Week 2: Calculus II Notes

Christian Parkinson GRE Prep: Calculus II Notes 1

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 converges or diverges even when you cannot ﬁnd the value. This is especially true for improper integrals: those in which the integrand had a singularity or the range goes to inﬁnity. We begin the Calculus II notes by discussing some basic convergence results.

Proposition 1 (p-test). Let p ∈ R. The integral Z ∞ dx p 1 x converges if p > 1 and diverges otherwise. The integral

Z 1 dx p 0 x converges if p < 1 and diverges otherwise.

Note that as a consequence, the integral Z ∞ dx p 0 x never converges. These can all be veriﬁed by a straight-forward calculation and hint at some good intuition. The borderline case is 1/x where neither integrals converge, so as a general rule, for Z ∞ f(x)dx 1 to converge, we will need f(x) → 0 faster than 1/x as x → ∞ and if g(x) has a singularity at x = 0, then for Z 1 g(x)dx 0 to 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 it easy to verify if an integral converges or diverges. The following two theorems give more rigorous forms of these general rules.

Theorem 2 (Comparison Test). Suppose that f, g : [0, ∞) → [0, ∞) are continuous and that f(x) ≤ g(x) for all x ∈ [0, ∞). Then Z ∞ Z ∞ (a) if g(x)dx converges, then so does f(x)dx, 0 0 Christian Parkinson GRE Prep: Calculus II Notes 2

Z ∞ Z ∞ (b) if f(x)dx diverges, then so does g(x)dx. 0 0 Theorem 3 (Limit Comparison Test). Suppose that f, g : [0, ∞) → [0, ∞) are continuous and that f(x) lim x→∞ g(x) exists as a number in (0, ∞)[excluding the endpoints]. Then Z ∞ Z ∞ f(x)dx and g(x)dx both converge or both diverge. 0 0 Note, the previous two theorems also hold for improper integrals where the functions have singularities 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 of an integral with ∞ as a limit is the asymptotic behavior of the integrand. We have a special notation for this. If f, g are such that f/g → L as x → ∞ where L is a finite non-zero number, then we write f ∼ g [note: in most contexts, L is required to be 1 in order to say f ∼ g; for our purposes, it is fine to allow L to be finite and non-zero]. We typically read this as “f is asymptotic to g.” Note that if f ∼ g, then there are constants c, C > 0 such that for sufficiently large x, we have cg(x) ≤ f(x) ≤ Cg(x). This shows that we can easily derive the Limit Comparison Test from the ordinary Comparison Test. A relaxed version of this requirement leads to a new definition. If there is C > 0 such that f(x) ≤ Cg(x) for all x sufficiently large, then we write f = O(g) [and say “f is big-oh of g”]. Thus f ∼ g iff f = 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 of diﬀerent common functions at ∞. We discuss a few here.

Proposition 4 (Growth Rates of x, ex, log(x) at ∞). Let α, β, γ > 0. Then 1 (a) lim = 0, x→∞ log(x)α log(x)α (b) lim = 0, x→∞ xβ xβ (c) lim = 0. x→∞ eγx What these tells us is that asympotically as x → ∞, any power of x is smaller than any exponential and any power of a logarithm is smaller than any power of x. In asymptotic notation, we write 1 log(x)α xβ eγx where f g means that f/g → 0 at ∞. We can compound these into new relationships xβ xβ log(x)α. Many more such relationships can be proved by simply by taking the required limit. Note that if f g, then f = O(g); Christian Parkinson GRE Prep: Calculus II Notes 3 the converse need not be true.

Example 5. Determine the convergence/divergence of the following integrals Z ∞ 1 (a) √ dx. 2 x + x + log(x) Z ∞ x6 + 4x3 + 3x2 + 2 (b) 9 dx 1 7x + x + 10 Z 1 (c) csc(x)1/2dx. 0 Solution. √ √ (a) Since x, log(x) x, there are constants c1, c2 such that for x large enough, x ≤ c1x, log(x) ≤ c2x. Thus Z ∞ dx Z ∞ dx √ ≥ C + 2 x + log(x) + x M (1 + c1 + c2)x where 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 the limit, we see

x9 + 4x6 + 3x5 + 2x3 1 + 4/x3 + 3/x4 + 2/x6 lim = lim = 1/7. x→∞ 7x9 + x + 10 x→∞ 7 + 1/x8 + 10/x9

R ∞ dx Thus by the Limit Comparison Test, our integral converges iﬀ 1 x3 converges. The latter 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 integral converges iﬀ Z 1 dx √ 0 x converges which it does.

