6: SEQUENCES AND SERIES OF FUNCTIONS, CONVERGENCE

6: SEQUENCES AND SERIES OF FUNCTIONS, CONVERGENCE STEVEN HEILMAN Contents 1. Review 1 2. Sequences of Functions 2 3. Uniform Convergence nd Continuity 3 4. Series of Functions nd the Weierstrss M-test 5 5. Uniform Convergence nd Integrtion 6 6. Uniform Convergence nd Differentition 7 7. Uniform Approximtion by Polynomils 9 8. Power Series 10 9. The Exponentil nd Logrithm 15 10. Trigonometric Functions 17 11. Appendix: Nottion 20 1. Review Remrk 1.1. From now on, unless otherwise specified, R n refers to Eucliden spce R n with n 1 positive integer, nd where we use the metric d l2 on R n. In prticulr, R refers to the metric spce R equipped with the metric d(x, y) = x y. Proposition 1.2. Let (X, d) be metric spce. Let (x (j) ) j=k be sequence of elements of X. Let x, x be elements of X. Assume tht the sequence (x (j) ) j=k converges to x with respect to d. Assume lso tht the sequence (x (j) ) j=k converges to x with respect to d. Then x = x. Proposition 1.3. Let < b be rel numbers, nd let f : [, b] R be function which is both continuous nd strictly monotone incresing. Then f is bijection from [, b] to [f(), f(b)], nd the inverse function f 1 : [f(), f(b)] [, b] is lso continuous nd strictly monotone incresing. Theorem 1.4 (Inverse Function Theorem). Let X, Y be subsets of R. Let f : X Y be bijection, so tht f 1 : Y X is function. Let x 0 X nd y 0 Y such tht f(x 0 ) = y 0. If f is differentible t x 0, if f 1 is continuous t y 0, nd if f (x 0 ) 0, then f 1 is differentible t y 0 with (f 1 ) (y 0 ) = 1 f (x 0 ). Dte: Februry 14, 2015. 1

2. Sequences of Functions As we hve seen in nlysis, it is often desirble to discuss sequences of points tht converge. Below, we will see tht it is similrly desirble to discuss sequences of functions tht converge in vrious senses. There re mny distinct wys of discussing the convergence of sequences of functions. We will only discuss two such modes of convergence, nmely pointwise nd uniform convergence. Before beginning this discussion, we discuss the limiting vlues of functions between metric spces, which should generlize our notion of limiting vlues of functions on the rel line. 2.1. Limiting Vlues of Functions. Definition 2.1. Let (X, d X ) nd (Y, d Y ) be metric spces, let E be subset of X, let f : X Y be function, let x 0 X be n dherent point of E, nd let L Y. We sy tht f(x) converges to L in Y s x converges to x 0 in E, nd we write lim x x0 ;x E f(x) = L, if nd only if, for every ε > 0, there exists δ = δ(ε) > 0 such tht, if x E stisfies d X (x, x 0 ) < δ, then d Y (f(x), L) < ε. Remrk 2.2. So, f is continuous t x 0 if nd only if lim f(x) = f(x 0). x x 0 ;x X And f is continuous on X if nd only if, for ll x 0 X, ( ) holds. Remrk 2.3. When the domin of x of the limit lim x x0 ;x X f(x) is cler, we will often insted write lim x x0 f(x). The following equivlence is generlized from its nlogue on the rel line. Proposition 2.4. Let (X, d X ) nd (Y, d Y ) be metric spces, let E be subset of X, let f : X Y be function, let x 0 X be n dherent point of E, nd let L Y. Then the following sttements re equivlent. lim x x0 ;x E f(x) = L. For ny sequence (x (j) ) in E which converges to x 0 with respect to the metric d X, the sequence (f(x (j) )) converges to L with respect to the metric d Y. Exercise 2.5. Prove Proposition 2.4. Remrk 2.6. From Propositions 2.4 nd 1.2, the function f cn converge to t most one limit L s x converges to x 0. Remrk 2.7. The nottion lim x x0 ;x E f(x) implicitly refers to convergence of the function vlues f(x) in the metric spce (Y, d Y ). Strictly speking, it would be better to write d Y somewhere next to the nottion lim x x0 ;x E f(x). However, this omission of nottion should not cuse confusion. 2.2. Pointwise Convergence nd Uniform Convergence. Definition 2.8 (Pointwise Convergence). Let (X, d X ) nd (Y, d Y ) be metric spces. Let (f j ) be sequence of functions from X to Y. Let f : X Y be nother function. We sy tht (f j ) converges pointwise to f on X if nd only if, for every x X, we hve lim f j(x) = f(x). j ( ) 2

