Math 104: Final exam solutions

Transcription

1 Mth 14: Fil exm solutios 1. Suppose tht (s ) is icresig sequece with coverget subsequece. Prove tht (s ) is coverget sequece. Aswer: Let the coverget subsequece be (s k ) tht coverges to limit s. The there exists K such tht k > K implies s k s < 1, d hece for k > K, s k < s + 1. Suppose tht (s ) is ot bouded bove. The there exists N such tht s N > s + 1, d sice it is icresig sequece, s > s + 1 for ll > N. Choosig k such tht k > K d k > N gives s k < s + 1 d s k > s + 1 which is cotrdictio. Hece (s ) is bouded bove, d sice it is icresig it coverges.

2 x 2. () Let f (x) = be fuctios defied o R, for ll 2 N. By usig the Weierstrß 2 (1+x 2 ) M-test, or otherwise, prove tht  f coverges uiformly o R. Aswer: If x < 1, the f (x) < 1 2 (1 + x 2 ) pple 1 2 = 1 2. If x 1 the f (x) pple x 2 (x 2 ) pple 1 2 (x) pple 1 2. Hece f (x) < 1 / 2 for ll x 2 R. Sice  1 / 2 coverges, it follows tht  f coverges by the Weierstrß M-test. (b) Cosider the sequece of fuctios defied o R s g (x) = 1/ for 1 pple x <, otherwise, for ll 2 N. Defie M = sup{ g (x) : x 2 R}. Prove tht  g coverges uiformly to limit g, but tht  M diverges. Sketch g(x) for pple x < 5. Aswer: From the defiitio, it c be see tht M = 1 /, d hece  M diverges. Cosider x 2 [ 1, ) for some 2 N. The  f (x) = 1 /. Sice f k (x) = for ll k > it follows tht  f (x) = 1 /. Thus  g coverges poitwise to 1/m if x 2 [m 1, m) for some m 2 N, g(x) = otherwise. To show tht covergece is uiform, cosider y e >. By the Archimede property, there exists N 2 N such tht 1 /N < e. For > N, ote tht g(x) 1/m if x 2 [m 1, m) for some m 2 N with m >,  g k (x) = otherwise, d thus g(x)  g k(x) < 1 / < e, so the covergece is uiform. The fuctio g is plotted i Fig. 1.

3 3. Cosider the sequece defied recursively ccordig to s 1 = 1 /3 d s +1 = ls (1 s ) for 2 N, where l is rel costt. () Prove tht (s ) coverges for l = 1. Aswer: Suppose tht s 2 (, 1). The s +1 = s (1 s ) < s d s +1 >, so s +1 2 (, s ). Sice s 1 = 1 /3 it follows by iductio tht (s ) is decresig sequece tht is bouded below by, so it coverges. (b) Prove tht (s ) hs coverget subsequece for l = 4. Aswer: First, ote tht s +1 = 4(s s 2 )=4 1 4! s = 1 4 s. 2 2 Suppose tht s 2 [, 1]. The s 1/2 2 [ 1/2, 1 /2], so 4(s 1/2) 2 2 [, 1], d hece s +1 2 [, 1]. Sice s 1 2 [, 1] it follows by mthemticl iductio tht s 2 [, 1] for ll 2 N. Thus the sequece is bouded, so it hs coverget subsequece vi the Bolzo Weierstrß theorem. (c) Prove tht (s ) diverges for l = 12. Aswer: Now suppose tht for 2, s > 2 1. The s +1 = 12 s 1 s > (2 1 1) > = 2. so s +1 > 2 (+1) 1. Sice s 2 = 8 /3 > 2 it follows by mthemticl iductio tht s > 2 1 for ll 2 N with > 2. Sice 2! s!, it follows tht s is ubouded d hece diverges.

4 4. A rel-vlued fuctio f defied o set S is defied to be Lipschitz cotiuous if there exists K > such tht for ll x, y 2 S, f (x) f (y) pplek x y. () Prove tht if f is Lipschitz cotiuous o S, the it is uiformly cotiuous o S. Aswer: Choose e >. The set d = e K. If x, y 2 S such tht x y < d, the d hece f is uiformly cotiuous. f (x) f (y) pplek x y < Kd = e (b) Cosider the fuctio f (x) = p x o the itervl [, ). Prove tht it is uiformly cotiuous, but ot Lipschitz cotiuous. Aswer: To show tht f is uiformly cotiuous, choose e >, d ote tht f (x) f (y) = p x p y = ( p x p y)( p x + p y) p x + p y = x y p x + p y. Suppose tht x < e 2 d y < e 2. The < f (x) < e d < f (y) < e so f (x) f (y) < e. Otherwise, either x e 2 or y e 2, so p x + p y e. If x y < d where d = e 2, the x y f (x) f (y) = p p < e2 x + y e = e, d hece the fuctio is uiformly cotiuous. To see tht f is ot Lipschitz cotiuous, cosider x = 1 / 2, d y = : f (x) f (y) x y = f (1 / 2 ) 1/ 2 =. Sice c be mde rbitrrily lrge, there does ot exist y K such tht f (x) f (y) pplek x y for ll x, y 2 [, ).

