Notes12.10.pdf
Applied Math And Engineering 111 with Pisano at University of Virginia
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By: Anonymous
Textbook:
Calculus (Stewart's Calculus Series)
Created: 2009-11-30
File Size: 4 page(s)
Views: 0
Textbook:
Calculus (Stewart's Calculus Series)Created: 2009-11-30
File Size: 4 page(s)
Views: 0
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11/22/2009 1 Section 12.10 Taylor and Maclaurin Series Other Functions How can we find power series representations for functions that are not related to ? 1 1 x? Answer: Use Taylor or MacLaurin Series. Theorem If a function, f(x), has a power series about a, then the series will have the form, ?? 0 () n n n Tx c x a ? ? ?? ? where This is a Taylor Series. If a = 0, then the series is also known as a Maclaurin Series. ? ? ? ? ! n n f a c n ? Why? If f (x) = T (x), then all derivatives should be equal at a. ???????? 234 01 2 3 4 () ()fx Tx c cxa cxa cxa cxa? ?? ?? ?? ?? ??? 0 () () ()f aTa fac? ?? ? ? ? ? ? ? 23 12 3 4 '() 23Tx c cxa cxa cxa? ???????? ? ? ? ? ? ? 1 '' 'f aTa fac? ?? ?? ?? 2 23 4 "() 2 32 43Tx c cxa cxa? ?? ? ?? ? ?? ?? ?? ?? ? ? 22 " "" "2 2 f a fa Ta fa c c????? ? ? ? ? 3 34 () 32 432Tx c cxa? ???? ??? ?? ?? ?? ?? ?? ?? ? ? ?? 3 33 3 33 32 32 f a faTa fa cc?????? ? Coefficients ? ? ?? ?? ?? ?? 0 1 2 3 3 ' " 2 cfa cfa f a c f a c ? ? ? ? 32? In general: ?? ? ? ! n n f a c n ? These are the coefficients that insure that the derivatives of T(x) and of f(x) at a are equal. Finding a Taylor/Maclaurin Series for a function is a matter of finding these coefficients. Example Find the Maclaurin series for Solution: () x f xe? ?? n n Tx cx ? ? ? 0n? Now find coefficients? n 0 11 1 1 1 2 1 1/2 ? ? ? ? n f x ? ? ? ?0 n f ?? ? ?0/ ! n n cf n? x e x e x e 1 ! n c n ? 0 1 ! x n n ex n ? ? ? ? 11/22/2009 2 Summary It can be shown that R = ?. 0 ! n x n x eR n ? ? ??? ? Interpretation ?? 23 0 1 !23! n x n xxx eTx x n ? ? ? ?????? ? ?for all x ?? 23 0 1 !23!! inn x n i x xx x eTx x ? ???????? ? ? for x close to 0 ? ? 0 1Tx? ? ? 1 1Tx x?? ?? 2 2 1 2 x Tx x? ?? ?? 23 3 1 26 x x Tx x??? ? 234 4 () 1 2624 x xx Tx x??? ? ? Interpretation continued ?? 1 0 111 111 !23! 1 11 2 1 n n ee T n e e ? ? ?? ? ????? ? ??? ? ? 11 2.5 2 11 11 2.6 26 e e ??? ? ??? ? ? The more terms that are included, the better the approximation. Maclaurin Series for cos (x) i 0 11 ? ? ? ? i f x ? ? ? ?0 i f ?? ? ?0/! i i cf i? cos x 1 0 0 2 ?1 ?1/2 3 0 0 4 1 1/4! sin x? cos x? sin x cos x Coefficients for cos x 24 24 11 cos 1 0 0 0 ! 11 cos 1 24! xxx xxx ??? ?? ?? ?? ? ? ? ? Note that the Taylor series includes only even terms, so it is an even function, just as cos x is an even function. It can be shown that R = ?. ?? ?? 2 0 cos 1 2! n n n x xR n ? ? ?? ?? ? Graphical Interpretation () cosf xx? 0 () 1Tx? Partial sums are known as Taylor Polynomials. 