Tht is, for ll x X, we hve lim d Y (f j (x), f(x)) = 0. j Tht is, for every x X nd for every ε > 0, there exists J > 0 such tht, for ll j > J, we hve d Y (f j (x), f(x)) < ε. Remrk 2.9. Note tht, if we chnge the point x, then the limiting behvior of f j (x) cn chnge quite bit. For exmple, let j be positive integer, nd consider the functions f j : [0, 1] R where f j (x) = j for ll x (0, 1/j), nd f j (x) = 0 otherwise. Let f : [0, 1] R be the zero function. Then f j converges pointwise to zero, since for ny x (0, 1], we hve f j (x) = 0 for ll j > 1/x. (And f j (0) = 0 for ll positive integers j.) However, given ny fixed positive integer j, there exists n x such tht f j (x) = j. Moreover, 1 f 0 j = 1 for ll positive integers j, but 1 f = 0. So, we see tht pointwise convergence does not preserve 0 the integrl of function. Remrk 2.10. Pointwise convergence lso does not preserve continuity. For exmple, consider f j : [0, 1] R defined by f j (x) = x j, where j N nd x [0, 1]. Define f : [0, 1] R so tht f(1) = 1 nd so tht f(x) = 0 for x [0, 1). Then f j converges pointwise to f s j, nd ech f j is continuous, but f is not continuous. In summry, pointwise convergence doesn t relly preserve ny useful nlytic quntities. The bove remrks show tht some points re chnging t much different rtes thn other points s j. A stronger notion of convergence will then fix these issues, where ll points in the domin re controlled simultneously. Definition 2.11 (Uniform Convergence). Let (X, d X ) nd (Y, d Y ) be metric spces. Let (f j ) be sequence of functions from X to Y. Let f : X Y be nother function. We sy tht (f j ) converges uniformly to f on X if nd only if, for every ε > 0, there exists J > 0 such tht, for ll j > J nd for ll x X we hve d Y (f j (x), f(x)) < ε. Remrk 2.12. Note tht the difference between uniform nd pointwise convergence is tht we simply moved the quntifier for ll x X within the sttement. This chnge mens tht the integer J does not depend on x in the cse of uniform convergence. Remrk 2.13. The sequences of functions from Remrks 2.9 nd 2.10 do not converge uniformly. So, pointwise convergence does not imply uniform convergence. However, uniform convergence does imply pointwise convergence. 3. Uniform Convergence nd Continuity We sw tht pointwise convergence does not preserve continuity. However, uniform convergence does preserve continuity. Theorem 3.1. Let (X, d X ) nd (Y, d Y ) be metric spces. Let (f j ) be sequence of functions from X to Y. Let f : X Y be nother function. Let x 0 X. Suppose f j converges uniformly to f on X. Suppose tht, for ech j 1, we know tht f j is continuous t x 0. Then f is lso continuous t x 0. 3

Exercise 3.2. Prove Theorem 3.1. Hint: it is probbly esiest to use the ε δ definition of continuity. Once you do this, you my require the tringle inequlity in the form d Y (f(x), f(x 0 )) d Y (f(x), f j (x)) + d Y (f j (x), f j (x 0 )) + d Y (f j (x 0 ), f(x 0 )). Corollry 3.3. Let (X, d X ) nd (Y, d Y ) be metric spces. Let (f j ) be sequence of functions from X to Y. Let f : X Y be nother function. Suppose (f j ) converges uniformly to f on X. Suppose tht, for ech j 1, we know tht f j is continuous on X. Then f is lso continuous on X. Uniform limits of bounded functions re lso bounded. Recll tht function f : X Y between metric spces (X, d X ) nd (Y, d Y ) is bounded if nd only if there exists rdius R > 0 nd point y 0 Y such tht f(x) B (Y,dY )(y 0, R) for ll x X. Proposition 3.4. Let (X, d X ) nd (Y, d Y ) be metric spces. Let (f j ) be sequence of functions from X to Y. Let f : X Y be nother function. Suppose (f j ) converges uniformly to f on X. Suppose lso tht, for ech j 1, we know tht f j is bounded. Then f is lso bounded. Exercise 3.5. Prove Proposition 3.4. 3.1. The Metric of Uniform Convergence. We will now see one dvntge to our bstrct pproch to nlysis on metric spces. We cn in fct tlk bout uniform convergence in terms of metric on spce of functions, s follows. Definition 3.6. Let (X, d X ) nd (Y, d Y ) be metric spces. Let B(X; Y ) denote the set of functions f : X Y tht re bounded. Let f, g B(X; Y ). We define the metric d : B(X; Y ) B(X; Y ) [0, ) by d (f, g) := sup d Y (f(x), g(x)). x X This metric is known s the sup norm metric or the L metric. We lso use d B(X;Y ) s synonym for d. Note tht d (f, g) < since f, g re ssumed to be bounded. Exercise 3.7. Show tht the spce (B(X; Y ), d ) is metric spce. Exmple 3.8. Let X = [0, 1] nd let Y = R. Consider the functions f(x) = x nd g(x) = 2x where x [0, 1]. Then f, g re bounded, nd d (f, g) = sup x 2x = sup x = 1. x [0,1] x [0,1] Here is our promised chrcteriztion of uniform convergence in terms of the metric d. Proposition 3.9. Let (X, d X ) nd (Y, d Y ) be metric spces. Let (f j ) be sequence of functions in B(X; Y ). Let f B(X; Y ). Then (f j ) converges uniformly to f on X if nd only if (f j ) converges to f in the metric d B(X;Y ). Exercise 3.10. Prove Proposition 3.9. Definition 3.11. Let (X, d X ) nd (Y, d Y ) be metric spces. continuous functions from X to Y s C(X; Y ) := {f B(X; Y ): f is continuous}. Define the set of bounded 4

Note tht C(X; Y ) B(X; Y ) by the definition of C(X; Y ). Also, by Corollry 3.3, C(X; Y ) is closed in B(X; Y ) with respect to the metric d. In fct, more is true. Theorem 3.12. Let (X, d X ) be metric spce, nd let (Y, d Y ) be complete metric spce. Then the spce (C(X; Y ), d B(X;Y ) C(X;Y ) C(X;Y ) ) is complete subspce of B(X; Y ). Tht is, every Cuchy sequence of functions in C(X; Y ) converges to function in C(X; Y ). Exercise 3.13. Prove Theorem 3.12 4. Series of Functions nd the Weierstrss M-test For ech positive integer j, let f j : X R be function. We will now consider infinite series of the form f j. The most nturl thing to do now is to determine in wht sense the series f j is function, nd if it is function, determine if it is continuous. Note tht we hve restricted the rnge to be R since it does not mke sense to dd elements in generl metric spce. Power series nd Fourier series perhps give the most studied exmples of series of functions. If x [0, 1] nd if j re rel numbers for ll j 1, we wnt to mke sense of the series j cos(2πjx). We wnt to know in wht sense this infinite series is function, nd if it is function, do the prtil sums converge in ny resonble mnner? We will return to these issues lter on. Definition 4.1. Let (X, d X ) be metric spce. For ech positive integer j, let f j : X R be function, nd let f : X R be nother function. If the prtil sums J f j converge pointwise to f s J, then we sy tht the infinite series f j converge pointwise to f, nd we write f = f j. If the prtil sums J f j converge uniformly to f s J, then we sy tht the infinite series f j converge uniformly to f, nd we write f = f j. (In prticulr, the nottion f = f j is mbiguous, since the nture of the convergence of the series is not specified.) Remrk 4.2. If series converges uniformly then it converges pointwise. converse is flse in generl. However, the Exercise 4.3. Let x ( 1, 1). For ech integer j 1, define f j (x) := x j. Show tht the series f j converges pointwise, but not uniformly, on ( 1, 1) to the function f(x) = x/(1 x). Also, for ny 0 < t < 1, show tht the series f j converges uniformly to f on [ t, t]. Definition 4.4. Let f : X R be bounded rel-vlued function. We define the sup-norm f of f to be the rel number f := sup f(x). x X Exercise 4.5. Let X be set. Show tht is norm on the spce B(X; R). Theorem 4.6 (Weierstrss M-test). Let (X, d) be metric spce nd let (f j ) be sequence of bounded rel-vlued continuous functions on X such tht the series (of rel numbers) f j is bsolutely convergent. Then the series f j converges uniformly to some continuous function f : X R. 5