5 5. Cosider the fuctio defied o the closed itervl [, b] s 1 if x = c, h c (x) = if x 6= c, where < c < b. () Prove tht h c is itegrble o [, b] d tht R b h c =. Aswer: Defie h = mi{b c, c }. For < h, defie the prtitio P = { = t < t 1 < t 2 < t 3 = b} where t 1 = c d t 2 = c +. The d L(h c, P ) = 3  m(h c, [t k, t k+1 ])(t k+1 t k ) = + 2m(h c, [c, c + ]) + = U(h c, P ) = 3  M(h c, [t k, t k+1 ])(t k+1 t k ) = + 2M(h c, [c, c + ]) + = 2. Sice c be mde rbitrrily smll, it follows tht U(h c ) pple d L(h c ). Hece U(h c )=L(h c )=, d h c is itegrble with itegrl R b h c =. (b) Suppose f is itegrble o [, b], d tht g is fuctio o [, b] such tht f (x) =g(x) except t fiitely my x i (, b). Prove tht g(x) is itegrble d tht R b f = R b g. You c mke use of bsic properties of itegrble fuctios. Aswer: If g differs from f t fiitely my poits, the it is possible to write g(x) = f (x)+  b i h ci (x) where the c i re the poits where they differ d b i = g(c i ) of itegrble fuctios it is itegrble d hece f (c i ). Sice g is the sum Z b g = Z b f + Z b  b i h ci = Z b f.

6 6. Cosider the sequece of fuctios h o R ccordig to 8 < 2 if < x < 1 /, h (x) = 2 if 1/ < x <, : otherwise. () Sketch h 1 d h 2. Aswer: The fuctios re show i Fig. 2 (b) Prove tht h coverges poitwise to o R. Aswer: At x =, lim! h (x) =lim! =. For x 6=, there exists N 2 N such tht 1 /N < x. Hece for > N, h (x) = d thus lim! h (x) =. (c) Let f be rel-vlued fuctio o R tht is differetible t x =. Prove tht Z lim h f = f ().! Aswer: Cosider y e >. The there exists d > such tht x < d implies f (x) f () x f () = f (x) x f () < e. (1) By the Archimede property, there exists N 2 N such tht 1 N < d. For > N, Z Z Z d h f = lim h f + lim h f = c! c d! Z 1/ Z 1/ 2 f + 2 f. By usig Eq. 1, Z 1/ f Z 1/ By the sme procedure d hece f Z 1/ ( f () x( f () e))dx = f () e f () e = f () 2 Z 1/ f Z 1/ Z Z 1/ h f f () = 2 from which it follows tht lim! h f = f (). Z f pple f () 2 + e 2 f Z 1/ 1/ e 2. f ( f () x( f ()+e))dx f () pple e

7 7. () Fid exmple of set A R where the iterior A is o-empty, but tht sup A 6= sup A d if A 6= if A. Aswer: Cosider A = { 2}[( 1, 1) [{2}. The sup A = 2 d if A = 2. However, A =( 1, 1), sice if x = ±2, the there is o eighborhood N d (x) A for y d >. Hece if A = 1 d sup A = 1. (b) Let B 1,...,B be subsets of R. Prove tht! \ B i = \ B i. Aswer: Cll the LHS E d the RHS D. Suppose x 2 E. The there exists d > such tht the eighborhood N d (x) T B i. Hece N d (x) B i for ll i = 1,..., d so x 2 B i for ll i = 1,...,. Hece x 2 D. Now suppose tht x 2 D. The x 2 B i for ll i = 1,...,, so there exist d i > such tht N di (x) B i for ll i = 1,...,. Let d = mi{d 1,...,d } >. The N d (x) B i for ll i = 1,..., d hece N d (x) T B i, so x 2 E. (c) Suppose tht {C i } is ifiite sequece of subsets of R. Prove tht! \ \ C i but tht these two sets my ot be equl. Aswer: The first rgumet from prt (b) c be still be pplied whe there re ifiite umber of sets, to estblish tht ( T C i ) T C i. However, the secod rgumet my fil if the ifimum of the d i is equl to zero. Motivitted by this, cosider C i =( 1/i, 1 /i). Sice these re ope sets, C i = C i. Hece, \ C i = {}. C i However d thus \ C i = {}! \ C i =.

