Week 7 Riemann Stieltjes Integration: Lectures 19-21

Save this PDF as:
 WORD  PNG  TXT  JPG

Size: px
Start display at page:

Download "Week 7 Riemann Stieltjes Integration: Lectures 19-21"

Transcription

1 Week 7 Riemnn Stieltjes Integrtion: Lectures 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 < 1,, n = b} be prtition of [, b]. Put α i = α( i ) α( i 1 ). M i = sup{f(x) : i 1 x i }. m i = inf{f(x) : i 1 x i }. U(P, f) = M i α i ; L(P, f) = m i α i. i=1 fdα = inf{u(p, f) : P }; Definition 1 If fdα = i=1 fdα = sup{l(p, f) : P }. fdα, then we sy f is Riemnn-Stieltjes (R-S) integrble w.r.t. to α nd denote this common vlue by b fdα := f(x)dα(x) := fdα = fdα. 0

2 Let R(α) denote the clss of ll R-S integrble functions on [, b]. Definition 2 A prtition P of [, b] is clled refinement of nother prtition P of [, b] if, points of P re ll present in P. We then write P P. Lemm 1 If P P then L(P ) L(P ) nd U(P ) U(P ). Enough to do this under the ssumption tht P hs one extr point thn P. And then it is obvious becuse if < b < c then inf{f(x) : x c} min{inf{f(x) : x b}, inf{f(x) : b x c} etc. Theorem 1 fdα fdα. For first of ll, becuse for every prtition P we hve U(P, f) L(P, f). Let P nd Q be ny two prtitions of [, b]. By tking common refinement T = P Q, nd pplying the bove lemm we get U(P ; f) U(T ; f) L(T ; f) L(Q; f) Now vrying Q over ll possible prtitions nd tking the supremum, we get U(P ) fdα. Now vrying P over ll prtitions of [, b] nd tking the infimum, we get the theorem. Theorem 2 Let f be bounded function nd α be monotoniclly incresing function. Then the following re equivlent. (i) f R(α). (ii) Given ɛ > 0, there exists prtition P of [, b] such tht U(P, f) L(P, f) < ɛ. 1

3 (iii) Given ɛ > 0, there exists prtition P of [, b] such tht for ll refinements of Q of P we hve U(Q, f) L(Q, f) < ɛ. (iv) Given ɛ > 0, there exists prtition P = { 0 < 1 < cdots < n } of [, b] such tht for rbitrry points t i, s i [ i 1, i ] we hve f(s i ) f(t i ) α i < ɛ. i=1 (v) There exists rel number η such tht for every ɛ > 0, there exists prtition P = { 0 < 1, < < n } of [, b] such tht for rbitrry points t i [ i 1, i ], we hve n i=1 f(t i) α i η < ɛ. Proof: (i) = (ii): By definition of the upper nd lower integrls, there exist prtitions Q, T such tht U(Q) fdα < ɛ/2; fdα L(T ) < ɛ/2. Tke common refinement P to Q nd T nd replce Q, T by P in the bove inequlities, nd then dd the two inequlities nd use the hypothesis (i) to conclude (ii). (ii) = (i): Since L(P ) fdα b fdα U(P ) the conclusion follows. (ii) = (iii): This follows from the previous theorem for if P P then (iii) = (ii): Obvious. L(P ) L(P ) U(P ) U(P ). (iii) = (iv): Note tht f(s i ) f(t i ) M i m i. Therefore, f(s i ) f(t i ) α i i i (M i m i ) α i = U(P, f) L(P, f) < ɛ. 2

4 (iv) = (iii): Choose points t i, s i [ i 1, i ] such tht m i f(s i ) < ɛ 2n α i, M i f(t i ) < ɛ 2n α i. Then U(P, f) L(P, f) i (M i m i ) α i i [ M i f(t i ) + m i f(s i ) + f(t i ) f(s i )] α i < 2ɛ. Thus so fr, we hve proved tht (i) to (iv) re ll equivlent to ech other. (i) = (v): We first note tht hving proved tht (i) to (iv) re ll equivlent, we cn use ny one of them. We tke η = fdα. Given ɛ > 0 we choose prtition P such tht L(P ) η < ɛ/3. nd prtition Q such tht (iv) holds with ɛ replced by ɛ/3. We then tke common refinement T of these two prtitions for which gin the sme would hold becuse of (iii). We now choose s i [ i 1, i ] such tht m i f(s i ) < ɛ 3n α i whenever α i is non zero. (If α i = 0 we cn tke s i to be ny point.) Then for rbitrry points t i [ i 1, i ], we hve = i f(t i) α i η [(f(t i ) f(s i ) + (f(s i ) m i ) + m i ] α i η i i f(s i ) f(t i ) α i + f(s i ) m i α i + L(P ) η i ɛ/3 + ɛ/3 + ɛ/3 = ɛ. (v) = (iv): Given ɛ > 0 choose prtition s in (v) with ɛ replced by ɛ/2. 3

5 Lecture 20 Fundmentl Properties of the Riemnn-Stieltjes Integrl Theorem 3 Let f, g be bounded functions nd α be n incresing function on n intervl [, b]. () Linerity in f : This just mens tht if f, g R(α), λ, µ R then λf + µg R(α). Moreover, (λf + µg) = λ fdα + µ fdα. (b) Semi-Linerity in α. This just mens if f R(α j ), j = 1, 2 λ j > 0 then f R(λ 1 α 1 + λ 2 α 2 ) nd moreover, fd(λ 1 α 1 + λ 2 α 2 ) = λ fdα 1 + µ fdα 2. (c) Let < c < b. Then f R(α) on [, b] if f R(α) on [, c] s well s on [c, b]. Moreover we hve fdα = c fdα + c fdα. (d) f 1 f 2 on [, b] nd f i R(α) then f 1dα f 2dα. (e) If f R(α) nd f(x) M then fdα M[α(b) α()]. (f) If f is continuous on [, b] then f R(α). (g) f : [, b] [c, d] is in R(α) nd φ : [c, d] R is continuous then φ f R(α). (h) If f R(α) then f 2 R(α). (i) If, f, g R(α) then fg R(α). (j) If f R(α) then f R(α) nd fdα f dα. 4

6 Proof: () Put h = f + g. Given ɛ > 0, choose prtitions P, Q of [, b] such tht U(P, f) L(P, f) < ɛ/2, P (Q, g) L(Q, g) < ɛ/2 nd replce these prtitions by their common refinement T nd then ppel to L(T, f) + L(T, g) L(T, h) U(T, h) U(T, f) + U(T, g). For constnt λ since U(P, λf) = λu(p, f); L(P, λf) = λl(p.f) it follows tht proof of (). λfdα = λ (b) This is esier: In ny prtition P we hve fdα. Combining these two we get the from which the conclusion follows. (λ 1 α 1 + λ 2 α 2 ) = λ 1 α 1 + λ 2 α 2 (c) All tht we do is to stick to those prtitions of [, b] which contin the point c. (d) This is esy nd (e) is consequence of (d). ɛ (f) Given ɛ > 0, put ɛ 1 =. Then by uniform continuity of f, α(b) α() there exists δ > 0 such tht f(t) f(s) < ɛ 1 whenever t, s [, b] nd t s < δ. Choose prtition P such tht α i < δ for ll i. Then it follows tht M i m i < ɛ 1 nd hence U(P ) L(P ) < ɛ. (g) Given ɛ > 0 by uniform continuity of φ, we get ɛ > δ > 0 such tht φ(t) φ(s) < ɛ for ll t, s [c, d] with t s < δ. There is prtition P of [, b] such tht U(P, f) L(P, f) < δ 2. 5