Exercise 4.7. Prove Theorem 4.6. (Hint: first, show tht the prtil sums J f j form Cuchy sequence in C(X; R). Then, use Theorem 3.12 nd the completeness of the rel line R.) Remrk 4.8. The Weierstrss M-test will be useful in our investigtion of power series. 5. Uniform Convergence nd Integrtion Theorem 5.1. Let < b be rel numbers. For ech integer j 1, let f j : [, b] R be Riemnn integrble function on [, b]. Suppose f j converges uniformly on [, b] to function f : [, b] R, s j. Then f is lso Riemnn integrble, nd lim j f j = Remrk 5.2. Before we begin, recll tht we require ny Riemnn integrble function g to be bounded. Also, for Riemnn integrble function g, we denote g s the supremum of ll lower Riemnn sums of g over ll prtitions of [, b]. And we denote g s the infimum of ll upper Riemnn sums of g over ll prtitions of [, b]. Recll lso tht function g is defined to be Riemnn integrble if nd only if g = g. Proof. We first show tht f is Riemnn integrble. First, note tht f j is bounded for ll j 1, since tht is prt of the definition of being Riemnn integrble. So, f is bounded by Proposition 3.4. Now, let ε > 0. Since f j converges uniformly to f on [, b], there exists J > 0 such tht, for ll j > J, we hve f j (x) ε f(x) f j (x) + ε, Integrting this inequlity on [, b], we hve (f j (x) ε) f f Since f j is Riemnn integrble for ll j 1, we therefore hve f. x [, b]. (f j (x) + ε). (b )ε + In prticulr, we get f j f f (b )ε + 0 f f 2(b )ε. Since ε > 0 is rbitrry, we conclude tht f = f, so f is Riemnn integrble. Now, from ( ), we hve: for ny ε > 0, there exists J such tht, for ll j > J, we hve f f j (b )ε. Since this holds for ny ε > 0, we conclude tht lim j f j = f, s desired. f j. ( ) 6

Remrk 5.3. In summry, if sequence of Riemnn integrble functions (f j ) converges to f uniformly, then we cn interchnge limits nd integrls lim f j = lim f j. j j Recll tht this equlity does not hold if we only ssume tht the functions converge pointwise. An nlogous sttement holds for series. Theorem 5.4. Let < b be rel numbers. For ech integer j 1, let f j : [, b] R be Riemnn integrble function on [, b]. Suppose f j converges uniformly on [, b]. Then f j is lso Riemnn integrble, nd Exercise 5.5. Prove Theorem 5.4. f j = f j. Exmple 5.6. Let x ( 1, 1). We know tht xj = x/(1 x), nd the convergence is uniform on [ r, r] for ny 0 < r < 1. Adding 1 to both sides, we get x j = 1 1 x. And this sum lso converges uniformly on [ r, r] for ny 0 < r < 1. Applying Theorem 5.4 nd integrting on [0, r], we get r r r j+1 j + 1 = 0 x j = 0 1 1 x. The lst function is equl to log(1 r), though we techniclly hve not defined the logrithm function yet. We will define the logrithm further below. 6. Uniform Convergence nd Differentition We now investigte the reltion between uniform convergence nd differentition. Remrk 6.1. Suppose sequence of differentible functions (f j ) converges uniformly to function f. We first show tht f need not be differentible. Consider the functions f j (x) := x 2 + 1/j, where x [ 1, 1]. Let f(x) = x. Note tht x x 2 + 1/j x + 1/ j. These inequlities follow by tking the squre root of x 2 x 2 + 1/j x 2 + 1/j + 2 x / j. So, by the Squeeze Theorem, (f j ) converges uniformly to f on [ 1, 1]. However, f is not differentible t 0. In conclusion, uniform convergence does not preserve differentibility. Remrk 6.2. Suppose sequence of differentible functions (f j ) converge uniformly to function f. Even if f is ssumed to be differentible, we show tht (f j ) my not converge to f. Consider the functions f j (x) := j 1/2 sin(jπx), where x [ 1, 1]. (We will ssume some bsic properties of trigonometric functions which we will prove lter on. Since we re 7

