Mathematics 805 Final Examination Answers

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1 . 5 poins Se he Weiersrss M-es. Mhemics 85 Finl Eminion Answers Answer: Suppose h A R, nd f n : A R. Suppose furher h f n M n for ll A, nd h Mn converges. Then f n converges uniformly on A.. 5 poins Se Lebesgue s Domined Convergence Theorem. b Se Lebesgue s Monoone Convergence Theorem. c Define [, n] f n n oherwise Noe h f n d, bu n f n. Why does his conrdic neiher he Domined nor he Monoone Convergence Theorems? Answer: Suppose h f n L nd f n f lmos everyone. Suppose furher h f n g lmos everywhere, wih g L. Then f L nd f f n. b Suppose h f n L is monoone sequence, nd suppose furher h f n is bounded. Then f n converges lmos everywhere o funcion f L, nd f f n. c This emple does no viole he Domined Convergence Theorem, becuse here is no funcion g L wih f n g. The emple does no viole he Monoone Convergence Theorem becuse he sequence f n is no monoone. 3. poins Suppose h K is compc meric spce, nd g : K R coninuous funcion, wih g > for ll K. Suppose furher h g n : K R is sequence of coninuous funcions converging uniformly o g on K. Show h here is some ineger N so h if n > N, hen g n > for ll K. Answer: Becuse K is compc, we now h g ins is minimum vlue some K. In priculr, we cn find some ɛ so h g ɛ for ll K. Becuse g n g uniformly, we cn find n ineger N so h if n N, hen g g n ɛ/ for ll K. This implies h g n ɛ/ for ll K, so g n > poins Le M, d nd M, d be meric spces, nd f : M M funcion. Define wh is men by f is uniformly coninuous. b Suppose h M, d nd M, d re meric spces, nd f : M M is uniformly coninuous. Suppose h A, B M wih d A, B. Show h d fa, fb. c Give n emple o show h i is possible for f : M M o be coninuous, wih d A, B nd d fa, fb. Answer: Given ny ɛ >, here is δ > so h if d, y < δ, hen d f, fy < ɛ. b Given ny ɛ >, we cn find δ > so h if d, y < δ, hen d f, fy < ɛ. Becuse d A, B, we cn find A nd b B so h d, b < δ. This sys h d f, fb < ɛ, nd so d fa, fb < ɛ. Becuse ɛ is rbirry, we cn conclude h d fa, fb. c Consider he funcion f : R R R defined by f, y y. Le A {n, n : n Z, n > }. Le B {n, : n Z, n > }. Then d A, B. However, fa {}, while fb {}, so d fa, fb. 5. poins Show h sin d

2 cn be defined s n improper Riemnn inegrl bu no s Lebesgue inegrl. Hin: Le, nd sin imie he proof h d cn be defined s n improper Riemnn inegrl bu no Lebesgue inegrl. Answer: Le, so h, d d, nd d d/. If, hen. Therefore, s Riemnn inegrl, we hve sin d b b b sin sin d b d Se u /, dv sin d, du 3/ d, nd v cos. We hve b sin d cos ] b b b b The firs i is cos, nd he inegrl converges, becuse Riemnn inegrl is defined. cos d. 3/ On he oher hnd, if he inegrl converged s Lebesgue inegrl, so would sme subsiuions led us o consider b n+ n sin b sin d. Bu n+ d sin d n + n 3/ d converges. Therefore, he improper n +. Becuse n + diverges, we now h he Lebesgue inegrl mus no eis. sin d. The 6. 5 poins As usul, we define e e B n n n! nd B n B n. Recll h B, B, nd B + 6. Show h B n n B n. Answer: We hve B n n! n e e B n n! n e B n + B n n e n! e + e e + e + e e e + n B n n! e / e Equing coefficiens, we see h B n + B n n B n. n B n n n! 7. poins Suppose h f C. Le m be n ineger. Prove using inducion on m h f f d + m B! f f m B m f m d

3 Answer: We firs hve he cse m. We mus show h f f d + B f f! f d f f + f d f d To chec his, we inegre by prs, seing u, du d, dv f d, nd v f. We hve f d f f + f d f d f f + ] f f f + f + f f. Now, for he inducive sep, we mus prove h m B f f m! or m m+ B! f d B m f m d f f m+ B m+ m +! f m+ d B m f m d B m+ f m f m m+ B m+ m +! m +! f m+ d Muliply hrough by m+, nd his is he sme s B m f m d m+ B m+ m +! f m f m B m+ m +! f m+ d We cn prove his by using inegrion by pr on he lef-hnd side. Le u f m, du f m+ d, dv Bm d, nd v Bm+ m+!. We hve B m f m d B ] m+ m +! f m B m+ m +! f m+ d m +! B m+ m +! B m+ f m B m+ f m f m f m B m+ m +! f m+ d B m+ m +! f m+ d So i remins o show h m+ B m+ B m+. If m + is odd, his is rue becuse B m+. If m + is even, his is rivilly rue inequliy poins Here is ye noher wy o compue migh be needed o do his problem: D n : n n e i sin D n sinn + sin K n : n D n + d. Recll he following fcs, some of which

4 cosn + K n n + cos sinn + / n + sin/ D n d cos sin / K n d Suppose h g is Riemnn-inegrble on [, b] wih g M. Show h b cos b M g d nd conclude h b Show h b cos g d. cos 4 sin d c Le f 4 sin, < Show h i is possible o define f so h f is coninuous nd bounded on [, ]. d Show h cos f d. e Show h nd hen f Le y nd conclude h g Finlly, inegre by prs o show h cos d sin / d 4. / sin sin d d. sin sin d d. Answer: This problem is en from An elemenry mehod for evluing in infinie inegrl, by M.R. Spiegel, The Americn Mhemicl Monhly, 58:8, Oc. 95, pp This is rivil: b cos g d b cos b g d M Mb d.

5 The i follows immediely. b We hve K d K d cos d cos cos sin d / cos 4 sin d / c Recll h sin y y y This mens h sin/ nd sin / Therefore, f 4 sin 4 sin 4 sin / + 4 / Therefore, he funcion is cully nlyic, if we define f o be. d Becuse f sisfies he hypoheses of pr, his follows immediely from. e We hve cos 4 sin d cos cos d 4 sin d cos d cos d d sin d sin d 4 f Le y, so y, nd hen d dy. Noe h if, hen y. We hve 4 sin sin sin dy dy sin y y dy sin / dy sin y y dy g Inegrion by prs is no quie rivil. We se u sin, dv d, du sin cos d sin d, nd v. We hve sin ] d sin sin + d sin y y dy

6 sin Now, we cn compue h nd hen dy d, so d dy: sin, nd s well. For he res, we subsiue y, sin d sin dy sin y y dy