UNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE


 Shona Lewis
 1 years ago
 Views:
Transcription
1 UNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE 1. Pointwise Convergence of Sequence Let E be set nd Y be metric spce. Consider functions f n : E Y for n = 1, 2,.... We sy tht the sequence (f n ) converges pointwise on E if there is function f : E Y such tht f n (p) f(p) for every p E. Clerly, such function f is unique nd it is clled the pointwise limit of (f n ) on E. We then write f n f on E. For simplicity, we shll ssume Y = R with the usul metric. Let f n f on E. We sk the following questions: (i) If ech f n is bounded on E, must f be bounded on E? If so, must sup p E f n (p) sup p E f(p)? (ii) If E is metric spce nd ech f n is continuous on E, must f be continuous on E? (iii) If E is n intervl in R nd ech f n is differentible on E, must f be differentible on E? If so, must f n f on E? (iv) If E = [, b] nd ech f n is Riemnn integrble on E, must f be Riemnn integrble on E? If so, must f n(x)dx f(x)dx? These questions involve interchnge of two processes (one of which is tking the limit s n ) s shown below. (i) lim sup f n (p) = sup lim f n(p). n p E p E n (ii) For p E nd p k p in E, lim lim f n(p k ) = lim lim f n(p k ). k n n k d (iii) lim n dx (f n) = d ( ) lim dx f n. n (iv) lim n f n (x)dx = ( lim f n(x) n ) dx. Answers to these questions re ll negtive. Exmples 1.1. (i) Let E := (0, 1] nd define f n : E R by { 0 if 0 < x 1/n, f n (x) := 1/x if 1/n x 1. Then f n (x) n for ll x E, f n f on E, where f(x) := 1/x. Thus ech f n is bounded on E, but f is not bounded on E. (ii) Let E := [0, 1] nd define f n : E R by f n (x) := 1/(nx + 1). Then ech f n is continuous on E, f n f on E, where f(0) := 1 nd f(x) := 0 if 0 < x 1. Clerly, f is not continuous on E. (iii) () Let E := ( 1, 1) nd define f n : E R by f n (x) := 1/(nx 2 + 1). Then ech f n is differentible on E nd f n f on ( 1, 1), 1
2 2 MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE (iv) where f(0) := 1 nd f(x) := 0 if 0 < x < 1. Clerly, f is not differentible on E. (b) Let E := R nd define f n : R R by f n (x) := (sin nx)/n. Then ech f n is differentible, f n f on R, where f 0. But f n (x) = cos nx for x R, nd (f n ) does not converge pointwise on R. For exmple, (f n (π)) is not convergent sequence. (c) Let E := ( 1, 1) nd define f n (x) := { [2 (1 + x) n ]/n if 1 < x < 0, (1 x) n /n if 0 x < 1. Then ech f n is differentible on E. (In prticulr, we hve f n (0) = 1 by L Hôpitl s Rule.) Also, f n f on ( 1, 1), where f 0. Also, { f n(x) (1 + x) n 1 if 1 < x < 0, = (1 x) n 1 if 0 x < 1. Further, f n g on ( 1, 1), where g(0) := 1 nd g(x) := 0 for 0 < x < 1. Clerly, f g. () Let E := [0, 1] nd define f n : [0, 1] R by { 1 if x = 0, 1/n!, 2/n!,..., n!/n! = 1, f n (x) := 0 otherwise. Then ech f n is Riemnn integrble on [0, 1] since it is discontinuous only t finite number of points. { 1 if x is rtionl, Also, f n f on [0, 1], where f(x) := 0 if x is irrtionl. For if x = p/q with p {0, 1, 2,..., q} N, then for ll n q, we hve n!x {0, 1, 2,...} nd so f n (x) = 1, while if x is n irrtionl number, then f n (x) = 0 for ll n N. We hve seen tht the Dirichlet function f is not Riemnn integrble. (b) Let E := [0, 1], nd define f n : [0, 1] R by f n (x) := n 3 xe nx. Then ech f n is Riemnn integrble nd f n f on [0, 1], where f 0. (Use L Hôpitl s Rule repetedly to show tht lim t t 3 /e st = 0 for ny s R with s > 0.) However, using Integrtion by Prts, we hve 1 0 xe nx dx = 1 n 2 1 n 2 e n 1 ne n for ech n N, so tht 1 0 f n(x)dx = n (n/e n ) (n 2 /e n ). (c) Let E := [0, 1] nd define f n : [0, 1] R by f n (x) := n 2 xe nx. As bove, ech f n is Riemnn integrble, nd f n f on [0, 1], where f 0, nd 1 0 f n(x)dx = 1 (1/e n ) (n/e n ) 1, which is not equl to 1 0 f(x)dx = Uniform Convergence of Sequence In n ttempt to obtin ffirmtive nswers to the questions posed t the beginning of the previous section, we introduce stronger concept of convergence.
3 UNIFORM CONVERGENCE 3 Let E be set nd consider functions f n : E R for n = 1, 2,.... We sy tht the sequence (f n ) of functions converges uniformly on E if there is function f : E R such tht for every ɛ > 0, there is n 0 N stisfying n n 0, p E = f n (p) f(p) < ɛ. Note tht the nturl number n 0 mentioned in the bove definition my depend upon the given sequence (f n ) of functions nd on the given positive number ɛ, but it is independent of p E. Clerly, such function f is unique nd it is clled the uniform limit of (f n ) on E. We then write f n f on E. Obviously, f n f on E = f n f on E, but the converse is not true : Let E := (0, 1] nd define f n (x) := 1/(nx + 1) for 0 < x 1. If f(x) := 0 for x (0, 1], then f n f on (0, 1], but f n f on (0, 1]. To see this, let ɛ := 1/2, note tht there is no n 0 N stisfying 1 f n (x) f(x) = nx + 1 < 1 2 for ll n n 0 nd for ll x (0, 1], since 1/(nx + 1) = 1/2 when x = 1/n, n N. A sequence (f n ) of relvlued functions defined on set E is sid to be uniformly Cuchy on E if for every ɛ > 0, there is n 0 N stisfying m, n n 0, p E = f m (p) f n (p) < ɛ. Proposition 2.1. (Cuchy Criterion for Uniform Convergence of Sequence) Let (f n ) be sequence of relvlued functions defined on set E. Then (f n ) is uniformly convergent on E if nd only if (f n ) is uniformly Cuchy on E. Proof. = ) Let f n f. For ll m, n N nd p E, we hve f m (p) f n (p) f m (p) f(p) + f(p) f n (p). =) For ech p E, (f n (p)) is Cuchy sequence in R, nd so it converges to rel number which we denote by f(p). Let ɛ > 0. There is n 0 N stisfying m, n n 0, p E = f m (p) f n (p) < ɛ. For ny m n 0 nd p E, letting n, we hve f m (p) f(p) ɛ. We hve the following useful test for checking the uniform convergence of (f n ) when its pointwise limit is known. Proposition 2.2. (Test for Uniform Convergence of Sequence) Let f n nd f be relvlued functions defined on set E. If f n f on E, nd if there is sequence ( n ) of rel numbers such tht n 0 nd f n (p) f(p) n for ll p E, then f n f on E. Proof. Let ɛ > 0. Since n 0, there is n 0 N such tht n n 0 = n < ɛ, nd so f n (p) f(p) < ɛ for ll p E. Exmple 2.3. Let 0 < r < 1 nd f n (x) := x n for x [ r, r]. Thenf n (x) 0 for ech x [ r, r]. Since r n 0 nd f n (x) 0 r n for ll x [ r, r], (f n ) is uniformly convergent on [ r, r]. Let us now pose the four questions stted in the lst section with convergence replced by uniform convergence. We shll nswer them one by one, but not necessrily in the sme order.
