Taylor formula remainder: various expressions

Taylor formula remainder: various expressions

Let

f

:

ℝ

→

ℝ

be an

n

+

1

times

differentiable function

, and let

T

n

,

a

⁢

(

x

)

its

n

t

⁢

h

⁢

-degree

Taylor polynomial

;

Then the following

expressions

for the

remainder

R

n

,

a

⁢

(

x

)

=

f

⁢

(

x

)

-

T

n

,

a

⁢

(

x

)

hold:

1)

R

n

,

a

⁢

(

x

)

=

1

n

!

⁢

∫

a

x

f

(

n

+

1

)

⁢

(

t

)

⁢

(

x

-

t

)

n

⁢

𝑑

t

(Integral
form)

2)

R

n

,

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)

R

n

,

a

⁢

(

x

)

=

f

(

n

+

1

)

⁢

(

ξ

)

n

!

⁢

(

x

-

ξ

)

n

⁢

(

x

-

a

)

for a

ξ

⁢

(

x

)

∈

(

a

,

x

)

(Cauchy form)

4)

R

n

,

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)

∫

a

b

R

n

,

a

⁢

(

x

)

⁢

𝑑

x

=

∫

a

b

f

(

n

+

1

)

⁢

(

x

)

(

n

+

1

)

!

⁢

(

b

-

x

)

n

+

1

⁢

𝑑

x

Proof:

1) Let’s proceed by induction.

n

=

0

.

∫

a

x

f

′

⁢

(

t

)

⁢

𝑑

t

=

f

⁢

(

x

)

-

f

⁢

(

a

)

=

R

0

,

a

⁢

(

x

)

, since

T

0

,

a

⁢

(

x

)

=

f

⁢

(

a

)

.

Let’s take it for true that

R

n

-

1

,

a

⁢

(

x

)

=

1

(

n

-

1

)

!

⁢

∫

a

x

f

(

n

)

⁢

(

t

)

⁢

(

x

-

t

)

n

-

1

⁢

𝑑

t

,
and let’s compute

∫

a

x

f

(

n

+

1

)

⁢

(

t

)

n

!

⁢

(

x

-

t

)

n

⁢

𝑑

t

by parts.

∫

a

x

f

(

n

+

1

)

⁢

(

t

)

n

!

⁢

(

x

-

t

)

n

⁢

𝑑

t

=

=

f

(

n

)

⁢

(

t

)

n

!

⁢

(

x

-

t

)

n

|

a

x

+

∫

x

a

f

(

n

)

⁢

(

t

)

n

!

⁢

n

⁢

(

x

-

t

)

n

-

1

⁢

𝑑

t

=

-

f

(

n

)

⁢

(

a

)

n

!

⁢

(

x

-

a

)

n

+

∫

x

a

f

(

n

)

⁢

(

t

)

(

n

-

1

)

!

⁢

(

x

-

t

)

n

-

1

⁢

𝑑

t

=

R

n

-

1

,

a

⁢

(

x

)

-

f

(

n

)

⁢

(

a

)

n

!

⁢

(

x

-

a

)

n

=

f

⁢

(

x

)

-

T

n

-

1

,

a

-

f

(

n

)

⁢

(

a

)

n

!

⁢

(

x

-

a

)

n

=

f

⁢

(

x

)

-

T

n

,

a

⁢

(

x

)

=

R

n

,

a

⁢

(

x

)

.

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

R

n

,

a

⁢

(

x

)

=

1

n

!

⁢

∫

a

x

f

(

n

+

1

)

⁢

(

t

)

⁢

(

x

-

t

)

n

⁢

𝑑

t

=

1

n

!

⁢

∫

a

x

f

(

n

+

1

)

⁢

(

t

)

⁢

(

x

-

t

)

n

-

p

+

1

⁢

(

x

-

t

)

p

-

1

⁢

𝑑

t

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

R

n

,

a

⁢

(

x

)

=

1

n

!

⁢

∫

a

x

f

(

n

+

1

)

⁢

(

t

)

⁢

(

x

-

t

)

n

-

p

+

1

⁢

(

x

-

t

)

p

-

1

⁢

𝑑

t

=

f

(

n

+

1

)

⁢

(

η

)

n

!

⁢

(

x

-

η

)

n

-

p

+

1

⁢

∫

a

x

(

x

-

t

)

p

-

1

⁢

𝑑

t

=

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:

∫

a

b

f

(

n

+

1

)

⁢

(

x

)

(

n

+

1

)

!

⁢

(

b

-

x

)

n

+

1

⁢

𝑑

x

=

=

f

(

n

)

⁢

(

x

)

(

n

+

1

)

!

⁢

(

b

-

x

)

n

+

1

|

a

b

+

∫

a

b

f

(

n

)

⁢

(

x

)

(

n

+

1

)

!

⁢

(

n

+

1

)

⁢

(

b

-

x

)

n

⁢

𝑑

x

=

-

f

(

n

)

⁢

(

a

)

(

n

+

1

)

!

