Taylor Polynomials and Taylor Series

Transcription

1 Taylor Polyomials ad Taylor Series Cotets Taylor Polyomials... Eample.... Eample Eample Eercises... 6 Eercise Solutios... 8 Taylor s Iequality... Eample.... Eample.... Eercises... 3 Eercise Solutios Ifiite Series... 6 Itroductio... 6 Geometric Series... 9 Eample Eercises... 3 Eercise Solutios... Appedi A. Proof of the Sum Formula for a Geometric Series Taylor Series... 4 Eample Taylor Series... 4 Eample Eample Eample Eample Summary to this Poit... 9 Taylor Polyomials... 9 Taylor Series... 9 Radius of Covergece... 3 Eample Eercises Eercise Solutios Commo Taylor Series & Radii of Covergece Operatios with Taylor Series Eample Eample Eample Eample Eample Eercises Eercise Solutios... 4

2 Math 5 - Taylor Polyomials & Taylor Series Sectio, Page of 43 TAYLOR POLYNOMIALS AND TAYLOR SERIES The followig otes are based i part o material developed by Dr. Ke Bube of the Uiversity of Washigto Departmet of Mathematics i the Sprig, 5. Modificatios made by the math faculty at Seattle Cetral College. Taylor Polyomials The taget lie to the graph of y= f ( ) at the poit = a is the lie goig through the poit ( a, fa ( )) that has slope f '( a ). By the poit-slope form of the equatio of a lie, its equatio is y fa ( ) = f'( a)( a) or y= fa ( ) + f'( a)( a) As you have already leared, the taget lie is a very good approimatio to y= f ( ) ear = a, as show i Figure. FIGURE. The lie y= fa ( ) + f'( a)( a) is taget to the graph of y= f ( ) at the poit ( a, fa ( )). Prited 6//6, :7 PM LU Jue 6; 3/4/; 3//; 5//9; 4/8/9; /4/8; 3/4/8; /5/8; /3/7; /8/7; //7;/9/7; ;/6/7;7/4/7; /9/7b,/9/7

3 Math 5 - Taylor Polyomials & Taylor Series Sectio, Page of 43 We will give a ame, T ( ), to the fuctio correspodig to the taget lie: T ( ) = fa ( ) + f '( a )( a ) For ear = a, we have f ( ) T( ). The taget lie fuctio T ( ) is called the Taylor polyomial of degree oe for f, ( ) cetered at = a. Notice that it satisfies the two coditios T ( a) = f( a) ad T '( a) = f'( a). I other words, T ( ) is the polyomial of degree oe that has the same fuctio value at = a ad the same first derivative value at = a as the origial fuctio f. ( ) We ca get a better approimatio, T ( ) ear a =, usig a parabola (a polyomial of degree two). The formula for T ( ) is f''( a) T ( ) = f( a) + f'( a)( a) + ( a) T ( ) is called the Taylor polyomial of degree two for f, ( ) cetered at a =. Sice T '( ) = f'( a) + f''( a)( a) ad T ''( ) = f''( a), T ( ) satisfies the three coditios: T ( a) = f( a), T' ( a) = f'( a) ad T ''( a) = f''( a). I other words, T ( ) is the polyomial of degree two that has the same fuctio value at = a, the same first derivative value at = a, ad the same secod derivative value at = a as the origial fuctio f. ( ) Eample. Fid the Taylor polyomials of degrees oe ad two for f ( ) = e, cetered at a =. Solutio: Sice f( ) = f'( ) = f''( ) = e, we have f() = f'() = f''() = e =, so the Taylor polyomial of degree oe (the taget lie to y= e at the poit (, )) is T ( ) = f() + f'()( ) = + The Taylor polyomial of degree two (the parabola that best fits y = e ear = ) is f ''() T ( ) = f() + f'()( ) + ( ) = + + Prited 6//6, :7 PM LU Jue 6; 3/4/; 3//; 5//9; 4/8/9; /4/8; 3/4/8; /5/8; /3/7; /8/7; //7;/9/7; ;/6/7;7/4/7; /9/7b,/9/7

4 Math 5 - Taylor Polyomials & Taylor Series Sectio, Page 3 of 43 FIGURE. The Taylor polyomials T ( ) ad T ( ) for ( ) f= e, cetered at a =. Notice that T ( ) does a better job of matchig f ( ) ear =. We ca get a eve better approimatio, T ( ) ear a 3 =, usig a cubic (a polyomial of degree three). The formula for T ( ) is 3 f''( a) f'''( a) T3 ( ) = f( a) + f'( a)( a) + ( a) + ( a) 6 3 T ( ) 3 is called the Taylor polyomial of degree three for f, ( ) cetered at = a. A short computatio (Eercise 3 o page 6) shows that T ( ) 3 satisfies the four coditios: T ( ) = f( a), T '( ) = f'( a), T ''( ) = f''( a), ad T '''( ) = f'''( a) I other words, T ( ) 3 is the polyomial of degree three that has the same fuctio value at = a, the same first derivative value at = a, the same secod derivative value at = a, ad the same third derivative value at = a, as the origial fuctio f. ( ) Prited 6//6, :7 PM LU Jue 6; 3/4/; 3//; 5//9; 4/8/9; /4/8; 3/4/8; /5/8; /3/7; /8/7; //7;/9/7; ;/6/7;7/4/7; /9/7b,/9/7

5 Math 5 - Taylor Polyomials & Taylor Series Sectio, Page 4 of 43 Eample. Fid the Taylor polyomial of degree three for f ( ) = si, cetered at a = 5π. 6 Solutio: f ( ) si f 5 π = 6 = ( ) = ( 5 3 ) f'( ) cos f''( ) si f ' π = 6 f '' 5π = 6 = ( ) = ( 5 3 ) f'''( ) cos f ''' π = 6 The Taylor polyomial of degree three (the cubic that best fits f ( ) = si ear a = 5π ) is 6 f'' 5π f''' 5π T 6 6 3( ) = f( 5π ) + f' ( 5π )( 5π ) + 5π + 5π ( 5π ) ( 5π ) ( 5π ) 3 3 = ( ) ( ) ( ) ( ) 3 3 FIGURE 3. The Taylor polyomial T ( ) 3 for f ( ) = si, cetered at a = 5π. 6 Prited 6//6, :7 PM LU Jue 6; 3/4/; 3//; 5//9; 4/8/9; /4/8; 3/4/8; /5/8; /3/7; /8/7; //7;/9/7; ;/6/7;7/4/7; /9/7b,/9/7

6 Math 5 - Taylor Polyomials & Taylor Series Sectio, Page 5 of 43 I geeral, the Taylor polyomial of degree for f, ( ) cetered at = a, is ( ) ( ) f''( a) f'''( a) 3 f ( a) ( ) f ( a) T ( ) = f( a) + f'( a)( a) + ( a) + ( a) ( a) + ( a) 6 ( )!! where! (read factorial ) is the product of the first positive itegers:! = 3...( ) ( ) For coveiece, we defie! = ad let f just mea the fuctio f itself. The the formula for T ( ) ca be writte usig summatio otatio: ( ) f ( a) T ( ) = ( a)! By a similar computatio to that for T ( ) or for T ( ) 3, it ca be show that T ( ) satisfies the + coditios T a = f a T a = f a T a = f a T a = f a T a = f a ( ) ( ) ( ) ( ) ( ) ( ), '( ) '( ), ''( ) ''( ),..., ( ) ( ), ad ( ) ( ) I other words, T ( ) is the polyomial of degree that has the same fuctio value, first derivative value, secod derivative value, etc., ad th derivative value at = a as the origial fuctio f. ( ) Eample.3 Fid the Taylor polyomial of degree for f ( ) =, cetered at a =. Solutio: = f () = f ( ) ( ) f'( ) ( ) ( ) ( ) f''( ) ( ) f = = = '() 3 = f ''() =! 4 = f '''() = 3! f'''( ) 3 ( ) etc. ( ) ( ) f ( ) =! ( ) + etc. ( f ) () =! The Taylor polyomial for degree for f ( ) =, cetered at a =, is ( ) ''() '''( ) 3 () f f a f T ( ) = f() + f'()( ) + ( ) + ( ) ( ) 6! 3 = Prited 6//6, :7 PM LU Jue 6; 3/4/; 3//; 5//9; 4/8/9; /4/8; 3/4/8; /5/8; /3/7; /8/7; //7;/9/7; ;/6/7;7/4/7; /9/7b,/9/7