Next, we discuss applications of the integral. We have already established that it can be used to calculate area under curves but there are a few other applications that show up often on the math subject GRE; most prominently arc-length calculations and surfaces/volumes of revolution. We discuss the formulas for these brieﬂy.

Proposition 6 (Arc Length). Let f :[a, b] → R be continuously diﬀerentiable. Then the arc-length of the graph (x, f(x)) between a and b is given by

Z b L = p1 + f 0(x)2dx. a Christian Parkinson GRE Prep: Calculus II Notes 4

More generally, if x, y :[a, b] → R2 are continuously diﬀerentiable, then the length of the curve parameterized by (x(t), y(t)) for t ∈ [a, b] is given by

Z b L = px0(t)2 + y0(t)2dt. a

Proposition 7 (Surface / Volume of Revolution). Suppose that f :[a, b] → R is continuous. If we were tor rotate the graph (x, f(x)) about the x-axis this will create a surface. The surface area of this surface is Z b A = 2π f(x)p1 + f 0(x)2dx a and the volume encapsulated by this surface is

Z b V = π f(x)2dx. a These formulas are easy to derive when you have a good picture in your mind; otherwise you 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 the formula, we see

Z 1 r x2 Z 1 dx L = 1 + dx = √ = arcsin(1) − arcsin(−1) = π/2 − (−π/2) = π, 2 2 −1 1 − x −1 1 − x where 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 rotating the graph of f about the x-axis?

Solution. Intuitively, this will create a right, circular cone of “height” 1 and radius 1. You π 2 π may recall from geometry, the volume is 3 r h = 3 . According to our formula, Z 1 π V = π x2dx = . 0 3

Likewise, the surface area√ (not including√ the “bottom” of the cone; i.e. the unit circle which forms the base) will be πr r2 + h2 = π 2 and our formula gives

Z 1 √ √ A = 2π x 1 + 12 dx = π 2. 0 Note: much like the arc-length formula, the surface area and volume formulas can be adjusted to account for parameterized coordinates or to account for rotation around the Christian Parkinson GRE Prep: Calculus II Notes 5 y-axis or any other line ax + by = c. Also, you may be asked for the volume created when the area between two graphs (x, f(x)) and (x, g(x)) is rotated around the x-axis. This will be given by Z b 2 2 π f(x) − g(x) dx. a The 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 number to each natural number. The value that a assigns to n ∈ N is typically denoted by an rather than a(n) though both may be used. In a slight abuse of notation, we then often refer to an as a sequence or otherwise denote sequences by (an), sometimes accompanied by a range for ∞ n, 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. Se- quences themselves (i.e. independent of series) account for many questions on the math subject GRE. A common question will give a recursive formula (i.e. a formula for an in terms 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 closed form 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,....

2 From here it is easy to surmise that the sequence is given by an = n . To prove this, one could use mathematical induction; however, proofs are not necessary on the GRE so this would be a waste of time. We will talk more about general methods for identifying sequences when 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 to L ∈ R if for all ε > 0, there is N ∈ N such that for all n ≥ N, we have |an − L| < ε. If no such 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, we write lim an = L. n→∞

Note that sometimes we drop the n → ∞ part since it is implied: we may say, lim an = L or an → L. Christian Parkinson GRE Prep: Calculus II Notes 6

The rules for limits are sequences are the same as those for limits of functions. We repeat them here.

Proposition 13 (Rules for limits of sequences). Suppose (an) and (bn) are two convergent sequences. Then

(a) lim (an + bn) = lim an + lim bn. n→∞ n→∞ n→∞ (b) lim (anbn) = lim an lim bn . n→∞ n→∞ n→∞ an limn→∞ an (c) lim = provided that lim bn 6= 0. n→∞ bn limn→∞ bn n→∞ (d) lim [αan] = α lim an for any constant α ∈ . n→∞ n→∞ R (e) If h : → is continuous, then lim h(an) = h lim an . That is, limits can slide R R n→∞ n→∞ 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 continuous iﬀ 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 of great interest since, for example, they can be used to approximate functions which cannot be explicity calculated in a finite number of steps; e.g. ex or cos(x).