7 The differences M i m i my behve in two different wys: Accordingly let us define A = {1 i n : M i m i < δ}, B = {1, 2,..., n} \ A. Put h = φ f. It follows tht M i (h) m i (h) < ɛ, i A. Therefore we hve δ( α i ) i m i ) α i < U(P, f) L(P, f) < δ i B i B(M 2. Therefore we hve i B α i < δ. Now let K be bound for φ(t) on [c, d]. Then U(P, h) L(P, h) = i (M i(h) m i (h)) α i = i A (M i(h) m i (h)) α i + i B (M i(h) m i (h)) α i ɛ(α(b) α()) + 2Kδ < ɛ(α(b) α() + 2K). Since ɛ > 0 is rbitrry, we re done. (h) Follows from (g) by tking φ(t) = t 2. (i) Write fg = 1 4 [(f + g)2 (f g) 2 ]. (j) Tke φ(t) = t nd pply (g) to see tht f R(α). Now let λ = ±1 so tht λ fdα 0. Then fdα = λ fdα = λfdα f dα. This completes the proof of the theorem. Theorem 4 Suppose f is monotonic nd α is continuous nd monotoniclly incresing. Then f R(α). Proof: Given ɛ > 0, by uniform continuity of α we cn find prtition P such tht ech α i < ɛ. 6

8 Now if f is incresing, then we hve M i = f( i ), m i = f( i 1 ). Therefore, U(P ) L(P ) = i [f( i ) f( i 1 )] α i < f(b) f())ɛ. Since ɛ > 0 is rbitrry, we re done. Lecture 21 Theorem 5 Let f be bounded function on [, b] with finitely mny discontinuities. Suppose α is continuous t every point where f is discontinuous. Then f R(α). Proof: Becuse of (c) of theorem 3, it is enough to prove this for the cse when c [, b] is the only discontinuity of f. Put K = sup f(t). Given ɛ > 0, we cn find δ 1 > 0 such tht α(c + δ 1 ) α(c δ 1 ) < ɛ. By uniform continuity of f on [, b] \ (c δ, c + δ) we cn find δ 2 > 0 such tht x y < δ 2 implies f(x) f(y) < ɛ. Given ny prtition P of [, b] choose prtition Q which contins the points c nd whose mesh is less thn min{δ 1, δ 2 }. It follows tht U(Q) L(Q) < ɛ(α(b) α()) + 2Kɛ. Since ɛ > 0 is rbitrry this implies f R(α). Remrk 1 The bove result leds one to the following question. Assuming tht α is continuous on the whole of [, b], how lrge cn be the set of discontinuities of function f such tht f R(α)? The nswer is not within R-S theory. Lebesgue hs to invent new powerful theory which not only nswers this nd severl such questions rised by Riemnn integrtion theory but lso provides sound foundtion to the theory of probbility. Exmple 1 We shll denote the unit step function t 0 by U which 7

9 is defined s follows: U(x) = { 0, x 0; 1, x > 0. By shifting the origin t other points we cn get other unit step function. For exmple, suppose c [, b]. Consider α(x) = U(x c), x [, b]. For ny bounded function f : [, b] R, let us try to compute fdα. Consider ny prtition P of [, b] in which c = k. The only non zero α i is α k = 1. Therefore U(P ) L(P ) = M k (f) m k (f). Now ssume tht f is continuous t c. Then by choosing k+1 close to k = c, we cn mke M k m k 0. This mens tht f R(α). Indeed, it follows tht M k f(c) nd m k f(c). Therefore, fdα = f(c). Now suppose f hs discontinuity t c of the first kind i.e, in prticulr, f(c + ) exists. It then follows tht M k m k f(c) f(c + ). Therefore, f R(α) iff f(c + ) = f(c). Thus, we see tht it is possible to destroy integrbility by just disturbing the vlue of the function t one single point where α itself is discontinuous. In prticulr, tke f = α. It follows tht α R(α) on [, b]. We shll now prove prtil converse to (c) of Theorem 3. Theorem 6 Let f be bounded function nd α n incresing function on [, b]. Let c [, b] t which (t lest) f or α is continuous. If f R(α) on [, b] then f R(α) on both [, c] nd [c, b]; moreover, in tht cse, fdα = c fdα + c fdα. 8

10 Proof: Assume α is continuous t c. If T c is the trnsltion function T c (x) = x c then the functions g 1 = U T nd g 2 = 1 U T re both in R(α) since they re discontinuous only t c. Therefore fg 1, fg 2 R(α). But these respectively imply tht f R(α) on [c, b] nd on [, c]. We now consider the cse when f is continuous t c. We shll prove tht f R(α) on [, c], the proof tht f R(α) on [c, b] being similr. Recll tht the set of discontinuities of monotonic function is countble. Therefore there exist sequence of points c n in [, c] (we re ssuming tht < c) such tht c n c. By the erlier cse f R(α) on ech of the intervls [, c n ]. We clim tht the sequence s n := cn fdα converges to limit which is equl to c fdα. Let K > 0 be bound for α. Given ɛ > 0 we cn choose δ > 0 such tht for x, y [c δ, c + δ], f(x) f(y) < ɛ/2k. If n 0 is big enough then n, m n 0 implies tht s n s m < ɛ. This mens {s n } Cuchy nd hence is convergent with limit equl to sy, s. Now choose n so tht s s n < ɛ. Put = α(c) α(c ). Since c n c, from the left, it follows tht α(c n ) α(c ). Choose n lrge enough so tht where L is bound for f. α(c n ) α(c ) < ɛ/l Now, choose ny prtition Q of [, c n ] so tht U(Q, f) s n < ɛ. This is possible becuse f R(α) on [, c n ]. Put P = Q {c}, M = mx{f(x) : x [c n, c]}. Then s + f(c) U(P, f) s s n + s n U(Q, f) + f(c) (α(c) α(c n ))M ɛ + ɛ + (f(c) M) + (α(c n ) α(c ))M 2ɛ + ɛ 2K + M ɛ K 4ɛ. 9

11 Theorem 7 Let {c n } be sequence of non negtive rel numbers such tht n c n <. Let t n (, b) be sequence of distinct points in the open intervl nd let α = n c nu T tn. Then for ny continuous function f on [, b] we hve fdα = n c n f(t n ). Proof: Observe tht for ny x [, b], 0 n U(x t n) n c n nd hence α(x) mkes sense. Also clerly it is monotoniclly incresing nd α() = 0 nd α(b) = n c n. Given ɛ > 0 choose n 0 such tht n>n 0 c n < ɛ. Tke α 1 = n n 0 U T tn, α 2 = n>n 0 U T tn. By (b) of theorem 3, nd from the exmple bove, we hve fdα 1 = n n 0 c n f(t n ). If K is bound for f on [, b] we lso hve fdα 2 < K(α 2 (b) α 2 ()) = K c n = Mɛ. n>n 0 Therefore, This proves the clim. fdα c n f(t n ) < Kɛ. n n 0 Theorem 8 Let α be n incresing function nd α R(x) on [, b]. Then for ny bounded rel function on [, b], f R(α) iff fα R(x). Furthermore, in this cse, fdα = f(x)α (x)dx. 10

12 Proof: Given ɛ > 0, since α is Riemnn integrble, by (iv) of theorem 2, there exists prtition P = { = 0 < 1 < < n = b} of [, b] such tht for ll s i, t i [ i 1, i ] we hve, α (s i ) α (t i ) x i < ɛ. i=1 Apply MTV to α to obtin t i [ i 1, i ] such tht α i = α (t i ) x i. Put M = sup f(x). Then f(s i ) α i = i=1 f(s i )α (t i ) x i. i=1 Therefore, f(s i ) x i i=1 f(s i )α (s i ) x i < i i=1 f(s i ) α (s i ) α (t i ) x i > Mɛ. Therefore f(s i ) x i i=1 f(s i )α (s i ) x i + Mɛ U(P, fα ) + Mɛ. i=1 Since this is true for rbitrry s i [ i 1, i ], it follows tht U(P, f, α) U(P, fα ) + Mɛ. Likewise, we lso obtin U(P, fα ) U(P, f, α)) + Mɛ. Thus U(P, f, α) U(P, fα ) < Mɛ. Exctly in the sme mnner, we lso get L(P, f, α L(P, fα ) < Mɛ. 11