7 Math 5 - Taylor Polyomials & Taylor Series Sectio, Page 6 of 43 Eercises. Let ( ) T = + + be the Taylor polyomial of degree two for f ( ) = e, cetered at a = (See Eample.). Verify directly by taig their derivatives that T ( ) ad ( ) f satisfy the three coditios: T () = f(), T '() = f'() ad T ''() = f''() Let T3 ( ) π π π = + be the Taylor polyomial of degree three for π f ( ) = si, cetered at a =. (See Eample.). Verify directly by taig their derivatives that 3 6 T ( ) ad f ( ) satisfy the four coditios: T π 3 f π, T3' π f' π, T3'' π f'' π, ad T3''' π f''' π = = = = f''( a) f'''( a) 3 3. Let T3 ( ) = f( a) + f'( a)( a) + ( a) + ( a) be the Taylor polyomial of degree three for the 6 fuctio f, ( ) cetered at the poit = a. a) Fid T3'( ), T3''( ), ad T3'''( ). b) Evaluate T3'( ), T3''( ), ad T3'''( ) at = a to verify that: T ( a) = f( a), T '( a) = f'( a), T'' ( a) = f''( a), ad T '''( a) = f'''( a) The fuctio f ( ) is approimated ear = by the secod-degree Taylor polyomial T ( ) = Fid the value of a) f (), b) f '() ad c) f ''(). 5. Suppose g is a fuctio with cotiuous derivatives, ad that g () = 3, g '() =, g ''() =, ad g '''() = 3. a) What is the Taylor polyomial of degree for g, cetered at a =? b) What is the Taylor polyomial of degree 3 for g, cetered at a =? c) Use T ( ) ad T ( ) 3 to approimate g (.). 6. For the fuctio f ( ) = l : a) List the first four derivatives of f. ( ) b) What are the values of the first four derivatives of f ( ) evaluated at a =? c) Write T ( ) as a polyomial i the log form : 4 4 A+ B( a) + C( a) + + E( a). d) Optioal Challege questio: Write T ( ) 4 usig summatio otatio. e) Graph f( ), T( ), T( ), T3( ), ad T4( ) o the same set of aes, with.5 < < ad 5< y < For the fuctio f ( ) = : a) Fid T( ), T( ), T( ), T3( ), T4( ), ad T5( ) cetered at a =. b) What is the sigificace of T ( )? c) What is T ( )? 3 Prited 6//6, :7 PM LU Jue 6; 3/4/; 3//; 5//9; 4/8/9; /4/8; 3/4/8; /5/8; /3/7; /8/7; //7;/9/7; ;/6/7;7/4/7; /9/7b,/9/7

8 Math 5 - Taylor Polyomials & Taylor Series Sectio, Page 7 of For the fuctio f ( ) = l : a) Fid the equatio of the lie taget to f ( ) at =. b) Fid a fuctio T ( ) that has the followig two properties T() = f() ad T'() = f'(). c) Graph f ( ) ad T ( ) o the same graph. 9. For the fuctio f ( ) = l : a) Fid a fuctio T ( ) that has the followig three properties: T() = f(), T' () = f'() ad T''() = f''() b) Graph f ( ) ad T ( ) o the same graph. c) What is the COMPLETE special ame give to T ( )? d) Fid the value of each of the followig: f() ad T(), f() ad T (), ad fially, f(), ad T ().. a) Fid the Taylor polyomial of degree 4 for the fuctio f ( ) = l, cetered at a =. b) What is the differece betwee T (.) 4 ad f (.)? (i.e. What is the error? ). a) Fid the parabola that best approimates the uit circle + y = ear the poit (, ). b) Use your aswer to part (a) to estimate the y-coordiate of the poit o the upper half of the uit circle with the -coordiate equal to... a) Fid the Taylor polyomial of degree 4, cetered at a = for the fuctio f ( ) = e. b) Compare this result to the Taylor polyomial of degree for the fuctio f ( ) = e, cetered at a =. What do you otice? c) Use your observatio i part (b) to write out the Taylor polyomial of degree for f ( ) = e. d) What is the Taylor polyomial of degree 5 for the fuctio f ( ) = e, cetered at a =? e) Double-chec these o your calculator. 3. Fid the Taylor polyomial of the give degree for the give fuctio, cetered at the give poit. (Leave your aswers as sums of powers of ( a), as i Eample..) a) f ( ) = l, degree 3, cetered at a =. b) f ( ) = +, degree 3, cetered at a =. c) f ( ) = si, degree 3, cetered at a = π /3. d) f ( ) = cos, degree 4, cetered at a = π. e) f ( ) =, degree, cetered at a =. 4. Show how you ca use the Taylor polyomial of degree 3 cetered at a =, which says that si to eplai why lim =. 3 si, 3! Prited 6//6, :7 PM LU Jue 6; 3/4/; 3//; 5//9; 4/8/9; /4/8; 3/4/8; /5/8; /3/7; /8/7; //7;/9/7; ;/6/7;7/4/7; /9/7b,/9/7

9 Math 5 - Taylor Polyomials & Taylor Series Sectio, Page 8 of 43 Eercise Solutios. T( ) = + + T() = f ( ) = e f() = T "( ) = + T '() = f'( ) = e f'() = T"( ) = T ''() = f"( ) = e f"() = f''( a) f'''( a) 3.(a) T3 ( ) = f( a) + f ( a)( a) + ( a) + ( a) 6 f'''( a) T3 '( ) = f'( a) + f''( a)( a) + ( a) T ''( ) = f''( a) + f'''( a)( a) 3 3 T '''( ) = f'''( a) 3 f''( a) f'''( a) 6 f'''( a) T3 '( a) = f'( a) + f''( a)( a a) + ( a a) = f'( a) T ''( a) = f''( a) + f'''( a)( a a) = f''( a) 3 (b) T3 ( a) = f( a) + f'( a)( a a) + ( a a) + ( a a) = f( a) 3 3 T '''( a) = f'''( a) f'() f''() 5. (a) T ( ) = f() + ( ) + ( )!! = f'() f''() f'''() (b) T3 ( ) = f() + ( ) + ( ) + ( )!! 3! 3 = (c) g(.) T (.) = 3.5 (d) g(.) T (.) = Prited 6//6, :7 PM LU Jue 6; 3/4/; 3//; 5//9; 4/8/9; /4/8; 3/4/8; /5/8; /3/7; /8/7; //7;/9/7; ;/6/7;7/4/7; /9/7b,/9/7

