We define a linear operator ∆mp :K →K, m∈N∗ ∆1p = ∆pan:=an+1−pan, (1) ∆m+1p an = ∆p(∆mp an

Mathematics and Informatics ICTAMI 2005 - Alba Iulia, Romania

ON SOME P-CONVEX SEQUENCES

T. incu Ioan

Abstract. The aim of this paper is to give some properties of thep-convex sequences.

LetKbe the set of all real sequences,K₊the set of all real positive sequence and p∈R\ {0}. We define a linear operator ∆^m_p :K →K, m∈N^∗

∆¹_p = ∆_pa_n:=a_n+1−pa_n, (1)

∆^m+1_p a_n = ∆_p(∆^m_p a_n) , for every n ∈N.

Definition 1. A sequence (a_n)n∈N from K is said to be p-convex of order m∈N^∗ if and only if

∆^m_p a_n≥0 , for all n∈N. (2) We denote by K_m^p the set of all real sequences p-convex.

Proposition 1. For n∈N, m∈N^∗ and p∈R\ {0} the equalities

∆^m+1_p a_n= ∆^m_p a_n+1−p∆^m_p a_n (3)

∆^m_p a_n =

m

X

k=0

(−1)^m−k m

k

p^m−ka_n+k (4)

holds.

Remark 1. For p= 1, we obtain ∆^m₁ a_n = ∆^ma_n, where ∆^ma_n represents the m-th order difference of the sequence (an)n∈N.

Example 1. For p 6= 1, a sequences (a_n)n∈N, where a_n =pⁿ, is p-convex the m-th order, but is not a 1-convex the m-th order for every m ∈N^∗.

Example 2. Let (a_n)_n∈N from K, a_n =n^rpⁿ, r ∈ {0,1, ..., m−1} where mZ,m ≥2, ∆^m_p a_n= 0.

We have A={±pⁿ,±npⁿ,±n²pⁿ, ...,±n^m−1pⁿ} ⊂K_m^p.

Proposition 2. We have

∆^m+r_p a_n= ∆^m_p (∆^r_pa_n) = ∆^r_p(∆^m_p a_n), (5) for every m, r ∈N^∗, n∈N, p∈R.

Theorem 1. For n, m∈N, the equality a_n+m =

m

X

k=0

m k

p^m−k∆^k_pa_n, (6) is verified.

Proof:

a_n+m =

m

X

k=0

m k

p^m−k

" _k X

i=0

(−1)^k−i k

i

p^k−ia_n+i

#

=

=

m

X

k=0

b(m, k)

k

X

i=0

c(k, i)a_n+i where

b(m, k) = m

k

p^m−k , c(k, i) = (−1)^k−i k

i

p^k−i a_n+m =

m

X

k=0

a_n+k

m

X

i=k

b(m, i)c(i, k) =

=

m

X

k=0

a_n+k

m−k

X

i=0

b(m, i+k)c(i+k, k) =

=

m

X

k=0

a_n+kp^m−k

m−k

X

i=0

(−1)ⁱ m

k

m−k i

=a_n+m. Theorem 2. We have

∆^m_p a_nb_n=

m

X

k=0

m k

∆^k₁a_n∆^m−k_p b_n+k, (7) for all n ∈N and m∈N.

Proof: We proceed by mathematical introduction.

Remark 2. For p= 1, we obtain the results of T. Popoviciu (see [3]).

We consider a linear operator T :K_m^p →K defined by T(a;n)=

n

X

k=0

ρ_k(n)a_s+k, wheres ∈Nis arbitrary, ρ_k(n)∈R, n ∈N, k = 0, n.

(8) Next, we will give some necesary and sufficienty condition for a matrix ρ=kρ_k(n)k^n∈N

k=0,n

to verify

T(K_m^p)⊆K₊. (9)

Lemma 1. Let m, s∈N, m≥2, ρ=kρ_k(n)k^n∈N

k=0,n. If

n

X

k=0

p^kρ_k(n) = 0 ,

n

X

k=0

p^kρ_k(n)kⁱ = 0 , i = 1,2, ..., m−1 (10) then

T(a;n) =

n−m

X

k=0

q_k(n)∆^m_p a_s+k (11)

where q_k(n) = (−1)^m (m−1)!

k

X

i=0

ρ_i(n)p^i−k−m(k−i+ 1)m−1, (x)_l = x(x+ 1)...(x+ l−1).

Proof: We proceed by mathematical induction.

Theorem 3. Let T :K →K, T(a;n) =

n

X

k=0

ρ_k(n)a_s+k, s ∈N. T(K_m^p)⊆ K₊ if and only if

i)

n

X

k=0

p^kρ_k(n) = 0,

n

X

k=0

p^kρ_k(n)kⁱ = 0, for i= 1,2, ..., m−1

ii)

k

X

i=0

ρ_i(n)(i−(k+ 1))(i−(k+ 2))...(i−(k+m−1))p^i−k−m ≤0, k = 0, n−m.

