Parametric Two-Point Integral Inequalities For n -Time Differentiable Functions With Applications ∗

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Parametric Two-Point Integral Inequalities For n -Time Differentiable Functions With Applications ^∗

Aristides I. Kechriniotis

^†

, Bill A. Kotsos

^‡

, Christos A. Tsonos

^§

Received 12 December 2008

Abstract

New parametric two point integral inequalities for n-time differentiable functions are presented. These inequalities are used to obtain some new estimations for the remainder in Taylor’s formula. New inequalities for the expectation and variance of a random variable defined on a finite interval are also given.

1 Introduction

In the literature on numerical integration, the following estimation is well known as the trapezoid inequality:

f(b) +f(a)

2 − 1

b−a Z b

a

f(x)dx

≤ (b−a)²

12 sup

x∈(a,b)

|f⁰⁰(x)|, (1) where the mapping f : [a, b] → R is twice differentiable on the interval (a, b), with the second derivative bounded on (a, b).For more results on the trapezoid inequality and their applications we refer to [3], while in [2] we can find the following result: Let f ∈Cⁿ[a, b] be a function such thatf⁽ⁿ⁺¹⁾is integrable and bounded on (a, b).Then for any positive numberρthe following estimation holds:

(1 +ρ) (b−a)ⁿ⁺¹ (n+ 2)! (n+ 1) inf

x∈(a,b)f⁽ⁿ⁺¹⁾(x)

≤ −ρ+ (−1)ⁿ⁺¹ b−a

Z b a

f(x)dx+ρf(b) + (−1)ⁿ⁺¹f(a) n+ 1

+

n−1

X

k=0

(n−k) (b−a)^k (n+ 1) (k+ 1)!

ρf^(k)(a) + (−1)^n+k+1f^(k)(b)

≤ (1 +ρ) (b−a)ⁿ⁺¹ (n+ 2)! (n+ 1) sup

x∈(a,b)

f⁽ⁿ⁺¹⁾(x). (2)

∗Mathematics Subject Classifications: 26D15, 26D20, 60E15.

†Department of Electronics, Technological Educational Institute of Lamia, 35100 Lamia, Greece

‡Department of Electronics, Technological Educational Institute of Lamia, 35100 Lamia, Greece

§Department of Electronics, Technological Educational Institute of Lamia, 35100 Lamia, Greece

216

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This inequality is used in [2] in order to obtain several inequalities forn-time differentiable functions, as for example some generalizations of inequality (1).

In what follows, we present some two point integral inequalities forn-time differentiable functions involving two parameters which are generalizations of inequality (2).

As applications to that some new interesting integral inequalities are given which are being used to obtain some estimations for the expectation and the variance of a random variable defined on a finite interval. The results presented here are related to the ones obtained in paper [1]. New inequalities involving the remainder in Taylor’s formula are also presented which give better approximations compared to the classical Taylor’s expansion.

2 Main Results

We begin with the following result.

THEOREM 1. Assume (p, q) is a pair of conjugate exponents, that is 1 ≤p, q ≤

∞, _p¹+¹_q = 1. Letg : [a, b]→Rbe a (n+ 1)-time differentiable function on (a, b) for somen≥1 withg⁽ⁿ⁺¹⁾integrable on (a, b).Then forn≥2 and allα, β∈Rwe have the inequalities

α+β b−a

Z b a

g(x)dx−(αg(a) +βg(b))

−

n−1

X

k=1

n−k n²−1

(nα−β)g^(k)(a) + (−1)^k(nβ−α)g^(k)(b)(b−a)^k (k+ 1)!

≤











(|nα−β|+|nβ−α|)(b−a)ⁿ⁺¹ (n+2)!(n²−1)

g⁽ⁿ⁺¹⁾

_∞ ifg⁽ⁿ⁺¹⁾∈L∞[a, b]

(|nα−β|^q+|nβ−α|^q)

1

q(b−a)^{n+ 1}^q (n−1)(n+1)![(nq+1)(nq+2)]

1 q

g⁽ⁿ⁺¹⁾

p ifg⁽ⁿ⁺¹⁾∈Lp[a, b], 1< p <∞

(b−a)ⁿ⁻¹ (n−1)(n+1)!

g⁽ⁿ⁺¹⁾

1max{|nα−β|,|nβ−α|} ifg⁽ⁿ⁺¹⁾∈L1[a, b]

, (3)

and the first inequality in (3) is sharp in the following two cases: i) n is odd and

β α ∈₁

n, n

, ii)nis even andα= 0 or _α^β ∈/₁

n, n . PROOF. LetKn(u, x) be the kernel given by

Kn(u, x) :=







α−nβ

n−1 (a−u)ⁿ if u∈[a, x]

nα−β

n−1 (b−u)ⁿ if u∈(x, b]

, n >1.