10 Math 5 - Taylor Polyomials & Taylor Series Sectio, Page 9 of (a) T ( ) = f() = f '() T ( ) = f() + ( ) = + 3! f'() f''() T ( ) = f() + ( ) + ( ) = !! f'() f''() f'''() 3 3 T3 ( ) = f() + ( ) + ( ) + ( ) = !! 3! f'() f''() f'''() 3 f ''''() 4 T4 ( ) = f() + ( ) + ( ) + ( ) + ( ) =!! 3! 4! = = ( 5) f'() f''() f'''() f''''() f () T5 ( ) = f() + ( ) + ( ) + ( ) + ( ) + ( ) =!! 3! 4! 5! = = (b) T is the costat term i f; ( ) the y-itercept i the graph. (c) T ( ) = f ( ) = (a) f ( ) = l, f() = f'( ) = f'() = f''( ) = f''() = So, f ''() T ( ) = f() + f'()( ) + ( ) = + ( ) + ( ) = ( ) (c) The Taylor Polyomial of degree, cetered at a =. f() = T () = f() = l().693 T () =.5 f () = udefied T () = 3 / Prited 6//6, :7 PM LU Jue 6; 3/4/; 3//; 5//9; 4/8/9; /4/8; 3/4/8; /5/8; /3/7; /8/7; //7;/9/7; ;/6/7;7/4/7; /9/7b,/9/7

11 Math 5 - Taylor Polyomials & Taylor Series Sectio, Page of 43. (a) y f ( ) = = ad a parabola is secod degree, so we are looig for ( ) ( ) () f= f = f'( ) = f'() = f''( ) = f''() = 3/ ( ) T, cetered at a =. f ''() T ( ) = f() + f'() ( ) + ( ) = + + ( ) = (.) (b) T (.) = = (a) T3 ( ) = l() + ( ) ( ) + ( ) (b) T3 ( ) = π π π (c) T3 ( ) = (d) T4 ( ) = + ( π) ( π) 4 (e) T 3 ( ) = = + Prited 6//6, :7 PM LU Jue 6; 3/4/; 3//; 5//9; 4/8/9; /4/8; 3/4/8; /5/8; /3/7; /8/7; //7;/9/7; ;/6/7;7/4/7; /9/7b,/9/7

12 Math 5 - Taylor Polyomials & Taylor Series Sectio, Page of 43 Taylor s Iequality I this sectio, we estimate the remaider term R( ) = f ( ) T( ), which is the differece betwee f ( ) ad the Taylor polyomial T ( ) for f, ( ) cetered at = a. The size of R ( ) tells us how good a approimatio T ( ) is to f: ( ) the smaller R( ) is, the closer T ( ) is to f. ( ) To estimate the remaider term R( ), we eed a formula for R( ). The formula we eed is ot provided here, but a cosequece of this formula is: Taylor s Iequality Suppose that f ( ) has + cotiuous derivatives o a iterval cotaiig both a ad, ad that ( + ) f () t M + for all values of t betwee a ad. The M+ + R( ) a, ( + )! where R( ) = f ( ) T( ) is the remaider term ad T ( ) is the Taylor polyomial of degree for f, ( ) cetered at = a. The form of this upper boud for R( ) is easy to remember; if we were to add oe more term to T ( ) we would get T ( ) + (both cetered at = a). This last term would be ( + ) f ( a) + a ( + )! If we were to tae absolute values ad replace boud for R( ). ( + f ) ( a) by M +, we would get the epressio for the upper Cautio: This memoic device for rememberig the form of the upper boud for R( ) does NOT mea that ( + ) ( + ) R ( ) f ( a) equals ( a) + f () c. This eact value of R( ) is ( a) + for some value of c betwee a ad. ( + )! ( + )! Notice that whe is very close to a, a is very small, so higher powers of a are much smaller tha lower powers. For eample, whe a =., the a =. ad 3 a =.. The absolute value of the remaider term is bouded by a costat times a +, which is oe higher power of a tha the degree. Amog all polyomials of degree, the Taylor polyomial T ( ) is the ONLY oe that has this property; ay other polyomial differs from f ( ) (for very close to a) by at least a costat times a, which is bigger tha a costat times a. This eplais why we call T ( ) the polyomial of degree that best fits f ( ) ear = a.

13 Math 5 - Taylor Polyomials & Taylor Series Sectio, Page of 43 Eample. The lie taget to the curve upper boud for R (.). y= si at (, ) is the lie y=. Use the Remaider Theorem to fid a Solutio: The lie y= is the Taylor polyomial T ( ) of degree oe for the fuctio f ( ) = si, cetered at a =. Because =, the differece betwee the curve ad its taget lie is R( ) = f ( ) T( ), ad Taylor s iequality taes the form = f T = R a si ( ) ( ) ( ) M! I the above formula we eed the value of M, which is a upper boud of f''( t ) for t values betwee a = ad =.. Sice f''( ) = si, we have f''( t) for all t, so we ca simply tae M =. This is a crude value for M! Substitutig M =, a = ad =., we determie a upper boud for the differece betwee the curve ad taget lie at =. : si.. = f(.) T(.) = R(.). =.5.! I the above eample we foud a upper boud o the error at eactly oe value of, that beig =.. I fact, this is a upper boud o the error for all values i the iterval... This is true because of the maer i which M is determied, alog with the fact that for all values i this iterval, We eplore this idea further i the et eample. Eample... Give a upper boud formula for the differece betwee the fuctio f ( ) = e ad its Taylor polyomial ( ) T = + + (cetered at a = ) for. (see Figure ). Solutio: Here, a = ad > a ad f'''( ) = e (which is positive ad icreasig for all ), so the maimum value of f'''( ) o the iterval. is tae o at the right edpoit: M f e. 3 = ma '''( ) =.. Usig the upper boud formula for R( ) give above, we obtai for. :.. M3 3 e 3 e 3 R ( ) =. <.84 3! 3! 6 Prited 6//6, :7 PM LU Jue 6; 3/4/; 3//; 5//9; 4/8/9; /4/8; 3/4/8; /5/8; /3/7; /8/7; //7;/9/7; ;/6/7;7/4/7; /9/7b,/9/7

14 Math 5 - Taylor Polyomials & Taylor Series Sectio, Page 3 of 43 Eercises. a) Determie a better value for M i Eample.. b) What is the ew error estimate?. For the Taylor polyomial of degree 4 for f ( ) = l, cetered at a = : a) What is the appropriate value of to use i the formula o the error oe maes whe usig T (.) 4 to estimate f (.)? ( ) M ( + )! + R a + to fid the upper boud st b) What is the ( + ) derivative i this situatio? st c) Is the ( + ) derivative a icreasig or decreasig fuctio? st d) Over what iterval must you graph the ( + ) derivative to fid a value for M +? e) Do you epect to fid the best (i.e., the smallest) value for M + o the left ed, o the right ed, or somewhere i the middle of the iterval? (Recall that M + must be at least as large as every value of the st ( + ) derivative i the iterval.) 3. Let f ( ) = +. (4) (4) (4) a) Show that the fuctio g ( ) = f ( ) is decreasig for ; coclude that ma f ( ) = f (). b) Use Taylor s Iequality to give a upper boud for the absolute value of the differece betwee the fuctio f ( ) = + ad its Taylor polyomial of degree 3, cetered at a =, for.. c) You foud T ( ) 3 i Eercise #3b of Sectio. Use your calculator to fid both f (.) =. ad T (.) 3 to as may decimal digits as your calculator shows. How does f(.) T3 (.) compare with your aswer to part (b)? 4. For the Taylor polyomial of degree 4, cetered at a = for f= ( ) si : a) Accordig to Taylor s Iequality, what is the largest possible differece betwee the Taylor polyomial ad the fuctio whe estimatig f (.75)? b) Would you epect R (.75) 4 to be larger, smaller or the same if we istead cetered our Taylor polyomial at π /4? Why? c) What is the upper boud o the error whe usig the Taylor polyomial of degree 4, cetered at a = π /4 to estimate f (.75)? 5. For g ( ) = e a) What is T ( ), cetered at a =? b) What is the maimum error we would epect i usig T ( ) to estimate g (.)? c) Is the error estimate larger, smaller or the same as Eample. s error estimate for T (.) for ( ) f= e? What is it about the graphs of e ad e that eplai this differece (or sameess)? d) Calculate the eact differece betwee f (.) ad its correspodig T (.). e) Calculate the eact differece betwee g (.) ad its correspodig T (.). Prited 6//6, :7 PM LU Jue 6; 3/4/; 3//; 5//9; 4/8/9; /4/8; 3/4/8; /5/8; /3/7; /8/7; //7;/9/7; ;/6/7;7/4/7; /9/7b,/9/7

