MATH 104 FINAL SOLUTIONS. 1. (2 points each) Mark each of the following as True or False. No justification is required. y n = x 1 + x x n n

Transcription

1 MATH 04 FINAL SOLUTIONS. ( poits ech) Mrk ech of the followig s True or Flse. No justifictio is required. ) A ubouded sequece c hve o Cuchy subsequece. Flse b) A ifiite uio of Dedekid cuts is Dedekid cut. Flse c) If f : R R is cotiuous o [, b], there is sequece of polyomils whose uiform limit o [, b] is f. True d) If f f uiformly o S, the f f uiformly o S. Flse e) If f is differetible o [, b] the it is itegrble o [, b]. True. Give sequece {x }, defie sequece {y } by settig y = x + x + + x. ) (6 poits) If x R, show tht y R. For y ɛ > 0, there exists N N such tht x < ɛ for ll > N. Also, sice x coverges, it is bouded d we hve tht x < M for ll N, for some M R. Let ɛ = (M + )ɛ. Note tht for > mx(n, N/ɛ) we hve Hece we hve tht y. y = (x ) + (x ) + + (x ) x + + x MN ɛ( N) + < (M + )ɛ = ɛ b) (4 poits) Give exmple of diverget sequece {x } (i.e. with o limit i R) for which {y } s defied bove coverges. Cosider the sequece {x } = {( ) }. This is diverget sequece, but {y } = {, 0, 3, 0, 5, 0,... } coverges to ) (8 poits) Fid exmple or prove tht the followig does ot exist: mootoe sequece tht hs o limit i R but hs subsequece covergig to rel umber. Such thig does ot exist. Ay bouded mootoe sequece hs limit i R, so mootoe sequece {s } tht hs o limit i R is ubouded. Suppose {s } is icresig (bouded below by

2 s ). Sice the sequece is ubouded, for y M R there exists some N such tht s N > M d, sice s is icresig, s > M for ll > N s well. Hece lim s = d y subsequece will hve the sme limit. Similrly, if s is decresig, it is ot bouded below, d for y M R there exists some N such tht s N < M d, sice s is decresig, s < M for ll > N s well. Hece lim s = d y subsequece will hve the sme limit. b) (7 poits) Let x = cos π 3. Fid coverget subsequece of {x } d compute lim sup x. Cosider {x 6 } = {}. This sequece is coverges to, d, sice cos x for ll x, we hve tht this is the lrgest possible subsequetil limit of {x }. Hece lim sup x =. 4) ) (6 poits) Cosider the series = 6, = + /. For ech of these, determie whether it coverges or diverges d justify your swer. For the first series, lim sup ( ) 6 / = lim 6 For the secod series, ote tht /, where / > = 0 < d so by the root test the series coverges. diverges d hs ll positive terms, d for ll. Hece by the compriso test this series diverges. +/ = b) (4 poits) Let {s } d {t } be sequeces of positive rel umbers. Show tht if s /t the s d t either both coverge or both diverge. Sice s /t, there is N N such tht d so d s t < < s t < 3, t < s < 3t for ll N. If t coverges, the N 3t coverges d by the bove d the compriso test, N s coverges s well. Sice s d N s differ by fiite umber of terms, oe coverges if d oly if the other does, d so our coclusio holds for s s well. Similrly, if s (d hece s ) coverges, the we hve tht sice t < s for ll > N, the series t coverges by the compriso test. Hece s d t either both coverge or diverge. 5) ) (5 poits) Suppose tht f coverges uiformly to f o set S R, d tht g is bouded

