Sequences and Series of Functions

Chapter 6 Sequeces ad Series of Fuctios 6.1. Covergece of a Sequece of Fuctios Poitwise Covergece. Defiitio 6.1. Let, for each N, fuctio f : A R be defied. If, for each x A, the sequece (f (x)) coverges (to a limit f(x)); that is, lim f (x) = f(x) x A, the we say that (f ) coverges poitwise to the limit fuctio f o A. I this case, we write f (x) f(x) or f f poitwise o A, or lim f (x) = f(x) to emphasize the limit is for. The poitwise covergece meas that, give each x A, ɛ > 0, N N such that f (x) f(x) < ɛ N. Note that the umber N here depeds o both x ad ɛ. Example 6.1. (i) Let f (x) = x2 +x, x R. For ay give x R, f (x) = x + x2 x as. Hece f (x) f(x) = x poitwise o R. (ii) Let g (x) = x o [0, 1]. Note that g (1) = 1 for all N; so (g (1)) 1. If 0 x < 1, the (g (x)) = (x ) 0. Hece the poitwise limit fuctio of g (x) o [0, 1] is give by { 0 x [0, 1) g(x) = 1 x = 1. Note that although each g is cotiuous o [0, 1], the poitwise limit fuctio g is ot cotiuous at x = 1. (iii) Let h (x) = x 1+ 1 2 1 o [ 1, 1]. The (h (x)) x poitwise o [ 1, 1]. (iv) (Exercise 5(b)) Let f (x) = { 1 x 1 x 5 2 x x < 1. 1

2 6. Sequeces ad Series of Fuctios The f (x) 1 x (x 0) ad f (0) 0. The limit fuctio o [ 5, 5] is ubouded. (v) Let h (x) = cos(x). The (h (x)) coverges if ad oly if x = 2kπ for k = 0, ±1, ±2,. Proof: Let a = cos(x) a. The a 2 = cos(2x) = 2 cos 2 (x) 1 = 2a 2 1 2a 2 1 = a; so a 0. From the formula cos[( + 1)x] + cos[( 1)x] = 2 cos(x) cos x, it follows that a +1 + a 1 = 2a cos x ad hece 2a = 2a cos x. Sice a 0, we must have cos x = 1; that is, x = 2kπ with k = 0, ±1, ±2,. For all such x s, h (x) = 1 ad hece (h (x)) 1 oly for such x s. Uiform Covergece. Assume that f is cotiuous o a set A for each N ad (f (x)) f(x) poitwise o A. Whe ca we coclude that f is cotiuous o A or at some poit c A? Give c A, ad ɛ > 0, we write by the triagle iequality, f(x) f(c) f (x) f(x) + f (x) f (c) + f (c) f(c). The last term o the right-had side ca be made < ɛ/3 by choosig large eough by the covergece of (f (c)) f(c) ad, oce is fixed, by choosig x close to c from the cotiuity of f at c the secod term ca be made < ɛ/3 as well. The first term, however, is the most troublesome sice, whe is chose, the term f (x) f(x) may ot be small as x c. To deal with this term, we eed the followig defiitio. Defiitio 6.2. (Uiform Covergece) Let f : A R ad f : A R be give fuctios. We say that the sequece (f ) coverges uiformly o A to fuctio f if, for every ɛ > 0, there exists a N N such that wheever x A ad N it follows that f (x) f(x) < ɛ. Remark 6.3. 1. For poitwise covergece, give ɛ > 0, the umber N is to be foud after x A is give (so N depeds o x), while for the uiform covergece, the umber N is to be foud that works for every x A (so N is idepedet of x). 2. From the defiitios, if (f ) coverges uiformly to f o A the (f (x)) f(x) poitwise o A. Therefore, the uiform limit fuctio must be the poitwise limit fuctio. Example 6.2. Cosider f (x) = x2 +x ad f(x) = x o R. We kow (f (x)) f(x) poitwise o R. However, give ɛ > 0, ca we fid a N N such that f (x) f(x) = x2 < ɛ N, x R? If such a N existed, we would take x = N ad = N to obtai 1 < ɛ, a cotradictio if our ɛ is chose < 1. Therefore, the sequece (f ) does ot coverge uiformly to f o R. I geeral, by egatig the defiitio, (f ) does ot coverge uiformly o A to a fuctio f if ad oly if there exists a ɛ 0 > 0 ad sequeces ( k ) i N with k k ad (x k ) i A such that f k (x k ) f(x k ) ɛ 0. Example 6.3. We also cosider f (x) = x2 +x ad f(x) = x but o bouded iterval [ b, b]. Sice f (x) f(x) = x2 b2 holds for all x [ b, b], i order to make this quatity < ɛ for all such x, we ca choose N N such that N > b2 ɛ. The, for all N ad x [ b, b], it does follow that