∞ th Deﬁnition 14 (Partial Sums & Series). Given a sequence (an)n=1, the N partial sum of (an) is given by N X sN = an. n=1

The series (or inﬁnite sum) of (an) is given (formally at least) by

s = lim sN N→∞ ∞ N X X an = lim an. N→∞ n=1 n=1

If the partial sums sn have a limit, then the series is said to converge; otherwise it is said to diverge.

In some cases, it is easy enough to look at the partial sums and explicitly take the limit.

1 1 Example 15. Calculate the inﬁnite sums of an = 2n and bn = n(n+1) . Christian Parkinson GRE Prep: Calculus II Notes 7

Solution. For an, we see 1 3 7 15 2N − 1 s = , s = , s = , s = , . . . , s = . 1 2 2 4 3 8 4 16 N 2N

Thus sN → 1 and so we say ∞ X 1 = 1. 2n n=1

For bn, we note that by partial fractions, 1 1 b = − . n n n + 1 Thus writing out the partial sum, we see 1 1 1 1 1 1 1 s = 1 − + − + − + ··· + − . N 2 2 3 3 4 N N + 1 We notice that the sum “telescopes”; that is, all intermediate terms are canceled and we are left with 1 s = 1 − → 1. N N + 1 Thus we also have ∞ X 1 = 1. n(n + 1) n=1 N(N+1) For examples of series that diverge, consider an = n, whence sN = 2 → ∞ or n+1 an = (−1) where s1 = 1, s2 = 0, s3 = 1, s4 = 0,... and the partial sums continue to oscillate and thus do not converge.

While it is fairly easy to tell if a sequence converges or diverges, it can be diﬃcult to tell whether a series converges or diverges. For example, it may be unclear initially whether

∞ ∞ ∞ X 1 X (−1)n X , or sin(n) n log(n) n=1 n=2 n=1 converge 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 the series is to converge, we at least need the summand to approach zero.

Theorem 16 (Divergence Test). Suppose that (an) is a sequence such that an 6→ 0. ∞ X Then an diverges. n=1 P∞ n P∞ Theorem 21 tells us, for example, that n=1 n+1 or n=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 Parkinson GRE Prep: Calculus II Notes 8

Proposition 17 (Harmonic Series). The Harmonic series given by

∞ X 1 n n=1 diverges. [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

∞ X 1 np n=1 converges 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 other integral convergence tests have analogs for sequence. The answer is yes.

Theorem 19 (Comparison Test for Series). Suppose that (an) and (bn) are sequences with non-negative terms such that an ≤ bn for all n ∈ N. Then ∞ ∞ X X (a) if bn converges then so does an. n=1 n=1

∞ ∞ X X (b) if an diverges then so does bn. n=1 n=1

Theorem 20 (Limit Comparison Test for Series). Suppose that (an) and (bn) are non-negative sequences such that bn > 0 for all n suﬃciently large. If a lim n = L ∈ (0, ∞) n→∞ bn then ∞ ∞ X X an and bn either both converge or both diverge. n=1 n=1 One feature that is more important for series than for integrals is sign changes. The re- ∞ (−1)n P √ sults above apply for non-negative sequences but they do not address series like n=1 n n [such series with summands of the form (−1) an where an are non-negative terms are called ∞ P √1 alternating series]. While n=1 n diverges (by the p-series test), with the alternating sign, there is cancellation occuring and perhaps there is enough cancellation that the alternating series converges. This is indeed the case; we state this as a proposition after a few other related deﬁnition and results. Christian Parkinson GRE Prep: Calculus II Notes 9

P∞ Deﬁnition 21 (Absolute Convergence). The series n=1 an is said to converge abso- P∞ lutely if the series n=1 |an| converges.

P∞ P∞ Theorem 22 (Absolute Convergence Theorem). If n=1 |an| converges, then n=1 an converges. That is, absolute convergence implies convergence. (This is a reﬂection of the fact 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 con- ∞ (−1)n P √ vergence), this still says nothing about n=1 n . For this we need another theorem.

