ICTAMI2007. 126KB Jun 04 2011 12:08:47 AM

ACTA UNIVERSITATIS APULENSIS

No 15/2008

ABOUT AN INTERMEDIATE POINT PROPERTY IN SOME
QUADRATURE FORMULAS
Ana Maria Acu
Abstract. In this paper we study a property of the intermediate point
from the quadrature formula of the Gauss-Jacobi type, the quadrature formula obtained in the paper [3] by using connection between the monospline
function and the numerical integration formula, and the generalized meanvalue formula of N. Cior˘anescu.
2000 Mathematics Subject Classification: 26D15, 65D30.
1. Introduction
In the specialized literature there are a lot of mean-value theorems. In
[1] and [8] the authors studied the property above for the intermediate point
from the quadrature formula of Gauss type.
The generalized quadrature formula of Gauss-Jacobi type has the form
([9])
Z b
m
X
α

β
(b − x) (x − a) f (x)dx =
Bm,k f (γk ) + Rm [f ],
(1)
a

k=0

the nodes γk , k = 0, m , with appears in (1) are given by
γk =

b−a
b+a
ak +
2
2
(α,β)

where ak , k = 0, m are the zeros of the Jacobi polynomial, Jm+1 and
Bm,k =

1 (b − a)α+β+1 (2m + α + β + 2)Γ(m + α + 1)Γ(m + β + 1)
.
d h (α,β) i
2
(α,β)
(m + 1)!Γ(m + α + β + 2)Jm (ak )
J
(x)
dx m+1
x=ak

For f ∈ C 2m+2 [a, b] the rest term is given by
Rm [f ] = (b−a)2m+α+β+3

f (2m+2) (ξ) (m+1)!Γ(m+α+2)Γ(m+β +2)Γ(m+α+β +2)
·
,
(2m+2)!
Γ(2m+α+β +3)Γ(2m+α+β +4)

19

A. M. Acu - About an intermediate point property in some...

a < ξ < b.
For α = β = 0 we have so-called Gauss-Legendre quadrature formula. If
f ∈ C 2m+2 [a, b], then for any x ∈ (a , b] there is cx ∈ (a, x) such that
Z

x

a

m

(x − a) X
f (t)dt =
m + 1 k=0

1

·f
h
i
d
(0,0)
(0,0)
Jm (ak )
J
(x)
dx m+1
x=ak
(x − a)2m+3 [(m + 1)!]4 (2m+2)
f
(cx ).
+
(2m + 3)[(2m + 2)!]3



x+a

x−a
ak +
2
2



(2)

Theorem 1.[1] If f ∈ C 2m+4 [a, b] and f (2m+3) (a) 6= 0, then for the interme1
cx − a
= .
diate point cx which appears in formula (2) we have lim
x→a x − a
2
In this paper we want to study the property of the intermediate point from
the quadrature formula of Gauss type with weight function w(x) = (b−x)(x−a).
In recent years a number of authors have considered generalization of
some known and some new quadrature rules. For example, P. Cerone and
S.S. Dragomir in [5] give a generalization of the midpoint quadrature rule:

Z

a

b

m−1
X

(b − a)k+1 (k)
f (t)dt =
f
[1+(−1) ] k+1
2
(k
+
1)!
k=0

where

k


(t − a)m


,

m!
Km (t) =
(t − b)m


,

m!


Z b

a+b
m
Km (t)f (m) (t)dt
+(−1)
2
a
(3)


a+b
t ∈ a,
2 
a+b
t∈(
,b
2



If f ∈ C m [a, b] and m is even , then for any x ∈ (a , b ] there is cx ∈ (a, x)

such that


Z x
m−2
k+1
X
(x − a)m+1 (m)
a+x
(k)
k (x − a)
+ m
f
f (cx )
f (t)dt =
[1 + (−1) ] k+1
2 (k + 1)!
2
2 (m + 1)!
a

k=0
(4)
In [2] was studied the following property of intermediate point from the
quadrature formula (4)
20

A. M. Acu - About an intermediate point property in some...
Theorem 2.[2] If f ∈ C 2m [a, b] , m is even and f (m+1) (a) 6= 0 , then for the
intermediate point cx that appears in formula (4), it follows:
cx − a
1
= .
x→a x − a
2
lim

In this paper we want to give a property of intermediate point from a
quadrature formula with the weight function w : [a, b] → (0, ∞), w(x) =
(b − x)(x − a).
N. Cior˘anescu demonstrated in [6] that the following formula is valid

Zb

f (m) (cb )
f (x)pm (x)w(x)dx =
m!

