Loading [MathJax]/jax/output/CommonHTML/jax.js

Taylor formula remainder: various expressions


Let f: be an n+1 times differentiable function, and let Tn,a(x) its nth -degree Taylor polynomialMathworldPlanetmath;

Then the following expressions for the remainder Rn,a(x)=f(x)-Tn,a(x) hold:

1)

Rn,a(x)=1n!xaf(n+1)(t)(x-t)n𝑑t

(Integral form)

2)

Rn,a(x)=f(n+1)(η)n!p(x-η)n-p+1(x-a)p

for a η(x)(a,x) and p>0 (Schlömilch form)

3)

Rn,a(x)=f(n+1)(ξ)n!(x-ξ)n(x-a)

for a ξ(x)(a,x)  (Cauchy form)

4)

Rn,a(x)=f(n+1)(ϑ)(n+1)!(x-a)n+1

for a ϑ(x)(a,x)  (Lagrange form)

Moreover the following result holds for the integral of the remainder from the center point  a to an arbitrary point b:

5)

baRn,a(x)𝑑x=baf(n+1)(x)(n+1)!(b-x)n+1𝑑x

Proof:

1) Let’s proceed by induction.

n=0. xaf(t)𝑑t=f(x)-f(a)=R0,a(x), since T0,a(x)= f(a).

Let’s take it for true that Rn-1,a(x)=1(n-1)!xaf(n)(t)(x-t)n-1𝑑t, and let’s compute xaf(n+1)(t)n!(x-t)n𝑑t by parts.

xaf(n+1)(t)n!(x-t)n𝑑t=
= f(n)(t)n!(x-t)n|xa+axf(n)(t)n!n(x-t)n-1𝑑t
= -f(n)(a)n!(x-a)n+axf(n)(t)(n-1)!(x-t)n-1𝑑t
= Rn-1,a(x)-f(n)(a)n!(x-a)n
= f(x)-Tn-1,a-f(n)(a)n!(x-a)n
= f(x)-Tn,a(x)=Rn,a(x).

2) Let’s write the remainder in the integral form this way:

Rn,a(x)=1n!xaf(n+1)(t)(x-t)n𝑑t=1n!xaf(n+1)(t)(x-t)n-p+1(x-t)p-1𝑑t

Now, since (x-t)p-1 doesn’t change sign between a and x, we can apply the integral Mean Value theorem (http://planetmath.org/IntegralMeanValueTheorem). So a point η(x)(a,x) exists such that

Rn,a(x)=1n!xaf(n+1)(t)(x-t)n-p+1(x-t)p-1𝑑t
= f(n+1)(η)n!(x-η)n-p+1xa(x-t)p-1𝑑t
= f(n+1)(η)n!p(x-η)n-p+1(x-a)p

(Note that the condition p>0 is needed to ensure convergence of the integral)

3) and 4) are obtained from Schlömilch form by plugging in p=1 and p=n+1 respectively.

5) Let’s start from the right-end side:

baf(n+1)(x)(n+1)!(b-x)n+1𝑑x=
= f(n)(x)(n+1)!(b-x)n+1|ba+baf(n)(x)(n+1)!(n+1)(b-x)n𝑑x
= -f(n)(a)(n+1)!(b-a)n+1+baf(n)(x)n!(b-x)n𝑑x
= =-nk=0f(k)(a)(k+1)!(b-a)k+1+baf(x)𝑑x
= -nk=0f(k)(a)k!ba(x-a)k𝑑x+baf(x)𝑑x
= ba(-nk=0f(k)(a)k!(x-a)k+f(x))𝑑x=baRn,a(x)𝑑x.

Note:

1) The proof of the integral form could also be stated as follow:

Let

ϕ(t)=nk=0f(k)(t)k!(x-t)k

Then ϕ(x)=f(x) and ϕ(a)=Tn,a(x), so that Rn,a(x)=ϕ(x)-ϕ(a)=xaϕ(t)𝑑t.

Let’s now compute ϕ(t).

ϕ(t)=nk=01k![f(k+1)(t)(x-t)k-f(k)(t)k(x-t)k-1]
= nk=0f(k+1)(t)k!(x-t)k-nk=1f(k)(t)(k-1)!(x-t)k-1
= nk=0f(k+1)(t)k!(x-t)k-n-1k=0f(k+1)(t)k!(x-t)k
= f(n+1)(t)n!(x-t)n.

2) From the integral form of the remainder it is possible to obtain the entire Taylor formula; indeed, repeatly integrating by parts, one gets:

xaf(n+1)(t)n!(x-t)n𝑑t=f(n)(t)n!(x-t)n|xa+xaf(n)(t)n!n(x-t)n-1𝑑t
= -f(n)(a)n!(x-a)n+xaf(n)(t)(n-1)!(x-t)n-1𝑑t
= -f(n)(a)n!(x-a)n-f(n-1)(a)(n-1)!(x-a)n-1+xaf(n-1)(t)(n-2)!(x-t)n-2𝑑t
= =-f(n)(a)n!(x-a)n-f(n-1)(a)(n-1)!(x-a)n-1--f(n-k+1)(a)(n-k+1)!(x-a)n-k+1+xaf(n-k+1)(t)(n-k)!(x-t)n-k𝑑t
= -nk=1f(k)(a)k!(x-a)k+xaf(t)𝑑t
= -nk=1f(k)(a)k!(x-a)k+f(x)-f(a)
= -nk=0f(k)(a)k!(x-a)k+f(x).

that is

f(x) = nk=0f(k)(a)k!(x-a)k+xaf(n+1)(t)n!(x-t)n𝑑t
= nk=0f(k)(a)k!(x-a)k+Rn,a(x).
Title Taylor formula remainder: various expressions
Canonical name TaylorFormulaRemainderVariousExpressions
Date of creation 2013-03-22 15:53:57
Last modified on 2013-03-22 15:53:57
Owner gufotta (12050)
Last modified by gufotta (12050)
Numerical id 19
Author gufotta (12050)
Entry type Result
Classification msc 41A58