Theorem 23 (Alternating Series Test). Suppose that (an) is a non-negative, decreasing P∞ n sequence 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 of Dirichlet, if (an) is a non-negative sequence decreasing to zero, and (bn) is any series whose P∞ partial sums are bounded, then n=1 anbn converges; the alternating series test is the case n that bn = (−1) . Note: it is a good exercise to prove by example that the assumption that (an) is decreasing is necessary; without this, the alternating series could diverge even if an → 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 decreasing function such that limx→∞ f(x) = 0. Then ∞ Z ∞ X f(x)dx converges if and only if f(n) converges. 0 n=0 That is, to check convergence of the sum, we simply need to check convergence of the integral. 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 all suﬃciently large n and assume that

an+1 lim = L ∈ [0, ∞). (i.e., assume the limit exists) n→∞ an Then ∞ X (a) if L < 1, then an converges, n=1 ∞ X (b) if L > 1, then an diverges. n=1 Christian Parkinson GRE Prep: Calculus II Notes 10

Theorem 26 (Root Test). Suppose that (an) is a sequence and assume that

pn lim |an| = L ∈ [0, ∞). (i.e., assume the limit exists) n→∞ Then ∞ X (a) if L < 1, then an converges, n=1

∞ X (b) if L > 1, then an diverges. n=1 Note that neither the ratio test or the root test can address the case that the given limit L = 1; in this case, these tests are inconclusive and another test must be used. Also, it is not strictly necessary for the limit to exist. We could replace L with the limit superior (which will 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 II curriculum but it can be very useful.

Theorem 27 (Cauchy Condensation Test). Let (an) be a non-negative sequence such that an → 0. Then ∞ ∞ X X n an converges if and only if 2 a2n converges. n=1 n=1 This test can be useful in series involving logarithms since log(2n) = n log(2).

Example 28. Determine the convergence/divergence of

∞ ∞ ∞ ∞ X n2 + 4 X (−1)n X X 1 (a) , (b) , (c) π−n, (d) . n3 + n3/2 n + log(n + log(n)) arctan(n)n n=1 n=1 n=0 n=1 Solution. There are several ways to check convergence or divergence. We suggest one way for 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 geometric progression; 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 evaluated explicitly.

Proposition 29 (Geometric Series). Let c ∈ R \{0}. Then ∞ X c , if |r| < 1, crn = 1−r divergent if |r| ≥ 1. n=0 Christian Parkinson GRE Prep: Calculus II Notes 11

This is easily proven by showing (using induction) that

N X c(1 − rN+1) crn = 1 − r n=1 and then taking the limit (or course, r = 1 needs to be dealt with separately, but this is no problem).

This last proposition is a nice segue into the final topic of Calculus II: series of functions and specifically, the Taylor series. Note that in the above proposition, the function f : (−1, 1) → R defined by 1 f(x) = , x ∈ (−1, 1) 1 − x could just as well be defined by

∞ X f(x) = xn, x ∈ (−1, 1). n=0 A natural question is then: what other functions has such representations as “infinite polynomials”? To understand this question we may think of approximating function locally by polynomials and allowing the degree of the approximating polynomial tend to ∞. Recall that differentiable functions look “locally linear”; thus it is natural to think that smooth functions (i.e., functions that are infinitely many times differentiable) may look locally like a polynomial of any degree that we choose. While this is not true of all smooth functions, it is true for most of the functions that one typically encounters in a Calculus course. In what follows, 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 a is given by 0 p1(x) = f(a) + f (a)(x − a).

You may notice that p1 is the unique ﬁrst degree polynomial which matches the value of f at a and the value of f 0 at a. If we want to also match the second derivative of f at a, we will need to up the order of the polynomial but some work shows that

f 00(a) p (x) = f(a) + f 0(a)(x − a) + (x − a)2 2 2 will work. Likewise, we can match the ﬁrst N derivatives of f at a using the polynomial

N X f (n)(a) p (x) = (x − a)n. N n! n=0

th These polynomials pN are sometimes called the N order Taylor approximations to f (at the point x = a). To see that these do actually approximate f near x = a, consider the following theorem. Christian Parkinson GRE Prep: Calculus II Notes 12

Theorem 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

N ! X f (n)(a) f(x) = p (x) + R (x) = (x − a)n + R (x), x ∈ N N n! N R n=0

N and RN (x) (x − a) as x → a.

th This shows that locally, pN approximates f to at least N order. There are several ways to explicity identify what form the function RN (x) actually takes but they are not of great importance for the math subject GRE; error bounds are more practical.

