Taylor_polynomials - fll Vloré About [I‘aylori...

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Unformatted text preview: fll Vloré About [I‘aylori Polynomials Supposfi has n 3: ll hontinnons derivatives and let [be the ma [Taylori polynomial oil fl (aboui‘l n 2 0]. Thd Estimate for the remainder £1 z — on pi 65$ ofl Salas—Hille—Etgenl (formula ME) can E restated as follows: Ifl | [("+1)@| S Q fori m E some interval 1] containing 07 then Cr—n—l—I 1 M ii :1: E n ( > n + 1 ! @(wll s zTHE U” m is afunctionl definedl nearmzm and therdis amta—nt Q mam g for a: near 0, we sayltlfat g(m] E (maxi?) (as m 4 Q m muchtherelationmgdoesto<1hatislzfljthatfij0 WMflasxjgwhereas“g(m)za(xk)”meansthatg(m]fi0hfileast as mk as$HDJ Withl this notation; according m (1] we [hale : U(:Jcn+]i)D on flail = [E20761 UCCMW. (2) W E; is the my polynomial E de re _ 0:2; with this property. [Indeedi Propositionl m. Suppose W has 0:); ill mntinnnus derivativesl and supposa m is a [poly- nomial of degree i M lsuch that = Q(:I:”+]i) as m —H (L Then Q is thd nth laflofl polfnomial bffi Proofl Let En be the nth [Faylori polynomial oil f. Subtracting the eguations f — WE] = len+1)fii M — Pnjgfl = Qlwn+1),we¢ oBtain PnKE) — Qngg) = kanflj. In othefi words, ifl Pn 2 gm and 2 gm, (Go—b0 tWI—Mflttmttlfl—szwl fiettingmzolmseethatw—bozp,or%=lm. Ihislbeingsoqifmdividebymm gfl (afi — j” (a2 — b2)a: (an — bn)a:"’1 2 0(33"). fiettingmzmmseethatgzlh. anmcandividdlgyfin W — i) (663 - b3>m Kai — banana = anvil. @ :1: 2 again, we get m 2 b2. W inductively, me am him; 2 A E all a so iBn = Q. I W position ll is @511 E calculating [Taylori polynomials. E Shows that w the formula (3 = is m the onlyi wayi F53 calculate Eng [ather2 ill by any means we canlfindlaj—[po ynomiaan ofWS'wmmM ang) QW),MQ7J must be P”. Here are lam nseflil applications of] this fact 2 (Taylor Polynomials bf Products. IE En]: land 3 be the nthl Taylor polynomials of f and g; respectively. Thenl f ($)g($l = [137560) Q($”+I)] [137560) Q($n+l)] : of degree S n in W Olmn+t). I hus2 E Hind the ntlt Taylot polynomial bf lfgl simply multiply lthd lntlt lFaylorl [polynomials of i and g togetherl discarding an] terms of degree E n. Example 1. What is the 6th Taylor polynomial of] E? Solutionl 2 3 5 6 Tan 2% 1 “3 ‘1" 4 = 7 4+ t “3 7 a: e :11 2 6 (Ma: ) a: a: ‘2 6 [2km 12 so the answer is $91 $41 E365 $336. Example 21 w is the 51h Taylor polynomial oil at san Solution: MESH m a a 004 W m] [36.51 n as] 21 61 24 112d 61 12d : 1_l 111—11 11—11 1 :1: 96% ML 6H Lem 6H xflfl m mlang), theislmflfl+§E—%E. lI‘aylor Polynomials ofl Compositions. H1 3” and g ham derinatnres gp to hrder m 1 and 9(0) : 92 we can Hind the nth Taylor polynomial 0t fl n g byl substituting lthd expans10n of g into the Taylot expansionl bf [£2 retainlng only the terms oil degree 3 fl. That isl suppose M = % ala: anmn Glynn“). Sincd 9(0) : and y is difierentiahle me ham 2 (Km) and hence film 1: a a + - - - + angi” a WW1]. Novvl plug in the Taylor expansion of p on the tight and multiplyl it @ discarding terms of degree E 01. Example 31 What is the 16th Taylor polynomial E Solution: 2 $12 611:]. a: m ngjj a emflzl 1:6 1 (2113181, so fife answen E 1 $61 Em . Example 41 What is the 4th Taylor polynomial of] flu”? Solutionl . 