Then, using twice the Taylor’s formula with an integral form of remainder we easily

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get

1 (n+ 1)! (b−a)

Z b a

Kn(u, x)g⁽ⁿ⁺¹⁾(u)dudx

= nβ−α

(n²−1) (b−a) Z b

a

Z a x

(a−u)ⁿ

n! g⁽ⁿ⁺¹⁾(u)dudx + nα−β

(n²−1) (b−a) Z b

a

Z b x

(b−u)ⁿ

n! g⁽ⁿ⁺¹⁾(u)dudx

= nβ−α

n²−1g(a) +nα−β

n−1 g(b)− (α+β) (n+ 1) (b−a)

Z b a

g(x)dx− 1 b−a

n

X

k=1

Ik, (4) where

Ik:= 1 k!

Z b a

g^(k)(x)

nβ−α

n²−1 (a−x)^k+nα−β

n²−1 (b−x)^k

dx, k≥0.

Now, using integration by parts we have, Ik+1−Ik=−

nα−β

n²−1g^(k)(a) + (−1)^knβ−α n²−1g^(k)(b)

(b−a)^k+1

(k+ 1)! , k≥0. (5) Moreover, it is easy to verify that any sequence (In), n≥0 the following identity holds

n

X

k=1

Ik =nI0+

n−1

X

k=0

(n−k) (Ik+1−Ik). (6) Combining (4) with (5) and (6) we easily get the following identity:

1 (n+ 1)! (b−a)

Z b a

= αg(a) +βg(b)−α+β b−a

Z b a

g(x)dx +

n−1

X

k=1

n−k n²−1

(nα−β)g^(k)(a) + (−1)^k(nβ−α)g^(k)(b)(b−a)^k (k+ 1)!. (7) Now we will use identity (7) to prove Theorem 1.

Ifg⁽ⁿ⁺¹⁾∈L∞[a, b], then it is not difficult to get:

1 (n+ 1)! (b−a)

Z b a

≤ (|α−nβ|+|nα−β|) (b−a)ⁿ⁺¹ (n+ 2)! (n²−1)

g⁽ⁿ⁺¹⁾

∞, (8) which combined with identity (7) lead to first estimation in (3).

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Let consider now that either n is an odd integer and ^β_α ∈ ₁

n, n

which means (nα−β) (nβ−α)≥0, ornis even and^β_α ∈/₁

n, n

(a6= 0), that is (nα−β) (nα−β)≤ 0. Then for g(x) = xⁿ⁺¹ we find out easily that inequality (8) holds as an equality.

That means, using the identity (7) as well, that the first inequality in (3) is sharp.

Now, ifg⁽ⁿ⁺¹⁾∈Lp[a, b], 1< p <∞,then, by using H¨older’s inequality, we have 1

(n+ 1)! (b−a)

Z b a

≤ 1

(n+ 1)! (b−a) Z b

a

Z b a

g⁽ⁿ⁺¹⁾(u)

p

dudx

!¹_p Z b

a

Z b a

|Kn(u, x)|^qdudx

!¹_q

= (|nα−β|^q+|nβ−α|^q)

1

q(b−a)ⁿ⁺¹^q (n−1) (n+ 1)! [(nq+ 1) (nq+ 2)]¹^q

g⁽ⁿ⁺¹⁾

p,

which combined with the identity (7) gives the second inequality in (3).

Finally, letg⁽ⁿ⁺¹⁾∈L1[a, b].Then, 1

(n+ 1)! (b−a)

Z b a

≤ 1

(n+ 1)! (b−a) sup

(u,x)∈[a,b]×[a,b]

|Kn(u, x)|

Z b a

g⁽ⁿ⁺¹⁾(u) dudx

= (b−a)ⁿ⁻¹ (n−1) (n+ 1)!

g⁽ⁿ⁺¹⁾

₁max{|nα−β|,|nβ−α|}, which combined with (7) leads to third inequality in (3).

