MATH 104: INTRODUCTORY ANALYSIS SPRING 2009/10 PROBLEM SET 8 SOLUTIONS. and x i = a + i. i + n(n + 1)(2n + 1) + 2a. (b a)3 6n 2
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1 MATH 104: INTRODUCTORY ANALYSIS SPRING 2009/10 PROBLEM SET 8 SOLUTIONS 6.9: Let f(x) { x 2 if x Q [, b], 0 if x (R \ Q) [, b], where > 0. Prove tht b. Solutio. Let P { x 0 < x 1 < < x b} be regulr prtitio of [, b] ito subitervls, i.e. x i b ( ) b d x i i for i 1,...,. Hee ( ) b U(P, f) M i x i x 2 i b [ ( )] b 2 i [ b ( ) ( ) ] b b 2 2 2i i 2 b ( ) b 2 ( ) b 2 2 i i 2 ( ) b 2 ( ) (b ) 2 ( 1) b ( 1)(2 1) b (b2 2 )(b ) 2 (b ) 6 2 b Note tht we hve used Exerises 1.1 d 1.14 for the sums. Sie f U(P, f) for every prtitio P, it follows tht f b for ll N. Therefore (b2 2 )(b ) 2 f b (b ) 6 2. (1) By Theorem 6.4, give y > 0, there exists δ > 0 suh tht U(P, f) f wheever P < δ. Sie for lrge eough, we will hve P (b )/ < δ, so Dte: My 6, 2010 (Versio 1.0). U(P, f) f 1
2 d so b (b2 2 )(b ) (b ) Therefore b f d by the rbitrriess of, it follows tht Hee by (1) d (2), b b f. f. (2) 6.12: Prove tht the followig futio is Riem-itegrble o [0, 1] eve though it hs ifiitely my disotiuities o [0, 1]: 1 if x 1 where 1, 2,,..., f(x) 0 otherwise. Solutio. Let > 0. There exists N N suh tht 1/ < /2 wheever > N. Let P be prtitio of [0, 1] where d suh tht So U(P, f) 0 x 0 < x 1 1 N 1 < x 2 < < x k 1 x i x i x i 1 < k 4N. for i 2,,..., k. M i x i 1 N 1 N 4N < 2 2 sie t most 2N of the umbers M 2,..., M k ould be 1 d the rest hve to be 0 (beuse [x 1, 1] [1/(N 1), 1] otis oly 1/N, 1/(N 1),..., 1/2, 1 where f tkes the vlue 1 the worst tht ould hppe is if ll these our mog x 2,..., x k ). Note tht L(P, f) m i x i 0 sie m i 0 for ll i 1, 2,..., k. Hee U(P, f) L(P, f) < d so f is Riem-itegrble by Theorem : Usig Defiitio 6. d Theorem 6.8, prove tht if is y rel umber the f provided f is Riem-itegrble o the lrgest of the itervls [, b], [, b], d [, ]. Solutio. If < < b, the proof of this is give i Theorem 6.8. For the se < b, Theorem 6.8 gi yields f b f f 2
3 d so f b by Defiitio 6.. Likewise for the se < b. 6.20: Prove Theorem 6.1. Hit: First prove tht the result holds whe f d g re oegtive itegble futios o [, b] d the use the idetity f fg (f f )(g g ) f g f g f g f g to estblish Theorem 6.1. Solutio. Suppose f d g re oegtive d Riem itegrble o [, b]. Let P { x 0 < x 1 < < x b} be prtitio. For y bouded futio F : [, b] R, deote m i (F ) For oegtive F, m(f ) Now sie f d g re oegtive, So U(P, fg) L(P, fg) if F (x) d M i(f ) sup F (x), x [x i 1,x i ] x [x i 1,x i ] if F (x) d M(F ) sup F (x). x [,b] x [,b] 0 m(f ) m i (F ) M i (F ) M(F ). M i (fg) M i (f)m i (g) d m i (f)m i (g) m i (fg). [M i (fg) m i (fg)] x i [M i (f)m i (g) m i (f)m i (g)] x i [M i (f)m i (g) M i (f)m i (g) M i (f)m i (g) m i (f)m i (g)] x i M i (f)[m i (g) m i (g)] x i M(f) f m i (g)[m i (f) m i (f)] x i [M i (g) m i (g)] x i M(g) [M i (f) m i (f)] x i M(f)[U(P, g) L(P, g)] M(g)[U(P, f) L(P, f)]. () Note tht if M(f) or M(g) 0, the f or g is the zero futio d therefore fg is the zero futio d therefore Riem itegrble. So we will ssume tht both M(f) > 0 d M(g) > 0. Let > 0 be give. Sie f d g re Riem itegrble, there exists P 1 suh tht U(P 1, g) L(P 1, g) < 2M(f) d there exists P 2 suh tht U(P 2, f) L(P 2, f) < 2M(g). Let P be ommo refiemet of P 1 d P 2. The by Lemm 2 o pp. 15, U(P, g) L(P, g) < U(P 1, g) L(P 1, g) < 2M(f)