3 fuctio o S. Prove tht the product g f coverges uiformly to g f. If g(x) = 0 o S, the {gf } = {0} which coverges uiformly to gf = 0. Otherwise, let g(x) < M > 0 for ll x S. For y ɛ > 0, there exists N R such tht f (x) f(x) < ɛ/m for ll x S, for ll > N. The g(x)f (x) g(x)f(x) = g(x) f (x) f(x) < ɛ for ll x S, for ll > N d hece gf coverges uiformly to gf. b) (5 poits) Let {f } be sequece of cotiuous fuctios o [, b] tht coverges uiformly to f o [, b]. Show tht if {x } is sequece i [, b] d if x x, the lim f (x ) = f(x). Sice f re cotiuous o the closed itervl [, b] d coverge uiformly to f, we hve tht f is cotiuous o [, b] s well. For y ɛ > 0 there exists N R such tht f (x ) f(x ) < ɛ/ for > N. Furthermore, sice x x d [, b] is closed, we hve x [, b]. Thus f is cotiuous t x d there is δ > 0 such tht if y S d x y < δ the f(x) f(y) < ɛ/. Note tht there is N R such tht x x < δ for ll > N, d hece f(x ) f(x) < ɛ/ for ll > N. Hece for > mx(n, N ), we hve tht f (x ) f(x) < f (x ) f(x ) + ɛ/ < ɛ d so f (x ) f(x) s desired. 6) ) (6 poits) Prove tht d(x, y) := mi( x y, ) is metric o R. First of ll, d(x, y) R for ll x, y R, so d is ideed fuctio from R R R. Next, ote tht d(x, y) 0 for ll x, y R, sice both d x y re oegtive for ll x, y R. Also, d(x, y) = 0 if d oly if d(x, y) = x y = 0, which is if d oly if x = y. Sice x y = y x, we lso hve tht d(x, y) = mi( x y, ) = mi( y x, ) = d(y, x). Filly, d(x, y) = mi( x y, ) mi( x z + z y, ) mi( x z, ) + mi( z y, ) = d(x, z) + d(z, y). So ll properties of metric re stisfied. b) (4 poits) Is the set ( 5, 5) ope with respect to this metric? Prove tht your swer is true. This set is ideed ope with respect to the bove metric. Let x ( 5, 5) d let δ = mi( x 5, x + 5, ). Cosider the eighborhood N(x) of rdius δ/ of x with respect to the bove metric. Sice δ/ <, we hve tht d(x, y) = x y for y N(x), d sice δ/ < mi( x 5, x + 5 ) we hve tht x y < mi( x 5, x + 5 ). Hece y ( 5, 5) d N(x) ( 5, 5) d x is iterior poit i this itervl. 7. ) (8 poits) Fid the Tylor series t 0 for f(x) = e x. Determie its rdius of covergece. You c use either the rtio or root test strtegy for this. Sice f () (x) = e x for ll, we hve tht f () (0) = for ll, d the Tylor series is =0 x!. Sice lim sup! (+)! = 0, we hve tht the rdius of covergece is.

4 b) (7 poits) Prove tht the Tylor series i prt () represets (is equl to) e x for ll x R. To show this, we use Tylor s theorem. I this cse, R (x) = e y ( + )! x+ where y is betwee 0 d x. If x > 0 the e y < e x d hece 0 < R (x) < e x x + ( + )! 0 s, d hece R (x) 0 by the squeeze lemm. If x < 0, the x < y < 0 d so e y < d R (x) < x + ( + )! 0 d so gi R (x) 0. For x = 0, e x = = 0 0!. 8. (0 poits) Let f be fuctio defied o R such tht f(x) f(y) (x y) for ll x, y R. Show tht f is differetible o R d tht f (x) = 0 for ll x R. Note tht for ll x, y R we hve y x y x. Sice y x 0 s y x, we hve by the squeeze lemm tht lso teds to 0 s y x d so y x lim = 0. y x y x So f(x) is differetible o R d its derivtive is 0 everywhere. 9) ) (6 poits) Let f d g be cotiuous fuctios o [, b] such tht b f = b g. Show tht there is x [, b] such tht f(x) = g(x). We hve tht b (f g) = 0. Suppose f(x) g(x) 0 for y x [, b]. Sice f d g re cotiuous, so is f g, d hece we hve either f(x) g(x) < 0 for ll x [, b] or f(x) g(x) > 0 for ll x [, b] (otherwise the itermedite vlue theorem would fil). I the first

5 cse, b (f g) > 0 sice y Riem sum correspodig to this itervl is bouded below by (b ) mi(f(x) g(x) x [, b]) which exists sice f g is cotiuous o [, b]. I the secod cse, b (f g) < 0 sice y Riem sum correspodig to this itervl is bouded below by (b ) mx(f(x) g(x) x [, b]) which exists sice f g is cotiuous o [, b]. Hece the itegrl cot be 0 which is cotrdictio, d there must be some x [, b] such tht f(x) = g(x). b) (4 poits) Costruct exmple of fuctios f, g, both itegrble o [, b], such tht b f = b g but f(x) g(x) for y x [, b]. Let f(x) = for x [, 0] d f(x) = for x [0, ]. Let g(x) = f(x) o [, ]. The f = g = 0 but f(x) g(x) for y x [, ].