INTRODUCTION In the paper [1], we have proved the following generalization of the classical Hermite- Hadamard inequality for convex functions, extended it to n nodes: Suppose that

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Volume 7, Issue 1, Article 24, 2006

APPLICATIONS OF THE EXTENDED HERMITE-HADAMARD INEQUALITY

Z. RETKES UNIVERSITY OFSZEGED

BOLYAIINSTITUTE

ARADI VÉRTANÚK TERE1 SZEGED, H-6720 HUNGARY

retkes@math.u-szeged.hu

Received 28 September, 2005; accepted 13 November, 2005 Communicated by P.S. Bullen

ABSTRACT. In this paper we give some combinatorial applications according to a new extension of the classical Hermite-Hadamard inequality proved in [1].

Key words and phrases: Convexity, Hermite-Hadamard inequality, Identities.

2000 Mathematics Subject Classification. 26A51.

1. INTRODUCTION

In the paper [1], we have proved the following generalization of the classical Hermite- Hadamard inequality for convex functions, extended it to n nodes: Suppose that−∞ ≤ a <

b ≤ ∞,f : (a, b)→Ris a strict convex function,xi ∈(a, b), i= 1, . . . , nsuch thatxi 6=xj if 1≤i < j ≤n. Then the following inequality holds:

n

X

i=1

F⁽ⁿ⁻¹⁾(x_i)

Π_i(x₁, . . . , x_n) < 1 n!

n

X

i=1

f(xi).

In the concave case<sign is changed to>. HereΠ_i(x₁, . . . , x_n) :=Qn

k=1 k6=i

(x_i−x_k)andF⁽⁰⁾(s), F⁽¹⁾(s), . . .,F⁽ⁿ⁻¹⁾(s), . . .is a sequence of functions defined recursively byF⁽⁰⁾(s) = f(s)and

d

dsF⁽ⁿ⁾(s) = F⁽ⁿ⁻¹⁾(s), n = 1,2, . . .. These sequences of functions of f are known as the iterated integrals of f. As an application of this main result we have proved the following identities:

(1.1)

n

X

k=1

xⁿ_k

Πk(x1, . . . , xn) =

n

X

k=1

x_k,

(1.2)

n

X

k=1

xⁿ⁻¹_k

Πk(x1, . . . , xn) = 1,

ISSN (electronic): 1443-5756 c

2006 Victoria University. All rights reserved.

292-05

(1.3)

n

X

k=1

1

x_k = (−1)ⁿ⁻¹

n

Y

k=1

x_k

n

X

k=1

1

x²_kΠ_k(x₁, . . . , x_n) (x_k 6= 0) and

(1.4)

n

Y

k=1

1

x_k = (−1)ⁿ⁻¹

n

X

k=1

1

x_kΠ_k(x₁, . . . , x_n) (x_k 6= 0).

In this paper we apply the formulae above to obtain closed combinatorial formulae and to inves- tigate the asymptotic behaviours of these sums. For the sake of the convenience of the reader we collect the transformational regulations for the quantityΠk(x1, . . . , xn)which plays a signifi- cant role in all of the formulae. Note that all of them are simple consequences of the definition ofΠ_k(x₁, . . . , x_n).

a) Ifα∈R, then

Π_k(α+x₁, . . . , α+x_n) = Π_k(x₁, . . . , x_n), that is,Π_kis shift invariant.

b)

Π_k(αx₁, . . . , αx_n) = αⁿ⁻¹Π_k(x₁, . . . , x_n).

c)

Π_k(α−x₁, . . . , α−x_n) = (−1)ⁿ⁻¹Π_k(x₁, . . . , x_n).

d) Ifx_k6= 0, k = 1, . . . , n, then Π_k

1

x₁, . . . , 1 x_n

= (−1)ⁿ⁻¹Π_k(x₁, . . . , x_n) xⁿ⁻²_k Πⁿ_j=1xj

.