15 Math 5 - Taylor Polyomials & Taylor Series Sectio, Page 4 of How may terms are ecessary to guaratee the Taylor polyomial, cetered at a =, is accurate to withi.5 of the eact value whe =.3 for each of the followig fuctios: a) f ( ) = e b) f ( ) = si c) f ( ) = cos 7. a) Draw a graph of the Taylor polyomial of degree three, cetered at a = for the fuctio f ( ) = l that illustrates the eact error i usig T ( ) 3 to estimate l. b) What problem do we ru ito i tryig to graph R ( )? 3 optioal: Discuss the pros ad cos of some reasoable way aroud the problem to show a graph of R ( ) This problem refies Eample.. Let T ( ) ad T ( ) be the Taylor polyomials of degrees oe ad two for f ( ) = si, cetered at a =. a) Show that T( ) = T( ). (So i this case, the taget lie is also the Taylor polyomial of degree.) b) Use Taylor s Iequality with = to give a eve better (i.e., smaller) upper boud o si for. tha is give i Eample.. 9. This problem cotiues the previous problem. Throughout this problem, f ( ) = si, ad all Taylor polyomials are cetered at a =. ( + a) Show that whe is odd, T( ) = T+ ( ). (Hit: Whe is odd, f ) () =.) b) Whe is odd, fid a upper boud o si T( ) = si T+ ( ) = R( ) for. by usig Taylor s Iequality for +. c) What is the smallest value of for which you ca guaratee that si T ( ) wheever.? (Hit: By part (a), the smallest will be odd. Use your aswer to part (b) to set up a equatio [actually, a iequality] for. Do ot try to solve this iequality for directly because appears i both a factorial ad a epoet. Istead, try guess ad chec: guess =, = 3, etc., ad see what your aswer to part (b) becomes.) d) For the you foud i part (c), use your calculator to fid both si(.) ad T (.). Be sure your calculator is set o radias, ot degrees, ad record as may decimal digits as your calculator shows. [Remar: Calculators use approimatios similar to Taylor polyomials to compute trigoometric, epoetial ad logarithmic fuctios.] Prited 6//6, :7 PM LU Jue 6; 3/4/; 3//; 5//9; 4/8/9; /4/8; 3/4/8; /5/8; /3/7; /8/7; //7;/9/7; ;/6/7;7/4/7; /9/7b,/9/7

16 Math 5 - Taylor Polyomials & Taylor Series Sectio, Page 5 of 43 Eercise Solutios. (a) M =. (b) error estimate =.5. ( + ) (5) 4! 4. (a) = 4 (b) f ( ) = f ( ) = = (c) decreasig (draw a graph) 5 5 (d) M + must be eamied over the iterval from a to ; i this case [,.]. ( + ) (5) 4 (e) A graph shows that f ( ) = f ( ) = is largest o the left ed (as is the case for ay decreasig 5 fuctio). (4) 5 3. (a) A graph shows that the largest value of f ( ) = 6( + ) value is 5/6. M / (b) R3( ) a + =. = (3 + )! 4! (c) 7/ occurs o the left ed ( = ), so its largest T (.).4885, so T (.) (.) calculator 3 = 3 f =, which is less tha R ( ) 3 calculated i part (b). 4. (a) The largest value we epect for ay derivative of si is, so we choose M + =. M5 5 5 R4 ( ) =.75 = ! 5! (b) We would epect R( ) to be smaller because.75 is closer to π /4 tha to zero, while othig else chages. (c) 5 5 M5 π π 4( ) =.75 = 4.7 R 5! 4 5! 4.3 M+ + e + 6. (a) R(.3).3 =.3.5 =. ( + )! ( + )! by trial ad error NOTE: M is chose at the right ed sice all derivatives are icreasig fuctios. M+ + + (b) R(.3).3 =.3.5 =. ( + )! ( + )! by trial ad error NOTE: The first-degree Taylor polyomial is already withi.5 of ay value <.3 i y= si! (c) Calculatios are the same as i part (b). 8 (a) f ( ) = si f() = f'( ) = cos f'() = f''( ) = si f''() = (b) T ( ) = f() + f'()( ) = f ''() = + + = M R ( ) =. =.6 3! 3! T ( ) f() f'()( ) ( ) Prited 6//6, :7 PM LU Jue 6; 3/4/; 3//; 5//9; 4/8/9; /4/8; 3/4/8; /5/8; /3/7; /8/7; //7;/9/7; ;/6/7;7/4/7; /9/7b,/9/7

17 Math 5 - Taylor Polyomials & Taylor Series Sectio 3, Page 6 of 43 3 Ifiite Series Itroductio We are all familiar with the fact that the fractio 3 is equivalet to the ifiite repeatig decimal So we could write: = The ifiite additio of fractios o the right had side above is a eample of what we call a ifiite series. Here is the formal defiitio of a ifiite series: Defiitio: A ifiite series is a epressio of the form a + a + a3 + a where a, a, a3, a 4,... is a ifiite sequece of real umbers. We ca epress the ifiite series usig summatio otatio: a + a + a3 + a = a. Notatio ote: Sometimes it will be coveiet to start the ifiite series with a, i which case the ifiite series will be epressed as: a + a + a3 + a = a. So far, all that we have doe is give a ame to what loos lie the sum of a ifiite umber of terms, however we still eed to mae clear what eactly it meas to fid such a sum. Some questios we have are: How do we add ifiitely may terms? Does every ifiite series have a sum? We cosider three eamples o the et page. After cosiderig these eamples we will mae some formal defiitios ad aswer the questios above.

18 Math 5 - Taylor Polyomials & Taylor Series Sectio 3, Page 7 of 43 Series #. Cosider the ifiite series: = We begi by looig at the partial sums for this series. + = = = = As you ca see, the partial sums seem to be approachig the umber. (Later i this sectio we will prove this rigorously, but for ow, we ca rely o our ituitio). I a situatio lie this where the partial sums are approachig a fiite umber, we say that the series is covergig to the fiite sum. We also say that the series is coverget. Series #. It is easy to come up with a series that is ot coverget. For eample, cosider the ifiite series which adds the positive itegers: as we add the terms, the partial sums: = Clearly, there is o fiite sum for this series because + = = = are icreasig without boud. Whe a ifiite series does ot have a fiite sum we say it is diverget. Series # 3. Fially let s cosider the series = This series is called the Harmoic Series. 3 4 Looig at the idividual terms that we are addig we might guess that this series also coverges. However, it turs out that the partial sums for this series are i fact ubouded. So this series does ot have a sum. It is diverget. (See Eercise 7 at the ed of this sectio where we prove that this series diverges). The harmoic series is importat because it represets a simple eample of a series whose idividual terms are approachig, but which evertheless does ot have a sum. This may seem couter ituitive but we have to remember that we are addig a ifiite umber of terms. So eve if they are very small they ca add up. I the case of the harmoic series, if we add eough terms we ca get the partial sums to eceed ay fiite umber! Prited 6//6, :7 PM LU Jue 6; 3/4/; 3//; 5//9; 4/8/9; /4/8; 3/4/8; /5/8; /3/7; /8/7; //7;/9/7; ;/6/7;7/4/7; /9/7b,/9/7