6.1. Covergece of a Sequece of Fuctios 3 f (x) f(x) b2 < ɛ. Hece (f ) uiformly coverges to f o [ b, b] (but ot o whole R, as see above). Cauchy Criterio for Uiform Covergece. Like other Cauchy criteria, this criterio gives the ecessary ad sufficiet coditio for a sequece of fuctios to coverge uiformly without kowig the limit fuctio. Theorem 6.1 (Cauchy Criterio for Uiform Covergece). Let f : A R for each N. The (f ) coverges uiformly o A if ad oly if, for each ɛ > 0, there exists a N N such that (6.1) f (x) f m (x) < ɛ x A,, m N. Proof. First assume (f ) coverges uiformly o A to a limit fuctio f. The, for each ɛ > 0, there exists a N N such that f (x) f(x) < ɛ/2 x A, N. Hece, wheever, m N ad x A, it follows that provig (6.1). f (x) f m (x) f (x) f(x) + f m (x) f(x) < ɛ/2 + ɛ/2 = ɛ, We ow assume coditio (6.1) holds. The, give each x A, the sequece (f (x)) is Cauchy; hece it coverges to a limit f(x) R. This defies a fuctio f : A R, which is the poitwise covergece limit of f (x). The coditio, used with ɛ replaced by ɛ/2, implies that there exists a N N such that for all x A ad, m N we have f (x) f m (x) < ɛ/2. Hece ɛ/2 < f (x) f m (x) < ɛ/2, m N. We ow fix x A ad N ad take the limit of sequece (f m (x)) i this iequality. Use the order limit theorem, we have ɛ/2 f (x) f(x) ɛ/2 N. Hece we have proved that f (x) f(x) ɛ/2 < ɛ holds for all x A ad N. This is othig but the defiitio of the uiform covergece of (f ) to f o A. Theorem 6.2 (Cotiuity of uiform limit fuctio). Let (f ) ad f be fuctios o A ad let (f ) coverge uiformly to f o A. Assume for each N the fuctio f is cotiuous at a poit c A. The f is cotiuous at c as well. Proof. Give ɛ > 0, by the uiform covergece of (f ) to f o A, there exists a N N such that f (x) f(x) < ɛ/3 x A, N. We fix = N ad cosider fuctio f N, which is cotiuous at c A; hece, there exists a umber δ > 0 such that f N (x) f N (c) < ɛ/3 x A, x c < δ. The, wheever x A ad x c < δ, it follows that f(x) f(c) f N (x) f(x) + f N (x) f N (c) + f N (c) f(c) This proves the cotiuity of f at c A. < ɛ/3 + ɛ/3 + ɛ/3 = ɛ.