13 Note tht the bove two inequlities hold for refinements of P s well. Now suppose f R(α). we cn then ssume tht the prtition P is chosen so tht It then follows tht U(P, f, α) L(P, fα) < Mɛ. U(P, fα ) L(P, fα ) < 3Mɛ. Since ɛ > 0 is rbitrry, this implies fα is Riemnn integrble. The other wy impliction is similr. Moreover, the bove inequlities lso estblish the lst prt of the theorem. Remrk 2 The bove theorems illustrte the power of Stieltjes modifiction of Riemnn theory. In the first cse, α ws stircse function (lso clled pure step function). The integrl therein is reduced to finite or infinite sum. In the ltter cse, α is differentible function nd the integrl reduced to the ordinry Riemnn integrl. Thus the R-S theory brings brings unifiction of the discrete cse with the continuous cse, so tht we cn tret both of them in one go. As n illustrtive exmple, consider thin stright wire of finite length. The moment of inerti bout n xis perpendiculr to the wire nd through n end point is given by l 0 x 2 dm where m(x) denotes the mss of the segment [0, x] of the wire. If the mss is given by density function ρ, then m(x) = x ρ(t)dt or 0 equivlently, dm = ρ(x)dx nd the moment of inerti tkes form l 0 x 2 ρ(x)dx. On the other hnd if the mss is mde of finitely mny vlues m i 12

14 concentrted t points x i then the inerti tkes the form x 2 i m i. i Theorem 9 Chnge of Vrible formul Let φ : [, b] [c, d] be strictly incresing differentible function such tht φ() = c, φ(b) = d. Let α be n incresing function on [c, d] nd f be bounded function on [c, d] such tht f R(α). Put β = α φ, g = f φ. Then g R(β) nd we hve gdβ = d c fdα. Proof: Since φ is strictly incresing, it defines one-one correspondence of prtitions of [, b] with those of [c, d], given by { = 0 < 1 < < n = b} {c = φ() < φ( 1 ) < < φ( n ) = d}. Under this correspondence observe tht the vlue of the two functions f, g re the sme nd lso the vlue of function α, β re lso the sme. Therefore, the two upper sums lower sums re the sme nd hence the two upper nd lower integrls re the sme. The result follows. 13

15 Week 8 Functions of Bounded Vrition Lectures Lecture 22 : Functions of bounded Vrition Definition 3 Let f : [, b] R be ny function. For ech prtition P = { = 0 < 1 < < n = b} of [, b], consider the vritions Let V (P, f) = f( k ) f( k 1 ). k=1 V f = V f [, b] = sup{v (P, f) : P is prtition of [, b]}. If V f is finite we sy f is of bounded vrition on [, b]. Then V f is clled the totl vrition of f on [, b]. Let us denote the spce of ll functions of bounded vritions on [, b] by BV[, b]. Lemm 2 If Q is refinement of P then V (Q, f) V (P, f). Theorem 10 () f, g BV[, b], α, β R = αf + βg BV[, b]. Indeed, we lso hve V αf+βg α V f + β V g. (b) f BV[, b] = f is bounded on [, b]. 14

16 (c) f, g BV[, b] = fg BV[, b]. Indeed, if f K, g L then V fg LV f + KV g. (d) f BV[, b] nd f is bounded wy from 0 then 1/f BV[, b]. (e) Given c [, b], f BV[, b] iff f BV[, c] nd f BV[c, b]. Moreover, we hve V f [, b] = V f [, c] + V f [c, b]. (f) For ny f BV[, b] the function V f : [, b] R defined by V f () = 0 nd V f (x) = V f [, x], < x b, is n incresing function. (g) For ny f BV[, b], the function D f = V f f is n incresing function on [, b]. (h) Every monotonic function f on [, b] is of bounded vrition on [, b]. (i) Any function f : [, b] R is in BV[, b] iff it is the difference of two monotonic functions. (j) If f is continuous on [, b] nd differentible on (, b) with the derivtive f bounded on (, b), then f BV[, b]. (k) Let f BV[, b] nd continuous t c [, b] iff V f : [, b] R is continuous t c. Proof: () Indeed for every prtition, we hve V (P, αf+βg) = αv (P, f)+ βv (P, g). The result follows upon tking the supremum. (b) Tke M = V f + f(). Then f(x) f(x) f() f() V (P, f) + f() M, where P is ny prtition in which, x re consecutive terms. (c) For ny two points x, y we hve, f(x)g(x) f(y)g(y) f(x) f(x) g(y) + g(y) f(x) f(y) Therefore, it follows tht V (P, f) KV g + LV f. 15

17 (d) Let 0 < m < f(x) for ll x [, b]. Then 1 f(x) 1 f(y) = f(x) f(y) f(x)f(y) It follows tht V (P, 1/f) V f for ll prtitions P nd hence the result. m 2 (e) Given ny prtition P of [, c] we cn extend it to prtition Q of [, b] by including the intervl [c, b]. Then V (Q, f) = V (P, f) + f(b) f(c) nd hence it follows tht if f BV[, b] then f BV[, c]; for similr reson, f BV[c, b] s well. Conversely suppose f BV[, c] BV[c, b]. Given prtion P of [, b] we first refine it to P by dding the point c nd then write Q = Q 1 Q 2 where Q i re the resstrictions of Q to [, c], [c, b] respectively. It follows tht V (P, f) V (Q, f) = V (Q 1, f) + V (Q 2, f) V f ([, c] + V f [c, b]. Therefore f BV[, b]. In either cse, the bove inequlity lso shows tht V f [, b] V f [, c + V f [c, b]. On the other hnd, since V (P, f) V (P, f) for ll P it follows tht V f [, b] = sup{v (P, f) : P is prtition of [, b]}. Since every prtition P is of the form P = Q 1 Q 2 where Q 1, Q 2 re rbitrry prtitions of [, c] nd [c, b] respectively, nd V (P, f) = V (Q 1, f) + V (Q 2, f) it follows tht V f [, b] = V f [, c] + V f [c, b]. (f) Follows from (e) (g) Let x < y b. Proving V f [, x] f(x) V f [, y] f(y) is the 16

18 sme s proving V f, x] + f(y) f(x) V f [, y]. For ny prtition P of [, x] let P = P {y}. Then V (P, f) + f(y) f(x) V (P, f) + f(y) f(x) = V (P, f) V f [, y]. Since this is true for ll prtitions P of [, x] we re through. (h) My ssume f is incresing. But then for every prtition P we hve V (P, f) = f(b) f() nd hence V f = f(b) f(). (i) If f BV[, b], from (f) nd (g), we hve f = V f (V f f) s difference of two incresing functions. The converse follows from () nd (h). (j) This is becuse then f stisfies Lipschitz condition f(x) f(y) M x y for ll x, y [, b]. Therefore for every prtition P we hve V (P, f) M(b ). (k) Observe tht V f is incresing nd hence V f (c ± ) exist. By (h) it follows tht sme is true for f. We shll show tht f(c) = f(c ± ) iff V f (c) = V f (c ± ) which would imply (k). So, ssume tht f(c) = f(c + ). Given ɛ > 0 we cn find δ 1 > 0 such tht f(x) f(c) < ɛ for ll c < x < c + δ 1, x [, b]. We cn lso choose prtition P = {c = x 0 < x 1 < < x n = b} such tht V f [c, b] ɛ < k f k. Put δ = min{δ 1, x 1 c}. Let now c < x < c + δ. Then V f (x) V f (c) = V f [c, x] = V f [c, b] V f [x, b] < ɛ + k f k V f [x, b] ɛ + f(x) f(c) + f(x 1 ) f(x) + k 2 f k V f [x, b] ɛ + ɛ + V f [x, b] V f [x, b] = 2ɛ. This proves tht V f (c + ) = V f (c) s required. 17