2. ASYMPTOTICALFORMULAE

a) Let1 < x₁ < · · · < x_n be variables ands₁ = x₁, s₂ = x₁ +x₂, . . ., s_n = x₁+· · ·+x_n. Thens_i 6=s_j if1 ≤i < j ≤n, hence formula (1.1) can be applied tos₁, . . . , s_n, consequently we have

(2.1)

n

X

j=1

s_j =

n

X

j=1

sⁿ_j

Πj(s1, . . . , sn). For the left-hand side of (2.1)

n

X

j=1

s_j =x₁ + (x₁+x₂) +· · ·+ (x₁+· · ·+x_n) =

n

X

j=1

j·xn−j+1.

The definition ofΠ_j implies Π_j(s₁, . . . , s_n)

= (s_j −s₁)· · ·(s_j −sj−1)(s_j−s_j+1)· · ·(s_j−s_n)

= (x₂+· · ·+x_j)· · ·(xj−1+x_j)·x_j(−1)^n−jx_j+1(x_j+1+x_j+2)· · ·(x_j+1+· · ·+x_n).

Using the above expressions we have the following identities for all1< x₁ <· · ·< x_n: (2.2)

n

X

j=1

j ·xn−j+1 =

n

X

j=1

(−1)^n−j(x₁+· · ·+x_j)ⁿ

(x₂+· · ·+x_j)· · ·(xj−1+x_j)x_jx_j+1· · ·(x_j+1+· · ·+x_n).

Now, ifx_k→1, k = 1, . . . , nthen (2.2) gives n+ 1

2

=

n

X

j=1

j =

n

X

j=1

(−1)^n−jjⁿ (j −1)!(n−j)!. Multiplying both sides by(n−1)!leads us to

n

X

j=1

(−1)^n−jjⁿ

n−1 j−1

= (n+ 1)!

2 . b) Apply now formula (1.3) tox_k = 1 + ^k_n, k = 1, . . . , n. Hence

n

X

k=1

1

1 + ^k_n = (−1)ⁿ⁻¹

n

Y

k=1

1 + k

n ⁿ

X

k=1

1

(1 + ^k_n)²Π_k(1 + _n¹, . . . ,1 + ⁿ_n)

= (−1)ⁿ⁻¹ 1 nⁿ

n

Y

k=1

(n+k)

n

X

k=1

n²

(n+k)²Π_k(_n¹,_n², . . . ,ⁿ_n)

= (−1)ⁿ⁻¹ 1 nⁿ

n

Y

k=1

(n+k)

n

X

k=1

n²

(n+k)²_nn−1¹ Π_k(1, . . . , n)

= (−1)ⁿ⁻¹

n

Y

k=1

(n+k)

n

X

k=1

n

(n+k)²(k−1)!(−1)^n−k(n−k)!

=n· 2n

n ⁿ

X

k=1

(−1)^k−1 (n+k)²

n−1 k−1

.

From this equality chain we conclude that (2.3)

2n n

n

X

k=1

(−1)^k−1 (n+k)²

n−1 k−1

= 1 n

n

X

k=1

1 1 + ^k_n =

n

X

k=1

1 n+k. Since

n→∞lim 1 n

n

X

k=1

1 1 + ^k_n =

Z 1 0

1

1 +udu= log 2, hence the left-hand side of (2.3) is convergent and

n→∞lim 2n

n n

X

k=1

(−1)^k−1 (n+k)²

n−1 k−1

= log 2.

Pursuant to the Stirling formula ²ⁿ_n

∼ ^√4_πnⁿ (n → ∞), consequently we have the following asymptotic expression for the sum on the left:

n

X

k=1

(−1)^k−1 (n+k)²

n−1 k−1

∼√ πlog 2

√n

4ⁿ (n→ ∞).

c) This example follows along the lines above but choosing x_k = 1 + ^k_n2

(k = 1, . . . , n).

Without the detailed computation we have

(2.4) 1

n

n

X

k=1

1

1 + (^k_n)² = 2n Qn

k=1(n²+k²) (n!)²

n

X

k=1

(−1)^k−1k² (n²+k²)²·

n k

n+k n

.

From this equality the limit of the right-hand side exists and

n→∞lim 2n· Qn

k=1(n²+k²) (n!)²

n

X

k=1

(−1)^k−1k² (n²+k²)²·

n k

n+k n

= lim

n→∞

1 n

n

X

k=1

1 1 + (^k_n)²

= Z 1

0

1 1 +u²du

= arctan 1 = π 4. Combining these observations we have

n

X

k=1

(−1)^k−1k² (n²+k²)²

n k

n+k n

∼ π

8n· (n!)²

Πⁿ_k=1(n²+k²) (n→ ∞).