Zb

xm pm (x)w(x)dx,

(5)

a

a

where f ∈ C m [a, b] and (pn )n≥0 is a sequence of orthogonal polynomials on
[a, b], in respect to a weight function w : [a, b] → (0, ∞).
In [4], the authors study a property of the intermediate point from the
mean-value formula of N. Cior˘anescu.
Theorem 3.[4] If f ∈ C m+1 [a, b] and f (m+1) (a) 6= 0, then the intermediate
point of the mean-value formula (5) satisfies the relation:
1
cb − a
=
b→a b − a
m+2
lim

In this paper we give a property of the intermediate point from the generalized mean-value formula of N. Cior˘anescu.
2. An intermediate point property in the quadrature
formulas of Gauss-Jacobi type
Let
Z

a

b

m
(b−a)3 X
(b−x)(x−a)f (x)dx =
m+3 k=0



b+a
b−a
1
f
ak +
d h (1,1) i
2
2
(1,1)
Jm+1 (x)
Jm (ak)
dx
x=ak
(m+1)!4 (m+2)(m+3)
f (2m+2) (ξ) (6)
+ (b−a)2m+5
4(2m+2)!3 (2m+3)2 (2m+5)
21

A. M. Acu - About an intermediate point property in some...
be the quadrature formula of Gauss-Jacobi type (1), which α = β = 1.
If f ∈ C 2m+2 [a, b], then for any x ∈ (a , b] there is cx ∈ (a, x) such that
Z

a

x

m
(x−a)3 X
(x−t)(t−a)f (t)dt =
m+3 k=0



1
x−a
x+a
ak +
f
d h (1,1) i
2
2
(1,1)
Jm+1 (x)
Jm (ak)
dx
x=ak
4
(m+1)! (m+2)(m+3)
+ (x−a)2m+5
f (2m+2) (cx ). (7)
4(2m+2)!3 (2m+3)2 (2m+5)

In this section we give a property of the intermediate point, cx , from the
quadrature formula of Gauss-Jacobi type (7). Here we prove a lemma which
help us in proving our theorem.
Lemma 1. If ak , k = 0, m are the
then the following relations hold:

i−3
ak + 1
m
X
2
=
d h (1,1) i
(1,1)
k=0 Jm (ak )
J
(x)
dx m+1
x=ak
2m+2

ak + 1
m
X
2
=
d h (1,1) i
(1,1)
k=0 Jm (ak )
J
(x)
dx m+1
x=ak
·
2m+3
ak + 1
m
X
2
h
i
d
(1,1)
(1,1)
k=0 Jm (ak )
Jm+1 (x)
dx
x=ak


=

−

(1,1)

zeroes of the Jacobi polynomials, Jm+1 ,

m+3
, for i = 3, 2m + 4,
i(i − 1)

(8)

m+3
(2m + 3)(2m + 4)(2m + 5)

(9)



(m + 1)!4 (m + 2)2 (m + 3)
,
(2m + 3) −
2(2m + 2)!2 (2m + 3)
1
22m+3



22m+2
2m + 5

22m (m + 1)!4 (m + 2)(m + 3)2
(2m + 2)!2 (2m + 3)(2m + 5)



.

(10)

Proof. If we choose a = 0, b = 1 and f (t) = ti−3 , i = 3, 2m + 5 in the
quadrature formula (6), then we obtain the relations (8) and (9).

22

A. M. Acu - About an intermediate point property in some...
If we choose a = −1, b = 1 and f (t) = ti , i = 0, 2m + 2 in the quadrature
formula (6), then we obtain the following relations:
m
X



aik
m+3
1+(−1)i , i = 0, 2m+1(11)
=
h
i
d
4(i+1)(i+3)
(1,1)
(1,1)
k=0 Jm (ak )
Jm+1 (x)
dx
x=ak
m
2m+2
X
ak
m+3
(12)
=
h
i
d
2(2m
+
3)(2m
+
5)
(1,1)
(1,1)
k=0 Jm (ak )
J
(x)
dx m+1
x=ak