4 6. Sequeces ad Series of Fuctios 6.2. Uiform Covergece ad Differetiatio Theorem 6.3. Let f (x) f(x) poitwise o [a, b] ad assume each f is differetiable o a ope iterval cotaiig [a, b]. If f coverges uiformly o [a, b] to a fuctio g, the f is differetiable ad f = g o [a, b]. Proof. Let c [a, b] be give. We wat to show Note that (6.2) f(x) f(c) x c f(x) f(c) lim = g(c). x c x c g(c) f(x) f(c) f (x) f (c) x c x c + f (x) f (c) f x c (c) + f (c) g(c). Give ɛ > 0, we wat to fid a δ > 0 such that each of the three terms is < ɛ/3 for all x c < δ ad x [a, b]. The third term is idepedet of x, but depeds o. Sice f (c) g(c), there exists a N 1 N such that f (c) g(c) < ɛ/3 N 1. The secod term for fixed ca be easily hadled. The first term requires the most work. Let s do it ow. Note that by applyig the MVT to fuctio f m (x) f (x) we have, for x c, (6.3) f m (x) f m (c) x c f (x) f (c) x c = (f m(x) f (x)) (f m (c) f (c)) x c = f m(α) f (α) for some α betwee x ad c (such a umber α depeds o may thigs: m,, x, c). However, sice (f ) coverges uiformly o [a, b], by the Cauchy Criterio for Uiform Covergece, there exists a N 2 N such that f (x) f m(x) < ɛ/4 x [a, b], m, N 2. We ow use (6.3) to coclude that f m (x) f m (c) f (x) f (c) x c x c = f m(α) f (α) < ɛ/4 for all x c ad all, m N 2. Take the limit as m ad use the order limit theorem, ad we have f(x) f(c) f (x) f (c) x c x c ɛ/4 < ɛ/3 x c, N 2. Now let N = max{n 1, N 2 }. Sice f N is differetiable at c with derivative f N (c), there exists a δ > 0 such that f N (x) f N (c) f x c N(c) < ɛ/3 x [a, b], x c < δ.

6.2. Uiform Covergece ad Differetiatio 5 For this δ > 0, wheever x [a, b] ad x c < δ, it follows from (6.2) with = N that f(x) f(c) g(c) x c f(x) f(c) f N(x) f N (c) x c x c + f N (x) f N (c) f x c N(c) + f N(c) g(c) < ɛ/3 + ɛ/3 + ɛ/3 = ɛ. So f (c) = g(c). We also have a stroger result uder a weaker assumptio. Theorem 6.4. Let (f ) be a sequece of differetiable fuctios defied o [a, b], ad assume (f ) coverges uiformly to a fuctio g o [a, b]. If there exists a poit c [a, b] for which the sequece (f (c)) coverges, the (f ) must coverge uiformly o [a, b] ad the limit fuctio f is differetiable o [a, b] ad satisfies f (x) = g(x) o [a, b]. Proof. Let us prove the uiform covergece of (f ) o [a, b]; the the differetiatio results follow from the previous theorem. We use the Cauchy criterio to show that (f ) coverges uiformly o [a, b]. Give ay ɛ > 0, by the uiform covergece of (f ), there exists a N 1 N such that f (x) f m(x) < ɛ/2(b a) x [a, b],, m N 1. Also, by the covergece of sequece (f (c)), there exists a N 2 N such that f (c) f m (c) < ɛ/2, m N 2. We use the MVT to fuctio f (x) f m (x) to obtai (f (x) f m (x)) (f (c) f m (c)) = (f (α) f m(α))(x c) for some α betwee x ad c ad hece α [a, b]. Therefore, wheever, m N 1, (f (x) f m (x)) (f (c) f m (c)) = (f (α) f m(α)) x c f (α) f m(α) (b a) < ɛ (b a) = ɛ/2. 2(b a) Fially, let N = max{n 1, N 2 }. The, wheever, m N ad x [a, b], it follows that f (x) f m (x) (f (x) f m (x)) (f (c) f m (c)) + (f (c) f m (c)) < ɛ/2 + ɛ/2 = ɛ. Hece, by the Cauchy Criterio, (f ) coverges uiformly o [a, b] to some fuctio f. This proves the theorem. Example 6.4. Let g (x) = x o x [0, 1]. Agai (g { ) uiformly coverges to g(x) = 0, but g (x) = x 1 0 0 x < 1 oly poitwise coverges to h(x) =, ot uiformly. 1 x = 1 Example 6.5. Let h (x) = ( 1) si(x) o R. The (h ) coverges uiformly to h(x) = 0 o R. However h (x) = ( 1) cos(x) does ot coverge for ay x R. The previous examples idicate that for a sequece of fuctios (f ) to coverge uiformly it is ot ecessary that (f ) coverge uiformly (eve poitwise).