4 4 MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE Uniform Convergence nd Boundedness. Proposition 2.4. Let f n nd f be relvlued functions defined on set E. If f n f on E nd ech f n is bounded on E, then f bounded on E. Proof. There is n 0 N such tht n n 0, p E = f n (p) f(p) < 1. Also, since f n0 is bounded on E, there is α 0 such tht p E = f n0 (p) α 0. Hence p E = f(p) f(p) f n0 (p) + f n0 (p) < 1 + α 0. The converse of the bove result is not true, tht is, ech f n s well s f bounded on E nd f n f f n f on E. For exmple, let E := (0, 1], f n (x) := 1/(nx + 1) nd f 0. Given set E, let B(E) denote the set of ll relvlued bounded functions defined on E. For f, g in B(E), define d(f, g) := sup{ f(p) g(p) : p E}. Then it is esy to see tht d is metric on B(E), known s the supmetric on B(E). Also, by Proposition 2.2, for f n nd f in B(E), we hve f n f on E if nd only if d(f n, f) 0, tht is, (f n ) converges to f in the supmetric on B(E). Similrly, (f n ) is uniformly Cuchy on E if nd only if d(f n, f m ) 0 s n, m, tht is, (f n ) is Cuchy sequence in the supmetric on B(E). Thus Propositions 2.1 nd 2.4 show tht B(E) is complete metric spce. Also, under the hypotheses of Proposition 2.4, we hve sup p E f n (p) sup p E nd so, sup p E f n (p) sup p E f(p). Uniform Convergence nd Integrtion. f(p) d(f n, f) 0, Proposition 2.5. Let (f n ) be sequence of relvlued functions defined on [, b]. If f n f on [, b] nd ech f n is Riemnn integrble on [, b], then f is Riemnn integrble on [, b] nd f n(x)dx f(x)dx. Proof. Since f n f nd ech f n is bounded, we see tht f is bounded on [, b] by Proposition 2.4. For n N, let α n := d(f n, f), where d denotes the supmetric on B([, b]). For ech n N nd x [, b], we hve f n (x) f(x) α n, tht is, f n (x) α n f(x) f n (x) + α n, nd so L(f n ) α n (b ) L(f) U(f) U(f n ) + α n (b ). But since f n is Riemnn integrble, we hve L(f n ) = U(f n ), nd hence 0 U(f) L(f) 2α n (b ) 0 s n. Thus L(f) = U(f), tht is, f is Riemnn integrble on [, b]. Also, tht is, f n (x)dx α n (b ) f(x)dx f n (x)dx + α n (b ), f n(x)dx f(x)dx αn (b ) 0 s n. The converse of the bove result is not true, tht is, ech f n s well s f Riemnn integrble on [, b], f n f on [, b] nd f n(x)dx f(x)dx f n f. For exmple, if f n (x) := 1/(nx + 1) for x [0, 1],
5 0 UNIFORM CONVERGENCE 5 f(0) := 1, f(x) := 0 for x (0, 1], then f n f on [0, 1], but f is integrble nd 1 ln(nx + 1) f n (x)dx = 1 ln(1 + n) 1 = 0 = f(x)dx. n 0 n Uniform Convergence nd Continuity. Proposition 2.6. Let (f n ) be sequence of relvlued functions defined on metric spce E. If f n f on E nd ech f n is continuous on E, then f is continuous on E. Proof. Let ɛ > 0. There is n 0 N such tht p E = f n0 (p) f(p) < ɛ/3. Consider p 0 E. Since f n0 is continuous t p 0, there is δ > 0 such tht p E, d(p, p 0 ) < δ = f n0 (p) f n0 (p 0 ) < ɛ/3. nd hence f(p) f(p 0 ) f(p) f n0 (p) + f n0 (p) f n0 (p 0 ) + f n0 (p 0 ) f(p 0 ) < ɛ, estblishing the continuity of f t p 0 E. The converse of the bove result is not true, tht is, ech f n s well s f continuous on metric spce E, f n f on E f n f. For exmple, let f n (x) := nxe nx nd f(x) := 0 for x [0, 1]. Since f n (0) = 0 nd for x (0, 1], f n (x) 0 s n by L Hôpitl s Rule, we see tht f n f. But there is no n 0 N such tht n n 0, x [0, 1] = nxe nx 0 < 1, since nxe nx = e 1 for x = 1/n, n N. However, the following prtil converse holds. Proposition 2.7. (Dini s Theorem) Let (f n ) be sequence of relvlued functions defined on compct metric spce E. If f n f on E, ech f n nd f re continuous on E, nd (f n ) is monotonic sequence (tht is, f n f n+1 for ll n N, or f n f n+1 for ll n N), then f n f on E. For proof, see Theorem 7.13 of [3]. The following exmples show tht neither the compctness of the metric spce E nor the continuity of the function f cn be dropped from Dini s Theorem: (i) E := (0, 1] nd f n (x) := 1/(nx + 1), x E, (ii) E := [0, 1] nd f n (x) := x n, x E. Uniform Convergence nd Differentition. Answers to the questions regrding differentition posed in the lst section re not ffirmtive even when f n f on n intervl of R. () Let f n (x) := x 2 + (1/n 2 ) nd f(x) := x for x [ 1, 1]. Since f n (x) f(x) = x 2 + (1/n 2 ) x 2 x 2 + (1/n 2 ) x 2 = 1 n for ll n N nd x [ 1, 1], Proposition 2.2 shows tht f n f on [ 1, 1]. Although ech f n is differentible on [ 1, 1], the limit function f is not. (b) In Exmple 1.1 (iii) (b), f n f on R, ech f n differentible, but (f n ) does not converge pointwise. (c) In Exmple 1.1 (iii) (c), f n f on ( 1, 1) nd ech f n s well s f is differentible on ( 1, 1), nd f n g on ( 1, 1), where g f. However, if we ssume the uniform convergence of the derived sequence (f n) long with the convergence of the sequence (f n ) t only one point of the intervl, we hve stisfctory nswer. 0
6 6 MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE Proposition 2.8. Let (f n ) be sequence of relvlued functions defined on [, b]. If (f n ) converges t one point of [, b], ech f n is continuously differentible on [, b] nd (f n) converges uniformly on [, b], then there is f : [, b] R such tht f is continuously differentible on [, b], f n f on [, b] nd in fct, f n f on [, b]. Proof. Let x 0 [, b] nd c 0 R be such tht f n (x 0 ) c 0. Also, let ech f n be continuously differentible nd g : [, b] R be such tht f n g on [, b]. By Proposition 2.6, the function g is continuous on [, b]. Define f : [, b] R by x f(x) := c 0 + g(t)dt for x [, b]. x 0 By prt (ii) of the Fundmentl Theorem of Clculus (FTC), f exists on [, b] nd f (x) = g(x) for x [, b]. Thus f is continuously differentible on [, b] nd f n g = f. Also, by prt (i) of the FTC, we hve f n (x) = f n (x 0 ) + x x 0 f n (t)dt for x [, b]. Hence for n N nd x [, b], we obtin x ( f n (x) f(x) f n (x 0 ) c 0 + f n (t) g(t) ) dt x 0 f n (x 0 ) c 0 + x x 0 sup f n (t) g(t) t [,b] Thus f n f on [, b] by Proposition 2.2. f n (x 0 ) c 0 + (b )d(f n, g). The converse of the bove result is not true, tht is, ech f n s well s f continuously differentible on [, b], f n f on [, b], f n f on [, b] f n f. For exmple, let f n (x) := (nx + 1)e nx /n nd f(x) := 0 for x [0, 1]. Since f n (x) = nxe nx for ech n N nd ll x [0, 1], ech f n is monotoniclly decresing on [0, 1]. As f n (0) = 1/n, we obtin f n (x) f(x) 1/n for ll x [0, 1], nd so f n f on [0, 1]. Also, we hve seen fter the proof of Proposition 2.6 tht f n f, but f n f on [0, 1]. Remrk 2.9. Proposition 2.8 holds if we drop the word continuously ppering (two times) in its sttement, but then the proof is much more involved. See Theorem 7.17 of [3]. The results in Propositions 2.4, 2.5, 2.6 nd 2.8 re summrized in the following theorem. Theorem (i) The uniform limit of sequence of relvlued bounded functions defined on set is bounded. (ii) The uniform limit of sequence of Riemnn integrble functions defined on [, b] is Riemnn integrble, nd its Riemnn integrl is the limit of the sequence of termwise Riemnn integrls, tht is, if (f n ) is uniformly convergent to f on [, b] nd ech f n is Riemnn integrble on [, b], then the function f is Riemnn integrble on [, b] nd f(x)dx = lim n f n(x)dx.
