CUBO A Mathematical Journal

Vol.10, N
o
¯ 01, (11–18). March 2008

Error Inequalities for a Taylor-like Formula

Nenad Ujević

Department of Mathematics

University of Split

Teslina 12/III, 21000 Split – CROATIA

email: ujevic@pmfst.hr

ABSTRACT

A Taylor-like formula is derived. Various error bounds for this formula are

established.

RESUMEN

Se deduce una formula de tipo Taylor. Se establecen varias cotas de error para

esta formula.

Key words and phrases: Error Inequalities, Taylor formula.

Math. Subj. Class.: 26D10.



12 Nenad Ujević CUBO
10, 1 (2008)

1 Introduction

In recent years a number of authors have considered the Taylor and generalized Taylor

formulas from an inequalities point of view. For example, this topic is considered in [1], [2],

[3], [4], [5], [6] and [8]. In [5] we can find the following generalization of Taylor formula:

f (x) = f (a) +

n
∑

k=1

(−1)k+1
[

Pk(x)f
(k)

(x) − Pk(a)f
(k)

(a)

]

+ Rn(f, a, x), (1)

Rn(f, a, x) = (−1)
n

∫ x

a

Pn(t)f
(n+1)

(t)dt,

where{Pk(t)}
∞
0 is a harmonic (or Appell) sequence of polynomials, that is

P
′
k(t) = Pk−1(t), P0(t) = 1.

If we substitute

Pk(t) =
(t − x)k

k!

in (1) then we get the classical Taylor formula:

f (x) = f (a) +

n
∑

k=1

(x − a)k

k!
f

(k)
(a) + R

C
n (f, a, x),

R
C
n (f, a, x) =

1

n!

∫ x

a

(x − t)nf (n+1)(t)dt.

In this paper we derive a Taylor-like formula. A way of obtaining this formula is similar to

the way described in [5]. However, here we do not use an Appell sequence of polynomials.

We use functions of the form

Sn(t) =

{

Pn(t), t ∈
[

a,
a+x

2

]

Qn(t), t ∈
(

a+x
2

, x
]

,

where Pn(t) and Qn(t) are Appell-like sequences of polynomials. We also establish various

error bounds for this formula. Similar error inequalities are established in [7] for some

quadrature rules.

Finally, we give an application of the mentioned Taylor-like formula to logarithmic

function.

2 Main results

Theorem 1 Let f : [a, x] → R be a function such that f (n) is absolutely continuous. Then

f (x) = f (a) −

n
∑

k=1

(−1)k(x − a)k

4kk!
(1 + k)

[

f
(k)

(x) − (−1)kf (k)(a)
]

(2)



CUBO
10, 1 (2008)

Error Inequalities for a Taylor-Like Formula 13

−

n
∑

k=2

(−1)k(x − a)k

4kk!
(1 − k)

[

1 − (−1)k
]

f
(k)

(
a + x

2
) + R(f ),

where

R(f ) = (−1)n
∫ x

a

Sn(t)f
(n+1)

(t)dt (3)

and

Sn(t) =







(t− 3a+x
4

)n−1

n!

[

t +
(n−3)a−(n+1)x

4

]

, t ∈
[

a,
a+x

2

]

(t− a+3x )n−1

n!

[

t +
(n−3)x−(n+1)a

4

]

, t ∈
(

a+x
2

, x
]

. (4)

Proof. We prove (2) by induction. We easily show that (2) holds for n = 1. Now

suppose that (2) holds for an arbitrary n. We have to prove that (2) holds for n → n + 1.

To simplify the proof we introduce the notations

Pn(t) =
(t − 3a+x

4
)
n−1

n!

[

t +
(n − 3)a − (n + 1)x

4

]

(5)

Qn(t) =
(t − a+3x

4
)
n−1

n!

[

t +
(n − 3)x − (n + 1)a

4

]

. (6)

We see that Pn and Qn form Appell sequences of polynomials, that is

P
′
n(t) = Pn−1(t), Q

′
n(t) = Qn−1(t), P0(t) = Q0(t) = 1.