19 Math 5 - Taylor Polyomials & Taylor Series Sectio 3, Page 8 of 43 Before cotiuig lets summarize some of the termiology that we have developed so far. Defiitio: For ay ifiite series So i geeral the th partial sum is s a, we defie the partial sums of the series as: s = a s = a + a s = a + a + a 3 3 s = a + a + a + a etc. = a. [Similarly, I this case the partial sums are s = a, s = a + a, s = a + a + a, ad so o]. = a for a series that starts with the ide s =. We are ow ready to state how we ca determie whether a series has a sum. Defiitio: If the partial sums for a series coverge to a fiite umber, which we will deote by S, we say that the series is coverget ad write: a = S. O the other had, if the partial sums do ot coverge to ay specific umber, we say that the ifiite series is diverget ad therefore does ot have a sum. Usig limit otatio we could summarize the relatioships stated i the defiitio above as follows. If lim s = S the a = S I words, if the partial sums coverge to a umber S, the S is the sum of the series. A thorough study of ifiite series is beyod the scope of these materials. If you are iterested i readig more o the geeral topic of ifiite series see Sectios i Calculus ad Cocepts by Stewart. The remaider of this sectio will focus o oe particular class of ifiite series, the so-called geometric series. Prited 6//6, :7 PM LU Jue 6; 3/4/; 3//; 5//9; 4/8/9; /4/8; 3/4/8; /5/8; /3/7; /8/7; //7;/9/7; ;/6/7;7/4/7; /9/7b,/9/7

20 Math 5 - Taylor Polyomials & Taylor Series Sectio 3, Page 9 of 43 Geometric Series A geometric series is a ifiite series that has the followig special form: 3 ar = a + ar + ar + ar +... (where a ) Here a is called the startig value, ad r is the costat multiplier. Note that each term is obtaied by multiplyig the precedig oe by r. The Sum of a Geometric Series 3 It ca be readily proved (see Appedi A o page 3) that the geometric series a + ar + ar + ar +... a will coverge to a fiite sum provided that < r <. Furthermore, the sum is eactly equal to r. O the other had, if either r or r, the the geometric series diverges. So we will write: a ar = if r < r a Note that r < is just aother way of writig < r <. Cautio: the formula for the sum r if r <. If r the the formula does ot apply sice there is o sum! oly wors Before movig o we poit out that the geometric series is very special. Most series do ot have a simple sum formula lie this. For geeral ifiite series, eve whe we ow that a sum eists, it is ofte difficult or impossible to fid a closed form epressio for the sum. Eample 3. Special case: a =. The geometric series whe a = is particularly simple i structure. For a =, we have 3 r = + r+ r + r +... The correspodig sum formula is r = if r <. r If we use i place of r the we ca write: 3 = = provided that < This is a importat result with some iterestig implicatios. Prited 6//6, :7 PM LU Jue 6; 3/4/; 3//; 5//9; 4/8/9; /4/8; 3/4/8; /5/8; /3/7; /8/7; //7;/9/7; ;/6/7;7/4/7; /9/7b,/9/7

21 Math 5 - Taylor Polyomials & Taylor Series Sectio 3, Page of 43 Here is a preview of what is comig up i the et sectio. Cosider the fuctio f ( ) =. This fuctio is defied for all real umbers ecept =. But otice that if we cofie to the iterval < < we ca say that f ( ) equals the sum of a coverget ifiite series. That is, we ca write: f 3 ( ) = = provided that < <. This meas that the partial sums etc. coverge to the fuctio y =. But these partial sums are the Taylor polyomials for this fuctio! 3 As we will see i the et sectio, the ifiite series is called the Taylor series for this fuctio. 3 Prited 6//6, :7 PM LU Jue 6; 3/4/; 3//; 5//9; 4/8/9; /4/8; 3/4/8; /5/8; /3/7; /8/7; //7;/9/7; ;/6/7;7/4/7; /9/7b,/9/7

22 Math 5 - Taylor Polyomials & Taylor Series Sectio 3, Page of 43 3 Eercises. Cosider the ifiite series ( ). Write dow the values for the first four partial sums for this series (i.e. s, s, s, s 3). Fid the sum of this series (if it eists). Justify your aswer.. The series = is a geometric series. What is its sum? Show that the ifiite series is a geometric series by writig it i the form ar. Calculate the sum. 4. Fid the sum of the series Epress the ifiite repeatig decimal as a sigle fractio by first writig it as a geometric series. 6. Determie whether the geometric series is coverget or diverget. If it is coverget, fid its sum: 4 a) b) c) ( ) π d) + 3 Hit: the first step is to put this ito the form ar. 7. The series = loos lie is should coverge but i fact diverges. 3 4 Prove that this series caot have a fiite sum. Hit: Use the figure below to coclude that s The use this iequality to show that the partial sums caot coverge. + > d. This series is called the Harmoic Series. It is the classic eample of the fact that a ifiite series may diverge eve though the idividual terms are covergig to zero. Prited 6//6, :7 PM LU Jue 6; 3/4/; 3//; 5//9; 4/8/9; /4/8; 3/4/8; /5/8; /3/7; /8/7; //7;/9/7; ;/6/7;7/4/7; /9/7b,/9/7

23 Math 5 - Taylor Polyomials & Taylor Series Sectio 3, Page of 43 3 Eercise Solutios. s s s s 3 = = ( ) = + = ( ) ( ) = + + = ( ) ( ) ( ) = = 3 ( ) ( ) ( ) ( ) Sice lim s does ot eist, this series does ot coverge to a value = = = = = = = = 9 7. I the drawig, the area iside the rectagles represets s. Sice each rectagle is above the fuctio f ( ) = / for all >, the area uder f ( ) = / is smaller tha the area iside the rectagles. (i.e. s + > d ) Sice we ow d = liml( t) = (is ubouded), we coclude that the larger area (iside the t rectagles) must also be ubouded. Prited 6//6, :7 PM LU Jue 6; 3/4/; 3//; 5//9; 4/8/9; /4/8; 3/4/8; /5/8; /3/7; /8/7; //7;/9/7; ;/6/7;7/4/7; /9/7b,/9/7

24 Math 5 - Taylor Polyomials & Taylor Series Sectio 3, Page 3 of 43 Appedi A. Proof of the Sum Formula for a Geometric Series Cosider the geometric series 3 ar = a + ar + ar + ar +... (where a ) Case. If r = the the series becomes a+ a+ a+... which clearly does ot have a fiite sum so the series diverges. Case. If r the we proceed as follows. Cosider the th partial sum: 3 s a ar ar ar... ar = [equatio ] Multiply both sides of equatio by r: 3 = [equatio ] rs ar ar ar ar + Subtract equatio from equatio ad do a little algebra to get a formula for s : s rs = a ar + rs = a r + ( ) ( ) a s = r + ( r ) Now to fid the sum of the geometric series, we must calculate the limit of the partial sums s as. So we eed to compute a a s = r = r r r + + lim lim ( ) lim( ) Now if r < the lim r + =. So, i that case, we fid the sum to be lim s a =. r O the other had if r >, the r + so lim s does ot eist. Fially, if r =, the the partial sums s will oscillate betwee ad a ad hece agai lim s does ot eist. Prited 6//6, :7 PM LU Jue 6; 3/4/; 3//; 5//9; 4/8/9; /4/8; 3/4/8; /5/8; /3/7; /8/7; //7;/9/7; ;/6/7;7/4/7; /9/7b,/9/7

