Advaced Placemet Specialty Coferece TEACHING THE IDEAS BEHIND POWER SERIES Preseted by LIN McMULLIN
Sequeces ad Series i Precalculus Power Series Itervals of Covergece & Covergece Tests Error Bouds Geometric Series New Series from Old Problems Q & A
From the Course Descriptio *IV. Polyomial Approimatios ad Series * Cocept of series. A series is defied as a sequece of partial sums, ad covergece is defied i terms of the limit of the sequece of partial sums. Techology ca be used to eplore covergece or divergece. * Series of costats. + Motivatig eamples, icludig decimal epasio. + Geometric series with applicatios. + The harmoic series. + Alteratig series with error boud. + Terms of series as areas of rectagles ad their relatioship to improper itegrals, icludig the itegral test ad its use i testig the covergece of p-series. + The ratio test for covergece ad divergece. + Comparig series to test for covergece or divergece.
* Taylor series. + Taylor polyomial approimatio with graphical demostratio of covergece. (For eample, viewig graphs of various Taylor polyomials of the sie fuctio approimatig the sie curve.) + Maclauri series ad the geeral Taylor series cetered at = a. + Maclauri series for the fuctios, e, si, cos, ad + Formal maipulatio of Taylor series ad shortcuts to computig Taylor series, icludig substitutio, differetiatio, atidifferetiatio, ad the formatio of ew series from kow series. + Fuctios defied by power series. + Radius ad iterval of covergece of power series. + Lagrage error boud for Taylor polyomials..
Precalculus Sequeces ad series: Sigma otatio, Recursive ad o-recursive defiitios of sequeces, Basic formulas for the sums of simple sequeces ( costat,, k k, etc.). k= k= k= Give a sequece they should be able to write the formula for the th term; give the th term they should be able to write the terms of the sequece. Defiitio of covergece of a series as the limit of the associated sequece of partial sums.
Types of Series Arithmetic series, Geometric series, Alteratig series, Harmoic series, Alteratig harmoic series, p-series
Decimals 0.3333... = 0.3 + 0.03 + 0.003 + 0.003 + ( ) ( ) ( ) ( ) 0 0.3 0 0.3 0 0.3 0 = + + + 0.3 0 3 + = + + + + 0.3 0.3 = = = 0.9 3 0 ( ) ( ) ( ) ( ) 0 3 0.999 = 0.9 0 + 0.9 0 + 0.9 0 + 0.9 0 + 0.9 0.9 = = = 0.9 0 ( ) 0.999... = 3 0.333... = 3 = 3 + 0.999....999... = = 0.999... =
a = To graph = ( ) = 0.7 t i the plae: Mode: Parametric Graph format: Dot Equatio Editor: t(t) = t yt(t) = ( 0.7)^t Widow: tmi =, tma = 30, tstep =, mi = 0, ma = 3, scl =, ymi = 0, yma =.5, yscl =.
a = To graph = ( ) = 0.7 t o a umber lie: Mode: Parametric Graph format: Dot Equatio Editor: t(t) = ( 0.7)^t ad yt(t) = Widow: tmi =, tma = 30, tstep =, mi = 0, ma =, scl =, The TRACE the graph ad watch it coverge. ymi = 0, yma =, yscl =.
Taylor Polyomial If f has derivatives at c the the th Taylor polyomial for f at = c is ( ) ( ) ( ) ( ) ( ) f c f c T ( ) = f ( c) + f ( c)( c) + c + + c!! If c = 0, the th Maclauri polyomial for f is ( ) ( ) ( ) ( ) ( ) ( ) f c f c T = f c + f c + + +!!
Let ( ) 3 f = + 4 4 a. Write the power series for f cetered at =. b. Epad the terms of the power series ad simplify. c. Is this a accidet or will this happe with ay polyomial? Eplai. ( ) = + ( ) + ( ) + ( ) 3 40 7 f Now epad the biomial terms ad see what you get!