only providing motivting exmple, we will not introduce ny circulr resoning.) Let f be the zero function. Since sin(jπx) 1, we hve d (f j, f) j 1/2, so (f j ) converges uniformly on [ 1, 1]. However, f j(x) = j 1/2 π cos(jπx). So, f j(0) = j 1/2 π. Tht is, (f j) does not converge pointwise to f. So, (f j) does not converge uniformly to f = 0. In conclusion, uniform convergence does not imply uniform convergence of derivtives. However, the converse sttement is true, s long s the sequence of functions converges t one point. Theorem 6.3. Let < b. For every integer j 1, let f j : [, b] R be differentible function whose derivtive (f j ) : [, b] R is continuous. Assume tht the derivtives (f j ) converge uniformly to function g : [, b] R s j. Assume lso tht there exists point x 0 [, b] such tht lim j f j (x 0 ) exists. Then the functions f j converge uniformly to differentible function f s j, nd f = g. Proof. Let x [, b]. From the Fundmentl Theorem of Clculus, for ech j 1, f j (x) f j (x 0 ) = x x 0 f j. By ssumption, L := lim j f j (x 0 ) exists. From Theorem 3.1, g is continuous, nd in x prticulr, g is Riemnn integrble on [, b]. Also, by Theorem 5.1, lim j x 0 f j exists nd is equl to x x 0 g. We conclude by ( ) tht lim j f j (x) exists, nd Define the function f on [, b] so tht lim f j(x) = L + j f(x) = L + We know so fr tht (f j ) converges pointwise to f. We now need to show tht this convergence is in fct uniform. We defer this prt to the exercises. x Exercise 6.4. Complete the proof of Theorem 6.3. Corollry 6.5. Let < b. For every integer j 1, let f j : [, b] R be differentible function whose derivtive f j : [, b] R is continuous. Assume tht the series of rel numbers f j is bsolutely convergent. Assume lso tht there exists x 0 [, b] such tht the series of rel numbers f j(x 0 ) converges. Then the series f j converges uniformly on [, b] to differentible function. Moreover, for ll x [, b], d dx Exercise 6.6. Prove Corollry 6.5. f j (x) = x x 0 g. x 0 g. d dx f j(x) The following exercise is nice counterexmple to keep in mind, nd it lso shows the necessity of the ssumptions of Corollry 6.5. ( ) 8

Exercise 6.7. (For this exercise, you cn freely use fcts bout trigonometry tht you lerned in your previous courses.) Let x R nd let f : R R be the function f(x) := 4 j cos(32 j πx). Note tht this series is uniformly convergent by the Weierstrss M-test (Theorem 4.6). So, f is continuous function. However, t every point x R, f is not differentible, s we now discuss. Show tht, for ll positive integers j, m, we hve f((j + 1)/32 m ) f(j/32 m ) 4 m. (Hint: for certin sequences of numbers ( j ), use the identity m 1 j = ( j ) + m + j=m+1 Also, use the fct tht the cosine function is periodic with period 2π, nd the summtion rj = 1/(1 r) for ll 1 < r < 1. Finlly, you should require the inequlity: for ll rel numbers x, y, we hve cos(x) cos(y) x y. This inequlity follows from the Men Vlue Theorem or the Fundmentl Theorem of Clculus.) Using the previous result, show tht, for ever x R, f is not differentible t x. (Hint: for every x R nd for every positive integer m, there exists n integer j such tht j 32 m x j + 1.) Explin briefly why this result does not contrdict Corollry 6.5. j. 7. Uniform Approximtion by Polynomils Definition 7.1 (Polynomil). Let < b be rel numbers nd let x [, b]. A polynomil on [, b] is function f : [, b] R of the form f(x) = k jx j, where k is nturl number nd 0,..., k re rel numbers. If k 0, then k is clled the degree of f. From the previous exercise, we hve seen tht generl continuous functions cn behve rther poorly, in tht they my never be differentible. Polynomils on the other hnd re infinitely differentible. And it is often beneficil to del with polynomils insted of generl functions. So, we mention below result of Weierstrss which sys: ny continuous function on n intervl [, b] cn be uniformly pproximted by polynomils. This fct seems to be relted to power series, but it is something much different. It my seem possible to tke generl (infinitely differentible) function, tke high degree Tylor polynomil of this function, nd then clim tht this polynomil pproximtes our originl function well. There re two problems with this pproch. First of ll, the continuous function tht we strt with my not even be differentible. Second of ll, even if we hve n infinitely differentible function, its power series my not ctully pproximte tht function well. Recll tht the function f(x) = e 1/x2 (where f(0) := 0) is infinitely differentible, but its Tylor polynomil is identiclly zero t x = 0. In conclusion, we need to use something other thn Tylor series to pproximte generl continuous function by polynomils. The proof of the Weierstrss pproximtion theorem introduces severl useful ides, but it is typiclly only proven in the honors clss. However, lter on, we will prove version of this theorem for trigonometric polynomils, nd this proof will be nlogous to the proof of the current theorem. 9

Theorem 7.2 (Weierstrss pproximtion). Let < b be rel numbers. Let f : [, b] R be continuous function, nd let ε > 0. Then there exists polynomil P on [, b] such tht d (P, f) < ε. (Tht is, f(x) P (x) < ε for ll x [, b].) Remrk 7.3. We cn lso stte this Theorem using metric spce terminology. Recll tht C([, b]; R) is the spce of continuous functions from [, b] to R, equipped with the sup-norm metric d. Let P ([, b]; R) be the spce of ll polynomils on [, b], so tht P ([, b]; R) is subspce of C([, b]; R), since polynomils re continuous. Then the Weierstrss pproximtion theorem sys tht every continuous function is n dherent point of P ([, b]; R). Put nother wy, the closure of P ([, b]; R) is C([, b]; R). P ([, b]; R) = C([, b]; R). Put nother wy, every continuous function on [, b] is the uniform limit of polynomils. 8. Power Series We now focus our discussion of series to power series. Definition 8.1 (Power Series). Let be rel number, let ( j ) be sequence of rel numbers, nd let x R. A forml power series centered t is series of the form j (x ) j, For nturl number j, we refer to j s the j th coefficient of the power series. Remrk 8.2. We refer to these power series s forml since their convergence is not gurnteed. Note however tht ny forml power series centered t converges t x =. It turns out tht we cn precisely identify where forml power series converges just from the symptotic behvior of the coefficients. Definition 8.3 (Rdius of Convergence). Let j(x ) j be forml power series. The rdius of convergence R 0 of this series is defined to be R := 1 lim sup j j 1/j. In the definition of R, we use the convention tht 1/0 = + nd 1/(+ ) = 0. Note tht it is possible for R to then tke ny vlue between nd including 0 nd +. Note lso tht R lwys exists s nonnegtive rel number, or s +, since the limit superior of positive sequence lwys exists s nonegtive number, or +. Exmple 8.4. The rdius of convergence of the series j( 2)j (x 3) j is 1 lim sup j j( 2) j = 1 1/j lim sup j 2j = 1 1/j 2. The rdius of convergence of the series 2j2 (x + 2) j is 1 lim sup j 2 = 1 j + = 0. 10