Proposition 31 (Error Bound for Taylor’s Theorem). Suppose that f : R → R is

N+1 times diﬀerentiable in a neighborhood of (a−δ, a+δ) of a ∈ R and that f (N+1)(x) ≤ M for all x ∈ (a − δ, a + δ). Further, let RN : R → R be as in Taylor’s Theorem. Then for all x ∈ (a − δ, a + δ), we have the bound

M |x − a|N+1 |R (x)| ≤ . N (N + 1)!

N+1 Intuitively this tells you that if f is smooth enough, then |f − pN | = O(|x − a| ) as x → a; that is, the error in the approximation should be on the order of |x − a|N+1 N+1 (RN ∼ |x − a| as x → a). This error bound can be used to calculate certain function within a given tolerance.

th x Example 32. Suppose that pN is the N order Taylor approximation to e centered at x = 0. How large does N need to be so that pN (1) approximates the value of e to two decimal places?

Solution. We see that

M |1 − 0|N+1 |e − p (1)| = |R (1)| ≤ N N (N + 1)!

x dn x x where M is the maximum of the N + 1 order derivative of e . Since dxn (e ) = e for any n, we easily find that M = e. To get the first two digits correct, we need |RN (1)| ≤ 0.01. Thus we 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 to write 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 at a ∈ R if there is some sequence (cn) and some open neighborhood I ⊆ R with a ∈ I such Christian Parkinson GRE Prep: Calculus II Notes 13

that ∞ X n f(x) = cn(x − a) , for all x ∈ I. n=0 In this case, we will have f (n)(a) c = n n! so that the sum above is exactly the limit of the Taylor approximations. We say that f is analytic in an open interval I ⊆ R if f is analytic at each point a ∈ I. In this case, we call the sum ∞ X f (n)(a) (x − a)n n! n=0 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 interval I ⊆ R, we have that the sequence (MN ) given by (N) MN = max f (x) x∈I is 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 uniformly in I].

This is easily proven using the error bound in Proposition 35. Note, this is not strictly a necessary condition. For example, ex is not bounded on R (nor are its derivatives), but it can 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 at a ∈ R and that (cn) is as in the deﬁnition of analyticity. Then we can take I = (a − r, a + r) where 1/n r = lim sup |cn| . n→∞ That is, if f is analytic at a, then it is analytic in an open neighborhood centered at a. The value r here is called the radius of convergence of the Taylor series of f centered at a and it can be +∞.

Example 36. Above we showed that

∞ 1 X = xn for x ∈ (−1, 1). 1 − x n=0 Christian Parkinson GRE Prep: Calculus II Notes 14

The above proposition shows that we cannot enlarge (−1, 1) at all; i.e., this is the maximum interval on which we have this equality. This is because the radius of convergence here is

r = lim sup |1|1/n = 1. n→∞ Note, this radius is found using the root test; due to some compatibility conditions with the root and ratio test, in most exercises it can be found just as easily using the ratio test, 1/n which is often simpler to apply. Also, in most practical example, the limit limn→∞ |cn| will actually exist and so we can do away with the lim sup. Finally, note that the proposition says 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 of convergence. By the above proposition, I = (a − r, a + r) will be the interior of the interval of convergence.

P∞ xn Example 37. Find the interval of convergence of the series n=1 n .

Solution. By the strict deﬁnition, we have that radius of convergence is 1 r = lim sup 1/n = 1. n→∞ n This shows that the series converges on (−1, 1). We can ﬁnd the same result using the ratio test. By the ratio test, the series will converge when

n |x|n+1 n lim n = |x| lim = |x| < 1 n→∞ (n + 1) |x| n→∞ n + 1

which 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 series which is divergent. At x = −1, we have the alternating harmonic series which converges by the alternating series test. Thus the interval of convergence is [−1, 1).

Note, this same convergence test can be performed for series that are not necessarily P∞ xn power 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 (al- ternatively, these can be derived by quickly taking the derivatives and using the formula).