4
2m+1 (m + 1)! (m + 2)(m + 3)
· 1−2
.
(2m + 2)!2 (2m + 3)
By using the following formulas (see [10]):
Γ(2m + α + 1)Γ(m + 1) (α,− 12 ) 2
(2x − 1),
Jm
Γ(m + α + 1)Γ(2m + 1)
Γ(2m + α + 2)Γ(m + 1) (α, 12 ) 2
(α,α)
J2m+1 (x) =
xJm (2x − 1),
Γ(m + α + 1)Γ(2m + 2)
d  (α,β) 1
(α+1,β+1)
Jm (x) = (m + α + β + 1)Jm−1
(x),
dx
2
(α,α)

J2m (x) =

we obtain
(1,1)

J2m (ak )

d h (1,1) i
2(2m+1)2 (1,−21) 2
(2,−1)
J2m+1 (x)
Jm (2ak −1)Jm 2 (2a2k −1), (13)
=
dx
m+1
x=ak
i
d h (1,1)
2(2m + 5)(2m + 3) 2
(1,1)
J2m+1 (ak )
J2m+2 (x)
· ak
(14)
=
dx
(m + 2)
x=ak
(2, 1 )
(1, 1 )
·Jm 2 (2a2k − 1)Jm 2 (2a2k − 1).

From the identity
(α,β)
(β,α)
Jm
(x) = (−1)m Jm
(−x)

it follows that
ak + am−k = 0,

(15)

where ak , k = 0, m are the zeroes of Jacobi polynomial of degree m + 1,
(1,1)
Jm+1 .
23

A. M. Acu - About an intermediate point property in some...
From (13), (14) and (15) we obtain
m
X
k=0

therefore 

a2m+3
k
= 0,
h
i
d
(1,1)
(1,1)
(x)
Jm (ak )
J
dx m+1
x=ak

2m+3
ak + 1
m
X
1
2
= 2m+3
h
i
d
2
(1,1)
(1,1)
k=0 Jm (ak )
Jm+1 (x)
dx
x=ak
m 2m+2
X
X  2m + 3 
+
i



m
X


k=0

a2m+3
k
d h (1,1) i
(1,1)
Jm+1 (x)
Jm (ak )
dx
x=ak



aik

h
i

d
(1,1)
(1,1)
k=0 i=0
Jm+1 (x)
Jm (ak )
dx
x=ak
X

m
2m+2
i
X
1
ak
2m + 3
= 2m+3
d h (1,1) i
i
2
(1,1)
i=0
k=0 Jm (ak )
J
(x)
dx m+1
x=ak
and by using (11) and (12) it follows the relation (10).

Theorem 4. If f ∈ C 2m+6 [a, b] and f (2m+3) 6= 0, then for the intermediate
point cx which appears in formula (7) we have
lim

x→a

1
cx − a
= .
x−a
2

Proof. Let us consider F, G : [a, b] → R defined by
Z x
(x − t)(t − a)f (t)dt
F (x) =

(16)

a

m
(x − a)3 X
−
m + 3 k=0



x +a
x−a
1
f
ak +
d h (1,1) i
2
2
(1,1)
Jm (ak )
Jm+1 (x)
dx
x=ak
4
(m
+
1)!
(m
+
2)(m
+
3)
f (2m+2) (a),
(17)
− (x − a)2m+5
3
2
4(2m + 2)! (2m + 3) (2m + 5)
G(x) = (x − a)2m+6 .
We have that F and G are (2m + 6) times derivable on [a, b],
G(i) (x) 6= 0, i = 1, 2m + 5 any x ∈ (a, b] ,
G(i) (a) = 0, i = 1, 2m + 5.
24

A. M. Acu - About an intermediate point property in some...
We observe that F (a) = F ′ (a) = F ′′ (a) = 0.
For i = 3, 2m + 4 we have
i−3
ak +1
m
i(i−1)(i−2) X
2
(i)
(i−3)
·f (i−3) (a)
F (a) = (i−2)f
(a)−
d h (1,1) i
m+3
(1,1)
k=0 Jm (ak )
J
(x)
dx m+1
x=ak


and by using relation (8) we obtain F (i) (a) = 0.
From relations (9) and (17) we obtain
(2m + 3)(2m + 4)(2m + 5)
F (2m+5) (a) = (2m + 3)f (2m+2) (a) −
m+3
2m+2

ak + 1
m
X
2
· f (2m+2) (a)
·
d h (1,1) i
(1,1)
k=0 Jm (ak )
(x)
J
dx m+1
x=ak
4
2
(m + 1)! (m + 2) (m + 3) (2m+2)
−
f
(a) = 0.
2(2m + 2)!2 (2m + 3)
By using successive l ′ Hospital rule and
F (2m+6) (a) = (2m + 4)f (2m+3) (a) − 2(2m + 4)(2m + 5)