Proof: Necessity: We considerT(K_m^p)⊆K₊. In example 2 it is shown that A={±pⁿ,±npⁿ, ...,±n^m−1pⁿ} ⊆K_m^p.

For (a_n)n∈N from A we obtain condition i) and ii).

Sufficienty: From lemma ?? we have T(a;n) =

n−m

X

k=0

qk(n)∆^m_p as+k,

where q_k=− 1 (m−1)!

k

X

i=0

ρ_i(n)(i−(k+ 1))...(i−(k+m−1))p^i−k−m.

From condition qk ≥ 0, k = 0,1, ..., n−m and (an)n∈N ∈ K_m^p it follows T(K_m^p)⊆K₊.

For p= 1, we obtain the results of T. Popoviciu (see [4]).

Next, let (A_n(a))_n∈Nbe the sequence of the means, that isA_n(a) = 1 n+ 1

n

X

k=0

a_k, n∈N.

Theorem 4. Let p < 1. If An(a) ∈ K_m−1^p , m ≥ 1, then operator An : K_m^p →K, A_n(a) = 1

n+ 1

n

X

k=0

a_k verify

A_n(K_m^p)⊆K_m^p. (12)

Proof: a_k= (k+ 1)A_k−kAk−1, k = 0,1, .... Then

∆^m_p a_k = ∆^m_p (k+ 1)A_k−∆^m_p kAk−1. From (??) we have

∆^m_p a_k =

m

X

i=0

m i

∆ⁱ₁(k+ 1)(∆^m−i_p A_k+i)−

m

X

i=0

m i

∆ⁱ₁k(∆^m−i_p Ak+i−1) =

= (k+ 1)∆^m_p A_k+m∆^m−1_p A_k+1−k∆^m_p A_k−1−m∆^m−1_p A_k

∆^m_p A_k = ∆^m−1_p (∆¹_pA_k) = ∆^m−1_p (A_k+1−pA_k) = ∆^m−1_p A_k+1−p∆^m−1_p A_k

∆^m−1_p Ak+1 = ∆^m_p Ak+p∆^m−1_p Ak.

After that

∆^m_p a_k= (m+k+ 1)∆^m_p A_k−k∆^m_p Ak−1+m(p−1)∆^m−1_p A_k. (13) (m+k)!

k! ∆^m_p a_k = (m+k+ 1)!

k! ∆^m_p A_k− (m+k)!

(k−1)! ∆^m_p Ak−1+ +m(p−1)(m+k)!

k! ∆^m−1_p A_k.

n

X

k=1

(m+k)!

k! ∆^m_p ak = (m+n+ 1)!

n! ∆^m_p An−(m+ 1)!∆^m_p A0+ +m(p−1)

n

X

k=1

(m+k)!

k! ∆^m−1_p A_k , n≥1

∆^m_p = n!

(m+n+ 1)!

" _n X

k=1

(m+k)!

k! ∆^m_p a_k+ +(m+ 1)!∆^m_p a₀+m(1−p)

n

X

k=1

(m+k)!

k! ∆^m−1_p A_k

# .

(14)

In (??) let k = 0, we obtain

∆^m_p a₀ = (m+ 1)∆^m_p A₀+m(p−1)∆^m−1_p A₀

∆^m_p A₀ = 1 m+ 1

∆^m_p a₀+m(1−p)∆^m−1_p A₀

. (15)

From (??) and (??) we obtain (??).

Remark 3. If we considerp= 1 in (??), we obtain the result of A. Lupa¸s (see [2]).

We consider following question: What are the conditions for a sequences (a_n)n∈N fromK to verify

A≤∆^m_p a_n≤B , for all n∈N , m∈N^∗, arbitrary (16) where A and B are constants that are not dependent by m and n.

Theorem 5. The sequence(a_n)_n∈N satisfied the condition (??) if and only if exists the real sequence (b_n)n∈N which verify

A ≤b_n≤B , for every n ≥m (17)

and a_n=

n

X

k=0

b_k(n−k+ 1)m−1

(m−1)! p^n−k , where (x)_l =x(x+ 1)...(x+l−1). (18)

Proof: Sufficienty:

∆^m_p a_n =

m

X

k=0

(−1)^m−k m

k

p^m−ka_n+k =

=

m

X

k=0

(−1)^m−k m

k

p^m−k

n+k

X

i=0

b_i(n+k−i+ 1)m−1

(m−1)! p^n+k−i =

=

m

X

k=0

c_m,k

n+k

X

i=0

d_n,k(m, i)b_i where

c_m,k = (−1)^m−k m

k

p^m−k, d_n,k(m, i) = (n+k−i+ 1)m−1

(m−1)! p^n+k−i

∆^m_p a_n =

n

X

j=0

b_j

m

X

r=0

c_m,rd_n,r(m, j) +

m

X

j=1

b_n+j

m

X

r=j

c_m,rd_n,r(m, n+j) =

=

n

X

j=0

b_jS_m,r(n, j) +

m

X

j=1

b_n+jS_m,r⁰ (n, j) with

S_m,r(n, j) =

m

X

r=0

c_m,rd_n,r(m, j) S_m,r⁰ (n, j) =

p

X

r=j

c_m,rd_n,r(m, n+j)