We have

(−1)n+1
∫ x

a

Sn+1(t)f
(n+2)

(t)dt

= (−1)n+1
∫

a+x
2

a

Pn+1(t)f
(n+2)

(t)dt + (−1)n+1
∫ x

a+x
2

Qn+1(t)f
(n+2)

(t)dt

= (−1)n+1
[

Pn+1(
a + x

2
)f

(n+1)
(
a + x

2
) − Pn+1(a)f

(n+1)
(a)

]

+(−1)n+1
[

Qn+1(x)f
(n+1)

(x) − Qn+1(
a + x

2
)f

(n+1)
(
a + x

2
)

]

+(−1)n
∫

a+x
2

a

Pn(t)f
(n+1)

(t)dt + (−1)n
∫ x

a+x
2

Qn(t)f
(n+1)

(t)dt

= (−1)n
∫ x

a

Sn(t)f
(n+1)

(t)dt + (−1)n+1
[

Pn+1(
a + x

2
) − Qn+1(

a + x

2
)

]

f
(n+1)

(
a + x

2
)

−(−1)n+1
[

Pn+1(a)f
(n)

(a) − Qn+1(x)f
(n)

(x)

]

= −

∫ x

a

f
′
(t)dt +

n
∑

k=1

(−1)k(x − a)k

4kk!

[

f
(k)

(x) − (−1)kf (k)(a)
]



14 Nenad Ujević CUBO
10, 1 (2008)

+

n
∑

k=2

(−1)k(x − a)k

4kk!
(1 − k)

[

1 − (−1)k
]

f
(k)

(
a + x

2
)

+(−1)n+1
[

Pn+1(
a + x

2
) − Qn+1(

a + x

2
)

]

f
(n+1)

(
a + x

2
)

−(−1)n+1
[

Pn+1(a)f
(n)

(a) − Qn+1(x)f
(n)

(x)

]

= −

∫ x

a

f
′
(t)dt +

n+1
∑

k=1

(−1)k(x − a)k

4kk!

[

f
(k)

(x) − (−1)kf (k)(a)
]

+

n+1
∑

k=2

(−1)k(x − a)k

4kk!
(1 − k)

[

1 − (−1)k
]

f
(k)

(
a + x

2
),

since

(−1)n+1
[

Pn+1(
a + x

2
) − Qn+1(

a + x

2
)

]

f
(n)

(
a + x

2
)

−(−1)n+1
[

Pn+1(a)f
(n)

(a) − Qn+1(x)f
(n)

(x)

]

=
(−1)n+1(x − a)n+1

4n+1(n + 1)!
(1 − n − 1)

[

1 − (−1)n+1
]

f
(n+1)

(
a + x

2
)

+
(−1)n+1(x − a)n+1

4n+1(n + 1)!

[

f
(n+1)

(x) − (−1)n+1f (n+1)(a)
]

.

This completes the proof.

Lemma 2 The functions Sn(t) satisfy:

∫ x

a

Sn(t)dt = 0, if n is odd, (7)

∫ x

a

|Sn(t)| dt =
(4n + 4)(x − a)n+1

4n+1(n + 1)!
, (8)

max
t∈[a,x]

|Sn(t)| =
(n + 1)(x − a)n

4nn!
. (9)

Proof. A simple calculation gives

∫ x

a

Sn(t)dt =
(x − a)n+1

4n(n + 1)!

[

1 − (−1)n+1
]

.

From the above relation we see that (7) holds, since 1 − (−1)n+1 = 0 if n is odd.



CUBO
10, 1 (2008)

Error Inequalities for a Taylor-Like Formula 15

We now consider some properties of the Appell sequences of polynomials Pn(t) and

Qn(t), given by (5) and (6), respectively. Since

t +
(n − 3)a − (n + 1)x

4
≤ 0, t ∈

[

a,
a + x

2

]

and

t +
(n − 3)x − (n + 1)a

4
≥ 0, t ∈

(

a + x

2
, x

]

we easily show that the following facts are valid.

If n is odd then Pn(t) ≤ 0 and Qn(t) ≥ 0. Furthermore, Pn(t) is an increasing function

for t ∈
[

a,
3a+x

4

)

and it is a decreasing function for t ∈
(

3a+x
4

,
a+x

2

]

. The function Qn(t) is

decreasing for t ∈
[

a+x
2

,
a+3x

4

)

and it is increasing for t ∈
(

3a+3x
4

, x
]

.

If n is even then Pn(t) is a decreasing function and Qn(t) is an increasing function.