2 . . 4 5mm Sinm 2 (IE. NM snhstitntel m — g3 a GE) £911 Sim: on the right and @ultiplfl fl; throwingallltermsfildegreq §4lintothe “C(azE’)” trashlcan; x3 1 x1 x3 £131 Sinmzl _ __ _ H m H-H + aagmfi], theisllrtm-HélE—ggfil. hi I 3 (Taylor Polynomials and] l’Hopital’g RuleJ Taylofi polynomiald can M be used efiectivelyi irfl computing limits ofl the tforni D/OJ Indeed2 supposg f J y, and theiri firsfl k: — 1 derivatives vanish fi :11 = Q; floufl their Vctli derivatives do W both Nanish. The Waylon expansions of i and g then look like (k) (k) M=V kfom ogfi, m: kfo)M:d 396+ 1. {aking the quotien’d and cancelling out EE, @ W : W®‘__W g asij M 9091(0) P001?» 9091(0) This is inJ accordance with T’Hopitai’d rule, but the devices discussed] above for computing H‘aylofi polynomials rrray [lead to the guicklfl than] kiireet bpplicationl of] l’Hopital. Example 51 What is [limggndksc2 — hind sing fl Solutioni :12: P—Qaagflfllfi—QtOQe/fi) Eflmzfifla@1,hnd :132—sin2d %x4+0(m51_%fi0(m) >1 :62 sin2m‘ $4+O($5)‘ — fl+‘0(x) ‘ 3 Exampleé? Evaluate m 1 : m . 20:1 logw m—M Solutioni HE me need to expand irfl powers E :1: — i]. [Eirst if agll2 1 :1: m—l—mwm m—1)—(m—1)logm—logm IZ—Z legal—IE-M C—Dlega: C—Dlega: Next, loge]: (:1:—1)—%l x—Tfiwafii—Wfi, andpluggingthisifintothetnumeratorand dermminatergiveg M—J—Q—P)2 [(13 1) $(33 1)2 DE—fl:—§+O($—1)j_g ‘ (w—1)2+O((a:—1)3) 1+Otx—1) ' 4 Higher Derivative Tests for Critical Pointsl Recall that H f’(a) : Q then has a local tninimum (respl maximum] at :0 = a if f”(a] 2 Q Kresp. f”(a] g 0]. Mil happens if f”(a] : {L7 LAnswer: The behavion oil f near a is controlled by its first nonvanishing Elerivativd at al Propositionl 2J W 3" has k continuous derixatizes mean E and flKa) 2 f”(a] : = fag—Um] : 0 but f(k)(a] 7E (L [flk is w f has a local minimum on Maximum a a according as f(k)(a] 2 Q or f(kl(a) Q (L Ifk: is M fl has hieithen a minimum non a m at (11 m [th (k — Taylor polynomial El fl about a is simply the constant f(a) [all the othen terms are zero)B so Taylor’s formula oil orden Vd — ll Mitlfl the [Lagrangd form of the remainaer Rk Becomes f‘Wc 2 f(a) W k! (a: m fon some 0 between a: and m Wm if :1; (and hence c) is m m E 130610;) is close m f(k)Ka). E particularl m is nonzero and has the same Sign as fWKa). on thd othen hand2 — a)? is always positive if k is even 1% changes SEQ at :1: = a if] E m Ihusl if] E even; flail — fig) is os1t1V on negativq for :13 a, according m the oflEEg but iflkz is oddl — E changes wasmcrossesa: mmfim I ...
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This note was uploaded on 10/12/2009 for the course MATH 134 taught by Professor Staff during the Fall '08 term at University of Washington.

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Taylor_polynomials - fll Vloré About [I‘aylori...

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