# Taylornotes - Estimation of functions using Taylor...

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Estimation of functions using Taylor polynomialscFrank Zorzitto, Faculty of Mathematics, University of WaterlooIf a power seriesa0+a1(x-p) +a2(x-p)2+· · ·+ak(x-p)k+. . .converges to a functionfon some interval(p-R, p+R)centered atp, then theseries’ coefficientsakmust be the Taylor coefficients off. Namely,ak=f(k)(p)k!,wheref(k)denotes the derivative offtakenktimes.This prompts us to pose the converse question. If a functionfhas derivatives ofall ordersf(k)(x), wherexruns through some interval(p-R, p+R), must theresulting Taylor seriesf(p) +f0(p)(x-p) +f00(p)2(x-p)2+· · ·+f(k)(p)k!(x-p)k+. . .converge tof(x)for allxin(p-R, p+R)?In order to answer our question, we need to examine theerrorthat the partial sumsof the Taylor series forfmake in estimatingf(x).The Taylor polynomialWe start with a functionfdefined on an intervalIthat contains a numberp. Sup-pose that the firstnderivatives off, namelyf0, f00, f000, . . . , f(n), are also all de-fined. There is no shortage of such functions. For every one of thesef, the poly-nomialTn(x) =f(p) +f0(p)(x-p) +f00(p)2!(x-p)2+· · ·+f(n)(p)n!(x-p)nis called theTaylor polynomial off, of degree up ton, expanded aboutp.As we observed at the start, iffis represented by a power seriesf(x) =a0+a1(x-p)+a2(x-p)2+· · ·+ak(x-p)k+. . .on some interval aroundp,1
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then then’th partial sum of this series is nothing but the Taylor polynomial offofdegree up ton.In sigma-notation, the Taylor polynomial can also be written asTn(x) =nXk=0f(k)(p)k!(x-p)k.The numbersf(k)(p)k!that go with(x-p)kare called theTaylor coefficientsoffat the pointp.We start counting our Taylor polynomials atn= 0.Whenn= 0, the Taylorpolynomial of degree up to0is the constant polynomialT0(x) =f(p).Whenn= 1, the Taylor polynomial is the much more interesting linear functionT1(x) =f(a) +f0(a)(x-a).Of course,T1(x)is the well known tangent line, also known as thelinear approxi-mationor thelinearization, offatp. Whenn= 2, the second Taylor polynomialbecomesT2(x) =f(p) +f0(p)(x-p) +f00(p)2(x-p)2.Whenn= 3we getT3(x) =f(p) +f0(p)(x-p) +f00(p)2(x-p)2+f000(p)6(x-p)3,and so on.The Taylor polynomial is special forfbecauseit is the one and only polynomial ofdegree up tonthat agrees withfand all of the derivativesf0, f00, . . . , f(n)at thecentral pointp.That isTn(p) =f(p)T0n(p) =f0(p)T00n(p) =f00(p)...T(n)n(p) =f(n)(p).2
Let us demonstrate this key observation forn= 2. A general polynomial of degreeup to2can be written to look likeg(x) =r+s(x-p) +t(x-p)2,wherer, s, tare the coefficients.We can see quite readily thatg0(x) =s+ 2t(x-p),g00(x) = 2t,and thusg(p) =r, g0(p) =s,g00(p) = 2t.Then the requirementsg(p) =f(p), g0(p) =f0(p), g00(p) =f00(p)come down to saying thatr=f(p),s=f0(p),t=f00(p)/2which means precisely thatg(x) =f(p) +f0(p)(x-p) +f00(p)2(x-p)2.

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Chapter 16 / Exercise 14
Calculus
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