Moreover(n!)² ∼2πn(ⁿ_e)²ⁿand

n

Y

k=1

(n²+k²) = n²ⁿ

n

Y

k=1

"

1 + k

n 2#

. On the other hand

1 nlog

n

Y

k=1

"

1 + k

n 2#

= 1 n

n

X

k=1

log

"

1 + k

n 2#

→ Z 1

0

log(1 +x²)dx (n→ ∞), hence

n

Y

k=1

"

1 + k

n 2#

∼e^{nR01log(1+x2)dx}. Using integration by parts to evaluate the integral in the exponent gives

Z 1 0

log(1 +x²)dx= log 2 + π

2 −2 =α >0.

Putting together these results gives the following asymptotical expression:

n

X

k=1

(−1)^k−1k² (n² +k²)²

n k

n+k n

∼ π²

4 e^{−n(log 2+π2)} (n→ ∞).

3. FORMULAE FORPOWER SUMS

Letu₁, . . . , u_n ∈ Rsuch that u_i 6= 0,|u_i| < 1, u_i 6= u_j if1 ≤ i < j ≤ nandx_i = 1−u_i (i= 1, . . . , n). Then applying (1.3) gives

n

X

i=1

1

1−u_i = (−1)ⁿ⁻¹

n

Y

k=1

(1−u_k)

n

X

i=1

1

(1−u_i)²Π_i(1−u₁, . . . ,1−u_n)

=

n

Y

k=1

(1−u_k)

n

X

i=1

1

(1−u_i)²Π_i(u₁, . . . , u_n). Under the above conditions _1−u¹

i =P∞

`=0u<sup>`</sup>_i, hence (3.1)

n

X

i=1

1 1−u_i =

n

X

i=1

∞

X

`=0

u^`_i =

∞

X

`=0 n

X

i=1

u^`_i =

∞

X

`=0

P(`),

where we use the denotationP(`) :=u<sup>`</sup>₁+· · ·+u<sup>`</sup><sub>n</sub>for(` = 0,1,2, . . .), and thus we have the following identities foru₁, . . . , u_n:

(3.2)

∞

X

`=0

P(`) =

n

Y

k=1

(1−uk)

n

X

i=1

1

(1−u_i)²Π_i(u₁, . . . , u_n). Apply now (3.2) tou_k= _n+1^k (k= 1, . . . , n). For the left-hand side:

∞

X

`=0

P(`) =

∞

X

`=0 n

X

k=1

k n+ 1

`

=

∞

X

`=0

1 (n+ 1)^`

n

X

k=1

k^`=

∞

X

`=0

P˜(`) (n+ 1)<sup>`</sup>, whereP˜(`) = 1<sup>`</sup>+ 2^{`</sup>+· · ·+n<sup>`} (`= 0,1,2, . . .). On the other hand

n

Y

k=1

1− k n+ 1

= n!

(n+ 1)ⁿ and Π_i

1

n+ 1, . . . , n n+ 1

= (−1)ⁿ⁻ⁱ(i−1)!(n−i)!

(n+ 1)ⁿ⁻¹ . Putting these expressions into the right-hand side of (3.2) we obtain (3.3)

∞

X

`=0

P˜(`)

(n+ 1)^` =n(n+ 1)

n

X

j=1

(−1)^j−1 j²

n−1 j −1

. A simple rearrangemant of (3.3) and using ⁿ_j ⁿ⁻¹_j−1

= ⁿ_j gives

∞

X

`=0

P˜(`) (n+ 1)<sup>`+1</sup> =

n

X

j=1

(−1)^j−1 j

n j

, and since it is known that

n

X

j=1

(−1)^j−1 j

n j

= Z 1

0

1−(1−s)ⁿ

s ds=

n

X

k=1

1 k, consequently we have the following identity for all fixedn:

∞

X

`=0

P˜(`) (n+ 1)<sup>`+1</sup> =

n

X

k=1

1 k. Let us start now with (3.1) substituting u_k = _x¹

k, x_k > 1 (k = 1, . . . , n) and x_i 6= x_j if 1 ≤ i < j ≤ n. Using the same technique applied above and the transformational rule d) we get the following identity forx₁, . . . , x_n:

(3.4)

∞

X

`=0 n

X

k=1

1

x^`_k = (−1)ⁿ⁻¹

n

Y

k=1

(x_k−1)

n

X

i=1

xⁿ_i

(x_i−1)²Π_i(x₁, . . . , x_n).

