Directory UMM :Journals:Journal_of_mathematics:OTHER:
ACTA UNIVERSITATIS APULENSIS
No 11/2006
Proceedings of the International Conference on Theory and Application of
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 the p-convex
sequences.
Let K be 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
∆1p = ∆p an := an+1 − pan ,
∆m+1
an = ∆p (∆m
p
p an ) ,
(1)
for every n ∈ N.
Definition 1. A sequence (an )n∈N from K is said to be p-convex of order
m ∈ N∗ if and only if
∆m
p an ≥ 0 , for all n ∈ N.
(2)
p
We denote by Km
the set of all real sequences p-convex.
Proposition 1. For n ∈ N, m ∈ N∗ and p ∈ R \ {0} the equalities
holds.
m
∆m+1
an = ∆m
p
p an+1 − p∆p an
m
X
m
m−k m
∆p an =
(−1)
pm−k an+k
k
k=0
(3)
(4)
m
m
Remark 1. For p = 1, we obtain ∆m
1 an = ∆ an , where ∆ an represents
the m-th order difference of the sequence (an )n∈N .
Example 1. For p 6= 1, a sequences (an )n∈N , where an = 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 (an )n∈N from K, an = nr pn , r ∈ {0, 1, ..., m − 1} where
mZ, m ≥ 2, ∆m
p an = 0.
p
We have A = {±pn , ±npn , ±n2 pn , ..., ±nm−1 pn } ⊂ Km
.
249
I. T.incu - On some p-convex sequences
Proposition 2. We have
r
r
m
∆pm+r an = ∆m
p (∆p an ) = ∆p (∆p an ),
(5)
for every m, r ∈ N∗ , n ∈ N, p ∈ R.
Theorem 1. For n, m ∈ N, the equality
m
X
m m−k k
an+m =
p
∆p an ,
k
k=0
(6)
is verified.
Proof:
an+m =
m
X
m
k
k=0
=
m
X
pm−k
b(m, k)
k
X
#
k
pk−i an+i =
(−1)k−i
i
i=0
k
X
c(k, i)an+i
i=0
k=0
where
"
m m−k
b(m, k) =
p
,
k
an+m =
m
X
an+k
k=0
=
=
m
X
k=0
m
X
an+k
m
X
i=k
m−k
X
k−i
c(k, i) = (−1)
k k−i
p
i
b(m, i)c(i, k) =
b(m, i + k)c(i + k, k) =
i=0
an+k p
m−k
k=0
Theorem 2. We have
∆m
p an b n
=
m−k
X
m m−k
(−1)
= an+m .
k
i
i=0
i
m
X
m
k=0
k
∆k1 an ∆pm−k bn+k ,
for all n ∈ N and m ∈ N.
Proof: We proceed by mathematical introduction.
250
(7)
I. T.incu - On some p-convex sequences
Remark 2. For p = 1, we obtain the results of T. Popoviciu (see [3]).
p
We consider a linear operator T : Km
→ K defined by
n
X
T (a; n)=
ρk (n)as+k , where s ∈ N is arbitrary, ρk (n) ∈ R, n ∈ N, k = 0, n.
k=0
(8)
Next, we will give some necesary and sufficienty condition for a matrix
ρ=kρk (n)k n∈N to verify
k=0,n
p
T (Km
) ⊆ K+ .
(9)
Lemma 1. Let m, s ∈ N, m ≥ 2, ρ = kρk (n)k n∈N . If
k=0,n
n
X
k
p ρk (n) = 0 ,
k=0
n
X
pk ρk (n)k i = 0 , i = 1, 2, ..., m − 1
(10)
k=0
then
T (a; n) =
n−m
X
qk (n)∆m
p as+k
(11)
k=0
k
where qk (n) =
l − 1).
(−1)m X
ρi (n)pi−k−m (k − i + 1)m−1 , (x)l = x(x + 1)...(x +
(m − 1)! i=0
Proof: We proceed by mathematical induction.
Theorem 3. Let T : K → K, T (a; n) =
n
X
pk ρk (n) = 0,
k=0
ii)
k
X
p
ρk (n)as+k , s ∈ N. T (Km
)⊆
k=0
K+ if and only if
i)
n
X
n
X
pk ρk (n)k i = 0, for i = 1, 2, ..., m − 1
k=0
ρi (n)(i − (k + 1))(i − (k + 2))...(i − (k + m − 1))pi−k−m ≤ 0,
i=0
k = 0, n − m.
