( ) k ( ) 1 T n 1 x = xk. Geometric series obtained directly from the definition. = 1 1 x. See also Scalars 9.1 ADV-1: lim n.
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1 Sclrs-9.0-ADV- Algebric Tricks d Where Tylor Polyomils Come From A.docx Pge of Algebric tricks ivolvig x. You c use lgebric tricks to simplify workig with the Tylor polyomils of certi fuctios.. T x = xk Geometric series obtied directly from the defiitio. See lso Sclrs 9. ADV-: lim x k = x x < Quick wys to evlute Tylor polyomils for f (x) of the form EXAMPLE: T 2 = x 2 x 2 ( ) k u(x) : By defiitio the idex o T must mtch the highest power i the Tylor polyomil. EXAMPLE: T 2 = T + x 2 2 x 2 T 2 = + x 2 ( ) k ( ) x 2k ( ) k = x 2 Mdiso, Wiscosi Pge of
2 Sclrs-9.0-ADV- Algebric Tricks d Where Tylor Polyomils Come From A.docx Pge 2 of The followig two theorems my mke pproximtig certi derivtives d idefiite itegrls esier (or eve possible). THEOREM: Term by term differetitio of Tylor polyomil. ( ) T f (x) = T f (x) You c compute the Tylor polyomil of f (x) by doig term by term differetitio of the Tylor polyomil of f (x). EXAMPLE: Evlute the Tylor polyomil of First questio: Does Yup: ( x) 2 or ( x). 2 x dx u = x ( ) 2 = x dx = ( ) 2 So we see tht for f (x) = For exmple: T 4 x ( ) 2 = u 2 ( du) = u 2 du x + C f (x) = ( x) 2 T x ( ) 2 = ( x) 2 dx look ythig like u(x)? du dx = du = dx = u + C 0 = u + C T = kx k k= x x 4 kx k = x 0 + 2x + 3x 2 + 4x 3 k= = + 2x + 3x 2 + 4x 3 = x k term by term Mdiso, Wiscosi Pge 2 of
3 Sclrs-9.0-ADV- Algebric Tricks d Where Tylor Polyomils Come From A.docx Pge 3 of THEOREM: Term by term idefiite itegrtio of Tylor polyomil. T + f (x) = T f (x) dx You c compute the Tylor polyomil of f (x) by doig term by term idefiite itegrtio of the Tylor polyomil of f (x). EXAMPLE: Evlute the Tylor polyomil of rct x. First questio: Does rct x Yup. ( rct x) = ( ) or ( rct x) dx + x 2 So we see tht for f (x) = rct x f (x) = T + rct x = T + x 2 dx T 2 + rct x = ( ) k x 2k dx Derived bove. ( ) k x 2k dx ( ) = T 2 + rct x = C + Solve for C usig x = 0. ( ) k 2k + x2k+ 0 = T 2 + { rct(0) } = C + 0 = C T 2 + rct x = 0 + A cocrete exmple: term by term look ythig like + x 2 ( ) k ( 2k + 0)2k+ = C + 0 ( ) k 2k + x2k+ u(x)? T (2 3)+ rct x = x x3 3 + x5 5 x7 7 Mdiso, Wiscosi Pge 3 of
4 Sclrs-9.0-ADV- Algebric Tricks d Where Tylor Polyomils Come From A.docx Pge 4 of Exmple: Computig the vlue of π. This is totlly fscitig! We showed erlier how to use Tylor polyomil to pproximte the vlue of e. ow we ll do the sme thig for π. Iterestig vlues of t x t(0) = 0 t π 6 = 3 t π 4 = t π 3 = 3 t π 2 UDEF ( ) k T 2 + rct x = x 2k+ From bove. 2k + Evlutig these expressios t x = we get: ( ) T 2 + { rct ( ) } k = ( ) k 2k + ( ) k π T 2 + = 4 2k + So we mke the pproximtio π 4 π 4 ( ) k 2k + ( ) k 2k + Mdiso, Wiscosi Pge 4 of
5 Sclrs-9.0-ADV- Algebric Tricks d Where Tylor Polyomils Come From A.docx Pge 5 of At = 400 we mtch π oly i the first 2 deciml plces. This computtio pproches π very slowly s icreses but we c do better. Mdiso, Wiscosi Pge 5 of
6 Sclrs-9.0-ADV- Algebric Tricks d Where Tylor Polyomils Come From A.docx Pge 6 of A better pproximtio of π which requires fr fewer terms. t(0) = 0 t π 6 = t π 3 4 = t π 3 = 3 t π 2 UDEF From prior experiece we kow tht the followig idetity is useful. t A + t B t(a + B) = t A t B We try to use this iformtio to fid other expressio cotiig π. Fid A d B such tht t(a + B) equls iterestig vlue from bove. By ispectio we try: A = rct 2 so t A = 2 B = rct 3 so t B = 3 t( A + B) = t A + t B t A t B = t rct 2 + rct 3 = For π 2 < ( A + B) < π 2 = = rct t rct 2 3 = rct() = π 4 rct 2 3 = π 4 4 rct 2 3 = π T 2 + rct x = ( ) k x 2k+ 2k + From bove. So we mke the pproximtio: rct x ( ) k x 2k+ 2k + Mdiso, Wiscosi Pge 6 of