19 Conversely, suppose V f (c + ) = V f (c). Then given ɛ > 0 we cn find δ > 0 such tht for ll c < x < c + δ we hve V f (x) V f (c) < ɛ. But then given x, y such tht c < y < x < c + δ it follows tht f(y) f(c) + f(x) f(y) V f ([c, x] = V f (x) V f (c) < ɛ which definitely implies tht f(x) f(y) ɛ. This completes the proof tht V f (c + ) = V f (c) iff f(c + ) = f(c). Similr rguments will prove tht V f (c ) = V f (c) iff f(c ) = f(c). Exmple 2 Not ll continuous functions on closed nd bounded intervl re of bounded vrition. A typicl exmples is f : [0, π] R defined by f(x) = { For ech n consider the prtition P = {0, x cos ( π x), x 0 0, x = n, 1 2n 1,..., 1} Then V (P, f) = n 1 k=1. As n, we know this tends to. k However, the function g(x) = xf(x) is of bounded vrition. To see this observe tht g is differentible in [0, 1] nd the derivtive is bounded (though not continuous) nd so we cn pply (j) of the bove theorem. Also note tht even prtil converse to (j) is not true, i.e., differentible function of bounded vrition need not hve its derivtive bounded. For exmple h(x) = x 1/3, being incresing function, is of bounded vrition on [0, 1] but its derivtive is not bounded. Remrk 3 We re now going extend the R-S integrl with integrtors α not necessrily incresing functions. In this connection, it should be noted tht condition (v) of theorem 63 becomes the strongest nd hence we dopt tht s the definition. 18

20 Definition 4 Let f, α : [, b] R be ny two functions. We sy f is R-S integrble with respect to to α nd write f R(α) if there exists rel number η such tht for every ɛ > 0 there exists prtition P of [, b] such tht for every refinement Q = { + x 0 < x 1 < < x n = b} of P nd points t i [x i 1, x i ] we hve f(t i ) α i η < ɛ. i=1 We then write η = fdα nd cll it R-S integrl of f with respect to α. It should be noted tht, in this generl sitution, severl properties listed in Theorem 64 my not be vlid. However, property (b) of Theorem 64 is vlid nd indeed becomes better. Lemm 3 For ny two functions α, β nd rel numbers λ, µ, if f R(α) R(β), then f R(λα + µβ). Moreover, in this cse we hve fd(λ α + µβ) = λ fdα + µ fdβ. Proof: This is so becuse for ny fixed prtition we hve the linerity property of : (λα + µβ) i = (λα + µβ)(x i x i 1 ) = λ( α) i + µ( β) i And hence the sme is true of the R-S sums. Therefore, if η = fdα, γ = fdβ, then it follows tht λη + µγ = fd(λα + µβ). 19

21 Theorem 11 Let α be function of bounded vrition nd let V denote its totl vrition function V : [, b] R defined by V (x) = V α [, x]. Let f be ny bounded function. Then f R(α) iff f R(V α ) nd f R(V α). Proof: The if prt is esy becuse of (). Also, we need only prove tht if f R(α) then f R(V ). Given ɛ > 0 choose prtition P ɛ so tht for ll refinements P of P ɛ, nd for ll choices of t k, s k [ i 1, i ], we hve, k ) f(s k )) α k < ɛ, V f (b) < k=1(f(t k We shll estblish tht U(P, f, V ) L(P, f, V ) < ɛk α k + ɛ. for some constnt K. By dding nd subtrcting, this tsk my be broken up into estblishing two inequlities [M k (f) m k (f)][ V k α k ] < ɛk/2; k [M k (f) m k (f)] α k < ɛk/2. Now observe tht V k α k 0 for ll k. Therefore if M is bound for f, then k [M k(f) m k (f)][ V k α k ] 2M k ( V k α k ) 2M(V f (b) α k ) < 2Mɛ. To prove the second inequlity, let us put A = {k : α k 0}; B = {1, 2,..., n} \ A. For k A choose t k, s k [ k 1, k ] such tht f(t k ) f(s k ) > M k (f) m k (f) ɛ; 20 k

22 nd for k B choose them so tht We then hve f(s k ) f(t k ) > M k (f) m k (f) ɛ. k [M k(f) m k (f)] α k < k A (f(t k) f(s k )) α k + k B (f(s k) f(t k )) k + ɛ k α k = k [f(t k) f(s k )) α k + ɛv (b) = ɛ(1 + V (b)). Putting K = mx{2m, 1 + V (b)} we re done. Corollry 1 Let α : [, b] R be of bounded vrition nd f : [, b] R be ny function. If f R(α) on [, b] then it is so on every subintervl [c, d] of [, b]. Corollry 2 Let f : [, b] R be of bounded vrition nd α : [, b] R be continuous of bounded vrition. Then f R(α). Proof: By (k) of the bove theorem, we see tht V (α) nd V (α) α re both continuous nd incresing. Hence by previous theorem, V (f) nd V (f) f re both integrble with respect to V (α) nd V (α) α. Now we just use the dditive property. Lecture 24 Exmple 3 : 1. Consider the double sequence, s m,n = m, m, n 1. m + n 21

23 Compute the two iterted limits nd record your results. lim m lim n s m,n, lim n lim m s m,n x 2 2. Let f n (x) = (1 + x 2 ), x R, n 1 nd put f(x) = n n f n(x). Check tht f n is continuous. Compute f nd see tht f is not continuous. 3. Define g m (x) = lim n (cos m!πx) 2n nd put g(x) = lim m g m (x). Compute g nd see tht g is discontinuous everywhere. Directly check tht ech g m is Riemnn integrble wheres g is not Riemnn integrble. sin nx 4. Consider the sequence h n (x) = nd put h(x) = lim n f n (x). n Check tht h 0. On the other hnd, compute lim n h n(x). Wht do you conclude? 5. Put λ n (x) = n 2 x(1 x 2 ) n, 0 x 1. Compute the lim n λ n (x). On the other hnd check tht Therefore we hve 1 0 = lim n [ 1 0 λ n (x)dx = ] λ n (x)dx n2 2n [lim n λ n (x)]dx = 0. These re ll exmples wherein certin nice properties of functions fil to be preserved under point-wise limit of functions. And we hve seen enough results to show tht these properties re preserved under uniform convergence. We know tht if sequence of continuous functions converges uniformly to function, then the limit function is continuous. We cn 22

24 now sk for the converse: Suppose sequence of continuous functions f n converges pointwise to function f which is lso continuous. the convergence uniform? The nswer in generl is NO. But there is sitution when we cn sy yes s well. Theorem 12 Let X be compct metric spce f n : X R be sequence of continuous functions converging pointwise to continuous function f. Suppose further tht f n is monotone. Then the f n f uniformly on X. Proof: Recll tht f is monotone incresing mens f n (x) f n+1 (x) for ll n nd for ll x. Likewise f n is monotone decresing mens f n (x) f n+1 (x) for ll n nd for ll x. Therefore, we cn define g n (x) = f(x) f n (x) to obtin sequence of continuous functions which monotoniclly decreses to 0. It suffices to prove tht g n converges uniformly. Given ɛ > 0 we wnt to find n 0 such tht g n (x) < ɛ for ll n n 0 nd for ll x X. Put K n = {x X : g n (x) ɛ}. Then ech K n is closed subset of X. Also g n (x) g n+1 (x) it follows tht K n+1 K n. On the other hnd, sunce g n (x) 0 it follows tht n K n =. Since this is hppening in compct spce X we conclude tht K n0 = for n 0. Remrk 4 The compctness is crucil s illustrted by the exmple: f n (x) = 1 nx + 1, 0 < x < 1. Is 23

MAT612-REAL ANALYSIS RIEMANN STIELTJES INTEGRAL

MAT612-REAL ANALYSIS RIEMANN STIELTJES INTEGRAL MAT612-REAL ANALYSIS RIEMANN STIELTJES INTEGRAL DR. RITU AGARWAL MALVIYA NATIONAL INSTITUTE OF TECHNOLOGY, JAIPUR, INDIA-302017 Tble of Contents Contents Tble of Contents 1 1. Introduction 1 2. Prtition

More information

Math 554 Integration

Math 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 information

1 i n x i x i 1. Note that kqk kp k. In addition, if P and Q are partition of [a, b], P Q is finer than both P and Q.