988 BC 4: Determie all values of for which the series k k coverges. l k ( k ) = 0 + k+ k+ l ( 3 ) lim k + lim l k + = k k k k l + 3 l ( k + ) By L'Hopital's rule, ( k + ) ( k + ) ( ) ( k ) l ( k + ) 3 lim lim k k + = + = lim = k l k+ 3 k k k + k + 3 l lim = k l 3 < <
Coverges for < < At =, series becomes l k = 0 ( k + ) diverges, by compariso with harmoic series 0 k ( ) ( k + ) At =, series becomes l coverges, by the alteratig series test. k = Serie s coverges for < k = 0 k +
Memorize the Maclauri Series ad the iterval of covergece for 3 4 e = + + + + + for all! 3! 4! 3 5 7 si = + + for all 3! 5! 7! 4 6 cos = + + for all! 4! 6! = + + + + + < < 3 4 -
000 BC 3: The Taylor series about = 5 for a certai fuctio f coverges to f( ) for all i the iterval of covergece. The th derivative of f at ( ) ( )! = 5 is give by f ( 5 ) = + ( ), ad f (5) =. (a) Write the third-degree Taylor polyomial for f about = 5. P3 ( f,5)( ) = ( 5) + ( 5) ( 5) 6 6 40 3 (b) Fid the radius of covergece of the Taylor series for f about = 5. Ratio test gives 5 < 5 < Radius is
(c) Show that the sith-degree Taylor polyomial for f about = 5 approimates f ( 6 ) with error less tha 000. By the alteratig series test the error is less tha ( ) ( ) 7 7 7! 6 5 = = < ( ) 7 7 (7 + ) 7! 9 5 000 Is this true at f(4)? Why, or why ot?
Legrage Form of the Remaider ad the Legrage Error Boud Taylor s Theorem: If f has derivatives of all orders o a iterval cotaiig a, the for ay positive iteger ad for all i the iterval, there eist a umber c betwee ad a such that: f ( a) ( ) f ( a) f( ) = f( a) + f ( a)( a) + ( a) + + + ( a) + R!! ( ) f ( + ) ( c )( R ( ) ) = a + ( + )! This is called the Legrage Form of the Remaider.
Eample : applyig the theorem to the sie fuctio cetered at the origi, there eists a c betwee ad zero such that si = + + si( c) 3 5 6 6 0 6! si(0.) = 0. (0.) (0.) + si( c )(0.) 3 5 6 6 0 70 si(0.) 0. (0.) (0.) 0.986693 3 5 6 0 Notice that the remaider term is ot calculated at = 0, but at some = c i the iterval (0, 0.), so the sith power term is used, R c = 6 5 70 si( )(0.) is ot zero. I the ope iterval (0, 0.) the largest the si(c) ca be is : (Note: si(0.) 0.9866933... ), so the largest the error ca be is ()(0.) 6 8.89 0 8 (or 70 error is cosiderably less, about 70 6 8 (0.986693)(0.).77 0 ). The actual.54 0 9.
Eample : applyig the theorem to c betwee ad zero such that e cetered at the origi, there eists a e = + + + + e 3 c 3! 4! 4 The at, say, = 0.: 3 ( ) ( ) ( ) e c = + 0. + 0. + 0. + e (0.) 0. 6 4 4 I the iterval [ 00,. ] the largest that e c ca be is 0. e.40, so the largest the error ca be is ( )( ) 4 5 4.40 0. 8.47 0. This is the Lagrage Error Boud. The actual error is about 6.94 0 5. e e 3 ( ) ( ) 0. 5 0. + 0. + 0. + 0. =.333 ± 8.47 0 = 6.407586...
999 BC 4: The fuctio f has derivatives of all orders for all real umbers. Assume f() = 3, f '() = 5, f ''() = 3, ad f '''() = 8. (a) Write the third-degree Taylor polyomial for f about = ad use it to approimate f(.5). (, ) = 3 + 5 ( ) + ( ) ( ) (.5 ) 4.958 3 3 8 T f f 3 6 (b) The fourth derivative of f satisfies the iequality ( 4 ) ( ) f 3 for all i the closed iterval [.5, ]. Use the Lagrage error boud o the approimatio to f(.5) foud i part (a) to eplai why f (.5 ) 5.
LEB = 3 4.5 =0.00785, f (.5 ) > 4.9583 0.00785 = 4.966 > 5 4! (c) Write the fourth degree Taylor polyomial, P(), for g( ) = f ( + ) about = 0. Use P to eplai why g must have a relative miimum at = 0. (, )( ) = 3 + 5 ( ) + ( ) ( ) ( )( ) 3 3 8 T f 3 6 T f, + = 3 + 5 + 3 4 ad from the coefficiets ( ) Secod Derivative Test. g (0) = 0 ad g 0 > 0 therefore a miimum by the
Geometric Series Method : The epressio a is similar to r. If we had a geometric series with a first term of a = ad a commo ratio of, the Turig this aroud, the power series for series 3 4 = + + + + + + + S =. must be the geometric ad the iterval of covergece must be all such that < or < <.
