials”? To understand this question we may think of approximating function locally bypolynomials and allowing the degree of the approximating polynomial tend to . Recallthat differentiable functions look “locally linear”; thus it is natural to think that smoothfunctions (i.e., functions that are infinitely many times differentiable) may look locally likea polynomial of any degree that we choose. While this is not true of all smooth functions, itis true for most of the functions that one typically encounters in a Calculus course. In whatfollows, unless explicitly stated, we assume all functions that we introduce are smooth.For demonstrative purposes, recall that the tangent line to a function f (x) at a point ais given byp1 (x) f (a) f 0 (a)(x a).You may notice that p1 is the unique first degree polynomial which matches the value of fat a and the value of f 0 at a. If we want to also match the second derivative of f at a, wewill need to up the order of the polynomial but some work shows thatp2 (x) f (a) f 0 (a)(x a) f 00 (a)(x a)22will work. Likewise, we can match the first N derivatives of f at a using the polynomialpN (x) NXf (n) (a)n 0n!(x a)n .These polynomials pN are sometimes called the N th order Taylor approximations to f (atthe point x a). To see that these do actually approximate f near x a, consider thefollowing theorem.
Christian ParkinsonGRE Prep: Calculus II Notes12Theorem 30 (Taylor’s Theorem). Let f : R R be N times differentiable at a R.Then there is a function RN : R R such that!NXf (n) (a)(x a)n RN (x), x Rf (x) pN (x) RN (x) n!n 0and RN (x) (x a)N as x a.This shows that locally, pN approximates f to at least N th order. There are several waysto explicity identify what form the function RN (x) actually takes but they are not of greatimportance for the math subject GRE; error bounds are more practical.Proposition 31 (Error Bound for Taylor’s Theorem). Suppose that f : R R isN 1 times differentiable in a neighborhood of (a δ, a δ) of a R and that f (N 1) (x) Mfor all x (a δ, a δ). Further, let RN : R R be as in Taylor’s Theorem. Then for allx (a δ, a δ), we have the boundM x a N 1. RN (x) (N 1)!Intuitively this tells you that if f is smooth enough, then f pN O( x a N 1 )as x a; that is, the error in the approximation should be on the order of x a N 1(RN x a N 1 as x a). This error bound can be used to calculate certain functionwithin a given tolerance.Example 32. Suppose that pN is the N th order Taylor approximation to ex centered atx 0. How large does N need to be so that pN (1) approximates the value of e to twodecimal places?Solution. We see that e pN (1) RN (1) M 1 0 N 1(N 1)!ndxxwhere M is the maximum of the N 1 order derivative of ex . Since dxfor any n,n (e ) ewe easily find that M e. To get the first two digits correct, we need RN (1) 0.01. Thuswe simply choose N to accomplish this. We find that using the above bound, N 5 gives RN (1) . 0.003 which is good enough (one can also verify that N 4 is not good enough).These propositions don’t quite answer our question. We would like to be able to able towrite f as an “infinite polynomial,” but these only give finite polynomial approximations.We pass to the “infinite polynomial” by taking N . However, some care is required.Definition 33 (Analyticity). A smooth function f : R R is said to be analytic ata R if there is some sequence (cn ) and some open neighborhood I R with a I such
Christian ParkinsonGRE Prep: Calculus II Notesthatf (x) Xcn (x a)n ,13for all x I.n 0In this case, we will havef (n) (a)n!so that the sum above is exactly the limit of the Taylor approximations. We say that f isanalytic in an open interval I R if f is analytic at each point a I. In this case, we callthe sum Xf (n) (a)(x a)nn!n 0cn the Taylor series for f centered at a.This gives us the definition that we want but does not tell us what functions are analytic.For that, we use the above error bound.Proposition 34. Suppose that f : R R is smooth and that for some open intervalI R, we have that the sequence (MN ) given byMN max f (N ) (x)x Iis bounded. Then f is analytic in I; that is,f (x) lim pN (x), x I.N [Under these hypotheses, we can actually make the stronger claim that pN f uniformlyin I].This is easily proven using the error bound in Proposition 35. Note, this is not strictlya necessary condition. For example, ex is not bounded on R (nor are its derivatives), but itcan be verified that ex can be represented by its Taylor series on all of R.Proposition 35 (Radius of Convergence). Suppose that f : R R is analytic ata R and that (cn ) is as in the definition of analyticity. Then we can take I (a r, a r)wherer lim sup cn 1/n .n That is, if f is analytic at a, then it is analytic in an open neighborhood centered at a. Thevalue r here is called the radius of convergence of the Taylor series of f centered at a and itcan be .Example 36. Above we showed that X1 xn for x ( 1, 1).1 x n 0
Christian ParkinsonGRE Prep: Calculus II Notes14The above proposition shows that we cannot enlarge ( 1, 1) at all; i.e., this is the maximuminterval on which we have this equality. This is because the radius of convergence here isr lim sup 1 1/n 1.n Note, this radius is found using the root test; due to some compatibility conditions withthe root and ratio test, in most exercises it can be found just as easily using the ratio test,which is often simpler to apply. Also, in most practical example, the limit limn cn 1/nwill actually exist and so we can do away with the lim sup. Finally, note that the propositionsays that f is analytic on (a r, a r); it does not say anything about the boundary points.At these points, f may still be equal to its Taylor series or the series may fail to converge.They must be checked separately. The set where a series converges is called