This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 HH + aagmﬁ], theisllrtmHélE—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—IEM 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 ...
View
Full
Document
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.
 Fall '08
 Staff
 Polynomials, The Land

Click to edit the document details