6 6. Sequeces ad Series of Fuctios 6.3. Series of Fuctios A series of fuctios is a ifiite series of the form f (x) = f 1 (x) + f 2 (x) + f 3 (x) +, =1 where f 1, f 2, are fuctios defied o a commo set A R. Let the sequece of partial sums be defied by s (x) = f k (x) N. Defiitio 6.4. The series of fuctios =1 f (x) is said to coverge poitwise o A if the sequece (s (x)) coverges poitwise o A; if (s (x)) f(x) poitwise o A, the we write f (x) = f(x) x A. =1 If (s ) coverges uiformly o A the we say that the series =1 f (x) coverges uiformly o A. Theorem 6.5. Let f be cotiuous fuctios defied o a set A R, ad assume =1 f (x) coverges uiformly o A to a fuctio f. The f is cotiuous o A. Proof. Easy cosequece of the case for sequece. Theorem 6.6 (Term-by-Term Differetiatio Theorem). Let (f ) be a sequece of differetiable fuctios defied o [a, b], ad assume =1 f coverges uiformly to a fuctio g o [a, b]. If there exists a poit c [a, b] for which the series =1 f (c) coverges, the the series =1 f must coverge uiformly to a differetiable fuctio f o [a, b] satisfyig f (x) = g(x) o [a, b]. I other words, if =1 f (c) coverges at oe poit c [a, b], ad =1 f coverges uiformly o [a, b], the f(x) = f (x) =1 is well-defied for all x [a, b] ad is differetiable with f (x) = f (x) x [a, b]. Proof. Use Theorem 6.4. =1 Theorem 6.7 (Cauchy Criterio for Uiform Covergece of Series). A series =1 f coverges uiformly o a set A R if ad oly if for every ɛ > 0 there exists a N N such that for all > m N, Proof. Use Theorem 6.1. f m+1 (x) + f m+2 (x) + + f (x) < ɛ x A. Corollary 6.8 (Weierstrass M-Test). For each N, let f be a fuctio defied o A satisfyig f (x) M for all x A, where M > 0 is a real umber. If =1 M coverges, the =1 f coverges uiformly o A. Proof. Use the previous Cauchy s Criterio.

6.4. Power Series 7 Example 6.6. Let f(x) = si kx k 3. (a) Show f ad is differetiable ad f is cotiuous o R. (b) Ca we determie if f is twice-differetiable? Example 6.7. Let f(x) = =1 x = x + x2 2 + x3 3 +. The f(x) coverges o [0, 1), but f(1) =. This is a special case of the power series to be leared ext. 6.4. Power Series A power series is a ifiite series of power fuctios: f(x) = a (x c) = a 0 + a 1 (x c) + a 2 (x c) 2 + a 3 (x c) 3 +. =0 The poit c is called the ceter of the power series ad sequece (a ) is called the sequece of coefficiets of the power series. The power series also coverges at its ceter c. I what follows, we always assume c = 0. Theorem 6.9. If a power series =0 a x coverges at some poit x 0 0, the it coverges absolutely for all x satisfyig x < x 0. Proof. If =0 a x 0 coverges, the the sequece (a x 0 ) is bouded. Let M > 0 satisfy a x 0 M for all N. Assume x R satisfies x < x 0. The a x = a x 0 x M x. x 0 Sice x/x 0 < 1, by compariso with the geometric series, it follows easily that =0 a x coverges; hece =0 a x coverges absolutely. Radius of Covergece. Let =0 a x be a power series. Let S = {x R =0 a x coverges}. There are three cases for this set S: (a) S = {0}. I this case, the power series coverges oly at ceter 0. (b) S = R. I this case, the power series coverges at every x R. (c) S {0} ad S R. I Case (c), there exist two poits x 0, x 1 R such that x 0 0, x 0 S ad x 1 / S. By Theorem 6.9, for all x R with x > x 1, the power series diverges at x; otherwise, if it coverges at some x with x > x 1 the by the theorem, x 1 S. Therefore, the set S is icluded i the iterval [ x 1, x 1 ] ad hece S is a oempty bouded set i R. By the AoC, let R = sup S. This R satisfies x 0 R x 1 ; hece R (0, ). x 0