7 UNIFORM CONVERGENCE 7 (iii) The uniform limit of sequence of continuous functions defined on metric spce is continuous. (vi) If sequence of continuously differentible functions defined on [, b] is convergent t one point of [, b] nd if the derived sequence is uniformly convergent on [, b], then the given sequence converges uniformly on [, b], the uniform limit is continuously differentible on [, b] nd its derivtive is the limit of the sequence of termwise derivtives, tht is, if (f n ) converges t one point of [, b], ech f n is continuously differentible on [, b] nd (f n ) is uniformly convergent on [, b], then (f n ) converges uniformly to continuously differentible function f on [, b], nd f (x) = lim n f n(x) for ll x [, b]. 3. Uniform Convergence of series The reder is ssumed to be fmilir with the elementry theory of series of rel numbers. (See, for exmple, Chpter 9 of [1], or Chpter 3 of [3].) Let (f k ) be sequence of relvlued functions defined on set E. Consider the sequence (s n ) of relvlued functions on E defined by s n := f f n = n f k. Note: Just s the sequence (f k ) determines the sequence (s n ), so does (s n ) determine (f k ): If we let s 0 = 0, then we hve f k = s k s k 1 for ll k N. We sy tht the series f k converges pointwise on E if the sequence (s n ) converges pointwise on E, nd we sy tht the series f k converges uniformly on E if the sequence (s n ) converges uniformly on E. For n N, the function s n is clled the nth prtil sum of the series f k nd if s n s, then the function s is clled its sum. Results bout convergence / uniform convergence of sequences of functions crry over to corresponding results bout convergence / uniform convergence of series of functions. Proposition 3.1. (Cuchy Criterion for Uniform Convergence of Series) Let (f k ) be sequence of relvlued functions defined on set E. Then the series f k converges uniformly on E if nd only if for every ɛ > 0, there is n 0 N such tht m n n 0, p E = m f k (p) < ɛ. k=n Proof. Use Proposition 2.1 for the sequence (s n ) of prtil sums. Proposition 3.2. (Weierstrss MTest for Uniform Absolute Convergence of Series) Let (f k ) be sequence of relvlued functions defined on set E. Suppose there is sequence (M k ) in R such tht f k (p) M k for ll k N nd ll p E. If M k is convergent, then f k converges uniformly nd bsolutely on E.
8 8 MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE Proof. Note tht m f k (p) k=n nd use Proposition 3.1. m f k (p) k=n m M k for ll m n, Exmples 3.3. (i) Consider the series k=0 xk, where x ( 1, 1). If r < 1, then the series converges uniformly on {x R : x r} since the series k=0 M k is convergent, where M k := r k for k = 0, 1,... (ii) For k N, let f k (x) := ( 1) k (x + k)/k 2, where x [0, 1]. We show tht the series f k converges uniformly on [0, 1]. For k N, let g k (x) := ( 1) k x/k 2, where x [0, 1]. Letting M k := 1/k 2 for k N, we observe tht the series k=0 M k is convergent. Hence the series g k(x) converges uniformly on [0, 1]. Also, the series ( 1)k /k converges uniformly on [0, 1], being convergent series of constnts. Since f k (x) = g k (x) + ( 1) k /k for k N nd x [0, 1], the series f k(x) converges uniformly on [0, 1]. This exmple lso shows tht the converse of Weierstrss Mtest does not hold: If M k := sup x [0,1] f k (x) = (1 + k)/k 2 for k N, then M k does not converge, since 1/k2 converges, but 1/k diverges. Proposition 3.4. (Dirichlet s Test for Uniform Conditionl Convergence of Series) Let (f k ) be monotonic sequence of relvlued functions defined on set E such tht f k 0 on E. If (g k ) is sequence of relvlued functions defined on E such tht the prtil sums of the series g k re uniformly bounded on E, then the series f kg k converges uniformly on E. In prticulr, the series ( 1)k f k converges uniformly on E. Proof. For ech p E, the series f k(p)g k (p) converges in R by Dirichlet s Test for conditionl convergence of series of rel numbers. (See, for exmple, Proposition 9.20 of [1], or Theorem 3.42 of [3].) For p E, let H(p) := f k(p)g k (p). Also, for n N, let G n := n g k nd H n := n f kg k. Further, let β R be such tht G n (p) β for ll n N nd ll p E. Then by using the prtil summtion formul n n 1 f k g k = (f k f k+1 )G k + f n G n for ll n 2, we hve H(p) H n (p) 2β f n+1 (p) for ll p E. Since f n+1 0 on E, it follows tht H n H on E, tht is, the series f kg k converges uniformly on E. In prticulr, letting g k (p) := ( 1) k for ll k N nd p E, nd noting tht G n (p) 1 for ll n N nd ll p E, we obtin the uniform convergence of the series ( 1)k f k on E. Exmple 3.5. Let E := [0, 1] nd f k (x) := x k /k for k N nd x [0, 1]. Then (f k ) is momotoniclly decresing sequence nd since f k (x) 1/k for k N nd x [0, 1], we see tht f k 0 on [0, 1] by Proposition 2.2. Hence the series ( 1)k x k /k converges uniformly on [0, 1]. k=n