1 i n x i x i 1. Note that kqk kp k. In addition, if P and Q are partition of [a, b], P Q is finer than both P and Q. Chpter 6 Integrtion In this chpter we define the integrl. Intuitively, it should be the re under curve. Not surprisingly, fter mny exmples, counter exmples, exceptions, generliztions, the concept of the

More information

Properties of the Riemann Stieltjes Integral

Properties of the Riemann Stieltjes Integral Properties of the Riemnn Stieltjes Integrl Theorem (Linerity Properties) Let < c < d < b nd A,B IR nd f,g,α,β : [,b] IR. () If f,g R(α) on [,b], then Af +Bg R(α) on [,b] nd [ ] b Af +Bg dα A +B (b) If

More information

Lecture 1. Functional series. Pointwise and uniform convergence.

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 information

Riemann Stieltjes Integration - Definition and Existence of Integral

Riemann 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 information

Homework 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[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 information

Lecture 3. Limits of Functions and Continuity

Lecture 3. Limits of Functions and Continuity Lecture 3 Limits of Functions nd Continuity Audrey Terrs April 26, 21 1 Limits of Functions Notes I m skipping the lst section of Chpter 6 of Lng; the section bout open nd closed sets We cn probbly live

More information

The Riemann-Stieltjes Integral

The Riemann-Stieltjes Integral Chpter 6 The Riemnn-Stieltjes Integrl 6.1. Definition nd Eistene of the Integrl Definition 6.1. Let, b R nd < b. ( A prtition P of intervl [, b] is finite set of points P = { 0, 1,..., n } suh tht = 0

More information

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

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

More information

MATH 409 Advanced Calculus I Lecture 19: Riemann sums. Properties of integrals.

MATH 409 Advanced Calculus I Lecture 19: Riemann sums. Properties of integrals. MATH 409 Advnced Clculus I Lecture 19: Riemnn sums. Properties of integrls. Drboux sums Let P = {x 0,x 1,...,x n } be prtition of n intervl [,b], where x 0 = < x 1 < < x n = b. Let f : [,b] R be bounded

More information

The final exam will take place on Friday May 11th from 8am 11am in Evans room 60.

The 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 information

Chapter 6. Riemann Integral

Chapter 6. Riemann Integral Introduction to Riemnn integrl Chpter 6. Riemnn Integrl Won-Kwng Prk Deprtment of Mthemtics, The College of Nturl Sciences Kookmin University Second semester, 2015 1 / 41 Introduction to Riemnn integrl

More information

Math 324 Course Notes: Brief description

Math 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 information

Math 360: A primitive integral and elementary functions

Math 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 information

Week 10: Riemann integral and its properties

Week 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 information

For a continuous function f : [a; b]! R we wish to define the Riemann integral

For 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 information

Math 8 Winter 2015 Applications of Integration

Math 8 Winter 2015 Applications of Integration Mth 8 Winter 205 Applictions of Integrtion Here re few importnt pplictions of integrtion. The pplictions you my see on n exm in this course include only the Net Chnge Theorem (which is relly just the Fundmentl

More information

MATH34032: Green s Functions, Integral Equations and the Calculus of Variations 1

MATH34032: 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 information

NOTES AND PROBLEMS: INTEGRATION THEORY

NOTES AND PROBLEMS: INTEGRATION THEORY NOTES AND PROBLEMS: INTEGRATION THEORY SAMEER CHAVAN Abstrct. These re the lecture notes prepred for prticipnts of AFS-I to be conducted t Kumun University, Almor from 1st to 27th December, 2014. Contents

More information

38 Riemann sums and existence of the definite integral.

38 Riemann sums and existence of the definite integral. 38 Riemnn sums nd existence of the definite integrl. In the clcultion of the re of the region X bounded by the grph of g(x) = x 2, the x-xis nd 0 x b, two sums ppered: ( n (k 1) 2) b 3 n 3 re(x) ( n These

More information

The First Fundamental Theorem of Calculus. If f(x) is continuous on [a, b] and F (x) is any antiderivative. f(x) dx = F (b) F (a).

The First Fundamental Theorem of Calculus. If f(x) is continuous on [a, b] and F (x) is any antiderivative. f(x) dx = F (b) F (a). The Fundmentl Theorems of Clculus Mth 4, Section 0, Spring 009 We now know enough bout definite integrls to give precise formultions of the Fundmentl Theorems of Clculus. We will lso look t some bsic emples

More information

Mapping the delta function and other Radon measures

Mapping 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 information

MATH 409 Advanced Calculus I Lecture 18: Darboux sums. The Riemann integral.

MATH 409 Advanced Calculus I Lecture 18: Darboux sums. The Riemann integral. MATH 409 Advnced Clculus I Lecture 18: Drboux sums. The Riemnn integrl. Prtitions of n intervl Definition. A prtition of closed bounded intervl [, b] is finite subset P [,b] tht includes the endpoints

More information

Chapter 4. Lebesgue Integration

Chapter 4. Lebesgue Integration 4.2. Lebesgue Integrtion 1 Chpter 4. Lebesgue Integrtion Section 4.2. Lebesgue Integrtion Note. Simple functions ply the sme role to Lebesgue integrls s step functions ply to Riemnn integrtion. Definition.

More information

Line Integrals. Partitioning the Curve. Estimating the Mass

Line Integrals. Partitioning the Curve. Estimating the Mass Line Integrls Suppose we hve curve in the xy plne nd ssocite density δ(p ) = δ(x, y) t ech point on the curve. urves, of course, do not hve density or mss, but it my sometimes be convenient or useful to

More information

Functions of bounded variation

Functions of bounded variation Division for Mthemtics Mrtin Lind Functions of bounded vrition Mthemtics C-level thesis Dte: 2006-01-30 Supervisor: Viktor Kold Exminer: Thoms Mrtinsson Krlstds universitet 651 88 Krlstd Tfn 054-700 10

More information

a n+2 a n+1 M n a 2 a 1. (2)

a 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 information

MATH 174A: PROBLEM SET 5. Suggested Solution

MATH 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 rel-vlued function on I), nd let L 1 (I) denote the completion

More information

Integrals along Curves.

Integrals along Curves. Integrls long Curves. 1. Pth integrls. Let : [, b] R n be continuous function nd let be the imge ([, b]) of. We refer to both nd s curve. If we need to distinguish between the two we cll the function the

More information

2 Definitions and Basic Properties of Extended Riemann Stieltjes Integrals

2 Definitions and Basic Properties of Extended Riemann Stieltjes Integrals 2 Definitions nd Bsic Properties of Extended Riemnn Stieltjes Integrls 2.1 Regulted nd Intervl Functions Regulted functions Let X be Bnch spce, nd let J be nonempty intervl in R, which my be bounded or

More information

Polynomial Approximations for the Natural Logarithm and Arctangent Functions. Math 230

Polynomial Approximations for the Natural Logarithm and Arctangent Functions. Math 230 Polynomil Approimtions for the Nturl Logrithm nd Arctngent Functions Mth 23 You recll from first semester clculus how one cn use the derivtive to find n eqution for the tngent line to function t given

More information

MA Handout 2: Notation and Background Concepts from Analysis

MA 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 information

Introduction to Real Analysis (Math 315) Martin Bohner

Introduction to Real Analysis (Math 315) Martin Bohner ntroduction to Rel Anlysis (Mth 315) Spring 2005 Lecture Notes Mrtin Bohner Author ddress: Version from April 20, 2005 Deprtment of Mthemtics nd Sttistics, University of Missouri Roll, Roll, Missouri 65409-0020

More information

Lecture 14: Quadrature

Lecture 14: Quadrature Lecture 14: Qudrture This lecture is concerned with the evlution of integrls fx)dx 1) over finite intervl [, b] The integrnd fx) is ssumed to be rel-vlues nd smooth The pproximtion of n integrl by numericl

More information

T b a(f) [f ] +. P b a(f) = Conclude that if f is in AC then it is the difference of two monotone absolutely continuous functions.