8 6. Sequeces ad Series of Fuctios Lemma 6.10. I Case (c), the power series coverges at all x with x < R ad diverges at all x with x > R. Proof. 1. If x < R = sup S, the there exists x S such that x > x. Sice the power series coverges at x ad x < x = x, by Theorem 6.9, the power series coverges at x. 2. If x > R, we show that the power series diverges at x. If ot, assume the power series coverges at x. Let y = x +R 2. The R < y < x. Sice the power series coverges at x ad y < x, by Theorem 6.9 agai, the power series coverges at y; hece y S. But R = sup S; so y R, a cotradictio. Defiitio 6.5 (Radius of Covergece). Let =0 a x be a power series with the set S defied as above. The radius of covergece of this power series is the umber R [0, ] defied as follows: R = 0 i Case (a); R = i Case (b); R = sup S (0, ) i Case (c). Remark 6.6. I fact, there is a formula for the radius of covergece give by 1 R = lim sup a i all cases. However, this formula will ot be eeded for our lecture ad the homework. Example 6.8. For the power series! x, =0 the th-term is a =!x. If x 0, the a +1 a = ( + 1) x, ad hece a. Therefore, a diverges. So the give power series oly coverges at x = 0 ad diverges for all x 0. The radius of covergece for this power series is R = 0. Example 6.9. For power series =0 x! = 1 + x + x2 2! + x3 3! + the th-term is a = x!. Hece a +1 a = x +1 ( + 1)!! x = x + 1 0 for all x 0; thus, the power series coverges absolutely for all x 0, ad certaily it always coverges for x = 0. Therefore the give series coverges absolutely for all x. The radius of covergece for this power series is R =. Example 6.10. I Case (c) above, the set S could oly be oe of the itervals ad all cases ca happe. [ R, R], [ R, R), ( R, R], ( R, R), (a) The power series =1 x / 2 has the radius of covergece R = 1 ad coverges exactly o [ 1, 1]. (b) The power series =1 x / has the radius of covergece R = 1 ad coverges exactly o [ 1, 1).

6.4. Power Series 9 (c) The power series =1 ( x) / has the radius of covergece R = 1 ad coverges exactly o ( 1, 1]. (d) The power series =1 x2 / has the radius of covergece R = 1 ad coverges exactly o ( 1, 1). Theorem 6.11. If a power series =0 a x coverges absolutely at a poit x 0 0, the it coverges uiformly o the iterval [ x 0, x 0 ]. Proof. Use Weierstrass s M-Test. Abel s Theorem. Lemma 6.12 (Abel s Lemma). Let sequece (b ) satisfy b 1 b 2 b 3 0 ad sequece (a ) satisfy a 1 + a 2 + + a A, N, for a costat A > 0. The, for all N, a 1 b 1 + a 2 b 2 + a 3 b 3 + + a b Ab 1. Proof. Let s 0 = 0 ad s k = a 1 + a 2 + + a k for k N. The a k = s k s k 1 for all k N. Hece a k b k = (s k s k 1 )b k = s k b k s k 1 b k = 1 s k b k s k b k+1 = k=0 1 1 s k b k + s b Sice s k A ad b k b k+1 0, it follows that 1 a k b k s k (b k b k+1 ) + s b 1 s k b k+1 = s k (b k b k+1 ) + s b. 1 1 s k (b k b k+1 ) + s b A (b k b k+1 ) + Ab = Ab 1. Theorem 6.13 (Abel s Theorem). If a power series =0 a x coverges at x = R > 0, the it coverges uiformly o the iterval [0, R]. If a power series =0 a x coverges at x = R < 0, the it coverges uiformly o the iterval [ R, 0]. Proof. Let us assume =0 a R coverges. We wat to use the Cauchy criterio to show that =0 a x coverges uiformly o [0, R]. Give ɛ > 0, from the covergece of =0 a R, there exists a umber N N such that a m+1 R m+1 + a m+2 R m+2 + + a R < ɛ/2, > m N. Let x [0, R]. For ay fixed m N, cosider sequece B = ( x R )m+ ad sequece A = a m+ R m+ with N. The the sequeces (B ) ad (A ) satisfy the coditios i Abel s Lemma above with A = ɛ/2. Hece a m+1 R m+1 ( x R )m+1 + a m+2 R m+2 ( x R )m+2 + + a R ( x x R ) ( R )m+1 ɛ/2 < ɛ.