Furthermore, Pn(t) > 0 for t ∈
[

a,
3a+x

4

)

and Pn(t) < 0 for t ∈
(

3a+x
4

,
a+x

2

]

, while Qn(t) < 0

for t ∈
[

a+x
2

,
a+3x

4

)

and Qn(t) > 0 for t ∈
(

3a+3x
4

, x
]

.

We use these properties to prove (8) and (9).

If n is odd then we have

∫ x

a

|Sn(t)| dt =

∫
a+x

2

a

|Pn(t)| dt +

∫ x

a+x
2

|Qn(t)| dt

=

∣

∣

∣

∣

∣

∫
a+x

2

a

Pn(t)dt

∣

∣

∣

∣

∣

+

∣

∣

∣

∣

∣

∫ x

a+x
2

Qn(t)dt

∣

∣

∣

∣

∣

=
(4n + 4)(x − a)n+1

4n+1(n + 1)!
.

If n is even then we find that the same result is valid. Thus (8) holds.

Finally, we have

max
t∈[a,x]

|Sn(t)| = max

{

max
t∈[a,

a+x
2

]
|Pn(t)| , max

t∈[
a+x

2
,x]

|Qn(t)|

}

= max

{
∣

∣

∣

∣

Pn(
a + x

2
)

∣

∣

∣

∣

,

∣

∣

∣

∣

Qn(
a + x

2
)

∣

∣

∣

∣

, |Pn(a)| , |Qn(x)|

}

=
(n + 1)(x − a)n

4nn!
.



16 Nenad Ujević CUBO
10, 1 (2008)

We introduce the notation

F (x, a) = f (a) −

n
∑

k=1

(−1)k(x − a)k

4kk!
(1 + k)

[

f
(k)

(x) − (−1)kf (k)(a)
]

−

n
∑

k=2

(−1)k(x − a)k

4kk!
(1 − k)

[

1 − (−1)k
]

f
(k)

(
a + x

2
).

Theorem 3 Let f : [a, x] → R be a function such that f (n) is absolutely continuous and

there exist real numbers γn, Γn such that γn ≤ f
(n+1)

(t) ≤ Γn, t ∈ [a, x]. Then

|f (x) − F (x, a)| ≤
Γn − γn
(n + 1)!

(2n + 2)(x − a)n+1

4n+1
if n is odd (10)

and

|f (x) − F (x, a)| ≤
(4n + 4)(x − a)n+1

4n+1(n + 1)!

∥

∥

∥
f

(n+1)
∥

∥

∥

∞
if n is even. (11)

Proof. Let n be odd. From (3) and (7) we get

R(f ) = (−1)n
∫ x

a

Sn(t)f
(n+1)

(t)dt = (−1)n
∫ x

a

Sn(t)

[

f
(n+1)

(t) −
γn + Γn

2

]

dt

such that we have

|R(f )| = |f (x) − F (x, a)| ≤ max
t∈[a,x]

∣

∣

∣

∣

f
(n+1)

(t) −
γn + Γn

2

∣

∣

∣

∣

∫ x

a

|Sn(t)| dt. (12)

We also have

max
t∈[a,x]

∣

∣

∣

∣

f
(n+1)

(t) −
γn + Γn

2

∣

∣

∣

∣

≤
Γn − γn

2
. (13)

From (12), (13) and (8) we get

|f (x) − F (x, a)| ≤
Γn − γn
(n + 1)!

(2n + 2)(x − a)n+1

4n+1
.

Let n be even. Then we have

|R(f )| = |f (x) − F (x, a)| ≤

∫ x

a

|Sn(t)| dt
∥

∥

∥
f

(n+1)
∥

∥

∥

∞
=

(4n + 4)(x − a)n+1

4n+1(n + 1)!

∥

∥

∥
f

(n+1)
∥

∥

∥

∞
.

Theorem 4 Let f : [a, x] → R be a function such that f (n) is absolutely continuous and let

n be odd. If there exists a real number γn such that γn ≤ f
(n+1)

(t), t ∈ [a, x] then

|f (x) − F (x, a)| ≤ (Tn − γn)
(n + 1)(x − a)n+1

4nn!
, (14)



CUBO
10, 1 (2008)

Error Inequalities for a Taylor-Like Formula 17

where

Tn =
f

(n)
(x) − f (n)(a)

x − a
.