T b a(f) [f ] +. P b a(f) = Conclude that if f is in AC then it is the difference of two monotone absolutely continuous functions. Rel Vribles, Fll 2014 Problem set 5 Solution suggestions Exerise 1. Let f be bsolutely ontinuous on [, b] Show tht nd T b (f) P b (f) f (x) dx [f ] +. Conlude tht if f is in AC then it is the differene

More information

STURM-LIOUVILLE BOUNDARY VALUE PROBLEMS

STURM-LIOUVILLE BOUNDARY VALUE PROBLEMS STURM-LIOUVILLE BOUNDARY VALUE PROBLEMS Throughout, we let [, b] be bounded intervl in R. C 2 ([, b]) denotes the spce of functions with derivtives of second order continuous up to the endpoints. Cc 2

More information

Math 4200: Homework Problems

Math 4200: Homework Problems Mth 4200: Homework Problems Gregor Kovčič 1. Prove the following properties of the binomil coefficients ( n ) ( n ) (i) 1 + + + + 1 2 ( n ) (ii) 1 ( n ) ( n ) + 2 + 3 + + n 2 3 ( ) n ( n + = 2 n 1 n) n,

More information

Best Approximation in the 2-norm

Best Approximation in the 2-norm Jim Lmbers MAT 77 Fll Semester 1-11 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 information

Czechoslovak Mathematical Journal, 55 (130) (2005), , Abbotsford. 1. Introduction

Czechoslovak 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 HENSTOCK-KURZWEIL INTEGRAL, Abbotsford (Received Jnury 22, 2003) Abstrct. When rel-vlued

More information

Riemann Integrals and the Fundamental Theorem of Calculus

Riemann Integrals and the Fundamental Theorem of Calculus Riemnn Integrls nd the Fundmentl Theorem of Clculus Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University September 16, 2013 Outline Grphing Riemnn Sums

More information

MORE FUNCTION GRAPHING; OPTIMIZATION. (Last edited October 28, 2013 at 11:09pm.)

MORE FUNCTION GRAPHING; OPTIMIZATION. (Last edited October 28, 2013 at 11:09pm.) MORE FUNCTION GRAPHING; OPTIMIZATION FRI, OCT 25, 203 (Lst edited October 28, 203 t :09pm.) Exercise. Let n be n rbitrry positive integer. Give n exmple of function with exctly n verticl symptotes. Give

More information

Sections 5.2: The Definite Integral

Sections 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 information

Convex Sets and Functions

Convex Sets and Functions B Convex Sets nd Functions Definition B1 Let L, +, ) be rel liner spce nd let C be subset of L The set C is convex if, for ll x,y C nd ll [, 1], we hve 1 )x+y C In other words, every point on the line

More information

440-2 Geometry/Topology: Differentiable Manifolds Northwestern University Solutions of Practice Problems for Final Exam

440-2 Geometry/Topology: Differentiable Manifolds Northwestern University Solutions of Practice Problems for Final Exam 440-2 Geometry/Topology: Differentible Mnifolds Northwestern University Solutions of Prctice Problems for Finl Exm 1) Using the cnonicl covering of RP n by {U α } 0 α n, where U α = {[x 0 : : x n ] RP

More information

More Properties of the Riemann Integral

More Properties of the Riemann Integral More Properties of the Riemnn Integrl Jmes K. Peterson Deprtment of Biologil Sienes nd Deprtment of Mthemtil Sienes Clemson University Februry 15, 2018 Outline More Riemnn Integrl Properties The Fundmentl

More information

Calculus and linear algebra for biomedical engineering Week 11: The Riemann integral and its properties

Calculus and linear algebra for biomedical engineering Week 11: The Riemann integral and its properties Clculus nd liner lgebr for biomedicl engineering Week 11: The Riemnn integrl nd its properties Hrtmut Führ fuehr@mth.rwth-chen.de Lehrstuhl A für Mthemtik, RWTH Achen Jnury 9, 2009 Overview 1 Motivtion:

More information

63. Representation of functions as power series Consider a power series. ( 1) n x 2n for all 1 < x < 1

63. 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 information

Section 6.1 Definite Integral

Section 6.1 Definite Integral Section 6.1 Definite Integrl Suppose we wnt to find the re of region tht is not so nicely shped. For exmple, consider the function shown elow. The re elow the curve nd ove the x xis cnnot e determined

More information

Section 6.1 INTRO to LAPLACE TRANSFORMS

Section 6.1 INTRO to LAPLACE TRANSFORMS Section 6. INTRO to LAPLACE TRANSFORMS Key terms: Improper Integrl; diverge, converge A A f(t)dt lim f(t)dt Piecewise Continuous Function; jump discontinuity Function of Exponentil Order Lplce Trnsform

More information

The Wave Equation I. MA 436 Kurt Bryan

The Wave Equation I. MA 436 Kurt Bryan 1 Introduction The Wve Eqution I MA 436 Kurt Bryn Consider string stretching long the x xis, of indeterminte (or even infinite!) length. We wnt to derive n eqution which models the motion of the string

More information

Mathematical Analysis: Supplementary notes I

Mathematical Analysis: Supplementary notes I Mthemticl Anlysis: Supplementry notes I 0 FIELDS The rel numbers, R, form field This mens tht we hve set, here R, nd two binry opertions ddition, + : R R R, nd multipliction, : R R R, for which the xioms

More information

Handout: Natural deduction for first order logic

Handout: Natural deduction for first order logic MATH 457 Introduction to Mthemticl Logic Spring 2016 Dr Json Rute Hndout: Nturl deduction for first order logic We will extend our nturl deduction rules for sententil logic to first order logic These notes

More information

NUMERICAL INTEGRATION. The inverse process to differentiation in calculus is integration. Mathematically, integration is represented by.

NUMERICAL INTEGRATION. The inverse process to differentiation in calculus is integration. Mathematically, integration is represented by. NUMERICAL INTEGRATION 1 Introduction The inverse process to differentition in clculus is integrtion. Mthemticlly, integrtion is represented by f(x) dx which stnds for the integrl of the function f(x) with

More information

Lecture 19: Continuous Least Squares Approximation

Lecture 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 information

Section 4.8. D v(t j 1 ) t. (4.8.1) j=1

Section 4.8. D v(t j 1 ) t. (4.8.1) j=1 Difference Equtions to Differentil Equtions Section.8 Distnce, Position, nd the Length of Curves Although we motivted the definition of the definite integrl with the notion of re, there re mny pplictions

More information

(0.0)(0.1)+(0.3)(0.1)+(0.6)(0.1)+ +(2.7)(0.1) = 1.35

(0.0)(0.1)+(0.3)(0.1)+(0.6)(0.1)+ +(2.7)(0.1) = 1.35 7 Integrtion º½ ÌÛÓ Ü ÑÔÐ Up to now we hve been concerned with extrcting informtion bout how function chnges from the function itself. Given knowledge bout n object s position, for exmple, we wnt to know

More information

Best Approximation. Chapter The General Case

Best Approximation. Chapter The General Case Chpter 4 Best Approximtion 4.1 The Generl Cse In the previous chpter, we hve seen how n interpolting polynomil cn be used s n pproximtion to given function. We now wnt to find the best pproximtion to given