REMARK 1. If we apply the first inequality in (3) forα= ^nρ−(−1)_n+1 ⁿ,β= ^ρ−(−1)_n+1ⁿⁿ, we get inequality (2). Therefore inequality (3) can be regarded as a generalization of (2).

If we apply the first inequality in (3) forn= 2 we have the following result:

COROLLARY 1. Letg: [a, b]→Rbe a three time differentiable function on (a, b) with g⁰⁰⁰bounded on (a, b). Then for allα, β∈Rwe have the inequality

α+β b−a

Z b a

g(x)dx−(αg(a) +βg(b)) + ((2α−β)g⁰(a)−(2β−α)g⁰(b))(b−a) 6

≤ (|2α−β|+|2β−α|) (b−a)³

72 kg⁰⁰⁰k_∞. (9)

Ifα= 0 or _α^β ∈/₁

2,2

, inequality (9) is sharp.

PROPOSITION 1. Letgbe as in Corollary 1. Then,

g⁰(a) Z b

a

g(x)−g(a) + 2g(b) 3

dx+g⁰(b) Z b

a

g(x)−2g(a) +g(b) 3

dx

≤ (|g⁰(a)|+|g⁰(b)|) (b−a)⁴

72 kg⁰⁰⁰k_∞, (10)

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and inequality (10) is sharp.

PROOF. Applying (9) for α= ²₃g⁰(b) + ¹₃g⁰(a) andβ = ¹₃g⁰(b) + ²₃g⁰(a), we get (10). Further, an easy calculation yields that forg(x) = (x−a)³ the equality in (10) holds. Therefore inequality (10) is sharp.

COROLLARY 2. Letg be as in Corollary 1. Ifg⁰(a) = 0 and g⁰(b)6= 0,then the following holds

1 b−a

Z b a

g(x)dx−2g(a) +g(b) 3

≤(b−a)³

72 kg⁰⁰⁰k_∞, (11) while ifg⁰(a)6= 0 andg⁰(b) = 0,we have the inequality

1 b−a

Z b a

g(x)dx−g(a) + 2g(b) 3

≤(b−a)³

72 kg⁰⁰⁰k_∞. (12) Both inequalities are sharp.

PROOF. The above inequalities result in as an easy application from (10). In the case wheng(x) = (x−a)³ andg(x) = (b−x)³ the equalities in (11) and (12) hold as well. Therefore the inequalities (11) and (12) are sharp.

3 An Application for Taylor’s Remainder

As usual, Rn(f;x0, x) denotes the remainder in Taylor’s formula, that is, Rn(f;x0, x) =f(x)−

n

X

k=0

f^(k)(x0)(x−x0)^k k! .

THEOREM2 . Let I ⊂ R be an open interval, and x0 ∈ I. If f : I → R is a (n+ 2)−time differentiable function withf⁽ⁿ⁺²⁾ integrable onI,then for allx∈I,

Rn(f;x0, x)−x−x0

n+ 1 Rn−1(f⁰;x0, x)

≤











|x−x0|ⁿ⁺² (n+2)!(n+1)

f⁽ⁿ⁺²⁾

∞ iff⁽ⁿ⁺²⁾∈L∞[{x0, x}]

|x−x0|^{n+1+ 1}^q (n+1)![(nq+1)(nq+2)]¹^q

f⁽ⁿ⁺²⁾

p iff⁽ⁿ⁺²⁾∈Lp[{x0, x}], 1< p <∞

|x−x0|ⁿ (n+1)!

f⁽ⁿ⁺²⁾

1 iff⁽ⁿ⁺²⁾∈L1[{x0, x}]

,

where [{x0, x}] denotes the closed interval [min{x, x0},max{x, x0}]. The first inequality is sharp.

PROOF. We distinguish two cases:

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First case:Letx≥x0.Then applying inequalities (3) forα= _n+1¹ ,β= _n+1ⁿ ,g(x) = f⁰(x),a=x,b=x0, and multiplying the result by (x−x0), we get,

f(x)−f(x0)−(f⁰(x) +nf⁰(x0))

n+ 1 (x−x0)

−

n−1

X

k=1

n−k

n+ 1f^(k+1)(x0)(x−x0)^k+1 (k+ 1)!