4 d Hee by (), U(P, f) L(P, f) < U(P 2, f) L(P 2, f) < 2M(g). U(P, fg) L(P, fg) M(f)[U(P, g) L(P, g)] M(g)[U(P, f) L(P, f)] < 2 2 d so fg is Riem-itegrble by Theorem 6.1. Now for geerl f d g, we pply wht we hve just proved to dedue tht f g, f g, f g, f g (ote tht they re ll produts of two oegtive futios) re Riem-itegrble. The we pply Theorem 6.6 to dedue tht f g f g f g f g is lso Riem-itegrble. But by the hit, this is just fg. 6.22: Prove tht if f d g re otiuous o [, b] d if g does ot hge sig i [, b] the there is poit (, b) suh tht f(x)g(x) dx f() g(x) dx (4) Solutio. We will ssume wlog tht g(x) 0 for ll x [, b] (if ot, osider g). Sie f is otiuous o [, b], so f is bouded o [, b]. Let d so Sie g(x) 0, Hee by Theorem 6.9, m m if f(x) d M sup f(x) x [,b] x [,b] 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 (4) is trivilly stisfied by y. Otherwise, set f(x)g(x) dx k : g(x) dx d ote tht k [m, M]. By Theorem 4.6 (IVT), there exists (, b) suh tht f() k. 6.2: If f R[, b] defie F (x) x f for every x [, b]. Prove tht F is otiuous o [, b]. Solutio. By Theorem 6.7, sie f R[, b], therefore f R[, ] for y [, b]. Hee F () exists for ll [, b]. We lim tht for ll (, b), d lso lim F (x) F () lim F (x) x x lim F (x) F (), lim x F (x) F (b), x b 4
5 d therefore F is otiuous o [, b] by Theorem.7. Sie f B[, b], there exists M > 0 suh tht f(x) M for ll x [, b]. Let > 0 d set δ /M. If x < δ, the by Problem 6.15 d Theorem 6.12, 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. 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]. 6.25: Prove: (1) If f R[0, b] d f is eve futio the f R[, b] d 2 (2) If f R[0, b] d f is odd futio the f R[, b] d 0 0. f. Solutio. (sketh) We defie symmetri prtitio Q of [, b] s prtitio with the property tht (i) 0 Q; d the property tht (ii) x Q if d oly x Q. For y bouded futio f o [, b] (ot eessrily eve or odd), we show tht sup{l(q, f) Q symmetri prtitio of [, b]}, if{u(q, f) Q symmetri prtitio of [, b]}, sie give y osymmetri prtitio P of [, b], we refie it ito symmetri prtitio by simply tkig Q P ( P ). Now it remis to observe tht for eve futio f d symmetri prtitio Q of [, b], L(Q, f) 2L(P, f), U(Q, f) 2U(P, f) where P Q [0, b] is prtitio of [0, b]. symmetri prtitio Q of [, b], L(Q, f) 0, U(Q, f) 0. Riem itegrbility d the required equtios the follows. Wheres for odd futio f d 6.27: Prove the itegrtio-by-prts formul: if f, g re differetible o [, b] d if f, g re itegrble o [, b] the Solutio. fg f(b)g(b) f()g() By the produt rule, f g. (fg) f g fg. (5) 5
6 Sie f, g re differetible d therefore otious d therefore Riem itegrble o [, b] d sie f, g re ssumed to be Riem itegrble o [, b], it follows from Theorem 6.6 d Problem 6.20 tht (fg) is Riem itegrble o [, b]. Hee by Theorem 6.17, (fg) f(b)g(b) f()g(). (6) Now subsitute (5) ito (6) d pply Theorem 6.6 to get the required formul. 6
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