More information

Euler-Maclaurin Summation Formula 1

Euler-Maclaurin Summation Formula 1 Jnury 9, Euler-Mclurin 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 information

Big idea in Calculus: approximation

Big idea in Calculus: approximation Big ide in Clculus: pproximtion Derivtive: f (x) = df dx f f(x +h) f(x) =, x h rte of chnge is pproximtely the rtio of chnges in the function vlue nd in the vrible in very short time Liner pproximtion:

More information

8 Laplace s Method and Local Limit Theorems

8 Laplace s Method and Local Limit Theorems 8 Lplce s Method nd Locl Limit Theorems 8. Fourier Anlysis in Higher DImensions Most of the theorems of Fourier nlysis tht we hve proved hve nturl generliztions to higher dimensions, nd these cn be proved

More information

Improper Integrals. The First Fundamental Theorem of Calculus, as we ve discussed in class, goes as follows:

Improper Integrals. The First Fundamental Theorem of Calculus, as we ve discussed in class, goes as follows: Improper Integrls The First Fundmentl Theorem of Clculus, s we ve discussed in clss, goes s follows: If f is continuous on the intervl [, ] nd F is function for which F t = ft, then ftdt = F F. An integrl

More information

7.2 The Definition of the Riemann Integral. Outline

7.2 The Definition of the Riemann Integral. Outline 7.2 The Definition of the Riemnn Integrl Tom Lewis Fll Semester 2014 Upper nd lower sums Some importnt theorems Upper nd lower integrls The integrl Two importnt theorems on integrbility Outline Upper nd

More information

3.4 Numerical integration

3.4 Numerical integration 3.4. Numericl integrtion 63 3.4 Numericl integrtion In mny economic pplictions it is necessry to compute the definite integrl of relvlued function f with respect to "weight" function w over n intervl [,

More information

Continuous Random Variables

Continuous Random Variables STAT/MATH 395 A - PROBABILITY II UW Winter Qurter 217 Néhémy Lim Continuous Rndom Vribles Nottion. The indictor function of set S is rel-vlued function defined by : { 1 if x S 1 S (x) if x S Suppose tht

More information

7 - Continuous random variables

7 - Continuous random variables 7-1 Continuous rndom vribles S. Lll, Stnford 2011.01.25.01 7 - Continuous rndom vribles Continuous rndom vribles The cumultive distribution function The uniform rndom vrible Gussin rndom vribles The Gussin

More information

CS667 Lecture 6: Monte Carlo Integration 02/10/05

CS667 Lecture 6: Monte Carlo Integration 02/10/05 CS667 Lecture 6: Monte Crlo Integrtion 02/10/05 Venkt Krishnrj Lecturer: Steve Mrschner 1 Ide The min ide of Monte Crlo Integrtion is tht we cn estimte the vlue of n integrl by looking t lrge number of

More information

Math Calculus with Analytic Geometry II

Math Calculus with Analytic Geometry II orem of definite Mth 5.0 with Anlytic Geometry II Jnury 4, 0 orem of definite If < b then b f (x) dx = ( under f bove x-xis) ( bove f under x-xis) Exmple 8 0 3 9 x dx = π 3 4 = 9π 4 orem of definite Problem

More information

STEP FUNCTIONS, DELTA FUNCTIONS, AND THE VARIATION OF PARAMETERS FORMULA. 0 if t < 0, 1 if t > 0.

STEP FUNCTIONS, DELTA FUNCTIONS, AND THE VARIATION OF PARAMETERS FORMULA. 0 if t < 0, 1 if t > 0. STEP FUNCTIONS, DELTA FUNCTIONS, AND THE VARIATION OF PARAMETERS FORMULA STEPHEN SCHECTER. The unit step function nd piecewise continuous functions The Heviside unit step function u(t) is given by if t

More information

The problems that follow illustrate the methods covered in class. They are typical of the types of problems that will be on the tests.

The problems that follow illustrate the methods covered in class. They are typical of the types of problems that will be on the tests. ADVANCED CALCULUS PRACTICE PROBLEMS JAMES KEESLING The problems tht follow illustrte the methods covered in clss. They re typicl of the types of problems tht will be on the tests. 1. Riemnn Integrtion

More information

u(t)dt + i a f(t)dt f(t) dt b f(t) dt (2) With this preliminary step in place, we are ready to define integration on a general curve in C.

u(t)dt + i a f(t)dt f(t) dt b f(t) dt (2) With this preliminary step in place, we are ready to define integration on a general curve in C. Lecture 4 Complex Integrtion MATH-GA 2451.001 Complex Vriles 1 Construction 1.1 Integrting complex function over curve in C A nturl wy to construct the integrl of complex function over curve in the complex

More information

Numerical integration

Numerical integration 2 Numericl integrtion This is pge i Printer: Opque this 2. Introduction Numericl integrtion is problem tht is prt of mny problems in the economics nd econometrics literture. The orgniztion of this chpter

More information

Improper Integrals. Introduction. Type 1: Improper Integrals on Infinite Intervals. When we defined the definite integral.

Improper Integrals. Introduction. Type 1: Improper Integrals on Infinite Intervals. When we defined the definite integral. Improper Integrls Introduction When we defined the definite integrl f d we ssumed tht f ws continuous on [, ] where [, ] ws finite, closed intervl There re t lest two wys this definition cn fil to e stisfied:

More information

Variational Techniques for Sturm-Liouville Eigenvalue Problems

Variational Techniques for Sturm-Liouville Eigenvalue Problems Vritionl Techniques for Sturm-Liouville Eigenvlue Problems Vlerie Cormni Deprtment of Mthemtics nd Sttistics University of Nebrsk, Lincoln Lincoln, NE 68588 Emil: vcormni@mth.unl.edu Rolf Ryhm Deprtment

More information

different methods (left endpoint, right endpoint, midpoint, trapezoid, Simpson s).

different methods (left endpoint, right endpoint, midpoint, trapezoid, Simpson s). Mth 1A with Professor Stnkov Worksheet, Discussion #41; Wednesdy, 12/6/217 GSI nme: Roy Zho Problems 1. Write the integrl 3 dx s limit of Riemnn sums. Write it using 2 intervls using the 1 x different

More information

x = b a n x 2 e x dx. cdx = c(b a), where c is any constant. a b

x = 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 information

UniversitaireWiskundeCompetitie. Problem 2005/4-A We have k=1. Show that for every q Q satisfying 0 < q < 1, there exists a finite subset K N so that

UniversitaireWiskundeCompetitie. Problem 2005/4-A We have k=1. Show that for every q Q satisfying 0 < q < 1, there exists a finite subset K N so that Problemen/UWC NAW 5/7 nr juni 006 47 Problemen/UWC UniversitireWiskundeCompetitie Edition 005/4 For Session 005/4 we received submissions from Peter Vndendriessche, Vldislv Frnk, Arne Smeets, Jn vn de

More information

20 MATHEMATICS POLYNOMIALS

20 MATHEMATICS POLYNOMIALS 0 MATHEMATICS POLYNOMIALS.1 Introduction In Clss IX, you hve studied polynomils in one vrible nd their degrees. Recll tht if p(x) is polynomil in x, the highest power of x in p(x) is clled the degree of

More information

We know that if f is a continuous nonnegative function on the interval [a, b], then b

We know that if f is a continuous nonnegative function on the interval [a, b], then b 1 Ares Between Curves c 22 Donld Kreider nd Dwight Lhr We know tht if f is continuous nonnegtive function on the intervl [, b], then f(x) dx is the re under the grph of f nd bove the intervl. We re going

More information

Necessary and Sufficient Conditions for Differentiating Under the Integral Sign

Necessary and Sufficient Conditions for Differentiating Under the Integral Sign Necessry nd Sufficient Conditions for Differentiting Under the Integrl Sign Erik Tlvil 1. INTRODUCTION. When we hve n integrl tht depends on prmeter, sy F(x f (x, y dy, it is often importnt to know when

More information

7.6 The Use of Definite Integrals in Physics and Engineering

7.6 The Use of Definite Integrals in Physics and Engineering Arknss Tech University MATH 94: Clculus II Dr. Mrcel B. Finn 7.6 The Use of Definite Integrls in Physics nd Engineering It hs been shown how clculus cn be pplied to find solutions to geometric problems

More information

Summer MTH142 College Calculus 2. Section J. Lecture Notes. Yin Su University at Buffalo

Summer MTH142 College Calculus 2. Section J. Lecture Notes. Yin Su University at Buffalo Summer 6 MTH4 College Clculus Section J Lecture Notes Yin Su University t Bufflo yinsu@bufflo.edu Contents Bsic techniques of integrtion 3. Antiderivtive nd indefinite integrls..............................................