2 2 () 1 2 x Tx?? 24 4 () 1 24! xx Tx?? ? 246 6 () 1 24!6! x xx Tx?? ? ? 11/22/2009 3 Example ?? ?? ?? ?????? 2246 0 0.01 0.01 0.01 0.01 cos(0.01) 1 1 2! 2 4! 6! n n n n ? ? ?? ?? ? ? ? ? ? ? ? ? ? 0 2 cos 0.01 (0.01) 1 0.01 T?? ???? 2 24 4 cos(0.01) (0.01) 1 .99995 2 0.01 0.01 cos(0.01) (0.01) 1 .9999500004 224 T T ???? ????? Using T 4 (0.01), ?? 6 15 0.01 1.4 10 6! error ? ??? Example 1) Find the Maclaurin Series for 2) Find Solution: 23 () cosf xx x? ? ?44 (0)f 1) 2) ?? ? ? ?? ?? ?? 2 3 2 0 62 0 () 1 2! () 1 2! n n n n n n x Tx x n x Tx n ? ? ?? ? ?? ?? ? ? ?? ?? 7 44 624 7 11 1 27! 14! n n c ? ? ? ? ?? ? ? ? ? ?? 44 44 1(0) 14! 44! 44! (0) 14! f f ? ? ?? Example Find the Maclaurin Series for sin (x): Solution: Don?t start from scratch. Use differentiation. ? ? 2 1 n n x R ? ? Now, differentiate? ??0 cos 2! n xR n ? ?? ?? Example continued ?? ?? ?? ?? ?? ?? 2 0 21 1 21 cos 1 2! 1 12 2! n n n n n n n dd x x dx dx n nx n ? ? ? ? ? ?? ?? ?? ?? ?? ?? ? ? ? ? ?? ?? ?? ?? ?? ?? ?? 1 2( 1) 1 1 0 21 1 0 21 0 1 sin 21! 1 2( 1) 1 ! sin 1 1 21! 21! n n n n n n n n n n n x x x n x n x x n n R ? ??? ? ? ? ? ? ? ? ? ? ?? ?? ? ?? ? ?? ?? ??? ? ? ? ? ? Example Is it appropriate to use T 3 (1.5),centered at 0 to approximate sin(1.5)? l lil dAnswer: No use a Tay or polynomia centere at ?/2. Taylor Series at ?/2 i 0 sin x 11 ? ? ? ? i f x ?? 2 i f ??? ?? ?? ?? /! 2 i i cf i ??? ? ?? ?? 1 cos x 0 0 2 ? sin x ? 1 ? 1/2 3 ? cos x 0 0 4 sin x 1 1/4! 11/22/2009 4 Taylor Series at ?/2 continued 24 22 sin 1 0 0 24! xx xR ???? ?? ?? ?? ?? ?? ?? ??? ?? ? ??? 2n ??? ?? ??0 2 sin 1 2! n n x xR n ? ? ? ?? ?? ?? ?? ? This Taylor series is convergent to sin (x) for all x, but Taylor Polynomial approximations will be accurate near ?/2. Compare Using ,?? ?? 21 0 sin 1 21! n n n x x n ?? ? ?? ? ? ?? 35 3 1.5 1.5 1.5 1.5 0.9375 0.06328125 3! 5! Tero?? ? ? ? Using Compare to ?? ?? 2 0 2 sin 1 2! n n n x x n ? ? ? ?? ? ?? ?? ?? ? ?? 24 6 2 1.5 1.5 22 1.5 1 0.9974939401 1.04672274 10 ! Tero ?? ? ?? ?? ?? ?? ?? ?? ?? ?? ? ? ? ? sin(1.5) 0.9974949866? Stacie Microsoft PowerPoint - Section12.10TaylorSeries
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About this note
By: Anonymous
Textbook:
Calculus (Stewart's Calculus Series)
Created: 2009-11-30
File Size: 4 page(s)
Views: 0
Textbook:
Calculus (Stewart's Calculus Series)Created: 2009-11-30
File Size: 4 page(s)
Views: 0
About StudyBlue
STUDYBLUE makes things that make you better at school.
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Things like personalized quizzes and friendly reminders about when (and what) to study next.
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“I have been getting MUCH better grades on all my tests for school. Flash cards, notes, and quizzes are great on here. Thanks!”
Kathy
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