9 UNIFORM CONVERGENCE 9 Results regrding the boundedness, Riemnn integrbility, continuity nd differentibility of the sum function of convergent series of functions cn be esily deduced from the corresponding results for the sequence of its prtil sums. Theorem 3.6. (i) The sum function of uniformly convergent series of relvlued bounded functions defined on set is bounded. (ii) The sum function of uniformly convergent series of Riemnn integrble functions defined on [, b] is Riemnn integrble, nd the series cn be integrted term by term, tht is, if f k is uniformly convergent on [, b] nd ech f k is Riemnn integrble on [, b], then the function f k is Riemnn integrble on [, b] nd ( ) f k (x) dx = f k (x)dx. (iii) The sum function of uniformly convergent series of relvlued continuous functions defined on metric spce is continuous. (vi) If series of continuously differentible functions defined on [, b] is convergent t one point of [, b] nd if the derived series is uniformly convergent on [, b], then the given series converges uniformly on [, b], the sum function is continuously differentible on [, b] nd the series cn be differentited term by term, tht is, if f k converges t one point of [, b], ech f k is continuously differentible on [, b] nd f k is uniformly convergent on [, b], then f k converges uniformly to continuously differentible function, nd ( ) f k (x) = f k (x) for ll x [, b]. Proof. The results follow by pplying Theorem 2.10 to the sequence of prtil sums of the given series. 4. Two Celebrted Theorems on Uniform Approximtion We hve seen in Proposition 2.6 tht uniform limit of sequence of continuous functions on metric spce is continuous. In this section, we reverse the procedure nd sk whether every continuous function on closed nd bounded intervl of R is the uniform limit of sequence of some specil continuous functions. For function f : [0, 1] R nd n N, we define the nth Bernstein polynomil of f by n ( ) ( n k B n (f) := f x k n) k (1 x) n k. k=0 Theorem 4.1. (Polynomil Approximtion Theorem of Weierstrss) If f : [0, 1] R is continuous, then B n (f) f on [0, 1]. Consequently, every relvlued continuous function on [0, 1] is the uniform limit of sequence of relvlued polynomil functions.
10 10 MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE For proof, see Theorem 7.26 of [3], or Corollry 3.12 of [2]. Remrk 4.2. Theorem 4.1 cn be used to prove tht every relvlued continuous function on ny closed nd bounded intervl [, b] is the uniform limit of sequence of relvlued polynomil functions. Let φ : [0, 1] [, b] be defined by φ(x) := (1 x) + xb for x [0, 1]. Then φ is bijective continuous function nd its continuous inverse φ 1 : [, b] [0, 1] is given by φ 1 (t) = (t )/(b ) for t [, b]. Given continuous relvlued function g on [, b], consider the continuous function f := g φ defined on [0, 1]. If (P n ) is sequence of polynomil functions such tht P n f on [0, 1], nd if we let Q n := P n φ 1, then since Q n (t) = P n ( (t )/(b ) ) for t [, b], ech Q n is polynomil function, nd Q n f φ 1 = g on [, b]. Insted of polynomils, let us now consider trigonometric polynomils for pproximting function. They re given by n 0 + ( k cos kx + b k sin kx) for n N, where 0, 1, 2,..., b 1, b 2,... re rel numbers. For Riemnn integrble function f on [ π, π], we define the Fourier coefficients of f by k (f) := 1 π π 0 (f) := 1 2π π π π f(t) cos kt dt, b k (f) := 1 π f(t)dt, nd for k N, π π f(t) sin kt dt. The series 0 (f) + ( k (f) cos kx + b k (f) sin kx ) of functions defined on [ π, π] is clled the Fourier series of the function f. For n = 0, 1, 2,..., let s n (f) denote the nth prtil sum of this series, nd consider the rithmetic mens of these prtil sums given by σ n (f) := s 0(f) + s 1 (f) + s n (f) n + 1 for n = 0, 1, 2... Theorem 4.3. (Trigonometric Polynomil Approximtion Theorem of Fejér) If f : [ π, π] R is continuous nd f( π) = f(π), then σ n (f) f on [ π, π]. Consequently, every relvlued continuous function on [ π, π] hving the sme vlue t π nd π is the uniform limit of sequence of relvlued trigonometric polynomil functions. For proof, see Theorem 8.15 nd Exercise 8.15 of [3], or Theorem 3.13 of [2]. References [1] S. R. Ghorpde nd B. V. Limye, A Course in Clculus nd Rel Anlysis, Springer Interntionl Ed., New Delhi, [2] B. V. Limye, Functionl Anlysis, New Age Interntionl, 2nd Ed., New Delhi, [3] W. Rudin, Principles of Mthemticl Anlysis, 3rd Ed., McGrw Hill, New Delhi, 1976.
Lecture 1. Functional series. Pointwise and uniform convergence.
1 Introduction. Lecture 1. Functionl series. Pointwise nd uniform convergence. In this course we study mongst other things Fourier series. The Fourier series for periodic function f(x) with period 2π is
More informationReview of Riemann Integral
1 Review of Riemnn Integrl In this chpter we review the definition of Riemnn integrl of bounded function f : [, b] R, nd point out its limittions so s to be convinced of the necessity of more generl integrl.
More informationAdvanced Calculus: MATH 410 Uniform Convergence of Functions Professor David Levermore 11 December 2015
Advnced Clculus: MATH 410 Uniform Convergence of Functions Professor Dvid Levermore 11 December 2015 12. Sequences of Functions We now explore two notions of wht it mens for sequence of functions {f n
More informationThe Regulated and Riemann Integrals
Chpter 1 The Regulted nd Riemnn Integrls 1.1 Introduction We will consider severl different pproches to defining the definite integrl f(x) dx of function f(x). These definitions will ll ssign the sme vlue
More informationChapter 6. Riemann Integral
Introduction to Riemnn integrl Chpter 6. Riemnn Integrl WonKwng Prk Deprtment of Mthemtics, The College of Nturl Sciences Kookmin University Second semester, 2015 1 / 41 Introduction to Riemnn integrl
More informationA BRIEF INTRODUCTION TO UNIFORM CONVERGENCE. In the study of Fourier series, several questions arise naturally, such as: c n e int
A BRIEF INTRODUCTION TO UNIFORM CONVERGENCE HANS RINGSTRÖM. Questions nd exmples In the study of Fourier series, severl questions rise nturlly, such s: () (2) re there conditions on c n, n Z, which ensure
More informationChapter 6. Infinite series
Chpter 6 Infinite series We briefly review this chpter in order to study series of functions in chpter 7. We cover from the beginning to Theorem 6.7 in the text excluding Theorem 6.6 nd Rbbe s test (Theorem
More informationf(x)dx . Show that there 1, 0 < x 1 does not exist a differentiable function g : [ 1, 1] R such that g (x) = f(x) for all
3 Definite Integrl 3.1 Introduction In school one comes cross the definition of the integrl of rel vlued function defined on closed nd bounded intervl [, b] between the limits nd b, i.e., f(x)dx s the
More informationThe final exam will take place on Friday May 11th from 8am 11am in Evans room 60.