S_m,r(n, j) =

m

X

r=0

(−1)^m−r m

r

p^m−r(n+r−j+ 1)m−1

(m−1)! p^n+r−j =

= p^m+n−j (m−1)!

m

X

r=0

(−1)^m−r m

r

Γ(n+r+m−j)

Γ(n+r−j+ 1) · Γ(n−j+m) Γ(n−j+m) =

= p^m+n−j (m−1)!

m

X

r=0

(−1)^m−r m

r

(n−j+m)_r(n−j +r+ 1) (n−j +r+ 2)...(n−j+m−1) =

= p^m+n−j (m−1)!

m

X

r=0

m r

(j −n−r−1)(j−n−r−2)...

...(j−n−m+ 1)(n−j +m)_r =

= −p^m+n+j (m−1)!

m

X

r=0

m r

(n−j+m)_r(j−n−m)_m−r j−n−m =

=− p^m+n−j

(m−1)! · (0)_m

j−n−m = 0.

In the previously calculations we have use the Chy-Vandermonde formula (x+y)_m =

m

X

k=0

m k

)(x)_k(y)_m−k

S_m,r⁰ (n, j) =

m

X

r=j

(−1)^m−r m

r

p^m−r(r−j + 1)m−1

(m−1)! p^r−j =

= p^m−j (m−1)!

m−j

X

r=0

(−1)^m−r−j m

r+j

p^m−r−j(r+ 1)m−1

S_m,r⁰ = p^m−j (m−1)!

m−j

X

r=0

(−1)^m−j−r

m−j r

^m

r+j

m−j r

(r+ 1)m−1 =

= p^m−j (m−1)!

m−j

X

r=0

(−1)^m−j−r

m−j r

m!

(m−j)!· r!

(r+j)!(r+ 1)_m−1 =

= p^m−j

(m−1)! · m!

(m−j)!

m−j

X

r=0

(−1)^m−j−r

m−j r

Γ(m)

Γ(r+j+ 1) · Γ(m+r) Γ(m) =

= mp^m−j (m−j)!

m−j

X

r=0

(−1)^m−j−r

m−j r

(r+j+ 1)(r+j + 2)...

...(m−1)(m)r=

= p^m−j (m−j)!

m−j

X

r=0

m−

r

(−m)m−j−r(m)_r =

= p^m−j

(m−j)!(0)m−j. For j ≤m−1 we obtain

S_m,r⁰ (n, j) = 0.

For j =m, we have

S_m,r⁰ =c_m,nd_n,m(m, n+m) = (1)m−1

(m−1)! = 1.

Results ∆^p_man =bn+m.

Necessity: For all real sequence we may consider the sequence (b_n) which define by

b0 =a0, b_n=a_n−

n−1

X

k=0

b_k(n−k+ 1)m−1

(m−1)! p^n−k , n= 1,2, ...

Because ∆^m_p a_n =b_n+m we obtain (??).

Example 3. Let (a_n)_n∈N, (b_n)_n∈N be the sequences defined by b_n=pⁿ , p∈(0,1),

a_n=

n

X

k=0

b_k(n−k+ 1)m−1

(m−1)! p^n−k = pⁿ

n!(m+ 1)_n. From theorem ?? we obtain

0≤∆^m_p a_n ≤1 , for all n∈N , m∈N^∗ and p∈(0,1).

We observe that then sequence (a_n)n∈N defined by a_n= pⁿ

n!(m+ 1)_n is p- convex with the order m.

References

[1] A. Lupa¸s, On convexity matrix transformations, Univ. Beograd Publ.

Elektrotehn. Fak. Ser. Mat. Fiz., No. 634-377 (1979), 189-191.

[2] A. Lupa¸s,On the means of convex sequences, Gazeta Matematicˇa, Seria (A), Anul IV, Nr. 1-2 (1983), 90-93.

[3] A. Lupa¸s, C. Manole, Capitole de Analiz’a numeric’a, Ed. Univ. ”Lu- cian Blaga” Sibiu 1994.

[4] T. Popoviciu, Les founctions convexes, Actualit´es Sci. Ind. Nr. 992, Paris 1945.

[5] Gh. Toader, A measure of convexity of sequences, Revue d ⁰analyse num´erique et de th´eorie de l ⁰approximation, Tome 22, N^◦ 1, 1993.

Ioan T. incu

”Lucian Blaga” University of Sibiu Romania