MATH 104: INTRODUCTORY ANALYSIS SPRING 2008/09 PROBLEM SET 10 SOLUTIONS. f m. and. f m = 0. and x i = a + i. a + i. a + n 2. n(n + 1) = a(b a) +

Transcription

1 MATH 04: INTRODUCTORY ANALYSIS SPRING 008/09 PROBLEM SET 0 SOLUTIONS Throughout this problem set, B[, b] will deote the set of ll rel-vlued futios bouded o [, b], C[, b] the set of ll rel-vlued futios otiuous o [, b], d R[, b] the set of ll relvlued futios Riem itegrble o [, b] For exmple, we will write f R[, b] to me tht f is Riem itegrble o [, b] d f / R[, b] to me tht it is t Let 0 < < b d let f m : [, b] R be defied by { x m if x Q [, b], f m (x) 0 if x (R \ Q) [, b] For m,,, fid the vlues of b f m d Dedue tht f m / R[, b] i ll ses Solutio Note tht for y prtitio P of [, b], L(P, f m ) 0 sie the irrtiol umbers re dese, d so f m 0 Let P { x 0 < x < < x b} be regulr prtitio of [, b] ito subitervls, ie x i b ( ) b d x i + i for i,, Hee f m ( ) b U(P, f ) M i x i x i b [ ( )] b + i b ( ) b + i ( ) b ( + ) (b ) + ( b (b ) + + ) b + (b ) Dte: My, 009 (Versio 0)

2 Sie f U(P, f ) for every prtitio P, it follows tht for ll N Therefore f b + f b (b ) () By theorem i the letures, give y ε > 0, there exists δ > 0 suh tht U(P, f ) f + ε wheever P < δ Sie for lrge eough, we will hve P (b )/ < δ, so U(P, f) f + ε d so b (b b ) + f + ε Therefore b f + ε d by the rbitrriess of ε, it follows tht Hee by () d (), f / R[, b] sie b f b f 0 < b Similr rgumets with the regulr prtitio P yield f / (b/ / ) d f () f f b Let f, g B[, b] d let (, b) () Show tht f R[, b] if d oly if f R[, ] d f R[b, ] Show tht f f + Solutio This ppered s Theorem 6 i the textbook (b) Show tht if f hs oly fiitely my disotiuities i [, b], the f R[, b] Give exmple of f with ifitely my disotiuities o [, b] with f R[, b] d other exmple where f / R[, b] Solutio This ppered s Theorem 8 i the textbook Let f : [0, ] R be if x for some N, f (x) 0 otherwise The eh poit i {/ N} is poit of disotiuity of f but f R[0, ] Let f : [0, ] R be { if x Q [0, ], f (x) 0 otherwise f

3 The eh poit i Q [0, ] is poit of disotiuity but f / R[0, ] s we hve show i the letures () Suppose f R[, b] d there exists fiite subset E [, b] suh tht g(x) f(x) for ll x [, b] \ E Show tht g R[, b] d Solutio f By ssumptio, the futio f g is give { f(x) g(x) if x E, (f g)(x) 0 otherwise Sie this hs oly fiitely my disotiuity, f g R[, b] by (b) The Riem itegrl of y futio tht is o-zero t fiite umber of poits is 0 d so g (f g) 0 Let 0 < < b Evlute the followig itegrls usig Riem sums with pproprite hoie of itermedite poits x b dx, x dx, dx x Hit: use these ( x ξ i i + x i x + x ) /, ξ i ( xi + ) x x i x, ξ i Solutio Sie f(x) : x is mootoe o [, b] (0, ), it is i R[, b] By theorem i the letures, x dx lim S(P, f, ξ) P 0 for y hoie of ξ Let P { x 0 < x < < x b} be prtitio of [, b] d hoose ξ {ξ ξ ξ } with ( x ξ i i + x i x + x ) / Note tht we hve ξ i (x, x i ) sie ( x x (x ) / < i + x i x + x ) / < (x i ) / x i for i,, Now S(P, f, ξ) f(ξ i ) x i (x i + x i x + x )(x i x ) (x i x ) (x x 0) (b ) Sie this is true for y P, we hve lim P 0 S(P, f, ξ) (b ) Sie g(x) : /x is mootoe o [, b] (0, ), it is i R[, b] Let P { x 0 < x < < x b} be prtitio of [, b] d hoose ξ {ξ ξ ξ } with ξ i x i x