Mth 104: finl informtion The finl exm will tke plce on Fridy My 11th from 8m 11m in Evns room 60. The exm will cover ll prts of the course with equl weighting. It will cover Chpters 1 5, 7 15, 17 21, 23
More informationThe Banach algebra of functions of bounded variation and the pointwise Helly selection theorem
The Bnch lgebr of functions of bounded vrition nd the pointwise Helly selection theorem Jordn Bell jordn.bell@gmil.com Deprtment of Mthemtics, University of Toronto Jnury, 015 1 BV [, b] Let < b. For f
More informationIMPORTANT THEOREMS CHEAT SHEET
IMPORTANT THEOREMS CHEAT SHEET BY DOUGLAS DANE Howdy, I m Bronson s dog Dougls. Bronson is still complining bout the textbook so I thought if I kept list of the importnt results for you, he might stop.
More information1 The Riemann Integral
The Riemnn Integrl. An exmple leding to the notion of integrl (res) We know how to find (i.e. define) the re of rectngle (bse height), tringle ( (sum of res of tringles). But how do we find/define n re
More informationEntrance Exam, Real Analysis September 1, 2009 Solve exactly 6 out of the 8 problems. Compute the following and justify your computation: lim
1. Let n be positive integers. ntrnce xm, Rel Anlysis September 1, 29 Solve exctly 6 out of the 8 problems. Sketch the grph of the function f(x): f(x) = lim e x2n. Compute the following nd justify your
More informationFor a continuous function f : [a; b]! R we wish to define the Riemann integral
Supplementry Notes for MM509 Topology II 2. The Riemnn Integrl Andrew Swnn For continuous function f : [; b]! R we wish to define the Riemnn integrl R b f (x) dx nd estblish some of its properties. This
More information7.2 Riemann Integrable Functions
7.2 Riemnn Integrble Functions Theorem 1. If f : [, b] R is step function, then f R[, b]. Theorem 2. If f : [, b] R is continuous on [, b], then f R[, b]. Theorem 3. If f : [, b] R is bounded nd continuous
More informationMath 554 Integration
Mth 554 Integrtion Hndout #9 4/12/96 Defn. A collection of n + 1 distinct points of the intervl [, b] P := {x 0 = < x 1 < < x i 1 < x i < < b =: x n } is clled prtition of the intervl. In this cse, we
More informationW. We shall do so one by one, starting with I 1, and we shall do it greedily, trying
Vitli covers 1 Definition. A Vitli cover of set E R is set V of closed intervls with positive length so tht, for every δ > 0 nd every x E, there is some I V with λ(i ) < δ nd x I. 2 Lemm (Vitli covering)
More informationMath 360: A primitive integral and elementary functions
Mth 360: A primitive integrl nd elementry functions D. DeTurck University of Pennsylvni October 16, 2017 D. DeTurck Mth 360 001 2017C: Integrl/functions 1 / 32 Setup for the integrl prtitions Definition:
More informationPrinciples of Real Analysis I Fall VI. Riemann Integration
21355 Principles of Rel Anlysis I Fll 2004 A. Definitions VI. Riemnn Integrtion Let, b R with < b be given. By prtition of [, b] we men finite set P [, b] with, b P. The set of ll prtitions of [, b] will
More informationHomework 4. (1) If f R[a, b], show that f 3 R[a, b]. If f + (x) = max{f(x), 0}, is f + R[a, b]? Justify your answer.
Homework 4 (1) If f R[, b], show tht f 3 R[, b]. If f + (x) = mx{f(x), 0}, is f + R[, b]? Justify your nswer. (2) Let f be continuous function on [, b] tht is strictly positive except finitely mny points
More informationEulerMaclaurin Summation Formula 1
Jnury 9, EulerMclurin Summtion Formul Suppose tht f nd its derivtive re continuous functions on the closed intervl [, b]. Let ψ(x) {x}, where {x} x [x] is the frctionl prt of x. Lemm : If < b nd, b Z,
More informationNOTES AND PROBLEMS: INTEGRATION THEORY
NOTES AND PROBLEMS: INTEGRATION THEORY SAMEER CHAVAN Abstrct. These re the lecture notes prepred for prticipnts of AFSI to be conducted t Kumun University, Almor from 1st to 27th December, 2014. Contents
More informationMath 324 Course Notes: Brief description
Brief description These re notes for Mth 324, n introductory course in Mesure nd Integrtion. Students re dvised to go through ll sections in detil nd ttempt ll problems. These notes will be modified nd
More informationAdvanced Calculus I (Math 4209) Martin Bohner
Advnced Clculus I (Mth 4209) Spring 2018 Lecture Notes Mrtin Bohner Version from My 4, 2018 Author ddress: Deprtment of Mthemtics nd Sttistics, Missouri University of Science nd Technology, Roll, Missouri
More informationReview on Integration (Secs ) Review: Sec Origins of Calculus. Riemann Sums. New functions from old ones.
Mth 20B Integrl Clculus Lecture Review on Integrtion (Secs. 5.  5.3) Remrks on the course. Slide Review: Sec. 5.5.3 Origins of Clculus. Riemnn Sums. New functions from old ones. A mthemticl description
More informationII. Integration and Cauchy s Theorem
MTH6111 Complex Anlysis 200910 Lecture Notes c Shun Bullett QMUL 2009 II. Integrtion nd Cuchy s Theorem 1. Pths nd integrtion Wrning Different uthors hve different definitions for terms like pth nd curve.