25 Math 5 - Taylor Polyomials & Taylor Series Sectio 4, Page 4 of 43 4 Taylor Series We ow cosider the limit of the Taylor polyomials T ( ) for a fuctio f, ( ) cetered at a = goes to ifiity. Eample 4., as the degree I Eample.3, we showed that the Taylor polyomial T ( ) (cetered at a = ) for the fuctio f ( ) = is 3 T ( ) = For what values of is it true that T ( ) f ( ) as? Solutio: For each fied value of, 3 T ( ) = = is the th partial sum of the ifiite geometric series 3 = = lim. By Eample 3., this series does ot coverge if ad it coverges to We coclude thatt ( ) f ( ) = as, if <. if <. The ifiite series 3 lim ( T ( ) ) = lim = = is called the Taylor series for f ( ) =, cetered at a =. Taylor Series The Taylor series for a fuctio f, ( ) cetered at = a, is the ifiite series ( ) f ( a) f''( a) f'''( a) 3 ( a) = f( a) + f'( a)( a) + ( a) + ( a) +...!! 3! Remar o ceterig at a = : I these otes, we will be looig mostly at Taylor series cetered at a =. Whe we tal about the Taylor series for a fuctio ad do ot say eplicitly where it is cetered, we will assume that it is cetered at a =.

26 Math 5 - Taylor Polyomials & Taylor Series Sectio 4, Page 5 of 43 Eample 4. shows that the Taylor series for f ( ) = actually coverges to f ( ) for < : 3 = = for < The Taylor series i Eample 4. has a very special property. As we saw i Appedi A, there is a simple closedform epressio for T ( ): T ( ) = +, The above epressio eables us to evaluate the limit lim T ( ) for < without difficulty. For almost all other Taylor series, we caot come up with a closed-form epressio for T ( ) that is simple eough to eable us to evaluate the limit lim T ( ) easily. Istead, we ca ofte use Taylor s Iequality to show that T( ) f ( ). Eamples 4., 4.3, ad 4.4 will illustrate how to do this. We start with a useful limit. Eample 4. Show that for ay fied real umber, as!. Solutio: Choose a iteger N. The N, so for N (with both ad N beig held fied), N =! N! N+ N+ N+ 3 N N! N = N! N But N N as. Cosequetly, N!! also approaches as. Prited 6//6, :7 PM LU Jue 6; 3/4/; 3//; 5//9; 4/8/9; /4/8; 3/4/8; /5/8; /3/7; /8/7; //7;/9/7; ;/6/7;7/4/7; /9/7b,/9/7

27 Math 5 - Taylor Polyomials & Taylor Series Sectio 4, Page 6 of 43 Eample 4. Fid the Taylor series for f ( ) = si ad show that it coverges to f ( ) = si for all. Solutio: We start by looig at the derivatives of si whe evaluated at = : f ( ) = si f() = f'( ) = cos f'() = f''( ) = si f''() = f'''( ) = cos f'''() = f''''( ) = si f''''() = ( We see that the values of f ) () for =,,, 3,... repeat i cycles of legth four:,,, -,,,, -, etc. The Taylor series for f ( ) = si is ( ) f () = ! 3! 5! 7! 9! = ! 5! 7! 9! j+ =.! j= j ( ) ( j + ) For each fied, we wat to show that this Taylor series coverges to f ( ) = si. This is equivalet to sayig that T ( ) (which is the th partial sum of the Taylor series) coverges to f, ( ) which is equivalet to showig that f ( ) T( ) = R( ) as. Sice the derivatives of si are either ± cos or ± si, ( + f ) ( ) for all ad all. So we ca tae M + = i Taylor s Iequality for all ad all. Usig the upper boud for R ( ) give i Taylor s Iequality, we obtai + + M + f ( ) T( ) = R( ) = ( + )! ( + )! The fractio approaches as by Eample 4.. We coclude that T ( ) f ( ) as for ( + )! each fied, so that the Taylor series for f ( ) = si actually coverges to f ( ) = si: ( ) f () = = si for all.! 3! 5! 7! 9! Figure 4 illustrates the covergece of the Taylor series for f ( ) (cetered at a = ) to si. The partial sums of the Taylor series are the Taylor polyomials. Prited 6//6, :7 PM LU Jue 6; 3/4/; 3//; 5//9; 4/8/9; /4/8; 3/4/8; /5/8; /3/7; /8/7; //7;/9/7; ;/6/7;7/4/7; /9/7b,/9/7

28 Math 5 - Taylor Polyomials & Taylor Series Sectio 4, Page 7 of 43 FIGURE 4 The Taylor polyomials T ( ), T3 ( ), T5 ( ), T7 ( ), T9 ( ), T( ), T3( ) ad T5( ) for f ( ) = si, cetered at a =. Eample 4.3 Fid the Taylor series for f ( ) = cos ad show that it coverges to f ( ) = cos for all. Solutio: Agai we start by computig derivatives ad evaluatig them at a = : f ( ) = cos f() = f'( ) = si f'() = f''( ) = cos f''() = f'''( ) = si f'''() = f''''( ) = cos f''''() = ( We otice the values of f ) () for =,,, 3,... repeat i cycles of legth four:,, -,,,, -,, etc. The Taylor series for f ( ) = cos is ( ) f () = !! 4! 6! 8! = ! 4! 6! 8! j j = ( ) ( j )! j= For each fied, to show that the Taylor series for f ( ) = cos coverges to f ( ) = cos, we must show that R ( ) as. Sice the derivatives of cos are either ± cos or ± si, we ca tae M + = for all ad all. Usig the upper boud for R ( ) give i Taylor s Iequality, we obtai + M + f ( ) T( ) = R( ) = ( + )! ( + )! Prited 6//6, :7 PM LU Jue 6; 3/4/; 3//; 5//9; 4/8/9; /4/8; 3/4/8; /5/8; /3/7; /8/7; //7;/9/7; ;/6/7;7/4/7; /9/7b,/9/7

29 Math 5 - Taylor Polyomials & Taylor Series Sectio 4, Page 8 of 43 + But as by Eample 4.. We coclude that T ( ) f ( ) as for each fied, so the ( + )! Taylor series for f ( ) = cos actually coverges to f ( ) = cos : j j ( ) = = cos for all. ( j)!! 4! 6! 8! j= The et eample is a little more complicated tha Eamples 4. ad 4.3. I Eamples 4. ad 4.3, we could use the value M + = for all ad all. The value we use for M + i Eample 4.4 will have to deped o. Eample 4.4 Fid the Taylor series for f ( ) = ead show that it coverges to f ( ) = e for all. Solutio: Sice ( ) ( ) '( ) ''( ) '''( )... f f f f f ( ) e = = = = = =, we have ( ) f() = f'() = f''() = f'''() =... = f () = e =. So the Taylor polyomial of degree is () (3) ( ) ( ) ( ) f () f () 3 f () ( ) f () T ( ) = f() + f ()( ) + ( ) + ( ) ( ) + ( ) 3! ( )!! 3 = ! 3! ( )!! ad the Taylor series for f ( ) = e is ( ) 3 4 j f () = =.!! 3! 4! j! j= The ( + ) st derivative of f ( ) = e is agai ( + ) t f () t e ( + ) f ( ) = e. For each fied, the maimum value of ( + ) = o the iterval t is tae at the right edpoit: M = ma f () t = e. Usig the upper boud for R ( ) give i Taylor s Iequality, we obtai + t + M+ + e f ( ) T( ) = R( ) = ( + )! ( + )! But this last epressio approaches as by Eample 4. (because is held fied, so fied as ). We coclude that T( ) f ( ) as for. For each fied <, the maimum value of f t e ( + ) () t = o the iterval t e is also held is tae o at the right ( + ) edpoit: M = ma f () t = e =. Usig the upper boud for R ( ) give i Taylor s Iequality, we obtai + t + M+ + f ( ) T( ) = R( ) = ( + )! ( + )! Prited 6//6, :7 PM LU Jue 6; 3/4/; 3//; 5//9; 4/8/9; /4/8; 3/4/8; /5/8; /3/7; /8/7; //7;/9/7; ;/6/7;7/4/7; /9/7b,/9/7