2m+3
ak + 1
m
X
2
· f (2m+3) (a),
·
d h (1,1) i
(1,1)
k=0 Jm (ak )
J
(x)
dx m+1
x=ak
(2m+6)
G
(a) = (2m + 6)!
we obtain

f (2m+3) (a)
F (2m+6) (x)
F (x)
= lim (2m+6)
=
(18)
x→a G
x→a G(x)
(x)
(2m + 6)!



2m+3
ak +1
m


X
2


· (2m + 4)−2(2m+4)(2m+5) ·
,
h
i
d


(1,1)
(1,1)
k=0 Jm (ak )
Jm+1 (x)
dx
x=ak
lim

25

A. M. Acu - About an intermediate point property in some...
but
(2m+2)
4
F (x)
(cx ) − f (2m+2) (a)
2m+5 (m + 1)! (m + 2)(m + 3) f
lim
= lim (x − a)
x→a G(x)
x→a
4(2m+2)!3 (2m+3)2 (2m+5)
(x − a)2m+6
f (2m+2) (cx ) − f (2m+2) (a) cx − a
(m + 1)!4 (m + 2)(m + 3)
·
·
,
= lim
x→a 4(2m+2)!3 (2m+3)2 (2m+5)
cx − a
x−a
namely
(m + 1)!4 (m + 2)(m + 3)
cx − a
F (x)
=
· f (2m+3) (a) · lim
.
(19)
lim
3
2
x→a x − a
x→a G(x)
4(2m+2)! (2m+3) (2m+5)

From (10), (18) and (19) it follows that the intermediate point cx which
appears in formula (7) verifies the property (16).
3. An intermediate point property in a quadrature formula
with weight function w(x) = (b − x)(x − a)
In [3] was studied the following quadrature formula
Z

b

w(t)f (t)dt =

a

m−1
X
k=0



(−1)k +1 ·

(b−a)k+3
f (k)
2k+2 (k+1)!(k+3)




a+b
+R[f ] ,
2

where f ∈ C m [a, b], w(t) = (b − t)(t − a),
Z b
m
Mn (t)f (m) (t)dt
R[f ] = (−1)
a

and



m+1
m+2
(t
−
a)
(t
−
a)
a
+
b


−2
, t ∈ [a,
 (b − a)
(m + 1)!
(m + 2)!
2 

Mm (t) =
.
m+1
m+2
(t − b)
(t − b)
a+b


−2
, t∈
,b
 (a − b)
(m + 1)!
(m + 2)!
2

If f ∈ C m [a, b] and m is even , then for any x ∈ (a , b ] there is cx ∈ (a, x)
such that


Z x
m−2
X

a+x
(x − a)k+3
(k)
k
f
(x − t)(t − a)f (t)dt =
(−1) + 1 · k+2
2 (k + 1)!(k +3)
2
a
k=0
+

(x − a)m+3
f (m) (cx ).
2m+1 (m + 1)!(m + 3)
26

(20)

A. M. Acu - About an intermediate point property in some...
In the above condition we have the following theorem
Theorem 5. If f ∈ C 2m+2 [a, b] , m is even and f (m+1) (a) 6= 0 , then for the
intermediate point cx that appears in formula (20), it follows:
lim

x→a

cx − a
1
= .
x−a
2

Proof. Let F, G : [a, b] → R definite as follows
Z

F (x) =

x

(x−t)(t−a)f (t)dt−

a

m−2
X
k=0

m+3



(−1)k +1 ·



(x−a)k+3
(k) a+x
f
2k+2 (k+1)!(k+3)
2

(x − a)
f (m) (a),
+ 1)!(m + 3)
G(x) = (x − a)m+4 .
−

2m+1 (m

We observe that F (a) = F ′ (a) = F ′′ (a) = 0. For i = 3, m + 2 we have

i(i − 1)(i − 2)
(i)
(i−3)
F (a) = f
(a) (i − 2) −
2i−1
)


i−3
X


1
i
−
3
= 0.
·
(−1)k + 1
k
(k + 1)(k + 3)
k=0
We find
F

(m+3)