⁢

(

b

-

a

)

n

+

1

+

∫

a

b

f

(

n

)

⁢

(

x

)

n

!

⁢

(

b

-

x

)

n

⁢

𝑑

x

=

…

=

-

∑

k

=

0

n

f

(

k

)

⁢

(

a

)

(

k

+

1

)

!

⁢

(

b

-

a

)

k

+

1

+

∫

a

b

f

⁢

(

x

)

⁢

𝑑

x

=

-

∑

k

=

0

n

f

(

k

)

⁢

(

a

)

k

!

⁢

∫

a

b

(

x

-

a

)

k

⁢

𝑑

x

+

∫

a

b

f

⁢

(

x

)

⁢

𝑑

x

=

∫

a

b

(

-

∑

k

=

0

n

f

(

k

)

⁢

(

a

)

k

!

⁢

(

x

-

a

)

k

+

f

⁢

(

x

)

)

⁢

𝑑

x

=

∫

a

b

R

n

,

a

⁢

(

x

)

⁢

𝑑

x

.

Note:

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

Let

ϕ

⁢

(

t

)

=

∑

k

=

0

n

f

(

k

)

⁢

(

t

)

k

!

⁢

(

x

-

t

)

k

Then

ϕ

⁢

(

x

)

=

f

⁢

(

x

)

and

ϕ

⁢

(

a

)

=

T

n

,

a

⁢

(

x

)

, so that

R

n

,

a

⁢

(

x

)

=

ϕ

⁢

(

x

)

-

ϕ

⁢

(

a

)

=

∫

a

x

ϕ

′

⁢

(

t

)

⁢

𝑑

t

.

Let’s now compute

ϕ

′

⁢

(

t

)

.

ϕ

′

⁢

(

t

)

=

∑

k

=

0

n

1

k

!

⁢

[

f

(

k

+

1

)

⁢

(

t

)

⁢

(

x

-

t

)

k

-

f

(

k

)

⁢

(

t

)

⁢

k

⁢

(

x

-

t

)

k

-

1

]

=

∑

k

=

0

n

f

(

k

+

1

)

⁢

(

t

)

k

!

⁢

(

x

-

t

)

k

-

∑

k

=

1

n

f

(

k

)

⁢

(

t

)

(

k

-

1

)

!

⁢

(

x

-

t

)

k

-

1

=

∑

k

=

0

n

f

(

k

+

1

)

⁢

(

t

)

k

!

⁢

(

x

-

t

)

k

-

∑

k

=

0

n

-

1

f

(

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:

∫

a

x

f

(

n

+

1

)

⁢

(

t

)

n

!

⁢

(

x

-

t

)

n

⁢

𝑑

t

=

f

(

n

)

⁢

(

t

)

n

!

⁢

(

x

-

t

)

n

|

a

x

+

∫

a

x

f

(

n

)

⁢

(

t

)

n

!

⁢

n

⁢

(

x

-

t

)

n

-

1

⁢

𝑑

t

=

-

f

(

n

)

⁢

(

a

)

n

!

⁢

(

x

-

a

)

n

+

∫

a

x

f

(

n

)

⁢

(

t

)

(

n

-

1

)

!

⁢

(

x

-

t

)

n

-

1

⁢

𝑑

t

=

-

f

(

n

)

⁢

(

a

)

n

!

⁢

(

x

-

a

)

n

-

f

(

n

-

1

)

⁢

(

a

)

(

n

-

1

)

!

⁢

(

x

-

a

)

n

-

1

+

∫

a

x

f

(

n

-

1

)

⁢

(

t

)

(

n

-

2

)

!

⁢

(

x

-

t

)

n

-

2

⁢

𝑑

t

=

…

=

-

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

+

∫

a

x

f

(

n

-

k

+

1

)

⁢

(

t

)

(

n

-

k

)

!

⁢

(

x

-

t

)

n

-

k

⁢

𝑑

t

…

=

-

∑

k

=

1

n

f

(

k

)

⁢

(

a

)

k

!

⁢

(

x

-

a

)

k

+

∫

a

x

f

′

⁢

(

t

)

⁢

𝑑

t

=

-

∑

k

=

1

n

f

(

k

)

⁢

(

a

)

k

!

⁢

(

x

-

a

)

k

+

f

⁢

(

x

)

-

f

⁢

(

a

)

=

-

∑

k

=

0

n

f

(

k

)

⁢

(

a

)

k

!

⁢

(

x

-

a

)

k

+

f

⁢

(

x

)

.

that is

f

⁢

(

x

)

=

∑

k

=

0

n

f

(

k

)

⁢

(

a

)

k

!

⁢

(

x

-

a

)

k

+

∫

a

x

f

(

n

+

1

)

⁢

(

t

)

n

!

⁢

(

x

-

t

)

n

⁢

𝑑

t

=

∑

k

=

0

n

f

(

k

)

⁢

(

a

)

k

!

⁢

(

x

-

a

)

k

+

R

n

,

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

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