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# Calculus II Notes for Week #10 MATH 1220

Tulane

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This 6 page Class Notes was uploaded by Zachary Hill on Wednesday March 16, 2016. The Class Notes belongs to MATH 1220 at Tulane University taught by Benjamin Klaff in Spring 2016. Since its upload, it has received 34 views. For similar materials see Calculus II in Mathematics (M) at Tulane University.

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Date Created: 03/16/16

MATH 1220 14 March 2016 WarmUp Tim xiii31 quotBEE EWE L LJHJlI39lrfJiEJ SEEMEl KNEW t39F t39ih ri xt rmer Hit mtginf B Lil b 0 0 12 cos1 COSB2 Sin1 SinB2 6032A 2COSACOSB 6032B SinzA 23inASinB SiIZZB 6032A SiIZZA 6032B SiIZZB 2COSACOSB 23inASinB 1 1 2cosAcosB 23inASinB 2 2COSACOSB 23inASinB o 0122 COSA B 12 3mm 132 COS2A B 2COSA B 1 sin2A B 1 2COSA B1 2 260SA B 0 Then 2 2COSACOSB 23inASinB 2 2COSA B 2COSACOSB 23inASinB 2COSA B COSA B COSACOSB SinASinB Reformulation Figure out what the Cauchy Condition for convergence of series should be o 2 an converges if and only if lim SN exists n0 new 0 Definition of a Cauchy Sequence a sequence an is a Cauch sequence if for all s gt O there s an index N such that if m 2 N and n 2 N then lam anl lt a 0 Then AESN L if and only if for every 8 gt 0 there is an N so that if m 2 n 2 N then ISmSnl an1 am m 71 2 a1 2 6 1 i0 i0 o This is because Sma0a1a2 anan1 am Sna0a1a2 an Sm Snan1 am 0 Then the Cauchy Condition for convergence of series o If an where n 2 O is a sequence then 2 an converges if and only if for every n0 sgt0thereisan NsothatifmZnZNthen an1 am lts We want this for the Weierstrass Mtest for uniform convergence of sequences of functions 0 Let fn where n 2 0 be a sequence of functions on the interval 1 Let Mn be a 00 sequence of real numbers so that 2 Mn converges also written as lt 00 and n0 Wool 3 Mn for every x in I Then the sequence fn converges uniformly on the interval 1 15 March 2016 Concept MapsFlowcharts 0 Suggested way to order and understand our concepts 0 Ratio Test for Infinite Series Llt1then Zan 0 Statement If 2 an is an infinite series and if lim 7 n0 n0 an1 an converges Gemmire Series for are rywm Theorem Ratio Tear ziiquot 5431 T7 at Comparison Tamer Eerie O 00 0 Proof Suppose Z r and lt r lt 1 Then an1lt ran Then M lt r so quot0 an1 2 a 3 3 k an2 lt ran1lt r an Then i lt r SO an3 lt ran2 lt r an Then anklt r an Then k k k oo oo 2 lt2mhmwhm2 2 z j0 j0 k OOJ390 j0 j0 0 Root Test for Series 0 Statement 2 an converges if a n L lt 1 n0 El c39rirrrmdrm39iz m quot 7 Trle ill39i39gzg39 39i H E i39i3939fi39nai39ilr39 n g a a q V anlt rnzrj39iarmll Thur e mumr all 3 E iiflm 39L iquot Irma 3W g mei imirqf 3quot If n Fr l39 q r J It 2 1 quot 3 Autumn 1 tjijlirrrrml I Emmmw Emma Hg EM germs 5 u Frrarrr Kiku uglwuw O Af rarr rii39 smear El39liirmuw k k 0 Since anltr then anltr Then for all nZN ZaNJlt ZVNTJ J0 J0 o Weierstrass MTest 0 We want to know expx is continuous at every x n0 iDQMovWmtf mWT39JT S LJ Mme WWWg Ai 7 quot39I 1 Statement fn is defined on an interval 1 Ifthere is a sequence Mn such that Vnx S Mn and 2 Mn lt 00 then fn converges uniformly n0 16 March 2016 Motivation Suppose you want a solution a function y yx to the differential equation y39 y that also satisfies y0 1 and you are only interested in a close to O i i H 8 O a Mvsterv Solution yx Z cnx where cm are coefficients for this geometric sequence n0 y01gt chOn1 n 0 9 0 1 2 n COO 610 620 cn0 1 Then c01 Assume Since yx Z cnx n0 Then 3 0 x 00 gr 2 deem n0 d 0 a d d xCOx Z 17 n 00 0 Z nCnx 1 n1 1ltngtltcngtltxnlgt 0ltn1gtltcn1gtltxngt Since y y39 then E cnx n 1cn1xquot n0 n0 Assume Z anxquot Z bnx ltgt an Inn for every n so for every n n0 n0 Cn n 1Cn1 Cn1 n1 Then c0 1 co 6160 1 H gt 62 C11 1 MG Ir NIquot C3 2 621 21 U C 3 i C4 C31 31 4 Then em because the above is recursive Then yx Z cnx 2 x n0 n0 Weierstrass MTest N 00 Interested in Z gt Z can be seen as fn gtf n0 n0 Have fnfnI gtER Need Mn sequence of numbers where Vnx SM for all x in I and 2 Mn lt oo n0 N 00 gone a gone N 00 Apply this to 2x77 and n0 n0 Choose R R RgtO Need Mn 21 so that quot90 SIS f Mn Apply ratio test to see if 2 Mn converges n0 00 M ZMnltoo ltgt1im quot1Llt1 n gtoo n n0 Rn1 Mn1 Rn1 n39 Mn Liquot n1 R n1 Then 20 is continuous on R R R gt 0 gt expx is continuous everywhere n

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