≤











(x−x0)ⁿ⁺² (n+2)!(n+1)

f⁽ⁿ⁺²⁾

∞ iff⁽ⁿ⁺²⁾∈L∞[{x0, x}]

(x−x0)^{n+1+ 1}^q (n+1)![(nq+1)(nq+2)]¹^q

f⁽ⁿ⁺²⁾

p iff⁽ⁿ⁺²⁾∈Lp[{x0, x}], 1< p <∞

(x−x0)ⁿ (n+1)!

f⁽ⁿ⁺²⁾

₁ iff⁽ⁿ⁺²⁾∈L1[{x0, x}].

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Now, if we replace kbyk−1, we have f(x)−f(x0)−(f⁰(x) +nf⁰(x0))

n+ 1 (x−x0)−

n−1

X

k=1

n−k

n+ 1f^(k+1)(x0)(x−x0)^k+1 (k+ 1)!

= f(x)−f(x0)−(f⁰(x) +nf⁰(x0))

n+ 1 (x−x0)−

n

X

k=2

n+ 1−k

n+ 1 f^(k)(x0)(x−x0)^k k!

= f(x)−f(x0)−f⁰(x0) (x−x0)−

n

X

k=2

f^(k)(x0)(x−x0)^k k!

−x−x0

n+ 1 f⁰(x)−f⁰(x0)−

n

X

k=2

f^(k)(x0)(x−x0)^k−1 (k−1)!

!

= f(x)−

n

X

k=0

f^(k)(x0)(x−x0)^k

k! −x−x0

n+ 1 f⁰(x)−

n−1

X

k=0

f^(k+1)(x0)(x−x0)^k k!

! .

Therefore, we conclude that

f(x)−f(x0)−(f⁰(x) +nf⁰(x0))

n+ 1 (x−x0)−

n−1

X

k=1

n−k

n+ 1f^(k+1)(x0)(x−x0)^k+1 (k+ 1)!

= Rn(f;x0, x)−x−x0

n+ 1Rn−1(f⁰;, x0, x).

Finally, using this latter equality in (13) we get the desired result.

Second case: Letx < x0.Applying (3) forα=_n+1ⁿ ,β= _n+1¹ ,g(x) =f⁰(x),a=x0, b=xand working in a way similar to the first case, we get the desired result as well.

Finally, forf(x) =xⁿ⁺², we readily calculate:

Rn(f;x0, x) = (n+ 2) (x−x0)ⁿ⁺¹x0+ (x−x0)ⁿ⁺²,

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and

Rn−1(f⁰;x0, x) = (n+ 2)Rn−1 xⁿ⁺¹;x0, x

= (n+ 2) (n+ 1) (x−x0)ⁿx0+ (n+ 2) (x−x0)ⁿ⁺¹. So we finally have

Rn(f;x0, x)−x−x0

n+ 1Rn−1(f⁰;x0, x)

= 1

n+ 1|x−x0|ⁿ⁺². On the other hand the following equality holds

|x−x0|ⁿ⁺² (n+ 2)! (n+ 1)

f⁽ⁿ⁺²⁾

_∞= 1

n+ 1|x−x0|ⁿ⁺².

Combining those last two equalities we conclude that forf(x) =xⁿ⁺² the equality in the first estimation of Theorem 2 holds. Consequently this inequality is sharp.

EXAMPLE 1. If we apply the first inequality of Theorem 2 for f(x) = e^x and x0= 0, after some straightforward algebra it follows that

e^x− 1 n+ 1−x

n

X

k=0

(n+ 1−k)x^k k!

≤ xⁿ⁺²

(n+ 2)! (n+ 1−x)e^x+|x|² .

EXAMPLE 2. If we apply the first inequality of Theorem 2 forf(x) = lnx, x, x0>

0,and taking into account that,

Rn(f;x0, x) = lnx−lnx0−

n

X

k=1

(−1)^k+1(x−x0)^k kx^k₀ ,

and

Rn−1(f⁰;x0, x) = 1 x− 1

x0 n−1

X

k=0

(x0−x)^k x^k₀

= 1

x− 1 x0

1−^(x⁰_x^−x)n ⁿ 0

1−^x⁰_x^−x₀ =(x0−x)ⁿ xxⁿ₀ , we get,

lnx−lnx0−

n

X

k=1

(−1)^k+1(x−x0)^k

kx^k₀ +(x0−x)ⁿ⁺¹ (n+ 1)xxⁿ₀

≤ 1

(n+ 2) (n+ 1)

|x−x0|ⁿ⁺² (min{x, x0})ⁿ⁺².