251
I. T.incu - On some p-convex sequences
p
Proof: Necessity: We consider T (Km
) ⊆ K+ . In example 2 it is shown that
p
A = {±pn , ±npn , ..., ±nm−1 pn } ⊆ Km
.
For (an )n∈N from A we obtain condition i) and ii).
Sufficienty: From lemma ?? we have
T (a; n) =
n−m
X
qk (n)∆m
p as+k ,
k=0
k
X
1
where qk = −
ρi (n)(i − (k + 1))...(i − (k + m − 1))pi−k−m .
(m − 1)! i=0
p
From condition qk ≥ 0, k = 0, 1, ..., n − m and (an )n∈N ∈ Km
it follows
p
T (Km ) ⊆ K+ .
For p = 1, we obtain the results of T. Popoviciu (see [4]).
n
1 X
Next, let (An (a))n∈N be the sequence of the means, that is An (a) =
ak ,
n + 1 k=0
n ∈ N.
p
Theorem 4. Let p < 1. If An (a) ∈ Km−1
, m ≥ 1, then operator An :
n
X
1
p
ak verify
Km
→ K, An (a) =
n + 1 k=0
p
p
An (Km
) ⊆ Km
.
(12)
Proof: ak = (k + 1)Ak − kAk−1 , k = 0, 1, .... Then
m
m
∆m
p ak = ∆p (k + 1)Ak − ∆p kAk−1 .
From (??) we have
m
m
X
X
m
m
i
m−i
m
∆p ak =
∆1 (k + 1)(∆p Ak+i ) −
∆i1 k(∆pm−i Ak+i−1 ) =
i
i
i=0
i=0
m−1
m−1
= (k + 1)∆m
Ak+1 − k∆m
Ak
p Ak + m∆p
p Ak−1 − m∆p
m−1
∆m
(∆1p Ak ) = ∆pm−1 (Ak+1 − pAk ) = ∆pm−1 Ak+1 − p∆pm−1 Ak
p Ak = ∆p
m−1
∆pm−1 Ak+1 = ∆m
Ak .
p Ak + p∆p
252
I. T.incu - On some p-convex sequences
After that
m
m
m−1
∆m
Ak .
p ak = (m + k + 1)∆p Ak − k∆p Ak−1 + m(p − 1)∆p
(13)
(m + k)! m
(m + k + 1)! m
(m + k)! m
∆p ak =
∆p Ak −
∆ Ak−1 +
k!
k!
(k − 1)! p
(m + k)! m−1
+m(p − 1)
∆p Ak .
k!
n
X
(m + n + 1)! m
(m + k)! m
∆p ak =
∆p An − (m + 1)!∆m
p A0 +
k!
n!
k=1
n
X
(m + k)! m−1
∆p Ak , n ≥ 1
+m(p − 1)
k!
k=1
" n
X (m + k)!
n!
∆m
=
∆m
p
p ak +
(m + n + 1)! k=1
k!
#
(14)
n
X
(m
+
k)!
∆pm−1 Ak .
+(m + 1)!∆m
p a0 + m(1 − p)
k!
k=1
In (??) let k = 0, we obtain
m
m−1
∆m
A0
p a0 = (m + 1)∆p A0 + m(p − 1)∆p
1 m
∆p a0 + m(1 − p)∆pm−1 A0 .
m+1
From (??) and (??) we obtain (??).
∆m
p A0 =
(15)
Remark 3. If we consider p = 1 in (??), we obtain the result of A. Lupa¸s
(see [2]).
We consider following question: What are the conditions for a sequences
(an )n∈N from K to verify
A ≤ ∆m
p an ≤ B ,
for all n ∈ N , m ∈ N∗ , arbitrary
where A and B are constants that are not dependent by m and n.
253
(16)
I. T.incu - On some p-convex sequences
Theorem 5. The sequence (an )n∈N satisfied the condition (??) if and only
if exists the real sequence (bn )n∈N which verify
A ≤ bn ≤ B ,
for every n ≥ m
(17)
and
an =
n
X
bk
k=0
(n − k + 1)m−1 n−k
p
, where (x)l = x(x + 1)...(x + l − 1). (18)
(m − 1)!