Ratioal Epressios Method : Log divisio yields the same result: ( ) 3 + + + + + + 3
Biomial Theorem Method 3: Epadig ( ) by the Biomial Theorem also gives the same result. ( ) ( ) ( )( ) 3 3 a + b = a + a b + a b + a b + 3 etc. ( ) ( )( )( 3) ( ) ( ) ( )( ) ( ) ( ) 3 ( ) = + ( ) ( ) + 4 3 + + 3 3 = + +
Ratioal Epressios Other fuctios ca be hadled i the same ways. Oe way to fid the 5 Maclauri Series for ay ratioal epressio, such as, is to arrage the + 5 terms with the lowest power first ad perform a log divisio. 3 5+ 5 + + 5 + 3 3 3 3 5 3 7 5 5 5 3 3 3 3 3 3 5 5 + 3 5 3 7 5 5 3 5 7
5 3 + = ( 5 ) 5 This is a geometric series with a first term of 3 ad a ratio of r = 3 ( 5 ) ( 5 ) ( 5 ) ( 5 ) 3 3 3 5 3 7 5 5 5 3 = 3 + 3 + 3 + 3 + = 3 + + Ad the iterval of covergece is < < 5 < 5 5 < < 5 5
A Mistake 3 = + ( ) + ( ) + ( ) + = + + + + 3 4 5 k k = + + 4 + 8 + 6 + + = = k = The series is geometric ad coverges whe < or < <. So the iterval of covergece is < <.
y 5 4 3 Iterval of covergece: < < 6 5 4 3 3 4 5 6 7 fuctio Power Series 3 4 5
But what if you do this? 3 = = = = = 4 8 6 3 ( ) 3 4 = The series is geometric ad coverges whe or < <. So the iterval of covergece is( ) ( ),,.
y 3 Iterval of Covergece < 6 5 4 3 3 4 5 6 7 3 Iterval of Covergece > 4 5
New Series from Old Treatig + as a geometric series with r = = gives: 4 6 k k = + + = ( ) for < < + k = 0 d But ta = d +. Therefore ta k 3 5 7 ( ) k = d = C + 3 + 5 7 + = C + k + k = The iitial coditio ( ) ta is ta 0 = 0 tells us that C = 0. So the power series for k ( ) k k for < < k = < <.
993 BC 3 Let f be the fuctio give by f( ) = e. (a) Write the first four ozero terms ad the geeral term for the Taylor series epasio of f( ) about = 0. / / 3 e = + + + + + +! 3!! e e 3 ( / ) ( / ) ( / ) = + + + + + +! 3!! 3! 3 3!! = + + + + + +
(b) Use the result from part (a) to write the first three ozero terms ad the geeral term of the series epasio about for = 0 for g ( ) e =. e / 3! 3 3!! = + + + + + + 3 / + + + + + + 3! 3!! e = / e = + + 3 + + +! 3!!
(c) For the fuctio g i part (b), fid g ( ) ad use it to show that =. 4( + )! 4 = ( ) g = + + + + + 3! 3!! ( ) ( ) ( + ) ( ) ( ) g = + + + + g 3! 3!! = + + + + 8 4! ( ) = + + + + 3! 3!! = + + + + 8 4! = 4! = / e Also; g( ) = e e g ( ) = e ( e ) g = = ( ) ( + ) ( )() / / = 4! 4 = 4 4
Euler s Formula 3 4 5 e = + + + + + +! 3! 4! 5! 3 4 5 i ( i ) ( i ) ( i ) ( i ) e = + i + + + + +! 3! 4! 5! Epad, ad simplify the terms. Group those terms without i ad those with i. 3 3 4 4 5 5 i i i i i e = + i + + + + +! 3! 4! 5! 4 6 3 5 7 i e = + + + + + i! 4! 6! 3! 5! 7!
4 6 3 5 7 e i = + + + + + i! 4! 6! 3! 5! 7! Recogize the Maclauri Series for cos ad si. e i = cos + i si Fially substitute = π ad simplify agai. e i π i π e = cos π + isi π i π e = + 0i + = 0