The rdius of convergence of the series 2 j2 (x + 2) j is 1 lim sup j 2 j = 1 0 = +. As we now show, the rdius of convergence tells us exctly where the power series converges. Theorem 8.5. Let j(x ) j be forml power series, nd let R be its rdius of convergence. () (Divergence outside of the rdius of convergence) If x R stisfies x > R, then the series j(x ) j is divergent t x. (b) (Convergence inside the rdius of convergence) If x R stisfies x < R, then the series j(x ) j is convergent t x. For the following items (c),(d) nd (e), we ssume tht R > 0. Then, let f : ( R, + R) be the function f(x) = j(x ) j, which exists by prt (b). (c) (Uniform convergence on compct intervls) For ny 0 < r < R, we know tht the series j(x ) j converges uniformly to f on [ r, + r]. In prticulr, f is continuous on ( R, + R) (by Theorem 3.1.) (d) (Differentition of power series) The function f is differentible on ( R, + R). For ny 0 < r < R, the series j j(x ) j 1 converges uniformly to f on the intervl [ r, + r]. (e) (Integrtion of power series) For ny closed intervl [y, z] contined in ( R, +R), we hve z (z ) j+1 (y ) j+1 f = j. j + 1 y Exercise 8.6. Prove Theorem 8.5. (Hints: for prts (),(b), use the root test. For prt (c), use the Weierstrss M-test. For prt (d), use Theorem 6.3. For prt (e), use Theorem 5.4.) Remrk 8.7. A power series my converge or diverge when x = R. Exercise 8.8. Give exmples of forml power series centered t 0 with rdius of convergence R = 1 such tht The series diverges t x = 1 nd t x = 1. The series diverges t x = 1 nd converges t x = 1. The series converges t x = 1 nd diverges t x = 1. The series converges t x = 1 nd t x = 1. We now discuss functions tht re equl to convergent power series. Definition 8.9. Let R nd let r > 0. Let E be subset of R such tht ( r, +r) E. Let f : E R. We sy tht the function f is rel nlytic on ( r, + r) if nd only if there exists power series j(x ) j centered t with rdius of convergence R such tht R r nd such tht this power series converges to f on ( r, + r). Exmple 8.10. The function f : (0, 2) R defined by f(x) = j(x 1)j is rel nlytic on (0, 2). 11

From Theorem 8.5, if function f is rel nlytic on ( r, + r), then f is continuous nd differentible. In fct, f is cn be differentited ny number of times, s we now show. Definition 8.11. Let E be subset of R. We sy tht function f : E R is once differentible on E if nd only if f is differentible on E. More generlly, for ny integer k 2, we sy tht f : E R is k times differentible on E, or just k times differentible, if nd only if f is differentible nd f is k 1 times differentible. If f is k times differentible, we define the k th derivtive f (k) : E R by the recursive rule f (1) := f nd f (k) := (f (k 1) ), for ll k 2. We lso define f (0) := f. A function is sid to be infinitely differentible if nd only if f is k times differentible for every k 0. Exmple 8.12. The function f(x) = x 3 is twice differentible on R, but not three times differentible on R. Note tht f (x) = 6 x, which is not differentible t x = 0. Proposition 8.13. Let R nd let r > 0. Let f be function tht is rel nlytic on ( r, + r), with the power series expnsion f(x) = j (x ) j, x ( r, + r). Then, for ny integer k 0, the function f is k times differentible on ( r, + r), nd the k th derivtive is given by f (k) (x) = j+k (j + 1)(j + 2) (j + k)(x ) j, x ( r, + r). Exercise 8.14. Prove Proposition 8.13. Corollry 8.15 (Tylor s formul). Let R nd let r > 0. Let f be function tht is rel nlytic on ( r, + r), with the power series expnsion f(x) = j (x ) j, x ( r, + r). Then, for ny integer k 0, we hve f (k) () = k! k, where k! = 1 2 k, nd we denote 0! := 1. In prticulr, we hve Tylor s formul f (j) () f(x) = (x ) j, x ( r, + r). j! Exercise 8.16. Prove Corollry 8.15 using Proposition 8.13. Remrk 8.17. The series f (j) () (x ) j is sometimes clled the Tylor series of f j! round. Tylor s formul sys tht if f is rel nlytic, then f is equl to its Tylor series. In the following exercise, we see tht even if f is infinitely differentible, it my not be equl to its Tylor series. 12

