X x x x m s x x x m s x e x x x x x l l l thus 2 4 6

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xxxxM SxxxM Sxexxxxx-=-+-+-= +-++++=---LLLThus24623424122121cos2!4!6!limlim12!3!4!4!6!lim3!4!1xxxxxxxxxexxxxxxx￲ ￲￲ ￲￲ ￲-+--=+------+-=----=-LLLLPartial Sums and Remainders of Taylor SeriesDefinition:The nthpartial sum ( )0( )( )()!ininifaTxxai==-is called the nthdegree Taylorpolynomialof f(x) about a. The difference f(x) – Tn(x) Rn(x) is called the nthremainderof the Taylor series of f(x) about a.Taylor Convergence TestTheorem:Let f(x) = Tn(x) + Rn(x). If lim( )0 |, 0 or nnRxxxaR RR￲ ￲=-<>=￲, then the Taylorseries converges to f(x) on the interval |xa| < R.Taylor’s Inequality TheoremTheorem:If (1)( )nfxMxad+-, then Rn(x) for the Taylor series of f(x) about asatisfies1( )(1)!nnMRxxan+-+.135
Proof:For n= 1,[]()11( )( )( )()( )( )( )( )()( )( )by FTC( )( )( )( )( )by FTC( )xxaaxaxaxtaaxtaaxtaaTxf afaxaRxf xfafaxaft dtfa dtftfadtftfa dtfz dz dtfzdz dtMdz dt=+-=---=-=--￲￲=￲￲������22if ()22!xaxaxadM ta dttaMM xa-=--=-=Example: 210sin( 1)(21)!nnnxxxn+==-+()()11M.S. ( )sinsin( )cos( )111So ( ).(1)!Does lim0 ?(1)1nnnnnnf xxxfxxfxMRxxnxxn++￲ ￲===+=+136a y1= T1(x)(a, f(a))y′1= T′1(x)
Proof: By the ratio test, 0!nnxn=converges x.By the divergence test, lim0!nnxn￲ ￲=.Therefore, 1lim0 (1)!nnxxn+￲ ￲=+.56/771. Evaluate 220( 1)6(2 )!nnnnnp=-.()2262006( 1)( 1)(2 )!6(2 )!cos32nnnnnnnnnppp==-=-==(like)45/771. Approximate ½20cosx dxto 3 decimal places using Maclaurin series.()[]22½½2000½40041120410M.S. cos( 1)(2 )!( 1)(2 )!( 1)(2 )!41( 1)(2 )!(2)(41)nnnnnnnnnnnnxx dxdxnxdxnnnnn==+=+==--=-=+-=+From alternating series where bn+1bnn> n0| n0,[]()()()34583213631933051and 1021! 245111.304101076,677,120(6!)(2)(13)119.0421010110,592(4!)(2 )(9)113.1251010320(2!)(2 )(5)nnnRRnnRRR-+------<++==<=>B=BB137
Therefore, ½25011cos0.4970.001(2!)(1)(2!)(2 )(5)x dx=-B.52/771. Find the first three non-zero terms of the Maclaurin series for y= sec x.Method 1: Division of power series2024624246246246246461seccos1( 1)(2 )!112!4!6!5122411224720122472022472024485724360nnnxxxnxxxxxxxxxxxxxxxxxxx===-=-+-++++= -+-+-+-+-+--+--+LLLLLLLThe first three non-zero terms of the Maclaurin series for y= sec xare 1, ½, and 524.Method 2: Examination and Substitution:23333(4)432325( )sec( )sectan( )sectansec( )sectan2sectan3sectan( )sectan3sectan15sectan5secf xxfxxxfxxxxfxxxxxxxfxxxxxxxx==￲￲=+￲￲￲=++=+++(4)(0)1(0)0(0)1(0)0(0)5fffff==￲￲=￲￲￲==2435. .1002!4!

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