Proposition 38 (Speciﬁc Taylor Series). We have

∞ X xn (a) ex = for all x ∈ , n! R n=0

∞ X (−1)nx2n (b) cos(x) = for all x ∈ , (2n)! R n=0 Christian Parkinson GRE Prep: Calculus II Notes 15

∞ X (−1)nx2n+1 (c) sin(x) = for all x ∈ , (2n + 1)! R n=0

∞ 1 X (d) = xn for x ∈ (−1, 1). 1 − x n=0 Note that all these series are centered at x = 0; it is fairly uncommon at this level (or on the math subject GRE) to come across a Taylor series that is centered somewhere besides x = 0 though it does occasionally happen. Note that the series for cos(x) and sin(x) can be derived from the series for ex using Euler’s identity eix = cos(x) + i sin(x) and collecting real and imaginary parts.

From these, using the next two propositions and some clever functional composition, we can derive many other series for common functions without having to actually work out the derivatives.

Proposition 39 (Term-by-term Diﬀerentiation.). Suppose that f : R → R is analytic at a ∈ with series R ∞ X n f(x) = cn(x − a) n=0 converging inside the interval of convergence I ⊆ R. Then for all x ∈ I, we have

∞ 0 X n−1 f (x) = ncn(x − a) . n=1 That 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 at a ∈ with series R ∞ X n f(x) = cn(x − a) n=0 converging inside the interval of convergence I ⊆ R. Let F : R → R be any anti-derivative of f. Then for all x ∈ I, we have

∞ X (x − a)n+1 F (x) = F (a) + c . n n + 1 n=0 That is, to ﬁnd the integral of a series, we can integrate term-by-term. Note that in this case, 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. What are the intervals on convergence for these series? Christian Parkinson GRE Prep: Calculus II Notes 16

1 x −x Solution. Recall that cosh(x) = 2 (e + e ). To ﬁnd the series for cosh(x), we can simply add the series for ex and e−x. Since

∞ X xn ex = , for all x ∈ , n! R n=0 we see that ∞ X (−1)nxn e−x = , for all x ∈ . n! R n=0 When adding these together, the odd terms will cancel and the even terms will be doubled. Thus we have ∞ X x2n cosh(x) = . (2n)! n=0 Since this is just the addition of two series which converge everywhere, this series should also converge everywhere; this can be checked using the ratio test. The the interval of convergence is R. For log(1 + x2), we note that

Z x 2 2t log(1 + x ) = 2 dt. 0 1 + t Now we can expand the integrand in a Taylor series. Using the geometric series, we see that

∞ 1 X = (−1)nt2n for t ∈ (−1, 1). 1 + t2 n=0 Then for such t, ∞ 2t X = 2 (−1)nt2n+1. 1 + t2 n=0 Thus integrating gives

∞ X (−1)nx2n+2 log(1 + x2) = , x ∈ (−1, 1). n + 1 n=0 Note that at both x = 1 and x = −1, this series will converge (by the alternating series test), thus the interval of convergence is [−1, 1].

One large application of Taylor series is evaluation of inﬁnite sums. We can see for example, that plugging in certain values of x, the series listed in Proposition 42 will give us the value of certain inﬁnite sums. For example,

∞ ∞ X 1 1 3 X (−1)n = = or = e−1. 3n 1 − (1/3) 2 n! n=0 n=0 Christian Parkinson GRE Prep: Calculus II Notes 17

Taylor series can help with much more involved sums though this sometimes requires skillful manipulations and some fortuitous recognition of series.

Example 42. Evaluate the following inﬁnite sums.

∞ ∞ 1 2 3 4 X (−1)k2kπ2k X (−1)n (a) + + + + ··· , (b) , (c) . 4 16 64 256 (2k + 2)! n + 1 k=0 n=0 Solution. For (a), we can write the sum as

∞ X n . 4n n=1 Notice that we have a geomtric term 1/4n in the summand (though it is accompanied by another 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:

∞ ∞ 1 X 1 X = xn =⇒ = nxn−1. 1 − x (1 − x)2 n=0 n=1 Plugging in x = 1/4 gives

∞ ∞ 16 X n 2 X n = =⇒ = . 9 4n−1 3 4n n=1 n=1

For (b), we notice the factorial in the denominator: this indicated that the sum can likely be evaulated using the Taylor series for sin(x), cos(x) or ex. Call the sum S. Note that