Proof: Sufficienty:
∆m
p an
m
X
m m−k
=
p
an+k =
(−1)
k
k=0
m
n+k
X
X
(n + k − i + 1)m−1 n+k−i
m−k m
m−k
=
(−1)
p
bi
p
=
k
(m
−
1)!
i=0
k=0
=
m
X
m−k
cm,k
dn,k (m, i)bi
i=0
k=0
where
n+k
X
m m−k
cm,k = (−1)
p
,
k
(n + k − i + 1)m−1 n+k−i
p
dn,k (m, i) =
(m − 1)!
m−k
∆m
p an
=
=
n
X
j=0
n
X
j=0
with
bj
m
X
cm,r dn,r (m, j) +
r=0
bj Sm,r (n, j) +
m
X
j=1
m
X
bn+j
m
X
cm,r dn,r (m, n + j) =
r=j
′
bn+j Sm,r
(n, j)
j=1
Sm,r (n, j) =
m
X
cm,r dn,r (m, j)
r=0
p
′
Sm,r
(n, j) =
X
cm,r dn,r (m, n + j)
r=j
254
I. T.incu - On some p-convex sequences
m
X
m m−r (n + r − j + 1)m−1 n+r−j
Sm,r (n, j) =
(−1)
p
p
=
r
(m − 1)!
r=0
m
pm+n−j X
m−r m Γ(n + r + m − j) Γ(n − j + m)
=
·
=
(−1)
(m − 1)! r=0
r Γ(n + r − j + 1) Γ(n − j + m)
m
pm+n−j X
m−r m
=
(−1)
(n − j + m)r (n − j + r + 1)
(m − 1)! r=0
r
(n − j + r + 2)...(n − j + m − 1) =
m
pm+n−j X m
(j − n − r − 1)(j − n − r − 2)...
=
(m − 1)! r=0 r
...(j − n − m + 1)(n − j + m)r =
m
(j − n − m)m−r
−pm+n+j X m
=
(n − j + m)r
=
(m − 1)! r=0 r
j−n−m
=−
m−r
(0)m
pm+n−j
·
= 0.
(m − 1)! j − n − m
In the previously calculations we have use the Chy-Vandermonde formula
(x + y)m =
m
X
m
k=0
′
Sm,r
(n, j)
m
X
k
)(x)k (y)m−k
m m−r (r − j + 1)m−1 r−j
p
=
=
(−1)
p
(m
−
1)!
r
r=j
m−j
m
pm−j X
m−r−j
(−1)
pm−r−j (r + 1)m−1
=
r+j
(m − 1)! r=0
m−r
255
I. T.incu - On some p-convex sequences
′
Sm,r
m
m−j
pm−j X
m
−
j
r+j
(−1)m−j−r
=
m−j (r + 1)m−1 =
(m − 1)! r=0
r
r
m−j
pm−j X
m−j
m!
r!
=
·
(r + 1)m−1 =
(−1)m−j−r
(m − 1)! r=0
(m − j)! (r + j)!
r
m−j
X
m!
Γ(m + r)
Γ(m)
pm−j
m−j−r m − j
·
·
=
(−1)
=
(m − 1)! (m − j)! r=0
Γ(r + j + 1)
Γ(m)
r
m−j
mpm−j X
m−j−r m − j
(−1)
(r + j + 1)(r + j + 2)...
=
r
(m − j)! r=0
...(m − 1)(m)r =
m−j
pm−j X m−
=
(−m)m−j−r (m)r =
(m − j)! r=0
r
=
pm−j
(0)m−j .
(m − j)!
For j ≤ m − 1 we obtain
′
Sm,r
(n, j) = 0.
For j = m, we have
′
Sm,r
= cm,n dn,m (m, n + m) =
(1)m−1
= 1.
(m − 1)!
Results ∆pm an = bn+m .
Necessity: For all real sequence we may consider the sequence (bn ) which
define by
b 0 = a0 ,
b n = an −
n−1
X
k=0
bk
(n − k + 1)m−1 n−k
p
,
(m − 1)!
Because ∆m
p an = bn+m we obtain (??).
256
n = 1, 2, ...
I. T.incu - On some p-convex sequences
Example 3. Let (an )n∈N , (bn )n∈N be the sequences defined by
bn = pn ,
an =
n
X
k=0
p ∈ (0, 1),
bk
(n − k + 1)m−1 n−k pn
p
= (m + 1)n .
(m − 1)!
n!
From theorem ?? we obtain
0 ≤ ∆m
p an ≤ 1 ,
for all n ∈ N , m ∈ N∗ and p ∈ (0, 1).
pn
We observe that then sequence (an )n∈N defined by an = (m + 1)n is pn!
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. ”Lucian 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 ′ analyse
num´erique et de th´eorie de l ′ approximation, Tome 22, N◦ 1, 1993.