4 Note tht we hve ξ i (x, x i ) sie for i,, Now x S(P, g, ξ) x < x i x < g(ξ i ) x i ( Sie this is true for y P, we hve x x i ) x i x i x i x (x i x ) x 0 x b x dx lim S(P, g, ξ) P 0 b Sie h(x) : / x is mootoe o [, b] (0, ), it is i R[, b] Let P { x 0 < x < < x b} be prtitio of [, b] d hoose ξ {ξ ξ ξ } with ( xi + ) x ξ i Note tht we hve ξ i (x, x i ) sie ( x + ) x ( xi + ) x ( xi + ) x i x < < x i for i,, Now S(P, h, ξ) h(ξ i ) x i xi + x (x i x ) ( x i x ) ( x x 0 ) ( b ) Sie this is true for y P, we hve 4 Let, b R d < b () Show tht dx lim S(P, h, ξ) ( b ) x P 0 C[, b] R[, b] B[, b] Solutio R[, b] B[, b] sie we defied Riem itegrls oly for bouded futios (this eed ot be the se for more geerl defiitios of Riem itegrbility) C[, b] R[, b] sie every otiuous futio o [, b] is Riem itegrble by theorem i our letures (b) Let f R[, b] Let F : [, b] R be defied by F (x) x f(t) dt for y x [, b] Show tht F C[, b] Solutio By Problem (), sie f R[, b], therefore f R[, ] for y [, b] Hee F () exists for ll [, b] We lim tht for ll (, b), lim F (x) F () lim F (x) x + x 4

5 d lso lim F (x) F (), lim x F (x) F (b), x b + d therefore F is otiuous o [, b] by theorem i our letures Sie f B[, b], there exists M > 0 suh tht f(x) M for ll x [, b] Let ε > 0 d set δ ε/m If x < + δ, the by Problem () d Theorem 4 i the letures, x F (x) F () f(t) dt f(t) dt x f(t) dt + f(t) dt x f(t) dt x f(t) dt M(x ) < Mδ ε This shows tht lim x + F (x) F () for ll [, b) A essetilly idetil rgumet shows tht lim x F (x) F () for ll (, b] 5 Let f, g R[, b] Suppose either g(x) 0 for ll x [, b] or g(x) 0 for ll x [, b] Let m if f(x) d M sup f(x) x [,b] x [,b] () Show tht there exists k [m, M] suh tht Solutio Sie d g(x) 0, f(x)g(x) dx k g(x) dx We will ssume wlog tht g(x) 0 for ll x [, b] (if ot, osider g) Hee by Theorem 40 i the letures, m m f(x) M mg(x) f(x)g(x) Mg(x) g(x) dx f(x)g(x) dx M g(x) dx If g(x) dx 0, the f(x)g(x) dx 0 d so y k [m, M] would work Otherwise, set f(x)g(x) dx k : g(x) dx d ote tht k [m, M] (b) Show tht if f C[, b], the there exists (, b) suh tht f(x)g(x) dx f() g(x) dx Solutio We use the ottios i () If f is otiuous, the by the Itermedite Vlue Theorem, there exists (, b) suh tht f() k 5

6 () Show tht if f is mootoe iresig, the there exists (, b) suh tht f(x)g(x) dx m g(x) dx + M g(x) dx Solutio We will ssume wlog tht g(x) 0 for ll x [, b] (if ot, osider g) For x [, b], let y F (y) : m g(x) dx + M g(x) dx y Sie g R[, b], by Problem 4(b), F : [, b] R is otiuous o [, b] Sie f is mototoe iresig d g(x) 0, we hve mg(x) f(x)g(x) Mg(x) for ll x [, b] Hee by Theorem 40 i the letures, F (b) m g(x) dx f(x)g(x) dx M g(x) dx F () Sie F is otiuous, it follows from the Itermedite Vlue Theorem tht there exists (, b) suh tht F () whih gives the equtio we eed f(x)g(x) dx, 6