More information

l 2 p2 n 4n 2, the total surface area of the

l 2 p2 n 4n 2, the total surface area of the Week 6 Lectures Sections 7.5, 7.6 Section 7.5: Surfce re of Revolution Surfce re of Cone: Let C be circle of rdius r. Let P n be n n-sided regulr polygon of perimeter p n with vertices on C. Form cone

More information

Farey Fractions. Rickard Fernström. U.U.D.M. Project Report 2017:24. Department of Mathematics Uppsala University

Farey Fractions. Rickard Fernström. U.U.D.M. Project Report 2017:24. Department of Mathematics Uppsala University U.U.D.M. Project Report 07:4 Frey Frctions Rickrd Fernström Exmensrete i mtemtik, 5 hp Hledre: Andres Strömergsson Exmintor: Jörgen Östensson Juni 07 Deprtment of Mthemtics Uppsl University Frey Frctions

More information

Section 14.3 Arc Length and Curvature

Section 14.3 Arc Length and Curvature Section 4.3 Arc Length nd Curvture Clculus on Curves in Spce In this section, we ly the foundtions for describing the movement of n object in spce.. Vector Function Bsics In Clc, formul for rc length in

More information

Math 113 Exam 1-Review

Math 113 Exam 1-Review Mth 113 Exm 1-Review September 26, 2016 Exm 1 covers 6.1-7.3 in the textbook. It is dvisble to lso review the mteril from 5.3 nd 5.5 s this will be helpful in solving some of the problems. 6.1 Are Between

More information

(a) A partition P of [a, b] is a finite subset of [a, b] containing a and b. If Q is another partition and P Q, then Q is a refinement of P.

(a) A partition P of [a, b] is a finite subset of [a, b] containing a and b. If Q is another partition and P Q, then Q is a refinement of P. Chpter 7: The Riemnn Integrl When the derivtive is introdued, it is not hrd to see tht the it of the differene quotient should be equl to the slope of the tngent line, or when the horizontl xis is time

More information

A Convergence Theorem for the Improper Riemann Integral of Banach Space-valued Functions

A Convergence Theorem for the Improper Riemann Integral of Banach Space-valued Functions Interntionl Journl of Mthemticl Anlysis Vol. 8, 2014, no. 50, 2451-2460 HIKARI Ltd, www.m-hikri.com http://dx.doi.org/10.12988/ijm.2014.49294 A Convergence Theorem for the Improper Riemnn Integrl of Bnch

More information

PDE Notes. Paul Carnig. January ODE s vs PDE s 1

PDE Notes. Paul Carnig. January ODE s vs PDE s 1 PDE Notes Pul Crnig Jnury 2014 Contents 1 ODE s vs PDE s 1 2 Section 1.2 Het diffusion Eqution 1 2.1 Fourier s w of Het Conduction............................. 2 2.2 Energy Conservtion.....................................

More information

DIRECT CURRENT CIRCUITS

DIRECT CURRENT CIRCUITS DRECT CURRENT CUTS ELECTRC POWER Consider the circuit shown in the Figure where bttery is connected to resistor R. A positive chrge dq will gin potentil energy s it moves from point to point b through

More information

Lecture 17. Integration: Gauss Quadrature. David Semeraro. University of Illinois at Urbana-Champaign. March 20, 2014

Lecture 17. Integration: Gauss Quadrature. David Semeraro. University of Illinois at Urbana-Champaign. March 20, 2014 Lecture 17 Integrtion: Guss Qudrture Dvid Semerro University of Illinois t Urbn-Chmpign Mrch 0, 014 Dvid Semerro (NCSA) CS 57 Mrch 0, 014 1 / 9 Tody: Objectives identify the most widely used qudrture method

More information

arxiv: v1 [math.ca] 9 Jun 2011

arxiv: v1 [math.ca] 9 Jun 2011 Men vlue integrl inequlities rxiv:1106.1807v1 [mth.ca] 9 Jun 2011 June, 2011 Rodrigo López Pouso Deprtment of Mthemticl Anlysis Fculty of Mthemtics, University of Sntigo de Compostel, 15782 Sntigo de Compostel,

More information

1 Sequences. 2 Series. 2 SERIES Analysis Study Guide

1 Sequences. 2 Series. 2 SERIES Analysis Study Guide 2 SERIES Anlysis Study Guide 1 Sequences Def: An ordered field is field F nd totl order < (for ll x, y, z F ): (i) x < y, y < x or x = y, (ii) x < y, y < z x < z (iii) x < y x + z < y + z (iv) 0 < y, x

More information

Henstock Kurzweil delta and nabla integrals

Henstock Kurzweil delta and nabla integrals Henstock Kurzweil delt nd nbl integrls Alln Peterson nd Bevn Thompson Deprtment of Mthemtics nd Sttistics, University of Nebrsk-Lincoln Lincoln, NE 68588-0323 peterso@mth.unl.edu Mthemtics, SPS, The University

More information

MA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and the Fundamental Theorem of Calculus (pp.

MA123, 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. 27-233, Gootmn) Chpter Gols: Assignments: Understnd the sttement of the Fundmentl Theorem of Clculus.

More information

MATH 101A: ALGEBRA I PART B: RINGS AND MODULES 35

MATH 101A: ALGEBRA I PART B: RINGS AND MODULES 35 MATH 101A: ALGEBRA I PART B: RINGS AND MODULES 35 9. Modules over PID This week we re proving the fundmentl theorem for finitely generted modules over PID, nmely tht they re ll direct sums of cyclic modules.

More information

New Expansion and Infinite Series

New Expansion and Infinite Series Interntionl Mthemticl Forum, Vol. 9, 204, no. 22, 06-073 HIKARI Ltd, www.m-hikri.com http://dx.doi.org/0.2988/imf.204.4502 New Expnsion nd Infinite Series Diyun Zhng College of Computer Nnjing University

More information

ECO 317 Economics of Uncertainty Fall Term 2007 Notes for lectures 4. Stochastic Dominance

ECO 317 Economics of Uncertainty Fall Term 2007 Notes for lectures 4. Stochastic Dominance Generl structure ECO 37 Economics of Uncertinty Fll Term 007 Notes for lectures 4. Stochstic Dominnce Here we suppose tht the consequences re welth mounts denoted by W, which cn tke on ny vlue between

More information

Interpreting Integrals and the Fundamental Theorem

Interpreting Integrals and the Fundamental Theorem Interpreting Integrls nd the Fundmentl Theorem Tody, we go further in interpreting the mening of the definite integrl. Using Units to Aid Interprettion We lredy know tht if f(t) is the rte of chnge of

More information

AMATH 731: Applied Functional Analysis Fall Some basics of integral equations

AMATH 731: Applied Functional Analysis Fall Some basics of integral equations AMATH 731: Applied Functionl Anlysis Fll 2009 1 Introduction Some bsics of integrl equtions An integrl eqution is n eqution in which the unknown function u(t) ppers under n integrl sign, e.g., K(t, s)u(s)

More information