More informationLecture 3 ( ) (translated and slightly adapted from lecture notes by Martin Klazar)
Lecture 3 (5.3.2018) (trnslted nd slightly dpted from lecture notes by Mrtin Klzr) Riemnn integrl Now we define precisely the concept of the re, in prticulr, the re of figure U(, b, f) under the grph of
More informationMath 118: Honours Calculus II Winter, 2003 List of Theorems. Lemma 5.1 (Partition Refinement) If P and Q are partitions of [a, b] such that Q P, then
Mth 118: Honours Clculus II Winter, 2003 List of Theorems Lemm 5.1 (Prtition Refinement) If P nd Q re prtitions of [, b] such tht Q P, then L(P, f) L(Q, f) U(Q, f) U(P, f). Lemm 5.2 (Upper Sums Bound Lower
More informationConvergence of Fourier Series and Fejer s Theorem. Lee Ricketson
Convergence of Fourier Series nd Fejer s Theorem Lee Ricketson My, 006 Abstrct This pper will ddress the Fourier Series of functions with rbitrry period. We will derive forms of the Dirichlet nd Fejer
More informationMath 118: Honours Calculus II Winter, 2005 List of Theorems. L(P, f) U(Q, f). f exists for each ǫ > 0 there exists a partition P of [a, b] such that
Mth 118: Honours Clculus II Winter, 2005 List of Theorems Lemm 5.1 (Prtition Refinement): If P nd Q re prtitions of [, b] such tht Q P, then L(P, f) L(Q, f) U(Q, f) U(P, f). Lemm 5.2 (Upper Sums Bound
More informationFUNDAMENTALS OF REAL ANALYSIS by. III.1. Measurable functions. f 1 (
FUNDAMNTALS OF RAL ANALYSIS by Doğn Çömez III. MASURABL FUNCTIONS AND LBSGU INTGRAL III.. Mesurble functions Hving the Lebesgue mesure define, in this chpter, we will identify the collection of functions
More informationUNIFORM CONVERGENCE. Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3
UNIFORM CONVERGENCE Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3 Suppose f n : Ω R or f n : Ω C is sequence of rel or complex functions, nd f n f s n in some sense. Furthermore,
More informationINTRODUCTION TO INTEGRATION
INTRODUCTION TO INTEGRATION 5.1 Ares nd Distnces Assume f(x) 0 on the intervl [, b]. Let A be the re under the grph of f(x). b We will obtin n pproximtion of A in the following three steps. STEP 1: Divide
More informationEuler, Ioachimescu and the trapezium rule. G.J.O. Jameson (Math. Gazette 96 (2012), )
Euler, Iochimescu nd the trpezium rule G.J.O. Jmeson (Mth. Gzette 96 (0), 36 4) The following results were estblished in recent Gzette rticle [, Theorems, 3, 4]. Given > 0 nd 0 < s
More informationProperties of the Riemann Integral
Properties of the Riemnn Integrl Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University Februry 15, 2018 Outline 1 Some Infimum nd Supremum Properties 2
More informationWeek 10: Riemann integral and its properties
Clculus nd Liner Algebr for Biomedicl Engineering Week 10: Riemnn integrl nd its properties H. Führ, Lehrstuhl A für Mthemtik, RWTH Achen, WS 07 Motivtion: Computing flow from flow rtes 1 We observe the
More informationOverview of Calculus I
Overview of Clculus I Prof. Jim Swift Northern Arizon University There re three key concepts in clculus: The limit, the derivtive, nd the integrl. You need to understnd the definitions of these three things,
More information11 An introduction to Riemann Integration
11 An introduction to Riemnn Integrtion The PROOFS of the stndrd lemms nd theorems concerning the Riemnn Integrl re NEB, nd you will not be sked to reproduce proofs of these in full in the exmintion in
More informationSTUDY GUIDE FOR BASIC EXAM
STUDY GUIDE FOR BASIC EXAM BRYON ARAGAM This is prtil list of theorems tht frequently show up on the bsic exm. In mny cses, you my be sked to directly prove one of these theorems or these vrints. There
More informationa n+2 a n+1 M n a 2 a 1. (2)
Rel Anlysis Fll 004 Tke Home Finl Key 1. Suppose tht f is uniformly continuous on set S R nd {x n } is Cuchy sequence in S. Prove tht {f(x n )} is Cuchy sequence. (f is not ssumed to be continuous outside
More informationProblem Set 4: Solutions Math 201A: Fall 2016
Problem Set 4: s Mth 20A: Fll 206 Problem. Let f : X Y be onetoone, onto mp between metric spces X, Y. () If f is continuous nd X is compct, prove tht f is homeomorphism. Does this result remin true
More informationReview of Calculus, cont d
Jim Lmbers MAT 460 Fll Semester 200910 Lecture 3 Notes These notes correspond to Section 1.1 in the text. Review of Clculus, cont d Riemnn Sums nd the Definite Integrl There re mny cses in which some
More informationCzechoslovak Mathematical Journal, 55 (130) (2005), , Abbotsford. 1. Introduction
Czechoslovk Mthemticl Journl, 55 (130) (2005), 933 940 ESTIMATES OF THE REMAINDER IN TAYLOR S THEOREM USING THE HENSTOCKKURZWEIL INTEGRAL, Abbotsford (Received Jnury 22, 2003) Abstrct. When relvlued
More information63. Representation of functions as power series Consider a power series. ( 1) n x 2n for all 1 < x < 1
3 9. SEQUENCES AND SERIES 63. Representtion of functions s power series Consider power series x 2 + x 4 x 6 + x 8 + = ( ) n x 2n It is geometric series with q = x 2 nd therefore it converges for ll q =
More informationDefinite integral. Mathematics FRDIS MENDELU
Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová Brno 1 Motivtion  re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function defined on [, b]. Wht is the re of the
More informationDefinite integral. Mathematics FRDIS MENDELU. Simona Fišnarová (Mendel University) Definite integral MENDELU 1 / 30
Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová (Mendel University) Definite integrl MENDELU / Motivtion  re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function
More informationPROBLEMS AND NOTES: UNIFORM CONVERGENCE AND POLYNOMIAL APPROXIMATION
PROBLEMS AND NOTES: UNIFORM CONVERGENCE AND POLYNOMIAL APPROXIMATION SAMEER CHAVAN Abstrct. These re the lecture notes prepred for the prticipnts of IST to be conducted t BP, Pune from 3rd to 15th November,
More informationSections 5.2: The Definite Integral
Sections 5.2: The Definite Integrl In this section we shll formlize the ides from the lst section to functions in generl. We strt with forml definition.. The Definite Integrl Definition.. Suppose f(x)
More informationAdvanced Calculus: MATH 410 Notes on Integrals and Integrability Professor David Levermore 17 October 2004
Advnced Clculus: MATH 410 Notes on Integrls nd Integrbility Professor Dvid Levermore 17 October 2004 1. Definite Integrls In this section we revisit the definite integrl tht you were introduced to when
More informationMAA 4212 Improper Integrals
Notes by Dvid Groisser, Copyright c 1995; revised 2002, 2009, 2014 MAA 4212 Improper Integrls The Riemnn integrl, while perfectly welldefined, is too restrictive for mny purposes; there re functions which
More informationLECTURE. INTEGRATION AND ANTIDERIVATIVE.
ANALYSIS FOR HIGH SCHOOL TEACHERS LECTURE. INTEGRATION AND ANTIDERIVATIVE. ROTHSCHILD CAESARIA COURSE, 2015/6 1. Integrtion Historiclly, it ws the problem of computing res nd volumes, tht triggered development
More informationRiemann is the Mann! (But Lebesgue may besgue to differ.)