7 Sclrs-9.0-ADV- Algebric Tricks d Where Tylor Polyomils Come From A.docx Pge 7 of This is job for computer. At = 8 we mtch π i the first 6 deciml plces. Lter o we ll show how to put error brs o this pproximtio. So cool! Mdiso, Wiscosi Pge 7 of
8 Sclrs-9.0-ADV- Algebric Tricks d Where Tylor Polyomils Come From A.docx Pge 8 of A closed form for R f (x) is rre but for f (x) = x T x T x R x = x = x = def = x + x f (x) T x R x = x x + x R x = x+ x We verify the followig: x =? x + x x Directly from the defiitio. You c verify this s show below = ( ) x+ x ( x) = x + x x x x x ( x 0 + x + x 2 + x x ) = x + ( ) ( ) = x + x + x 2 + x x + x + x + = x + Mdiso, Wiscosi Pge 8 of
9 Sclrs-9.0-ADV- Algebric Tricks d Where Tylor Polyomils Come From A.docx Pge 9 of Where does the formul for Tylor Polyomils come from? Here re some ituitive exmples to get us strted. Suppose tht we do t kow the computtiol rule (geericlly represeted by the symbol f (x) ) for some fuctio f but we do kow the vlues of multiple derivtives of f (x) evluted t the sigle poit x =. Here is how oe might do polyomil pproximtios of f (x) er (d sometimes ot so er) x =. The stdrd lier pproximtio of f (x) er x = with Tylor ottio. T ottio: T { f (x)}(b) mes the Tylor Polyomil T ote tht: T { f (x)}() = f () So { f (x)} def = T f (x) def = f () + f '()(x ). ( T { f (x)}) () = T { f (x)} mimics f (x) pretty well t x =. f () { f (x)} evluted t x = b. T f (x)(b) def = T { f (x)}(b) T ( b) def = T { f (x)}(b) whe the cotext is cler. q.v. p. 82 We typiclly do t use the ottio T f () becuse it c resobly be iterpreted to me f () ote tht 0 0 def =. Mdiso, Wiscosi Pge 9 of
10 Sclrs-9.0-ADV- Algebric Tricks d Where Tylor Polyomils Come From A.docx Pge 0 of Here s possible qudrtic pproximtio of f (x) er x =. f () T 2 { f (x)} = f () + f ()(x ) + (x ) 2. It ws chose so tht: 2 T 2 f (x) So { }( ) = f () ( T 2 { f (x)}) () = f () ( T 2 { }) () = T 2 f (x) f () { f (x)} mimics f (x) pretty well t x =. The detiled derivtio is show lter i the chpter. Observtio: If the 2 d derivtive is positive t the the st derivtive is icresig so we icrese our estimte. If the 2 d derivtive is egtive the we decrese our estimte. See ed of chpter for more detil o how we derived this formul. Here s possible cubic pproximtio of f (x) er x =. 3 T 3 { f (x)} = f () (x f () )0 + (x f () ) + 2 (x )2 + It ws chose so tht: T 3 f (x) { } () = f () ( T 3 { f (x)}) () = f () ( T 3 { f (x)}) () = f () f () 3 2 (x )3. ( ) () = T 3 { f (x)}) f () So T 3 f (x) mimics pretty well t { } f (x) x =. Observtio: If the 3 rd derivtive is positive t the the 2 d derivtive is icresig so we icrese our estimte. If the 3 rd derivtive is egtive the we decrese our estimte. Mdiso, Wiscosi Pge 0 of
11 Sclrs-9.0-ADV- Algebric Tricks d Where Tylor Polyomils Come From A.docx Pge of Proof of theorem from Sclrs 9.0 ADV- THEOREM: For P (x) = def T f (x) P () = f (), P () = f (),... P () () = f () () d P (x) = def T f (x) is the oly polyomil tht hs this property. Proof: By defiitio ( ) =, ( ) k = 0 for k > 0 f (k ) ( ) is costt for every pir of costts d k. k! P () = f (k ) ( ) k! ( ) k = f (0) () 0! + = f () = f () + 0 k= f (k ) ( ) k! ( ) k The first term i the summtio bove is costt so whe we tke the derivtive below it becomes 0 d the summtio oly goes from to. etc. This hppes for ech successive derivtive. f P () (k ) ( ) = k! k( ) k = f () () f (k ) ( ) +! k! P () = P () = k= = f () = f () + 0 k=2 f (k ) ( ) k! k( k ) ( ) k 2 = f (2) ()! + = f () = f () + 0 k=3 f (k ) ( ) k! k( k ) ( k 2) ( ) k 3 = f (3) () = f () = f () + 0! k=2 k=3 f (k ) ( ) k! ( ) k( ) k k(k ) ( ) k 2 f (k ) + k! k(k ) k 2 Ad this ptter cotiues for ll the derivtives (wheever they re defied). k=4 ( )( ) k 3 Mdiso, Wiscosi Pge of
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