Ioan T.incu
”Lucian Blaga” University of Sibiu
Romania
257
No 11/2006
Proceedings of the International Conference on Theory and Application of
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 the p-convex
sequences.
Let K be 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
∆1p = ∆p an := an+1 − pan ,
∆m+1
an = ∆p (∆m
p
p an ) ,
(1)
for every n ∈ N.
Definition 1. A sequence (an )n∈N from K is said to be p-convex of order
m ∈ N∗ if and only if
∆m
p an ≥ 0 , for all n ∈ N.
(2)
p
We denote by Km
the set of all real sequences p-convex.
Proposition 1. For n ∈ N, m ∈ N∗ and p ∈ R \ {0} the equalities
holds.
m
∆m+1
an = ∆m
p
p an+1 − p∆p an
m
X
m
m−k m
∆p an =
(−1)
pm−k an+k
k
k=0
(3)
(4)
m
m
Remark 1. For p = 1, we obtain ∆m
1 an = ∆ an , where ∆ an represents
the m-th order difference of the sequence (an )n∈N .
Example 1. For p 6= 1, a sequences (an )n∈N , where an = 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 (an )n∈N from K, an = nr pn , r ∈ {0, 1, ..., m − 1} where
mZ, m ≥ 2, ∆m
p an = 0.
p
We have A = {±pn , ±npn , ±n2 pn , ..., ±nm−1 pn } ⊂ Km
.
249
I. T.incu - On some p-convex sequences
Proposition 2. We have
r
r
m
∆pm+r an = ∆m
p (∆p an ) = ∆p (∆p an ),
(5)
for every m, r ∈ N∗ , n ∈ N, p ∈ R.
Theorem 1. For n, m ∈ N, the equality
m
X
m m−k k
an+m =
p
∆p an ,
k
k=0
(6)
is verified.
Proof:
an+m =
m
X
m
k
k=0
=
m
X
pm−k
b(m, k)
k
X
#
k
pk−i an+i =
(−1)k−i
i
i=0
k
X
c(k, i)an+i
i=0
k=0
where
"
m m−k
b(m, k) =
p
,
k
an+m =
m
X
an+k
k=0
=
=
m
X
k=0
m
X
an+k
m
X
i=k
m−k
X
k−i
c(k, i) = (−1)
k k−i
p
i
b(m, i)c(i, k) =
b(m, i + k)c(i + k, k) =
i=0
an+k p
m−k
k=0
Theorem 2. We have
∆m
p an b n
=
m−k
X
m m−k
(−1)
= an+m .
k
i
i=0
i
m
X
m
k=0
k
∆k1 an ∆pm−k bn+k ,
for all n ∈ N and m ∈ N.
Proof: We proceed by mathematical introduction.
250
(7)
I. T.incu - On some p-convex sequences
Remark 2. For p = 1, we obtain the results of T. Popoviciu (see [3]).
p
We consider a linear operator T : Km
→ K defined by
n
X
T (a; n)=
ρk (n)as+k , where s ∈ N is arbitrary, ρk (n) ∈ R, n ∈ N, k = 0, n.
k=0
(8)
Next, we will give some necesary and sufficienty condition for a matrix
ρ=kρk (n)k n∈N to verify
k=0,n
p
T (Km
) ⊆ K+ .
(9)
Lemma 1. Let m, s ∈ N, m ≥ 2, ρ = kρk (n)k n∈N . If
k=0,n
n
X
k
p ρk (n) = 0 ,
k=0
n
X
pk ρk (n)k i = 0 , i = 1, 2, ..., m − 1
(10)
k=0
then
T (a; n) =
n−m
X
qk (n)∆m
p as+k
(11)
k=0
k
where qk (n) =
l − 1).
(−1)m X
ρi (n)pi−k−m (k − i + 1)m−1 , (x)l = x(x + 1)...(x +
(m − 1)! i=0
Proof: We proceed by mathematical induction.
Theorem 3. Let T : K → K, T (a; n) =
n
X
pk ρk (n) = 0,
k=0
ii)
k
X
p
ρk (n)as+k , s ∈ N. T (Km
)⊆
k=0
K+ if and only if
i)
n
X
n
X
pk ρk (n)k i = 0, for i = 1, 2, ..., m − 1
k=0
ρi (n)(i − (k + 1))(i − (k + 2))...(i − (k + m − 1))pi−k−m ≤ 0,
i=0
k = 0, n − m.