Riemnn is the Mnn! (But Lebesgue my besgue to differ.) Leo Livshits My 2, 2008 1 For finite intervls in R We hve seen in clss tht every continuous function f : [, b] R hs the property tht for every ɛ >
More informationFundamental Theorem of Calculus and Computations on Some Special HenstockKurzweil Integrals
Fundmentl Theorem of Clculus nd Computtions on Some Specil HenstockKurzweil Integrls WeiChi YANG wyng@rdford.edu Deprtment of Mthemtics nd Sttistics Rdford University Rdford, VA 24142 USA DING, Xiofeng
More informationMAT612REAL ANALYSIS RIEMANN STIELTJES INTEGRAL
MAT612REAL ANALYSIS RIEMANN STIELTJES INTEGRAL DR. RITU AGARWAL MALVIYA NATIONAL INSTITUTE OF TECHNOLOGY, JAIPUR, INDIA302017 Tble of Contents Contents Tble of Contents 1 1. Introduction 1 2. Prtition
More informationRegulated functions and the regulated integral
Regulted functions nd the regulted integrl Jordn Bell jordn.bell@gmil.com Deprtment of Mthemtics University of Toronto April 3 2014 1 Regulted functions nd step functions Let = [ b] nd let X be normed
More informationCalculus in R. Chapter Di erentiation
Chpter 3 Clculus in R 3.1 Di erentition Definition 3.1. Suppose U R is open. A function f : U! R is di erentible t x 2 U if there exists number m such tht lim y!0 pple f(x + y) f(x) my y =0. If f is di
More informationODE: Existence and Uniqueness of a Solution
Mth 22 Fll 213 Jerry Kzdn ODE: Existence nd Uniqueness of Solution The Fundmentl Theorem of Clculus tells us how to solve the ordinry differentil eqution (ODE) du = f(t) dt with initil condition u() =
More informationMATH34032: Green s Functions, Integral Equations and the Calculus of Variations 1
MATH34032: Green s Functions, Integrl Equtions nd the Clculus of Vritions 1 Section 1 Function spces nd opertors Here we gives some brief detils nd definitions, prticulrly relting to opertors. For further
More informationMA 124 January 18, Derivatives are. Integrals are.
MA 124 Jnury 18, 2018 Prof PB s oneminute introduction to clculus Derivtives re. Integrls re. In Clculus 1, we lern limits, derivtives, some pplictions of derivtives, indefinite integrls, definite integrls,
More informationarxiv: v1 [math.ca] 7 Mar 2012
rxiv:1203.1462v1 [mth.ca] 7 Mr 2012 A simple proof of the Fundmentl Theorem of Clculus for the Lebesgue integrl Mrch, 2012 Rodrigo López Pouso Deprtmento de Análise Mtemátic Fcultde de Mtemátics, Universidde
More informationMA Handout 2: Notation and Background Concepts from Analysis
MA350059 Hndout 2: Nottion nd Bckground Concepts from Anlysis This hndout summrises some nottion we will use nd lso gives recp of some concepts from other units (MA20023: PDEs nd CM, MA20218: Anlysis 2A,
More information1 The Lagrange interpolation formula
Notes on Qudrture 1 The Lgrnge interpoltion formul We briefly recll the Lgrnge interpoltion formul. The strting point is collection of N + 1 rel points (x 0, y 0 ), (x 1, y 1 ),..., (x N, y N ), with x
More informationCalculus II: Integrations and Series
Clculus II: Integrtions nd Series August 7, 200 Integrls Suppose we hve generl function y = f(x) For simplicity, let f(x) > 0 nd f(x) continuous Denote F (x) = re under the grph of f in the intervl [,x]
More informationINDIAN INSTITUTE OF TECHNOLOGY BOMBAY MA205 Complex Analysis Autumn 2012
Lecture 6: Line Integrls INDIAN INSTITUTE OF TECHNOLOGY BOMBAY MA205 Complex Anlysis Autumn 2012 August 8, 2012 Lecture 6: Line Integrls Lecture 6: Line Integrls Lecture 6: Line Integrls Integrls of complex
More informationx = b a n x 2 e x dx. cdx = c(b a), where c is any constant. a b
CHAPTER 5. INTEGRALS 61 where nd x = b n x i = 1 (x i 1 + x i ) = midpoint of [x i 1, x i ]. Problem 168 (Exercise 1, pge 377). Use the Midpoint Rule with the n = 4 to pproximte 5 1 x e x dx. Some quick
More informationAnalysis Comp Study Guide
Anlysis Comp Study Guide The Rel nd Complex Number Systems nd Bsic Topology Theorem 1 (CuchySchwrz Inequlity). ( n ) 2 k b k b 2 k. 2 k As ( k tb k ) 2 0, s qudrtic in t it hs t most one root. So the
More informationRiemann Stieltjes Integration  Definition and Existence of Integral
 Definition nd Existence of Integrl Dr. Adity Kushik Directorte of Distnce Eduction Kurukshetr University, Kurukshetr Hryn 136119 Indi. Prtition Riemnn Stieltjes Sums Refinement Definition Given closed
More informationIntegrals  Motivation
Integrls  Motivtion When we looked t function s rte of chnge If f(x) is liner, the nswer is esy slope If f(x) is nonliner, we hd to work hrd limits derivtive A relted question is the re under f(x) (but
More informationMATH 174A: PROBLEM SET 5. Suggested Solution
MATH 174A: PROBLEM SET 5 Suggested Solution Problem 1. Suppose tht I [, b] is n intervl. Let f 1 b f() d for f C(I; R) (i.e. f is continuous relvlued function on I), nd let L 1 (I) denote the completion
More informationThe HenstockKurzweil integral
fculteit Wiskunde en Ntuurwetenschppen The HenstockKurzweil integrl Bchelorthesis Mthemtics June 2014 Student: E. vn Dijk First supervisor: Dr. A.E. Sterk Second supervisor: Prof. dr. A. vn der Schft
More informationDefinite Integrals. The area under a curve can be approximated by adding up the areas of rectangles = 1 1 +
Definite Integrls 5 The re under curve cn e pproximted y dding up the res of rectngles. Exmple. Approximte the re under y = from x = to x = using equl suintervls nd + x evluting the function t the lefthnd
More informationTOPICS IN FOURIER ANALYSISII. Contents
TOPICS IN FOURIER ANALYSISII M.T. NAIR Contents. Trigonometric series nd Fourier series 2 2. Riemnn Lebesgue Lemm 4 3. Dirichlet kernel 6 4. DirichletDini criterion for convergence 8 5. Ce`sro summblity
More informationRiemann Sums and Riemann Integrals
Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 2013 Outline 1 Riemnn Sums 2 Riemnn Integrls 3 Properties
More informationChapter 8: Methods of Integration
Chpter 8: Methods of Integrtion Bsic Integrls 8. Note: We hve the following list of Bsic Integrls p p+ + c, for p sec tn + c p + ln + c sec tn sec + c e e + c tn ln sec + c ln + c sec ln sec + tn + c ln
More informationThe Riemann Integral
Deprtment of Mthemtics King Sud University 20172018 Tble of contents 1 Antiderivtive Function nd Indefinite Integrls 2 3 4 5 Indefinite Integrls & Antiderivtive Function Definition Let f : I R be function
More informationarxiv: v1 [math.ca] 11 Jul 2011
rxiv:1107.1996v1 [mth.ca] 11 Jul 2011 Existence nd computtion of Riemnn Stieltjes integrls through Riemnn integrls July, 2011 Rodrigo López Pouso Deprtmento de Análise Mtemátic Fcultde de Mtemátics, Universidde
More information0.1 Properties of regulated functions and their Integrals.