If there exists a real number Γn such that f
(n+1)

(t) ≤ Γn, t ∈ [a, x] then

|f (x) − F (x, a)| ≤ (Γn − Tn)
(n + 1)(x − a)n+1

4nn!
. (15)

Proof. We have

|R(f )| = |f (x) − F (x, a)| =

∣

∣

∣

∣

∫ x

a

(f
(n+1)

(t) − γn)Sn(t)dt

∣

∣

∣

∣

,

since (7) holds. Then we have

∣

∣

∣

∣

∫ x

a

(f
(n+1)

(t) − γn)Sn(t)dt

∣

∣

∣

∣

≤ max
t∈[a,x]

|Sn(t)|

∫ x

a

(f
(n+1)

(t) − γn)dt

=
(n + 1)(x − a)n

4nn!

[

f
(n)

(x) − f (n)(a) − γn(x − a)
]

=
(n + 1)(x − a)n+1

4nn!
(Tn − γn) .

In a similar way we can prove that (15) holds.

Remark 5 Note that we can apply the estimations (10) and (11) only if f (n+1) is bounded.

On the other hand, we can apply the estimation (14) if f (n+1) is unbounded above and we

can apply the estimation (15) if f (n+1) is unbounded below.

3 An application to logarithmic function

We now apply the formula (2) to logarithmic function. We have

f
(j)

(t) =
(−1)j (j − 1)!

(1 + t)j
if f (t) = ln(1 + t). (16)

From (2), (16) and a = 0, f (t) = ln(1 + t) we get

F (x) = −

n
∑

k=1

(−1)kxk

4kk

[

(1 + k)

(

(−1)k+1

(1 + x)k
+ 1

)

+
(−1)k+1(1 − k)(1 − (−1)k)

(1 +
x
2
)k

]

(17)

≈ ln(1 + x), x ∈

(

−
4

5
, 4

)

.



18 Nenad Ujević CUBO
10, 1 (2008)

The standard formula for this function is given by

S(x) =

m
∑

k=1

(−1)k+1xk

k
≈ ln(1 + x), x ∈ (−1, 1) . (18)

Many numerical examples show that n can be much less than m if we wish to obtain a prior

given accuracy and if x is close to 1 (x < 1).

Let us choose x = 0.99 and give the accuracy of order E − 14. The ”exact” value

is ln(1 + 0.99) = 0.688134643528734. If we use (17) with n = 22 then we get F (0.99) ≈

0.688134643528725. If we use (18) with m = 5000 then we get S(0.99) ≈ 0.688134643528737.

All calculations are done in double precision arithmetic. The first approximate result is

obtained faster than the second one. Similar results are obtained when we chose x = 0.9,

x = 0.95, etc.

Received: May 2006. Revised: August 2006.

References

[1] G.A. Anastassiou and S.S. Dragomir, On some estimates of the remainder in Tay-

lor’s formula, J. Math. Anal. Appl., 263 (2001), 246–263.

[2] P. Cerone, Generalized Taylor’s formula with estimates of the remainder, in Inequality

Theory and Applications, Vol 2, Y. J. Cho, J. K. Kim and S. S. Dragomir (Eds.), Nova

Sicence Publ., New York, (2003), 33-52.

[3] S.S. Dragomir, New estimation of the remainder in Taylor’s formula using Grüss type

inequalities and applications, Math. Inequal. Appl., 2(2) (1999), 183–193.

[4] H. Gauchman, Some integral inequalities involving Taylor’s remainder I, J. Inequal.

Pure Appl. Math., 3(2), Article 26, (2002).

[5] M. Matić, J. Pečarić and N. Ujević, On new estimation of the remainder in gen-

eralized Taylor’s formula, Math. Inequal. Appl., 2(3), (1999), 343–361.

[6] E. Talvila, Estimates of the remainder in Taylor’s theorem using the Hentstock-

Kurzweil integral, Czechoslovak Math. J., 55(4), (2005), 933–940.

[7] N. Ujević and A.J. Roberts, A corrected quadrature formula and applications,

ANZIAM J., 45 (E), (2004), E41-E56.

[8] N. Ujević, A new generalized perturbed Taylor’s formula, Nonlin. Funct. Anal. Appl.,

7(2), (2002), 255-267.


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