Exercise 8.18. Define function f : R R by f(0) := 0 nd f(x) := e 1/x2 for x 0. Show tht f is infinitely differentible, but f (k) (0) = 0 for ll k 0. So, being infinitely differentible does not imply tht f is equl to its Tylor series. (You my freely use properties of the exponentil function tht you hve lerned before.) Corollry 8.19 (Uniqueness of power series). Let R nd let r > 0. Let f be function tht is rel nlytic on ( r, + r), with two power series expnsions f(x) = j (x ) j, x ( r, + r). Then j = b j for ll j 0. f(x) = b j (x ) j, x ( r, + r). Proof. By Corollry 8.15, we hve k! k = f (k) () = k!b k for ll k 0. Since k! 0 for ll k 0, we divide by k! to get k = b k for ll k 0. Remrk 8.20. Note however tht power series cn hve very different expnsions if we chnge the center of the expnsion. For exmple, the function f(x) = 1/(1 x) stisfies f(x) = x j, x ( 1, 1). However, t the point 1/2, we hve the different expnsion f(x) = 1 1 x = 2 1 2(x 1/2) = 2(2(x 1/2)) j = 2 j+1 (x 1/2) j, x (0, 1). Note lso tht the first series hs rdius of convergence 1 nd the second series hs rdius of convergence 1/2. 8.1. Multipliction of Power Series. Lemm 8.21 (Fubini s Theorem for Series). Let f : N N R be function such tht (j,k) N N f(j, k) is bsolutely convergent. (Tht is, for ny bijection g : N N N, the sum l=0 f(g(l)) is bsolutely convergent.) Then ( f(j, k)) = f(j, k) = ( f(j, k)). k=1 (j,k) N N Proof Sketch. We only consider the cse f(j, k) 0 for ll (j, k) N. The generl cse then follows by writing f = mx(f, 0) min(f, 0), nd pplying this specil cse to mx(f, 0) nd min(f, 0), seprtely. Let L := (j,k) N N f(j, k). For ny J, K > 0, we hve J K k=1 J, K, we conclude tht k=1 such tht J K k=1 > L ε. Since (j,k) N N finite set X N N such tht (j,k) X k=1 f(j, k) L. Letting f(j, k) L. Let ε > 0. It remins to find J, K f(j, k) converges bsolutely, there exists f(j, k) > L ε. But then we cn choose J, K sufficiently lrge such tht {(j, k) X} {(j, k): 1 j J, 1 k K}. Therefore, J K k=1 f(j, k) (j,k) X f(j, k) > L ε, s desired. 13

Theorem 8.22. Let R nd let r > 0. Let f nd g be functions tht re rel nlytic on ( r, + r), with power series expnsions f(x) = j (x ) j, x ( r, + r). g(x) = b j (x ) j, x ( r, + r). Then the function fg is lso rel nlytic on ( r, + r). For ech j 0, define c j := j k=0 kb j k. Then fg hs the power series expnsion f(x)g(x) = c j (x ) j, x ( r, + r). Proof. Fix x ( r, + r). By Theorem 8.5, both f nd g hve rdius of convergence R r. So, both j(x ) j nd b j(x ) j re bsolutely convergent. Define C := j (x ) j, D := bj (x ) j. Then both C, D re finite. For ny N 0, consider the prtil sum N N j (x ) j b k (x ) k. k=0 We cn re-write this sum s N j (x ) j N bk (x ) k N j (x ) j D CD. Since this inequlity holds for ll N 0, the series j (x ) j b k (x ) k k=0 (j,k) N N is convergent. Tht is, the following series is bsolutely convergent. j (x ) j b k (x ) k. Now, using Lemm 8.21, (j,k) N N j (x ) j b k (x ) k = Rewriting this equlity, f(x)g(x) = (j,k) N N (j,k) N N j (x ) j k=0 b k (x ) k = j (x ) j b k (x ) k = j (x ) j g(x) = f(x)g(x). (j,k) N N j b k (x ) j+k. 14

Since the sum is bsolutely convergent, we cn rerrnge the order of summtion. For ny fixed positive integer l, we sum over ll positive integers j, k such tht j + k = l. Tht is, we hve l f(x)g(x) = j b k (x ) l = (x ) l s b s l. l=0 (j,k) N N: j+k=l l=0 9. The Exponentil nd Logrithm We cn now use the mteril from the previous sections to define nd investigte vrious specil functions. Definition 9.1. For every rel number x, we define the exponentil function exp(x) to be the rel number x j exp(x) := j!. Theorem 9.2 (Properties of the Exponentil Function). () For every rel number x, the series x j exists nd is rel number for every x R, the power series convergence R = +, nd exp is n nlytic function on (, + ). s=0 is bsolutely convergent. So, exp(x) j! x j hs rdius of j! (b) exp is differentible on R, nd for every x R, we hve exp (x) = exp(x). (c) exp is continuous on R, nd for ll rel numbers < b, we hve exp = exp(b) exp(). (d) For every x, y R, we hve exp(x + y) = exp(x) exp(y). (e) exp(0) = 1. Also, for every x R, we hve exp(x) > 0, nd exp( x) = 1/ exp(x). (f) exp is strictly monotone incresing. Tht is, whenever x, y re rel numbers with x < y, we hve exp(x) < exp(y). Exercise 9.3. Prove Theorem 9.2. (Hints: for prt (), use the rtio test. For prts (b) nd (c), use Theorem 8.5. For prt (d), you my need the binomil formul (x + y) k = k k! j!(k j)! xj y k j. For prt (e), use prt (d). For prt (f), use prt (d) nd show tht exp(x) > 1 for ll x > 0.) Definition 9.4. We define the rel number e by e := exp(1) = Proposition 9.5. For every rel number x, we hve exp(x) = e x. Exercise 9.6. Prove Proposition 9.5. (Hint: first prove the proposition for nturl numbers x. Then, prove the proposition for integers. Then, prove the proposition for rtionl numbers. Finlly, use the density of the rtionls to prove the proposition for rel numbers. You should find useful identifies for exponentition by rtionl numbers.) 1 j! 15