Putting x_k = k + 1 (k = 1, . . . , n) in (3.4) and simplifying the right-hand side we have the following equality:

(3.5)

∞

X

`=0

1

2^{`</sup> +· · ·+ 1 (n+ 1)<sup>`}

=

n

X

i=1

(−1)ⁱ⁻¹

i (i+ 1)ⁿ n

i

.

Observe that ₂¹` +· · ·+ <sub>(n+1)</sub><sup>1</sup> ` = ζ^[n+1](`)−1, where ζ<sup>[n+1]</sup>(`)denotes the (n+ 1)th partial sum of the series ζ(`) = P∞

k=1 1

k<sup>`</sup>, if ` ≥ 2. Moreover let Hn denote the nth partial sum of

the harmonic series, that is H_n = Pn k=1

1

k. Using this quantity and separating the summands corresponding to`= 0and`= 1gives

(3.6) n+H_n+1−1 +

∞

X

`=2

[ζ^[n+1](`)−1] =

n

X

i=1

(−1)ⁱ⁻¹

i (i+ 1)ⁿ n

i

.

Rearranging (3.6) and taking the limit asn→ ∞yields that:

(3.7)

∞

X

`=2

[ζ(`)−1] = 1 + lim

n→∞

" _n X

1=1

(−1)ⁱ⁻¹

i (i+ 1)ⁿ n

i

−n−H_n+1

# .

This relation prompted us to investigate the quantities arising in both sides of the equality.

On one hand the sum of the series is equal to 1 as an easy computation shows below:

∞

X

`=2

[ζ(`)−1] =

∞

X

`=2

∞

X

k=2

1 k^` =

∞

X

k=2

1 k²

∞

X

j=0

1 k^j

=

∞

X

k=2

1 k²

1 1− ¹_k

=

∞

X

k=2

1

k(k−1) = 1.

Hence

n→∞lim

" _n X

i=1

(−1)ⁱ⁻¹

i (i+ 1)ⁿ n

i

−n−H_n+1

#

= 0.

In fact we prove more in the following lemma.

Lemma 3.1. For alln≥1the identity below holds

n

X

i=1

(−1)ⁱ⁻¹

i (i+ 1)ⁿ n

i

=n+H_n.

Proof. In the paper [1] we proved that ifx₁, . . . , x_n ∈ Rsuch thatx_i 6= x_j if1 ≤ i < j ≤ n then

n

X

i=1

x^j_i

Π_i(x_i, . . . , x_n) =









 Pn

k=1x_k if j =n

1 if j =n−1

0 if 0≤j ≤n−2.

Applying this result tox_i = ¹_i (i= 1, . . . , n)gives that

n

X

i=1

(−1)ⁱ⁻¹ i i^n−j

n i

=











H_n if j =n

1 if j =n−1 0 if 0≤j ≤n−2.

Moreover

n

X

i=1

(−1)ⁱ⁻¹

i (i+ 1)ⁿ n

i

=

n

X

i=1

(−1)ⁱ⁻¹ i

" _n X

k=0

i^n−k n

k #

n i

=

n

X

k=0

" _n X

i=1

(−1)ⁱ⁻¹ i i^n−k

n i

# n

k

= n

n−1

+H_n n

n

=n+H_n

as we stated.

REFERENCES

[1] Z. RETKES, An extension of the Hermite-Hadamard inequality (submitted)

[2] S.S. DRAGOMIR AND C.E.M. PEARCE, Selected Topics on Hermite-Hadamard Inequalities, RGMIA Monographs, Victoria University, 2000. [ONLINE: http://rgmia.vu.edu.au/

monographs/index.html]

[3] W. RUDIN, Real and Complex Analysis, McGraw-Hill Book Co., 1966.

[4] G.H. HARDY, J.E. LITTLEWOOD AND G. PÓLYA, Inequalities, Cambridge University Press, 1934.