Introduction - UCONN

1 THE remainder IN TAYLOR SERIESKEITH (x) be infinitely differentiable on an intervalIaround a numbera. On the intervalI,Taylor s inequalitybounds the difference betweenf(x) and itsnth degree Taylor polynomialcentered ataTn,a(x) =f(a) +f (a)(x a) +f (a)2!(x a)2+ +f(n)(a)n!(x a)nin terms of the magnitude of the (n+ 1)th derivative off: if|f(n+1)(x)| Mfor allxinIthen Taylor s inequality saysifbis inIthen|f(b) Tn,a(b)| M|b a|n+1(n+ 1)!.We will derive this inequality in two ways, using exact formulas forf(x) Tn,a(x) involvingderivatives and involving (Differential form of the remainder (Lagrange, 1797)).With notation asabove, forn 0andbin the intervalI,f(b) =n k=0f(k)(a)k!(b a)k+f(n+1)(c)(n+ 1)!(b a)n+1=Tn,a(b) +f(n+1)(c)(n+ 1)!(b a)n+1for somecstrictly numbercdepends ona,b, andn. Whenn= 0 the theorem saysf(b) =f(a) +f (c)(b a) for somecstrictly betweenaandb, which is the Mean Value (Integral form of the remainder (Cauchy, 1821)).

2 With notation as above,forn 0andbin the intervalI,f(b) =n k=0f(k)(a)k!(b a)k+ baf(n+1)(t)n!(b t)ndt=Tn,a(b) + + baf(n+1)(t)n!(b t) 0 this isf(b) =f(a) + baf (t)dt, which is the Fundamental Theorem of the differential form of the remainder , the integral form of the remainder involvesno additional parameters (Lagrange) form of the remainderTo prove Theorem we will use an extension of Rolle s theorem to higher (x)be infinitely differentiable on an interval containing the numbersaandb, wherea6=b. IfF(a) =F(b)andF(j)(a) = 0for1 j n, then there is somecstrictly betweenaandbsuch thatF(n+1)(c) = Rolle s theorem forF(x), sinceF(a) =F(b) we haveF (c1) = 0 for somec1strictly betweenaandb. Then fromF (a) = 0 andF (c1) = 0, by Rolle s theorem for thefunctionF (x) we haveF (c2) = 0 for somec2strictly betweenaandc1(soc2is strictlybetweenaandb). Next, fromF (a) = 0 andF (c2) = 0 we have by Rolle s theorem forF (x) thatF (c3) = 0 for somec3strictly betweenaandc2(soc3is strictly betweenaandb).

3 Continuing in this way, we eventually get somecnstrictly betweenaandbsuch thatF(n)(cn) = 0. Combining this withF(n)(a) = 0 there is somecn+1strictly betweenaandcn(so also strictly betweenaandb) such thatF(n+1)(c) = 0. Proof of SetE(x) =f(x) Tn,a(x) =f(x) n k=0f(k)(a)k!(x a) to the choice of coefficients inTn,a(x) we haveE(a) =f(a) f(a) = 0, E (a) =f (a) f (a) = 0, .. , E(n)(a) =f(n)(a) f(n)(a) = if we set( )F(x) =E(x) E(b)(b a)n+1(x a)n+1then we haveF(j)(a) = 0 forj= 0, .. , nandF(b) =E(b) E(b) = 0. By Theorem ,there is somecstrictly betweenaandbsuch thatF(n+1)(c) = 0. SinceF(n+1)(x) =E(n+1)(x) E(b)(n+ 1)!(b a)n+1=f(n+1)(x) E(b)(n+ 1)!(b a)n+1,we get0 =F(n+1)(c) =f(n+1)(c) E(b)(n+ 1)!(b a)n+1,soE(b)(b a)n+1=f(n+1)(c)(n+ 1)!.Substituting this into ( ),F(x) =E(x) f(n+1)(c)(n+ 1)!(x a)n+1,THE remainder IN TAYLOR SERIES3so atx=bwe get0 =F(b) =E(b) f(n+1)(c)(n+ 1)!(b a)n+1=f(b) Tn,a(b) f(n+1)(c)(n+ 1)!

4 (b a)n+1. Taylor s inequality is an immediate consequence of this differential form of the remainder :if|f(n+1)(x)| Mfor allxfromatob, then|f(n+1)(c)| M, so|f(b) Tn,a(b)|=|f(n+1)(c)(b a)n+1/(n+ 1)!| M|b a|n+1/(n+ 1)!. (Cauchy) form of the remainderProof of Start with the Fundamental Theorem of Calculus in the formf(b) =f(a) + baf (t) integration by parts withu=f (t) anddv=dt, sodu=f (t)dtand takev=t b(notv=t) to getf(b) =f(a) +f (t)(t b) ba ba(t b)f (t)dt=f(a) f (a)(a b) ba(t b)f (t)dt=f(a) +f (a)(b a) + ba(b t)f (t) integration by parts again withu=f (t) anddv= (b t)dt, sodu=f (t)dtandtakev= (b t)2/2. Thenf(b) =f(a) +f (a)(b a) + ba(b t)f (t)dt=f(a) +f (a)(b a) f (t)2(b t)2 ba+ ba(b t)22f (t)dt=f(a) +f (a)(b a) +f (a)2(b a)2+ ba(b t)22f (t) another integration by parts withu=f (t) anddv=12(b t)2dtwe getf(b) =f(a) +f (a)(b a) +f (a)2(b a)2+f (a)6(b a)3+ ba(b t)36f(4)(t) repeated integration by parts we eventually getf(b) =n k=0f(k)(a)k!

5 (b a)k+ ba(b t)nn!f(n+1)(t)dt. We will derive Taylor s inequality from Theorem in two CONRADM ethod 1: Assume|f(n+1)(t)| Mfor alltfromatob. Fora < b,|f(b) Tn,a(b)| ba|b t|nn!|f(n+1)(t)|dt ba(b t)nn!M dt=M(b a)n+1(n+ 1)!,where the last calculation comes from the Fundamental Theorem of Calculus. Forb < a, weget in a similar way|f(b) Tn,a(b)| M(a b)n+1/(n+ 1)!. Putting both cases together,|f(n+1)(t)| Mfortfromatob= |f(b) Tn,a(b)| M|b a|n+1(n+ 1)!.Method 2: We make a change of variables in the integral to bypass the need for separatecases as in the first method. The integral is taken fromatob(whethera < bora > b),and numbers fromatobcan be written in parametric form asa+ (b a)uasuruns from 0to 1. Therefore with the change of variablest=a+ (b a)uthe integral remainder equals ba(b t)nn!f(n+1)(t)dt= 10((b a)(1 u))nn!f(n+1)(a+ (b a)u)(b a)du=(b a)n+1n! 10(1 u)nf(n+1)(a+ (b a)u)du,so if|f(n+1)(t)| Mfor alltfromatobthen the absolute value of the integral remainderis at most|b a|n+1n!

6 10(1 u)nM du=M|b a|n+1n! 10(1 u)ndu=M|b a|n+1n!1n+ 1=M|b a|n+1(n+ 1)!.Remark s inequality is not due to Taylor. In fact, Taylor s treatment of powerseries (in his bookMethodus Incrementorum Directa et Inversa, written in 1715) was notconcerned with justifications of convergence or error estimates, and preceded by almost 80years the work of Lagrange and by over 100 years the work of Cauchy.