From now on, we use exp(x) nd e x interchngebly. Remrk 9.7. Since e > 1 by the definition of e, we hve e x + s x + nd e x 0 s x. So, from the Intermedite Vlue Theorem, the rnge of exp is (0, ). Since exp is strictly incresing on R, exp is therefore injective on R, so exp is bijection from R to (0, ). Therefore, exp hs n inverse function from (0, ) to R. Definition 9.8. We define the nturl logrithm function log: (0, ) R (which is lso clled ln) to be the inverse of the exponentil function. So, exp(log(x)) = x for every x (0, ), nd log(exp(x)) = x for every x R. Remrk 9.9. Since exp is continuous nd strictly monotone incresing, log is lso continuous nd strictly monotone incresing by Proposition 1.3. Since exp is differentible nd its derivtive is never zero, the Inverse Function Theorem (Theorem 1.4) implies tht log is lso differentible. Theorem 9.10. () For every x (0, ), we hve log (x) = 1/x. So, by the Fundmentl Theorem of Clculus, for ny 0 < < b, we hve (1/t)dt = log(b) log(). (b) For ll x, y (0, ), we hve log(x) + log(y) = log(xy). (c) For ll x (0, ), we hve log(1/x) = log x. In prticulr, log(1) = 0. (d) For ny x (0, ) nd y R, we hve log(x y ) = y log x. (e) For ny x ( 1, 1), we hve x j log(1 x) = j. In prticulr, log is nlytic on (0, 2) with the power series expnsion ( 1) j+1 log(x) = (x 1) j, x (0, 2). j Exercise 9.11. Prove Theorem 9.10. (Hints: for prt (), use the Inverse Function Theorem or Chin Rule. For prts (b),(c) nd (d), use Theorem 9.2 nd the lws of exponentition. For prt (e), let x ( 1, 1), use the geometric series formul 1/(1 x) = xj nd integrte using Theorem 8.5.) 9.1. A Digression concerning Complex Numbers. Our investigtion of trigonometric functions below is significntly improved by the introduction of the complex number system. We will lso use the complex exponentil in our discussion of Fourier series. Definition 9.12 (Complex Numbers). A complex number is ny expression of the form + bi where, b re rel numbers. Formlly, the symbol i is plceholder with no intrinsic mening. Two complex numbers + bi nd c + di re sid to be equl if nd only if = c nd b = d. Every rel number x is considered complex number, with the identifiction x = x+0i. The sum of two complex numbers is defined by (+bi)+(c+di) := (+c)+(b+d)i. The difference of two complex numbers is defined by (+bi) (c+di) := ( c)+(b d)i. The product of two complex numbers is defined by (+bi)(c+di) := (c bd)+(d+bc)i. If c+di c 0, the quotient of two complex numbers is defined by (+bi)/(c+di) := (+bi)( d i). c 2 +d 2 c 2 +d 2 16

The complex conjugte of complex number + bi is defined by + bi := bi. The bsolute vlue of complex number + bi is defined by + bi := 2 + b 2. The spce of ll complex numbers is clled C. Remrk 9.13. We write i s shorthnd for 0 + i. Note tht i 2 = 1. Remrk 9.14. The complex numbers obey ll of the usul rules of lgebr. For exmple, if v, w, z re complex numbers, then v(w+z) = vw+vz, v(wz) = (vw)z, nd so on. Specificlly, the complex numbers C form field. Also, the rules of complex rithmetic re consistent with the rules of rel rithmetic. Tht is, 3 + 5 = 8 whether or not we use ddition in R or ddition in C. The opertion of complex conjugtion preserves ll of the rithmetic opertions. If w, z re complex numbers, then w + z = w + z, w z = w z, w z = w z, nd w/z = w/z for z 0. The complex conjugte nd bsolute vlue stisfy z 2 = zz. Remrk 9.15. If z C, then z = 0 if nd only if z = 0. If z, w C, then it cn be shown tht zw = z w, nd if w 0, then z/w = z / w. Also, the tringle inequlity holds: z + w z + w. So, C is metric spce if we use the metric d(z, w) := z w. Moreover, C is complete metric spce. The theory we hve developed to del with series of rel functions lso covers complexvlued functions, with lmost no chnge to the proofs. For exmple, we cn define the exponentil function of complex number z by z j exp(z) := j!. The rtio test then cn be proven in exctly the sme mnner for complex series, nd it follows tht exp(z) converges for every z C. Mny of the properties of Theorem 9.2 still hold, though we cnnot del with ll of these properties in this clss. However, the following identity is proven in the exct sme wy s in the setting of rel numbers: for ny z, w C, we hve exp(z + w) = exp(z) exp(w). Also, we should note tht exp(z) = exp(z), which follows by conjugting the prtil sums J zj /j!, nd then letting J. We briefly mention tht the complex logrithm is more difficult to define, minly becuse the exponentil function is not invertible on C. This topic is deferred to the complex nlysis clss. 10. Trigonometric Functions Besides the exponentil nd logrithmic functions, there re mny different kinds of specil functions. Here, we will only mention the sine nd cosine functions. One s first encounter with the sine nd cosine functions probbly involved their definition in terms of the edge lengths of right tringles. However, we will show below n nlytic definition of these functions, which will lso fcilitte the investigtion of the properties tht they possess. The complex exponentil plys crucil role in this development. 17