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4 Applications for Expectation and Variance

Many authors using trapezoid and Ostrowski type inequalities have produced estimations for the expectation and variance of a random variableX defined on a finite interval. For example in [4] we can see some related inequalities using an Ostrowski type inequality.

We want use some inequalities of the second section to obtain new inequalities for the expectation and the variance of a random variableX defined on a finite interval.

Some similar results are presented in [1].

Let f : [a, b] → (0,∞) be a differentiable function on (a, b) with f⁰ bounded on (a, b).Assume thatf is a probability density function of a random variableX, that is Rb

af(x)dx= 1. Denote byµand σ² respectively the expectation and the variance of X.

PROPOSITION 2. The following inequalities hold,

σ²+ (b−µ)

2a+b

3 −µ

≤ (b−a)⁴

36 kf⁰k_∞, (14)

and

σ²+ (a−µ)

2b+a

3 −µ

≤ (b−a)⁴

36 kf⁰k_∞. (15)

Both inequalities are sharp.

PROOF. Letg: [a, b]→R+be the function given byg(x) =Rx a

Ru

a f(t)dtdu.Then we have, that g⁰(a) = 0, g⁰(b) = 16= 0. Now, if we apply (11) for g,and taking into account that we have

g(a) = 0,

g(b) = Z b

a

(u−b)⁰ Z u

a

f(t)dtdu= Z b

a

(b−u)f(u)du=b−µ, and

Z b a

g(x)dx = Z b

a

(x−b)⁰ Z x

a

Z u a

f(t)dtdudx=− Z b

a

(x−b) Z x

a

f(t)dtdx

= −1 2

Z b a

(x−b)²0Z x a

f(t)dtdx= 1 2

Z b a

(x−b)²f(x)dx

= 1

2

σ²+ (b−µ)² ,

we directly get (14). An easy calculation yields that forf(x) = ^2(x−a)_(b−a)2, (14) holds as an equality. So inequality (14) is sharp.

Let now g(x) = Rx b

Ru

b f(t)dtdu, x ∈ [a, b]. Then, in a similar way as above we could state that

g(b) = g⁰(b) = 0, g⁰(a) =−1, g(a) =µ−a, Z b

a

g(x)dx = 1 2

σ²+ (µ−a)² .

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So, we can apply (12) to obtain (15),and is easy to verify that forf(x) = ^2(b−x)_(b−a)2 the equality in (15) holds.

Now, applying Corollary 1 first forg(x) = Rx a

Ru

a f(t)dtdu and then for g(x) = Rx

b

Ru

b f(t)dtdu,in similar way as above, we can prove the following proposition.

PROPOSITION 3. For allα, β∈R, we have the inequalities,

3 (α+β)

σ²+ (b−µ)²

−6β(b−µ) (b−a)−(2β−α) (b−a)²

≤ (|2α−β|+|2β−α|) (b−a)⁴

12 kf⁰k_∞, (16)

and

3 (α+β)

σ²+ (µ−a)²

−6α(µ−a) (b−a)−(2α−β) (b−a)²

≤ (|2α−β|+|2β−α|) (b−a)⁴

12 kf⁰k_∞. (17)

REMARK 2. Settingβ =−αin inequalities (16) and (17) we find that

µ−3b−a 2

≤(b−a)³

12 kf⁰k_∞, and

µ−3a−b 2

≤(b−a)³

12 kf⁰k_∞.

References

[1] N. S. Barnett and P. Cerone, S. S. Dragomir, Further inequalities for the expectation and variance of a random variable defined on a finite interval, Math. Inequal. and Appl.,6(2003), 23–36.

[2] A. I. Kechriniotis and N. D. Assimakis, Generalizations of the trapezoid inequalities based on a new mean value Theorem for the remainder in Taylor’s formula, J.

Inequal. Pure and Appl. Math.,7(3, Art. 90)(2006).

[3] D. S. Mitrinovic, J. E. Pecaric and A. M. Fink, Inequalities for Functions and Their Integrals and Derivatives, Kluwer Academic, Dordrecht, 1994.

[4] A. Rafiq, N. A. Mir and F. Zafar, A generalized Ostrowski type inequality for a random variable whose probability density function belongs toLp[a, b], Appl. Math.

E-Notes, 8(2008), 246–253.