Taylor formula remainder: various expressions
Let f:ℝ→ℝ be an n+1 times differentiable function, and let Tn,a(x) its nth -degree Taylor polynomial;
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+1∫xa(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 | |||
= | …=-n∑k=0f(k)(a)(k+1)!(b-a)k+1+∫baf(x)𝑑x | |||
= | -n∑k=0f(k)(a)k!∫ba(x-a)k𝑑x+∫baf(x)𝑑x | |||
= | ∫ba(-n∑k=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)=n∑k=01k![f(k+1)(t)(x-t)k-f(k)(t)k(x-t)k-1] | ||||
= | n∑k=0f(k+1)(t)k!(x-t)k-n∑k=1f(k)(t)(k-1)!(x-t)k-1 | |||
= | n∑k=0f(k+1)(t)k!(x-t)k-n-1∑k=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 | |||
… | ||||
= | -n∑k=1f(k)(a)k!(x-a)k+∫xaf′(t)𝑑t | |||
= | -n∑k=1f(k)(a)k!(x-a)k+f(x)-f(a) | |||
= | -n∑k=0f(k)(a)k!(x-a)k+f(x). |
that is
f(x) | = | n∑k=0f(k)(a)k!(x-a)k+∫xaf(n+1)(t)n!(x-t)n𝑑t | ||
= | n∑k=0f(k)(a)k!(x-a)k+Rn,a(x). |
Title | Taylor formula remainder: various expressions |
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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 |