251
I. T.incu - On some p-convex sequences
p
Proof: Necessity: We consider T (Km
) ⊆ K+ . In example 2 it is shown that
p
A = {±pn , ±npn , ..., ±nm−1 pn } ⊆ Km
.
For (an )n∈N from A we obtain condition i) and ii).
Sufficienty: From lemma ?? we have
T (a; n) =
n−m
X
qk (n)∆m
p as+k ,
k=0
k
X
1
where qk = −
ρi (n)(i − (k + 1))...(i − (k + m − 1))pi−k−m .
(m − 1)! i=0
p
From condition qk ≥ 0, k = 0, 1, ..., n − m and (an )n∈N ∈ Km
it follows
p
T (Km ) ⊆ K+ .
For p = 1, we obtain the results of T. Popoviciu (see [4]).
n
1 X
Next, let (An (a))n∈N be the sequence of the means, that is An (a) =
ak ,
n + 1 k=0
n ∈ N.
p
Theorem 4. Let p < 1. If An (a) ∈ Km−1
, m ≥ 1, then operator An :
n
X
1
p
ak verify
Km
→ K, An (a) =
n + 1 k=0
p
p
An (Km
) ⊆ Km
.
(12)
Proof: ak = (k + 1)Ak − kAk−1 , k = 0, 1, .... Then
m
m
∆m
p ak = ∆p (k + 1)Ak − ∆p kAk−1 .
From (??) we have
m
m
X
X
m
m
i
m−i
m
∆p ak =
∆1 (k + 1)(∆p Ak+i ) −
∆i1 k(∆pm−i Ak+i−1 ) =
i
i
i=0
i=0
m−1
m−1
= (k + 1)∆m
Ak+1 − k∆m
Ak
p Ak + m∆p
p Ak−1 − m∆p
m−1
∆m
(∆1p Ak ) = ∆pm−1 (Ak+1 − pAk ) = ∆pm−1 Ak+1 − p∆pm−1 Ak
p Ak = ∆p
m−1
∆pm−1 Ak+1 = ∆m
Ak .
p Ak + p∆p
252
I. T.incu - On some p-convex sequences
After that
m
m
m−1
∆m
Ak .
p ak = (m + k + 1)∆p Ak − k∆p Ak−1 + m(p − 1)∆p
(13)
(m + k)! m
(m + k + 1)! m
(m + k)! m
∆p ak =
∆p Ak −
∆ Ak−1 +
k!
k!
(k − 1)! p
(m + k)! m−1
+m(p − 1)
∆p Ak .
k!
n
X
(m + n + 1)! m
(m + k)! m
∆p ak =
∆p An − (m + 1)!∆m
p A0 +
k!
n!
k=1
n
X
(m + k)! m−1
∆p Ak , n ≥ 1
+m(p − 1)
k!
k=1
" n
X (m + k)!
n!
∆m
=
∆m
p
p ak +
(m + n + 1)! k=1
k!
#
(14)
n
X
(m
+
k)!
∆pm−1 Ak .
+(m + 1)!∆m
p a0 + m(1 − p)
k!
k=1
In (??) let k = 0, we obtain
m
m−1
∆m
A0
p a0 = (m + 1)∆p A0 + m(p − 1)∆p
1 m
∆p a0 + m(1 − p)∆pm−1 A0 .
m+1
From (??) and (??) we obtain (??).
∆m
p A0 =
(15)
Remark 3. If we consider p = 1 in (??), we obtain the result of A. Lupa¸s
(see [2]).
We consider following question: What are the conditions for a sequences
(an )n∈N from K to verify
A ≤ ∆m
p an ≤ B ,
for all n ∈ N , m ∈ N∗ , arbitrary
where A and B are constants that are not dependent by m and n.
253
(16)
I. T.incu - On some p-convex sequences
Theorem 5. The sequence (an )n∈N satisfied the condition (??) if and only
if exists the real sequence (bn )n∈N which verify
A ≤ bn ≤ B ,
for every n ≥ m
(17)
and
an =
n
X
bk
k=0
(n − k + 1)m−1 n−k
p
, where (x)l = x(x + 1)...(x + l − 1). (18)
(m − 1)!