MA244 Anlysis III Solutions. Sheet 2. NB. THESE ARE SKELETON SOLUTIONS, USE WISELY!. Properties of regulted functions nd their Integrls.. (Q.) Pick ny ɛ >. As f, g re regulted, there exist φ, ψ S[, b]:
More informationWeek 7 Riemann Stieltjes Integration: Lectures 1921
Week 7 Riemnn Stieltjes Integrtion: Lectures 1921 Lecture 19 Throughout this section α will denote monotoniclly incresing function on n intervl [, b]. Let f be bounded function on [, b]. Let P = { = 0
More informationRiemann Sums and Riemann Integrals
Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 203 Outline Riemnn Sums Riemnn Integrls Properties Abstrct
More information7 Improper Integrals, Exp, Log, Arcsin, and the Integral Test for Series
7 Improper Integrls, Exp, Log, Arcsin, nd the Integrl Test for Series We hve now ttined good level of understnding of integrtion of nice functions f over closed intervls [, b]. In prctice one often wnts
More informationON THE CINTEGRAL BENEDETTO BONGIORNO
ON THE CINTEGRAL BENEDETTO BONGIORNO Let F : [, b] R be differentible function nd let f be its derivtive. The problem of recovering F from f is clled problem of primitives. In 1912, the problem of primitives
More informationIndefinite Integral. Chapter Integration  reverse of differentiation
Chpter Indefinite Integrl Most of the mthemticl opertions hve inverse opertions. The inverse opertion of differentition is clled integrtion. For exmple, describing process t the given moment knowing the
More informationTHE EXISTENCEUNIQUENESS THEOREM FOR FIRSTORDER DIFFERENTIAL EQUATIONS.
THE EXISTENCEUNIQUENESS THEOREM FOR FIRSTORDER DIFFERENTIAL EQUATIONS RADON ROSBOROUGH https://intuitiveexplntionscom/picrdlindeloftheorem/ This document is proof of the existenceuniqueness theorem
More information1. On some properties of definite integrals. We prove
This short collection of notes is intended to complement the textbook Anlisi Mtemtic 2 by Crl Mdern, published by Città Studi Editore, [M]. We refer to [M] for nottion nd the logicl stremline of the rguments.
More informationChapter 28. Fourier Series An Eigenvalue Problem.
Chpter 28 Fourier Series Every time I close my eyes The noise inside me mplifies I cn t escpe I relive every moment of the dy Every misstep I hve mde Finds wy it cn invde My every thought And this is why
More informationMapping the delta function and other Radon measures
Mpping the delt function nd other Rdon mesures Notes for Mth583A, Fll 2008 November 25, 2008 Rdon mesures Consider continuous function f on the rel line with sclr vlues. It is sid to hve bounded support
More informationn f(x i ) x. i=1 In section 4.2, we defined the definite integral of f from x = a to x = b as n f(x i ) x; f(x) dx = lim i=1
The Fundmentl Theorem of Clculus As we continue to study the re problem, let s think bck to wht we know bout computing res of regions enclosed by curves. If we wnt to find the re of the region below the
More informationf(x i ) x. i=1 In section 4.2, we defined the definite integral of f from x = a to x = b as f(x) dx = lim f(x i ) x; i=1
The Fundmentl Theorem of Clculus As we continue to study the re problem, let s think bck to wht we know bout computing res of regions enclosed by curves. If we wnt to find the re of the region below the
More informationLecture 1: Introduction to integration theory and bounded variation
Lecture 1: Introduction to integrtion theory nd bounded vrition Wht is this course bout? Integrtion theory. The first question you might hve is why there is nything you need to lern bout integrtion. You
More informationFundamental Theorem of Calculus for Lebesgue Integration
Fundmentl Theorem of Clculus for Lebesgue Integrtion J. J. Kolih The existing proofs of the Fundmentl theorem of clculus for Lebesgue integrtion typiclly rely either on the Vitli Crthéodory theorem on
More information1. GaussJacobi quadrature and Legendre polynomials. p(t)w(t)dt, p {p(x 0 ),...p(x n )} p(t)w(t)dt = w k p(x k ),
1. GussJcobi qudrture nd Legendre polynomils Simpson s rule for evluting n integrl f(t)dt gives the correct nswer with error of bout O(n 4 ) (with constnt tht depends on f, in prticulr, it depends on
More informationMath& 152 Section Integration by Parts
Mth& 5 Section 7.  Integrtion by Prts Integrtion by prts is rule tht trnsforms the integrl of the product of two functions into other (idelly simpler) integrls. Recll from Clculus I tht given two differentible
More information7.2 The Definite Integral
7.2 The Definite Integrl the definite integrl In the previous section, it ws found tht if function f is continuous nd nonnegtive, then the re under the grph of f on [, b] is given by F (b) F (), where
More informationLecture 19: Continuous Least Squares Approximation
Lecture 19: Continuous Lest Squres Approximtion 33 Continuous lest squres pproximtion We begn 31 with the problem of pproximting some f C[, b] with polynomil p P n t the discrete points x, x 1,, x m for
More informationAntiderivatives/Indefinite Integrals of Basic Functions
Antiderivtives/Indefinite Integrls of Bsic Functions Power Rule: In prticulr, this mens tht x n+ x n n + + C, dx = ln x + C, if n if n = x 0 dx = dx = dx = x + C nd x (lthough you won t use the second
More informationAnalysis III. Ben Green. Mathematical Institute, Oxford address:
Anlysis III Ben Green Mthemticl Institute, Oxford Emil ddress: ben.green@mths.ox.c.uk 2000 Mthemtics Subject Clssifiction. Primry Contents Prefce 1 Chpter 1. Step functions nd the Riemnn integrl 3 1.1.
More informationLecture 19: Continuous Least Squares Approximation
Lecture 19: Continuous Lest Squres Approximtion 33 Continuous lest squres pproximtion We begn 31 with the problem of pproximting some f C[, b] with polynomil p P n t the discrete points x, x 1,, x m for
More informationMA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and the Fundamental Theorem of Calculus (pp.
MA123, Chpter 1: Formuls for integrls: integrls, ntiderivtives, nd the Fundmentl Theorem of Clculus (pp. 27233, Gootmn) Chpter Gols: Assignments: Understnd the sttement of the Fundmentl Theorem of Clculus.
More informationBest Approximation in the 2norm
Jim Lmbers MAT 77 Fll Semester 111 Lecture 1 Notes These notes correspond to Sections 9. nd 9.3 in the text. Best Approximtion in the norm Suppose tht we wish to obtin function f n (x) tht is liner combintion
More informationCalculus III Review Sheet
Clculus III Review Sheet 1 Definitions 1.1 Functions A function is f is incresing on n intervl if x y implies f(x) f(y), nd decresing if x y implies f(x) f(y). It is clled monotonic if it is either incresing
More information