Definition 10.1. Let x be rel number. We then define cos(x) := eix + e ix. 2 sin(x) := eix e ix. 2i We refer to cos s the cosine function, nd we refer to sin s the sine function. Remrk 10.2. Using the power series expnsion for the exponentil, we cn then derive power series expnsions for sine nd cosine s follows. Let x R. Then e ix = 1 + ix x 2 /2! ix 3 /3! + x 4 /4! + e ix = 1 ix x 2 /2! + ix 3 /3! + x 4 /4! Therefore, using the definitions of sine nd cosine, cos(x) = 1 x 2 /2! + x 4 ( 1) j x 2j /4! =. (2j)! sin(x) = x x 3 /3! + x 5 /5! = ( 1) j x 2j+1. (2j + 1)! So, if x R then cos(x) R nd sin(x) R. Also, sine nd cosine rel nlytic on (, ), e.g. since their power series converge on (, ) by the rtio test. In prticulr, the sine nd cosine functions re continuous nd infinitely differentible. Theorem 10.3 (Properties of Sine nd Cosine). () For ny rel number x we hve cos(x) 2 + sin(x) 2 = 1. In prticulr, sin(x) [ 1, 1] nd cos(x) [ 1, 1] for ll rel numbers x. (b) For ny rel number x, we hve sin (x) = cos(x), nd cos (x) = sin(x). (c) For ny rel number x, we hve sin( x) = sin(x) nd cos( x) = cos(x). (d) For ny rel numbers x, y we hve cos(x + y) = cos(x) cos(y) sin(x) sin(y) nd sin(x + y) = sin(x) cos(y) + cos(x) sin(y). (e) sin(0) = 0 nd cos(0) = 1. (f) For every rel number x, we hve e ix = cos(x) + i sin(x) nd e ix = cos(x) i sin(x). Exercise 10.4. Prove Theorem 10.3. (Hints: whenever possible, write everything in terms of exponentils.) Lemm 10.5. There exists positive rel number x such tht sin(x) = 0. Proof. We rgue by contrdiction. Suppose sin(x) 0 for ll x > 0. We conclude tht cos(x) 0 for ll x > 0, since cos(x) = 0 implies tht sin(2x) = 0, by Theorem 10.3(d). Since cos(0) = 1, we conclude tht cos(x) > 0 for ll x > 0 by the Intermedite Vlue Theorem. Since sin(0) = 0 nd sin (0) = 1 > 0, we know tht sin is positive for smll positive x. Therefore, sin(x) > 0 for ll x > 0 by the Intermedite Vlue Theorem. Define cot(x) := cos(x)/ sin(x). Then cot is positive on (0, ), nd cot is differentible for x > 0. From the quotient rule nd Theorem 10.3(), we hve cot (x) = 1/ sin 2 (x). So, cot (x) 1 for ll x > 0. Then, by the Fundmentl Theorem of Clculus, for ll x, s > 0, we hve cot(x + s) cot(x) s. Letting s shows tht cot eventully becomes negtive on (0, ), contrdiction. 18

Let E be the set E := {x (0, ): sin(x) = 0}, so tht E is the set of zeros of the sine function. By Lemm 10.5, E is nonempty. Also, since sin is continuous, E is closed set. (Note tht E = sin 1 (0).) In prticulr, E contins ll of its dherent points, so E contins inf(e). Definition 10.6. We define π to be the number π := inf{x (0, ): sin(x) = 0}. Then π > 0 nd sin(π) = 0. Since sin is nonzero on (0, π) nd sin (0) = 1 > 0, we conclude tht sin is positive on (0, π). Since cos (x) = sin(x), we see tht cos is decresing on (0, π). Since cos(0) = 1, we therefore hve cos(π) < 1. Since sin 2 (π) + cos 2 (π) = 1 nd sin(π) = 0, we conclude tht cos(π) = 1. We therefore deduce Euler s fmous formul e iπ = cos(π) + i sin(π) = 1. Here re some more properties of sine nd cosine. Theorem 10.7. () For ny rel x we hve cos(x+π) = cos(x) nd sin(x+π) = sin(x). In prticulr, we hve cos(x + 2π) = cos(x) nd sin(x + 2π) = sin(x), so tht sin nd cos re 2πperiodic. (b) If x is rel, then sin(x) = 0 if nd only if x/π is n integer. (c) If x is rel, then cos(x) = 0 if nd only if x/π is n integer plus 1/2. Exercise 10.8. Prove Theorem 10.7. 19

11. Appendix: Nottion Let A, B be sets in spce X. Let m, n be nonnegtive integers. Z := {..., 3, 2, 1, 0, 1, 2, 3,...}, the integers N := {0, 1, 2, 3, 4, 5,...}, the nturl numbers Z + := {1, 2, 3, 4,...}, the positive integers Q := {m/n: m, n Z, n 0}, the rtionls R denotes the set of rel numbers R = R { } {+ } denotes the set of extended rel numbers C := {x + y 1 : x, y R}, the complex numbers denotes the empty set, the set consisting of zero elements mens is n element of. For exmple, 2 Z is red s 2 is n element of Z. mens for ll mens there exists R n := {(x 1,..., x n ): x i R, i {1,..., n}} A B mens A, we hve B, so A is contined in B A B := {x A: x / B} A c := X A, the complement of A A B denotes the intersection of A nd B A B denotes the union of A nd B Let (X, d) be metric spce, let x 0 X, let r > 0 be rel number, nd let E be subset of X. Let (x 1,..., x n ) be n element of R n, nd let p 1 be rel number. B (X,d) (x 0, r) = B(x 0, r) := {x X : d(x, x 0 ) < r}. E denotes the closure of E int(e) denotes the interior of E E denotes the boundry of E n (x 1,..., x n ) lp := ( x i p ) 1/p i=1 (x 1,..., x n ) l := mx i=1,...,n x i 20

Let f, g : (X, d X ) (Y, d Y ) be mps between metric spces. Let V X, nd let W Y. f(v ) := {f(v) Y : v V }. f 1 (W ) := {x X : f(x) W }. d (f, g) := sup d Y (f(x), g(x)). x X B(X; Y ) denotes the set of functions f : X Y tht re bounded. C(X; Y ) := {f B(X; Y ): f is continuous}. Let X be set, nd let f : X C be complex-vlued function. f := sup f(x). x X 11.1. Set Theory. Let X, Y be sets, nd let f : X Y be function. The function f : X Y is sid to be injective (or one-to-one) if nd only if: for every x, x V, if f(x) = f(x ), then x = x. The function f : X Y is sid to be surjective (or onto) if nd only if: for every y Y, there exists x X such tht f(x) = y. The function f : X Y is sid to be bijective (or one-to-one correspondence) if nd only if: for every y Y, there exists exctly one x X such tht f(x) = y. A function f : X Y is bijective if nd only if it is both injective nd surjective. Two sets X, Y re sid to hve the sme crdinlity if nd only if there exists bijection from X onto Y. UCLA Deprtment of Mthemtics, Los Angeles, CA 90095-1555 E-mil ddress: heilmn@mth.ucl.edu 21