∞ ∞ X (−1)k(2k + 2)π2k X (−1)kπ2k S = − 2 = S + S . (2k + 2)! (2k + 2)! 1 2 k=0 k=0 We see ∞ ∞ X (−1)kπ2k 1 X (−1)kπ2k+1 sin(π) S = = = = 0. 1 (2k + 1)! π (2k + 1)! π k=0 k=0 Now ∞ ∞ X (−1)kπ2k 2 X (−1)k+1π2k+2 S = −2 = . 2 (2k + 2)! π2 (2k + 2)! k=0 k=0 This looks exactly like the Taylor series for cos(x) with x = π plugged in except that k replaced in the summad by k + 1. This has the eﬀect of omitting the ﬁrst term inthe Taylor series which is cos(0) = 1. Thus

∞ ! 2 X (−1)kπ2k 2 4 S = −1 + = (−1 + cos(π) = − . 2 π2 (2k)! π2 π2 k=0 Christian Parkinson GRE Prep: Calculus II Notes 18

Thus 4 S = S + S = − . 1 2 π2 For (c), we can imagine this may have arisen from plugging in x = −1 to a Taylor series. Indeed, for x ∈ (−1, 1),

∞ ∞ ∞ ! X xn+1 X Z x Z x X Z x dt = tndt = tn dt = = − log(1 − x). n + 1 1 − t n=0 n=0 0 0 n=0 0 Now substituting x = −1, we see

∞ X (−1)n log(2) = . n + 1 n=0 Note, at the end here, we sort of cheated: we substituted in x = −1 when the original manipulation was only valid for x ∈ (−1, 1). This is justiﬁed by Abel’s Theorem.

Theorem 43 (Abel’s Theorem). Suppose that f is analytic in (a − r, a + r) ⊆ R with series ∞ X n f(x) = cn(x − a) . n=0 If f is continous from the left at a + r (resp. continuous from the right at a − r), then P∞ n P∞ n P∞ n n=0 cnr converges (resp. n=0 cn(−r) converges) and f(a + r) = n=0 cnr (resp. P∞ n f(a − r) = n=0 cn(−r) ).

P∞ (−1)n This theorem validates that log(1 − x) = n=0 n+1 since log(1 − x) remains continuous as x → −1. Another use for Taylor series is in evaluating limits since Taylor series decide local behavior of a function. For example, we saw previously using l’Hˆopital’srule that

sin(x) lim = 1. x→0 x This can easily be computed using Taylor approximations as well. Writing out the ﬁrst few terms of the Taylor series, we see that near x = 0,

x3 sin(x) ≈ x − . 6 Thus sin(x) x2 ≈ 1 − → 1 as x → 0. x 6 This can greatly simplify limits which appear tricky at ﬁrst.

sin(x − sin(x)) Example 44. Compute lim . x→0 x3 Christian Parkinson GRE Prep: Calculus II Notes 19

Solution. This limit can be performed using l’Hˆopital’srule and manipulating the result, but with knowledge of Taylor series it is very easy. From the above, we see that near x = 0, we have x3 x − sin(x) = + O(x4). 6 In the limit, the higher order terms disappear, so we see

sin(x − sin(x)) sin(x3/6) lim = lim x→0 x3 x→0 x3 and since sin(x) ≈ x for x near 0, we have sin(x3/6) ≈ x3/6, so

sin(x − sin(x)) x3/6 1 lim = lim = . x→0 x3 x→0 x3 6 As a final note, in light of our viewing Taylor series as “infinite polynomials,” there should be some compatibility with finite polynomials. Indeed, there is.

Proposition 45. For any polynomial p : R → R and any a ∈ R, we see

∞ X p(n)(a) p(x) = (x − a)n, for all x ∈ . n! R n=0

Note that there is no issue checking convergence since p(n)(x) = 0 for all x, when n is larger than the degree of the polynomial; thus the sum is actually ﬁnite. This proposition tells us that any polynomial is equal to its Taylor series centered at any point. This fact is occasionally useful on the math subject GRE; for example, one problem on a previous math GRE asks to identify the values of a0, a1, a2, a3 so that

3 2 3 x − x + 1 − a0 + a1(x − 2) + a2(x − 2) + a3(x − a)

which is very simple if you view the right hand side as the Taylor series for the polynomial p(x) = x3 − x + 1 centered at x = 2.