10 6. Sequeces ad Series of Fuctios Therefore, a m+1 x m+1 + a m+2 x m+2 + + a x < ɛ, x [0, R], > m N. This proves the uiform covergece of =0 a x o [0, R]. The Success of Power Series. Give a power series =0 a x, the differetiated series =1 a x 1 is also a power series. Theorem 6.14. If a power series =0 a x coverges for all x ( R, R), the the differetiated series =1 a x 1 coverges for all x ( R, R) as well. Proof. Let t be such that x < t < R. The coclusio follows from the idetity a x 1 = 1 ( x 1 ) t t 1 a t. Sice r = x /t < 1, it follows that (r 1 ) 0. This ca be show by the covergece of series r usig the ratio-test. Theorem 6.15. If a power series coverges poitwise o a set A, the it coverges uiformly o ay compact subset of A. Proof. Whe A cotais oe of R or R, where R is the radius of covergece, we eed the Abel s theorem. Other cases, the theorem ca be proved without it. Theorem 6.16. Assume the power series g(x) = a x =0 coverges o a set A. The, g is cotiuous o A ad differetiable o ay ope iterval ( R, R) A with the derivative give by the term-by-term differetiatio g (x) = a x 1, x ( R, R). =1 Moreover, g is ifiitely differetiable o ( R, R), ad the higher-order derivatives ca be obtaied via the term-by-term differetiatio of the previous differetiated power series. Proof. Oly for the cotiuity of g at possibly the ed-poits of the iterval of covergece is the previous theorem eeded. All other coclusios ca be proved without usig the Abel s theorem. Example 6.11. Let f(x) = =0 x! = 1 + x + x2 2! + x3 + (x R). 3! The, from the term-by-term differetiatio, f x 1 (x) = ( 1)! = 1 + x + x2 2! + x3 3! + = f(x). =1 Also f(0) = 1. Let g(x) = f(x)e x. The g(0) = 1 ad g (x) = f (x)e x f(x)e x = 0 for all x R. So g(x) 1; that is, f(x) = e x for all x R. I particular, with x = 1, 1 + 1 + 1 2! + 1 3! + 1 4! + = e = 2.718281828459045....