the interval ofconvergence. By the above proposition, I (a r, a r) will be the interior of the intervalof convergence.Example 37. Find the interval of convergence of the seriesP xnn 1 n .Solution. By the strict definition, we have that radius of convergence isr lim supn 1n1/n 1.This shows that the series converges on ( 1, 1). We can find the same result using the ratiotest. By the ratio test, the series will converge whenn x n 1nlim x 1n x limn (n 1) x n n 1which again shows convergence on ( 1, 1). The endpoints need to be checked separately(this is where the ratio test and root test will fail). At x 1, we have the harmonic serieswhich is divergent. At x 1, we have the alternating harmonic series which converges bythe alternating series test. Thus the interval of convergence is [ 1, 1).Note, this same convergence test can be performed for series thatnecessarilyP are xnotnpower series. For example, we could use the ratio test to find where n 0 1 xn coverges.There are several Taylor series which are ubiquitous enough to merit memorization (alternatively, these can be derived by quickly taking the derivatives and using the formula).Proposition 38 (Specific Taylor Series). We have(a)xe Xxnn 0(b)cos(x) n!for all x R, X( 1)n x2nn 0(2n)!for all x R,
Christian Parkinson(c)sin(x) GRE Prep: Calculus II Notes X( 1)n x2n 1n 0(2n 1)!15for all x R, (d)X1 xn for x ( 1, 1).1 x n 0Note that all these series are centered at x 0; it is fairly uncommon at this level (or onthe math subject GRE) to come across a Taylor series that is centered somewhere besidesx 0 though it does occasionally happen. Note that the series for cos(x) and sin(x) can bederived from the series for ex using Euler’s identity eix cos(x) i sin(x) and collecting realand imaginary parts.From these, using the next two propositions and some clever functional composition, wecan derive many other series for common functions without having to actually work out thederivatives.Proposition 39 (Term-by-term Differentiation.). Suppose that f : R R is analyticat a R with series Xcn (x a)nf (x) n 0converging inside the interval of convergence I R. Then for all x I, we have0f (x) Xncn (x a)n 1 .n 1That is, to find the derivative of a convergent series, we can differentiate term-by-term. Note,this proposition states f 0 will be analytic with the same radius of convergence as f .Proposition 40 (Term-by-term Integration.). Suppose that f : R R is analytic ata R with series Xcn (x a)nf (x) n 0converging inside the interval of convergence I R. Let F : R R be any anti-derivativeof f . Then for all x I, we haveF (x) F (a) Xn 0cn(x a)n 1.n 1That is, to find the integral of a series, we can integrate term-by-term. Note that in thiscase, the radius of convergence also does not change.Example 41. Find the Taylor series for cosh(x) and log(1 x2 ) centered at x 0. Whatare the intervals on convergence for these series?
Christian ParkinsonGRE Prep: Calculus II Notes16Solution. Recall that cosh(x) 12 (ex e x ). To find the series for cosh(x), we can simplyadd the series for ex and e x . Sincexe Xxnn 0we see thate x n!for all x R,, X( 1)n xnn!n 0for all x R.,When adding these together, the odd terms will cancel and the even terms will be doubled.Thus we have Xx2ncosh(x) .(2n)!n 0Since this is just the addition of two series which converge everywhere, this series should alsoconverge everywhere; this can be checked using the ratio test. The the interval of convergenceis R.For log(1 x2 ), we note thatZ x2t2dt.log(1 x ) 20 1 tNow we can expand the integrand in a Taylor series. Using the geometric series, we see that X1 ( 1)n t2n for t ( 1, 1).1 t2n 0Then for such t, X2t( 1)n t2n 1 . 21 t2n 0Thus integrating giveslog(1 x2 ) X( 1)n x2n 2n 0n 1,x ( 1, 1).Note that at both x 1 and x 1, this series will converge (by the alternating seriestest), thus the interval of convergence is [ 1, 1].One large application of Taylor series is evaluation of infinite sums. We can see forexample, that plugging in certain values of x, the series listed in Proposition 42 will give usthe value of certain infinite sums. For example, X113 3n1 (1/3)2n 0or X( 1)nn 0n! e 1 .
Christian ParkinsonGRE Prep: Calculus II Notes17Taylor series can help with much more involved sums though this sometimes requires skillfulmanipulations and some fortuitous recognition of series.Example 42. Evaluate the following infinite sums. X ( 1)k 2kπ 2kX ( 1)n2341 · · · , (b), (c).(a) 4 16 64 256(2k 2)!n 1n 0k 0Solution. For (a), we can write the sum as Xn.n4n 1Notice that we have a geomtric term 1/4n in the summand (though it is accompanied byanother term); this should hint that (a) comes from the geometric series somehow. Indeed,to make that n appear in the numerator, we can differentiate the geometric series: X1 xn1 x n 0 X1 nxn 1 .(1 x)2n 1Plugging in x 1/4 gives 16 X n 94n 1n 1 2 X n .3 n 1 4nFor (b), we notice the factorial in the denominator: this indicated that the sum can likelybe evaulated using the Taylor series for sin(x), cos(x) or ex . Call the sum S. Note thatS X( 1)k (2k 2)π 2k(2k 2)!k 0We seeS1 2 X( 1)k π 2kk 0(2k 2)! S1 S2 . X( 1)k π 2kk 0 1 X ( 1)k π 2k 1sin(π) 0.(2k 1)!π k 0 (2k 1)!πNowS2 2 2 X ( 1)k 1 π 2k 2 2.(2k 2)!π k 0 (2k 2)! X( 1)k π 2kk 0This looks exactly like the Taylor series for cos(x) with x π plugged in except that kreplaced in the summad by k 1. This has the effect of omitting the first term inthe Taylorseries which is cos(0) 1. Thus! X2( 1)k π 2k24S2 2
Christian Parkinson GRE Prep: Calculus II Notes 1 Week 2: Calculus II Notes We concluded the Calculus I notes with Riemann integration, Fundamental Theorem of Calculus and some helpful integration techniques. However, ofter times, you will be asked to identify whether an integral conv
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