Proof: Sufficienty:
∆m
p an
m
X
m m−k
=
p
an+k =
(−1)
k
k=0
m
n+k
X
X
(n + k − i + 1)m−1 n+k−i
m−k m
m−k
=
(−1)
p
bi
p
=
k
(m
−
1)!
i=0
k=0
=
m
X
m−k
cm,k
dn,k (m, i)bi
i=0
k=0
where
n+k
X
m m−k
cm,k = (−1)
p
,
k
(n + k − i + 1)m−1 n+k−i
p
dn,k (m, i) =
(m − 1)!
m−k
∆m
p an
=
=
n
X
j=0
n
X
j=0
with
bj
m
X
cm,r dn,r (m, j) +
r=0
bj Sm,r (n, j) +
m
X
j=1
m
X
bn+j
m
X
cm,r dn,r (m, n + j) =
r=j
′
bn+j Sm,r
(n, j)
j=1
Sm,r (n, j) =
m
X
cm,r dn,r (m, j)
r=0
p
′
Sm,r
(n, j) =
X
cm,r dn,r (m, n + j)
r=j
254
I. T.incu - On some p-convex sequences
m
X
m m−r (n + r − j + 1)m−1 n+r−j
Sm,r (n, j) =
(−1)
p
p
=
r
(m − 1)!
r=0
m
pm+n−j X
m−r m Γ(n + r + m − j) Γ(n − j + m)
=
·
=
(−1)
(m − 1)! r=0
r Γ(n + r − j + 1) Γ(n − j + m)
m
pm+n−j X
m−r m
=
(−1)
(n − j + m)r (n − j + r + 1)
(m − 1)! r=0
r
(n − j + r + 2)...(n − j + m − 1) =
m
pm+n−j X m
(j − n − r − 1)(j − n − r − 2)...
=
(m − 1)! r=0 r
...(j − n − m + 1)(n − j + m)r =
m
(j − n − m)m−r
−pm+n+j X m
=
(n − j + m)r
=
(m − 1)! r=0 r
j−n−m
=−
m−r
(0)m
pm+n−j
·
= 0.
(m − 1)! j − n − m
In the previously calculations we have use the Chy-Vandermonde formula
(x + y)m =
m
X
m
k=0
′
Sm,r
(n, j)
m
X
k
)(x)k (y)m−k
m m−r (r − j + 1)m−1 r−j
p
=
=
(−1)
p
(m
−
1)!
r
r=j
m−j
m
pm−j X
m−r−j
(−1)
pm−r−j (r + 1)m−1
=
r+j
(m − 1)! r=0
m−r
255
I. T.incu - On some p-convex sequences
′
Sm,r
m
m−j
pm−j X
m
−
j
r+j
(−1)m−j−r
=
m−j (r + 1)m−1 =
(m − 1)! r=0
r
r
m−j
pm−j X
m−j
m!
r!
=
·
(r + 1)m−1 =
(−1)m−j−r
(m − 1)! r=0
(m − j)! (r + j)!
r
m−j
X
m!
Γ(m + r)
Γ(m)
pm−j
m−j−r m − j
·
·
=
(−1)
=
(m − 1)! (m − j)! r=0
Γ(r + j + 1)
Γ(m)
r
m−j
mpm−j X
m−j−r m − j
(−1)
(r + j + 1)(r + j + 2)...
=
r
(m − j)! r=0
...(m − 1)(m)r =
m−j
pm−j X m−
=
(−m)m−j−r (m)r =
(m − j)! r=0
r
=
pm−j
(0)m−j .
(m − j)!
For j ≤ m − 1 we obtain
′
Sm,r
(n, j) = 0.
For j = m, we have
′
Sm,r
= cm,n dn,m (m, n + m) =
(1)m−1
= 1.
(m − 1)!
Results ∆pm an = bn+m .
Necessity: For all real sequence we may consider the sequence (bn ) which
define by
b 0 = a0 ,
b n = an −
n−1
X
k=0
bk
(n − k + 1)m−1 n−k
p
,
(m − 1)!
Because ∆m
p an = bn+m we obtain (??).
256
n = 1, 2, ...
I. T.incu - On some p-convex sequences
Example 3. Let (an )n∈N , (bn )n∈N be the sequences defined by
bn = pn ,
an =
n
X
k=0
p ∈ (0, 1),
bk
(n − k + 1)m−1 n−k pn
p
= (m + 1)n .
(m − 1)!
n!
From theorem ?? we obtain
0 ≤ ∆m
p an ≤ 1 ,
for all n ∈ N , m ∈ N∗ and p ∈ (0, 1).
pn
We observe that then sequence (an )n∈N defined by an = (m + 1)n is pn!
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. ”Lucian 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 ′ analyse
num´erique et de th´eorie de l ′ approximation, Tome 22, N◦ 1, 1993.
Ioan T.incu
”Lucian Blaga” University of Sibiu
Romania
257