6.5. Taylor Series 11 Example 6.12. (Exercise 6.5.1.) Note that the power series g(x) = x x2 2 + x3 3 = =1 1 x ( 1) has radius of covergece R = 1 ad coverges exactly for x ( 1, 1]. We ca prove this by just usig the fact g( 1) diverges ad g(1) coverges. (Explai why.) Hece g is cotiuous o ( 1, 1] ad is differetiable o ( 1, 1). The derivative is give by g (x) = ( x) 1, 1 < x < 1. =1 So, by the geometric series, g (x) = 1 that is, g(x) = 1+x x 0 for 1 < x < 1. But g(0) = 0. Hece g (t) dt = x 0 1 1 + t dt = [l 1 + t ] x 0 = l 1 + x = l(1 + x); ( 1) =1 1 x = l(1 + x), 1 < x < 1. This idetity also holds whe x = 1 sice g is cotiuous at x = 1. So we have that the value of the alterative harmoic series is give by ( 1) 1 g(1) = = 1 1 2 + 1 3 1 + = l 2. 4 6.5. Taylor Series =1 The power series defies a ifiitely differetiable fuctio o its ope iterval of covergece. Give a ifiitely differetiable fuctio o a ope iterval, ca we express the fuctio as a power series cetered at a iterior poit of the iterval? Such a series is called the Taylor series of the fuctio cetered at the give poit. Let f be a ifiitely differetiable fuctio defied o ( R, R). Suppose f equals a power series cetered at 0 as f(x) = a x = a 0 + a 1 x + a 2 x 2 +, x ( R, R). =0 The a 0 = f(0). By the term-by-term differetiatio, we have, for all k = 1, 2, 3,, f (k) (x) = ( 1) ( k + 1)a x k, x ( R, R). =k Hece f (k) (0) = k!a k. Therefore, with 0! = 1, we have a k = f (k) (0) k!, k = 0, 1, 2,. This shows that if a fuctio f equals a power series ear 0 the the coefficiets a k of the power series must be give by the above formula.

12 6. Sequeces ad Series of Fuctios Defiitio 6.7. Let f be a ifiitely differetiable fuctio ear a poit a R. The the power series f () (a) (6.4) (x a) ear a! =0 is called the Taylor series of f about a. Whe a = 0, the Taylor series iis also called the Maclauri series; but we will still call it the Taylor series. Lagrage s Remaider Theorem. Does the Taylor series of f coverge to f ear a? We assume a = 0. The followig result is useful to aswer such a questio. Theorem 6.17 (Lagrage s Remaider Theorem). Let f be ifiitely differetiable o ( R, R) ad defie N f (k) (0) S N (x) = x k x ( R, R). k! k=0 The, give ay 0 < x < R, there exists a umber c with c < x such that the error term E N (x) = f(x) S N (x) satisfies E N (x) = f (N+1) (c) (N + 1)! xn+1. Proof. Without loss of geerality, assume 0 < x < R. Cosider fuctio E N (t) = f(t) S N (t). The E () N (0) = 0 for all = 0, 1, 2,, N, ad E (N+1) N (t) = f (N+1) (t) for all t ( R, R). So, by the Geeralized Mea-Value Theorem repeatedly, we have x > x 1 > x 2 > > x N+1 > 0 such that E N (x) x N+1 = E N (x 1) (N + 1)x N 1 = E N (x 2) (N + 1)Nx N 1 2 ad this proves the theorem with c = x N+1. = = E(N+1) N (x N+1 ) (N + 1)N 2 1 = f (N+1) (c) (N + 1)!, Example 6.13. Show si x = =0 ( 1) (2 + 1)! x2+1 x R. Proof. Let f(x) = si x. The f (2) (x) = ( 1) si x, for all = 0, 1, 2,.... Therefore the power series ( 1) (2 + 1)! x2+1 =0 f (2+1) (x) = ( 1) cos x is exactly the Taylor series of si x about 0. Let S N (x) be the partial sum up to power x N. The k ( 1) S 2k (x) = S 2k+1 (x) = (2 + 1)! x2+1. =0

6.5. Taylor Series 13 By Lagrage s Remaider Theorem, f (2k+2) (c) si x S 2k+1 (x) = E 2k+1 (x) = (2k + 2)! x2k+2 1 (2k + 2)! x 2k+2 0 as k for all x R. This proves the covergece; moreover it gives the error estimate k ( 1) (2 + 1)! x2+1 si x x 2k+2 (2k + 2)! =0 for all x R ad all k = 0, 1,. A Couterexample. The Taylor series of f may ot be equal to f ear a. Example 6.14. Let f(x) = { e 1/x2 for x 0, 0 for x = 0. The f is ifiitely differetiable at all x 0. It is a good exercise to show that f has all orders of derivatives at 0 ad f () (0) = 0 for all = 0, 1, 2,. Therefore, the Taylor series of f about 0 is idetically zero; obviously f(x) 0 wheever x 0. This shows that f is ot equal to its Taylor series ear 0.