30 Math 5 - Taylor Polyomials & Taylor Series Sectio 4, Page 9 of 43 As before, the last epressio approaches as by Eample 4.. We coclude that T( ) f ( ) as for <. So the Taylor series for f ( ) = e actually coverges to f ( ) = e: 3 4 = = e for all.!! 3! 4! Summary to this Poit Taylor Polyomials Let f be a fuctio with derivatives of order, for,, 3,..., N throughout some iterval cotaiig a as a iterior poit. The for N, the Taylor polyomial of order, geerated by f at = a is: () (3) ( ) () f ( a) f ( a) 3 f ( a) T ( ) = f( a) + f ( a)( a) + ( a) + ( a) ( a) 3!! ( ) f ( a) T ( ) = ( a)! I the special case where a =, the Taylor polyomial is: () (3) ( ) () f () f () 3 f () T ( ) = f() + f () !! T ( ) = ( ) f ()! Taylor Series Let f be a fuctio with derivatives of all orders throughout some iterval cotaiig a as a iterior poit. The the Taylor series geerated by f at = a is: ( ) () (3) ( ) f ( a) () f ( a) f ( a) 3 f ( a) ( a) = f( a) + f ( a)( a) + ( a) + ( a) ( a) +...! 3!! I the special case where a =, the Taylor series is: ( ) () (3) ( ) f () () f () f () 3 f () = f() + f () ! 3!! The Taylor series cetered at a = is also called the MacLauri series. Prited 6//6, :7 PM LU Jue 6; 3/4/; 3//; 5//9; 4/8/9; /4/8; 3/4/8; /5/8; /3/7; /8/7; //7;/9/7; ;/6/7;7/4/7; /9/7b,/9/7

31 Math 5 - Taylor Polyomials & Taylor Series Sectio 4, Page 3 of 43 CAUTION: Just because we ca compute the Taylor series for a fuctio usig the above formula, does ot i itself guaratee that the series actually coverges to the fuctio. So, i geeral, we caot equate a fuctio with its Taylor series. We have to establish this equality i each case before we ca assert it. ( ) f ( a) That is why we did ot write f ( ) = ( a) above.! Radius of Covergece As we have see i Eample 4., Taylor series do ot always coverge for all. Figure 5 below illustrates the covergece of the Taylor series for f ( ) = to f ( ) = for <, ad the divergece for. The partial sums of the Taylor series are the Taylor polyomials. Figure 5 shows oly T ( ), T ( ), T ( ), T ( ) 4 7 ad T ( ) 3 to avoid crowdig. FIGURE 5. The Taylor polyomials T ( ), T4 ( ), T7 ( ), T( ) ad T3( ) for f ( ) =, cetered at a =. Prited 6//6, :7 PM LU Jue 6; 3/4/; 3//; 5//9; 4/8/9; /4/8; 3/4/8; /5/8; /3/7; /8/7; //7;/9/7; ;/6/7;7/4/7; /9/7b,/9/7

32 Math 5 - Taylor Polyomials & Taylor Series Sectio 4, Page 3 of 43 The vertical asymptote at = clearly poses a problem for the covergece of this Taylor series for. Notice, however, that there is also a problem i covergece for, eve though f ( ) does t have a vertical asymptote at =. As we will see below, it is impossible for a Taylor series cetered at a = to diverge for > ad also coverge for <. So i some sese it is true that the vertical asymptote at = ot oly causes divergece of this Taylor series for >, it also causes divergece of this Taylor series for <. The behavior that we saw i Eample 4. (the series coverges for < ad diverges for > ) is surprisigly geeral for all Taylor series: Fact: Radius of Covergece For the Taylor series ( ) f ( a) ( a) for a fuctio f ( ) cetered at! = a, there are oly three possibilities: (i) The series coverges oly at = a. (I this case, we say R =.) (ii) The series coverges absolutely for all. (I this case, we say R =.) (iii) There is a positive umber R for which the series coverges absolutely wheever a < R ad diverges wheever a > R. The umber R (which may be or ifiity) is called the radius of covergece for the Taylor series for f, ( ) cetered at = a. Remar: Notice that there is o coclusio about covergece whe a = R, i.e. whe = a R or = a+ R. It is possible that the series could coverge at both, oe, or either of these poits. We will ot study what happes whe a = R i these otes. Eample 4.5 Eample 4. shows that the radius of covergece of the Taylor series for f ( ) = is R =. Eamples 4., 4.3, ad 4.4 show that the radius of covergece of the Taylor series for each of (a) si (b) cos ad (c) e is R =. Prited 6//6, :7 PM LU Jue 6; 3/4/; 3//; 5//9; 4/8/9; /4/8; 3/4/8; /5/8; /3/7; /8/7; //7;/9/7; ;/6/7;7/4/7; /9/7b,/9/7

33 Math 5 - Taylor Polyomials & Taylor Series Sectio 4, Page 3 of 43 4 Eercises. Cosider the Taylor series for f ( ) = si( ) a) Epress the series i summatio otatio. b) Fid a epressio for R ( ). 3 c) Keepig i mid the proof (Eample 4.) that as, show that lim R( ) = for all (fied)! values of.. Fid the Taylor series for f ( ) = cos ad show that it coverges to cos for all values of. 3. Fid the Taylor series for f ( ) = e + ad show that it coverges to e + for all values of. 4. For what values of does the Taylor series for f ( ) coverge to f ( ) =? Prited 6//6, :7 PM LU Jue 6; 3/4/; 3//; 5//9; 4/8/9; /4/8; 3/4/8; /5/8; /3/7; /8/7; //7;/9/7; ;/6/7;7/4/7; /9/7b,/9/7

34 Math 5 - Taylor Polyomials & Taylor Series Sectio 4, Page 33 of 43 4 Eercise Solutios.(a) f ( ) = si( ) f() = f'( ) = cos( ) f'() = f''( ) = si( ) f''() = f'''( ) = cos( ) f'''() = ( ) ( ) f ( a) f () ( a) = ( )!! () () () (3) (4) f () f () f () f () 3 f () 4 = !!! 3! 4! 3 = ! + + ( ) = ( + )! R3 = (b) ( ) (c) ( ) M 4! 4! + M + + lim R lim = lim = ( + ) ( + )!!, sice is fied ad per Eample f ( ) = e f() = e + f'( ) = e f'() = e + f''( ) = e f''() = e + ( ) ' '' ''' ( 4) ( ) () () () () 3 () 4 f a f f f f f ( a) = !!!! 3! 4! e = e + e+ + = e!. ( + ) t+ We ow f () t = e, which is icreasig o the iterval t. Evaluatig of this iterval gives us M = e +. + t e + at the right edpoit + + M e lim R ( ) lim lim = e = ( + )! ( + )!, for all (fied) values of. Prited 6//6, :7 PM LU Jue 6; 3/4/; 3//; 5//9; 4/8/9; /4/8; 3/4/8; /5/8; /3/7; /8/7; //7;/9/7; ;/6/7;7/4/7; /9/7b,/9/7