8 8. () If f is cotiuous strictly icresig fuctio o R, prove tht d(x, y) = f (x) f (y) defies metric o R. Aswer: Cosider the three properties of beig metric M1. For ll x 2 R, d(x, x) = f (x) f (x) =. If d(x, y) = the f (x) = f (y), d if f is strictly icresig the x = y. M2. For ll x, y 2 R, d(x, y) = f (x) f (y) = f (y) f (x) = d(y, x) d thus the metric is symmetric. M3. For ll x, y, z 2 R, d(x, y)+d(y, z) = f (x) f (y) + f (y) f (z) pple f (x) f (z) = d(x, z) which follows from the usul trigle iequlity. (b) Prove tht d is equivlet to the Euclide metric d E (x, y) = x y. Aswer: Write N r (x) d Nr E (x) for the eighborhoods of rdius r with respect to d d d E respectively. To prove tht the two metrics re equivlet, cosider y x 2 R, d e >. By cotiuity, there exists d > such tht x y < d implies f (x) f (y) < e. Hece Nd E(x) N e(x). Sice f is cotiuous d strictly icresig, it hs cotious strictly icresig iverse f 1. Cosider y x 2 R d e >. By usig cotiuity of f 1 t f (x), there exists d > such tht f (x) z < d implies f 1 ( f (x)) f 1 (z) < e so x f 1 (z) < e. The Hece the two metrics re equivlet. N d (x) = {y 2 R : f (x) f (y) < d} {y 2 R : x f 1 ( f (y)) < e} = {y 2 R : x y < e} = N E e (x). (c) Suppose g is cotiuous strictly icresig fuctio o [, ) where g() =. Is the fuctio d 2 (x, y) =g( x y ) lwys metric? Either prove the result, or fid couterexmple. Aswer: This does ot lwys defie metric. Cosider g(x) =x 2. The d 2 (, 1)+ d 2 (1, 2) = = 2 but d 2 (, 2) =2 2 = 4 so the trigle iequlity is violted.

9 9. Cosider the cotiuous fuctio defied o R s x f (x) = e x 1 if x 6=, c if x =, where c 2 R. For this questio you c ssume bsic properties of the expoetil, such tht it is cotiuous, differetible, d hs the Tylor series e x = Â k= xk k!. () Use L Hôpitl s rule to compute lim x! f (x) d hece determie c. Aswer: By L Hopitl s rule, lim x! f (x) =lim x! 1/e x t if d oly if lim x! f (x) = f (), d thus c = 1. (b) Show tht f is differetible o R d compute f. = 1. A fuctio is cotiuous Aswer: The derivtive for x 6= is Usig the defiitio of the derivtive, f (x) = (ex 1) xe x (e x 1) 2. f () =lim! f () f () = lim! e 1 1 = e + 1 lim! (e 1) = 1 e lim! ( + 1)e 1 = e lim! e ( + 2) = 1 2 where L Hôpitl s rule hs bee pplied twice. (c) Clculte the fuctio limits lim f (x), x! lim (x + f (x)). x! Use the results to sketch f d f T o ( 1, 1). Aswer: The first limit is lim f (x) = lim x! x! sice e x /x! s x! ; this c be verified by usig the Tylor series to see tht e x > x2 /2 for ll x >. The secod limit is e x x 1 1 x = lim (x + f (x)) = lim x x(ex 1 + 1) xe x! x! e x = x lim 1 x! e x 1 = sice s xe x! s x!. The fuctios f d f T re plotted i Fig. 3.

10 1. Give fuctio f o [, b], defie the totl vritio of f to be (  ) Vf = sup f (t k ) f (t k 1 ) where the supremum is tke over ll prtitios P = { = t < t 1 <...< t = b} of [, b]. () Clculte Vf for the fuctio defied o [ 1, 1] s f (x) = 2 if x <, 3 if x. Aswer: Cosider y prtitio P = { = t < t 1 <... < t = b}. The there exists k such tht t k 1 < pple t k. For j < k, f (t j ) f (t j 1 = ( 2) ( 2) =. Similrly, if j > k, f (t j ) f (t j 1 = 3 3 =, d thus  f (t k ) f (t k 1 ) = + f (t k ) f (t k 1 ) + = 3 ( 2) = 5. Sice this is true for rbitrry prtitio, it follows tht Vf = 5. (b) Prove tht if f is differetible o itervl [, b], d tht f is cotiuous the Vf = R b f. Aswer: To prove tht Vf pple R b f, cosider y prtitio P of [, b]. The, by usig the Fudmetl Theorem of Clculus,  f (t k ) f (t k 1 ) =  Z tk t k 1 f pple Z tk Z b  f = f. t k 1 Sice f is itegrble o [, b], for ll e >, there exists prtitio P such tht U( f, P) L( f, P) < e, d hece L( f, P) > ( R b f ) e. By usig the Me Vlue Theorem, for ech k, there exists x k 2 (t k 1, t k ) such tht Hece  f (t k ) f (t k 1 ) = f (x k )= f (t k) f (t k 1 ) t k t k 1.  f (x k )(t k t k 1 )  m( f, [t k 1, t k ])(t k t k 1 ) Z b = L( f, P) > f e. R b It is therefore possible to fid prtitio such tht the sum is lrger th f e for R b ll e > d thus Vf f. Combiig this with the first result shows Vf = R b f.

11 g(x) x Figure 1: Grph of the fuctio cosidered i questio y -2-4 y = h 1 (x) y = h 2 (x) x Figure 2: Grph showig the fuctios h 1 d h 2 cosidered i questio 5.

12 12 1 y = f (x) y = f T (x) 8 6 y x Figure 3: Grph showig the fuctios cosidere i questio 9 o L Hôpitl s rule.