35 Math 5 - Taylor Polyomials & Taylor Series Sectio 4, Page 34 of 43 Commo Taylor Series & Radii of Covergece Here are a few series worth rememberig. [Yes that meas you should memorize them!] Fuctio Taylor series Restrictios Radius of Covergece u u+ u + u + u +... = u u < R = + u ( ) u+ u u + u = u u < R = cosu u u u u u = ( ) oe R =! 4! 6! 8! ( )! siu u u u u u u = ( ) oe R = 3! 5! 7! 9! ( + )! u e 3 4 u u u u u...! 3! 4!! = oe R =

36 Math 5 - Taylor Polyomials & Taylor Series Sectio 5, Page 35 of 43 5 Operatios with Taylor Series We coclude our discussio of Taylor series by itroducig some operatios that ca be doe with Taylor series. The first of these operatios is simple substitutio. Quic substitutio We ca fid the Taylor series for a fuctio by maig a u-substitutio i ay of the series we foud i the previous sectio. Eample 5. Fid the Taylor series for f ( ) = + 4, fid its radius of covergece, ad show that it coverges to f ( ) = + 4 wheever it coverges. Solutio: By Eample 4., the Taylor series for gu () = u is 3 + u+ u + u +. Sice f ( ) = g( 4 ), we mae the substitutio u= 4 i the Taylor series for gu () to obtai the Taylor series for f: ( ) ( ) ( ) ( 3 ) ( ) = 4 To fid the radius of covergece, ote that the Taylor series for f ( ) = is a geometric series with + 4 r = 4. By Eample 3., this series is coverget for r < ad diverget for r. j= j Sice r < 4 < 4 < < <, 4 we ow that the Taylor series for f ( ) coverges whe The radius of covergece is R =. < ad diverges whe. Prited 6//6, :7 PM LU Jue 6; 3/4/; 3//; 5//9; 4/8/9; /4/8; 3/4/8; /5/8; /3/7; /8/7; //7;/9/7; ;/6/7;7/4/7; /9/7b,/9/7

37 Math 5 - Taylor Polyomials & Taylor Series Sectio 5, Page 36 of 43 FIGURE 6. Taylor polyomials T ( ), T4 ( ), T8( ) ad T( ) for f ( ) = + 4, cetered at a =. Figure 6 illustrates that the Taylor series for f ( ) = (cetered at a = ) coverges to f ( ) + 4 = + 4 for <, ad the diverges for. The partial sums of the Taylor series are the Taylor polyomials. Figure 6 shows oly T ( ), T4 ( ), T8( ) ad T( ) to avoid crowdig. At first glace, the radius of covergece of the fuctio i Figure 6 is mysterious. There are o apparet vertical asymptotes to cause problems, but the radius of covergece is still oly R = /. It turs out, however, that there are vertical asymptotes (called poles) if we allow to be a comple umber; this topic is discussed i detail i more advaced courses (Math 47 at the UW for eample). Prited 6//6, :7 PM LU Jue 6; 3/4/; 3//; 5//9; 4/8/9; /4/8; 3/4/8; /5/8; /3/7; /8/7; //7;/9/7; ;/6/7;7/4/7; /9/7b,/9/7

38 Math 5 - Taylor Polyomials & Taylor Series Sectio 5, Page 37 of 43 Term-by-term differetiatio ad itegratio We ca fid derivatives ad itegrals of fuctios that are represeted by Taylor series by differetiatig or itegratig each term i the Taylor series. Eample 5. d Use Taylor series to compute ( e ). d Solutio: 3 4 d d ( e ) = d d! 3! 4! = ! 3! 4! 3 4 = ! 3! 4! = e Eample 5.3 d Use Taylor series to compute (si ). d Solutio: d d (si ) = + + = + + d d 3! 5! 7! 3! 5! 7! 4 6 = + +! 4! 6! = cos Eample 5.4 Fid the Taylor series for l( + ). Solutio: We could compute the Taylor series directly by taig derivatives ad evaluatig them at =, as we did i Eamples 4., 4.3, ad 4.4. Alteratively, we ote that l( + ) = l + = d whe + > or > + Usig the series epasio for, we ca evaluate the itegral term by term: + l( + ) = l + for > = d = + + d for < < ( ) = C To evaluate the costat C, substitute i = to get C =. We get l( + ) = d = + + = ( ) Prited 6//6, :7 PM LU Jue 6; 3/4/; 3//; 5//9; 4/8/9; /4/8; 3/4/8; /5/8; /3/7; /8/7; //7;/9/7; ;/6/7;7/4/7; /9/7b,/9/7

39 Math 5 - Taylor Polyomials & Taylor Series Sectio 5, Page 38 of 43 Oe useful fact is that, as log as the ceterig value is the same, the followig all have the same radius of covergece: The Taylor series for a fuctio f ( ) The Taylor series for its derivative f'( ) (obtaied from term-by-term differetiatio of the Taylor series for f) ( ) The Taylor series for its idefiite itegral f ( ) d (obtaied from term-by-term itegratio of the Taylor series for f) ( ) For eample, the Taylor series for has radius of covergece R =, so the Taylor series for l( + ) that + we obtaied i Eample 5.4 also has radius of covergece R =. Eample 5.5 Fid the atiderivative, F, ( ) of f ( ) = e which satisfies F () =. Solutio: We use the substitutio u= i the Taylor series to get e 3 4 e u = + u+ u + u + u +! 3! 4! = =! 3! 4! j! j j ( ) j= Itegrate term by term, ad set the costat C = to mae F () =, to obtai j+ j F ( ) = = ( ) 3! 5! 7 3! 9 4! ( j+ ) j! j= u Eample 4.4 shows that the Taylor series for e coverges to substitutio u= as i Eample 5.5, the Taylor series for coverges to for all, ad thus the radius of covergece of the Taylor series for e u e for all u. So by maig the e that we obtaied i Eample 5.5 e is R =. We coclude that the radius of covergece of the Taylor series for the fuctio F ( ) i Eample 5.5 is also R =. We could view this series as the formula which defies F. ( ) The error fuctio erf ( ), used widely i statistics, is give by the formula t erf ( ) = F( ) = e dt π π Prited 6//6, :7 PM LU Jue 6; 3/4/; 3//; 5//9; 4/8/9; /4/8; 3/4/8; /5/8; /3/7; /8/7; //7;/9/7; ;/6/7;7/4/7; /9/7b,/9/7

40 Math 5 - Taylor Polyomials & Taylor Series Sectio 5, Page 39 of 43 5 Eercises Directios for eercises #-9: Use the series o page 34 to fid the Taylor series (cetered at a = ) for the followig fuctios AND fid the ew radius of covergece: Simple substitutio. cos( ) e 4. 6 Multiplyig both sides by the same object si 8. [Careful!] Differetiatig both sides 9. ( ) d = d Hit: ( ). Fid the derivative of si, by differetiatig the Taylor series for si.. Fid the derivative of cos, by differetiatig the Taylor series for cos.. Fid the derivative of e, by differetiatig the Taylor series for e. Itegratig both sides 3. Usig Taylor series, fid the atiderivative, F, ( ) of si, which satisfies F () =. 4. Usig Taylor series, fid the atiderivative, F, ( ) of cos, which satisfies F () =. 5. Usig Taylor series, fid the atiderivative, F, ( ) of e, which satisfies F () =. Miig them up l( + 3) 9. arcta. Fid a Taylor series that represets the atiderivative, F, ( ) of [Importat questio: Which itegratio techiques would be used to help us fid e satisfyig F () =. e d?] Prited 6//6, :7 PM LU Jue 6; 3/4/; 3//; 5//9; 4/8/9; /4/8; 3/4/8; /5/8; /3/7; /8/7; //7;/9/7; ;/6/7;7/4/7; /9/7b,/9/7