·



(m + 1)(m + 2)(m + 3)
2m+2
)


m−2
X

m
+
2
1
m
− m+1 = 0
(−1)k + 1
k
(k + 1)(k + 3)
2
k=0

(a) = f

(m)

(a) (m + 1) −

and
F

(m+4)


(m + 2)(m + 3)(m + 4)
(a) = f
(a) (m + 2) −
2m+3
)


m−2
X

1
m
+
1
·
(−1)k + 1
k
(k + 1)(k + 3)
k=0
(m+1)

=

(m + 2)(m + 4) (m+1)
f
(a).
2m+2
27

A. M. Acu - About an intermediate point property in some...
We have that F and G are (m+4) times derivable on [a, b], F (i) (a) = G(i) (a) =
0, for i = 0, m + 3 and G(i) (x) 6= 0, i = 1, m + 3 any x ∈ (a, b]. By using
successive l ′ Hospital rule we obtain
F (x)
1
F m+4 (x)
= lim m+4
= m+2
f (m+1) (a),
x→a G(x)
x→a G
(x)
2
(m + 1)!(m + 3)

(21)

cx − a
1
F (x)
= m+1
f (m+1) (a) lim
.
x→a x − a
x→a G(x)
2
(m + 1)!(m + 3)

(22)

lim

but

lim

From relation (21) and (22) we have
1
cx − a
= .
x→a x − a
2
lim

4. An intermediate point property from the mean-value
˘nescu
formula of N. Ciora
The polynomial
Ps,m (x) = xm + am−1 xm−1 + · · · + a1 x + a0 ,
which satisfies the orthogonality conditions
Z b
[Ps,m (x)]2s+1 xk w(x)dx = 0, k = 0, 1, · · · , m − 1
a

is called s-orthogonal polynomial with respect to the weight function
w : [a, b] → (0, ∞).
The following equality is the generalized mean-value formula of N. Cior˘anescu
(see [9])
Zb

2s+1
f (x)Ps,m
(x)w(x)dx

f (m) (cb )
=
m!

Zb

2s+1
xm Ps,m
(x)w(x)dx.

(23)

a

a

We observe that for s = 0 we obtain the mean-value formula of N. Cior˘anescu
(5).
We construct the functions (Vk )k=0,m as follows
28

A. M. Acu - About an intermediate point property in some...

2s+1
V0 (x) = w(x)Ps,m
(x),

Vj (x) =

Z

x

Vj−1 (x)dx, j = 1, m.

a

Here we prove a lemma which help us in proving our theorem.
Lemma 2. We have the following equalities
Vj (a) = 0, Vj (b) = 0,
for any j = 1, m,
Z b
Z b
2s+1
m
f (m) (x)Vm (x)dx.
f (x)Ps,m (x)w(x)dx = (−1)

(24)
(25)

a

a

Proof. We have Vj (a) = 0, for any j = 1, m.
Z
Z b
2s+1
V0 (x)dx = Ps,m
(x)w(x)dx = 0.
V1 (b) =
a

For every k ∈ {2, 3, . . . , m} we have
Z b  k−1 (k−1)
Z b
x
Vk−1 (x)dx =
Vk−1 (x)dx
Vk (b) =
(k − 1)!
a
a
 k−1 (ν) b
k−2
X
x

=
(−1)k−ν−2 [Vk−1 (x)](k−ν−2)


(k − 1)!
ν=0
a
Z b
k−1
x
+ (−1)k−1
[Vk−1 (x)](k−1)
dx
(k − 1)!
a
b
Z
k−2
X
xk−ν−1 
(−1)k−1 b
k−ν−2
V0 (x)xk−1 dx
=
(−1)
Vν+1 (x)
 + (k − 1)!
(k
−
ν
−
1)!
a
a
ν=0
Z
b
(−1)k−1
2s+1
=
xk−1 Ps,m
(x)w(x)dx = 0.
(k − 1)! a
We have
Z b
m−1
X
b
2s+1
f (x)Ps,m
(x)w(x)dx =
(−1)m−ν−1 f (m−ν−1) (x)Vm−ν (x)a
a

ν=0

m

+ (−1)

Z

b

f (m) (x)Vm (x)dx,

a

and by using relation (24) we obtain the equality (25).
29

A. M. Acu - About an intermediate point property in some...
Theorem 6. If f ∈ C m+1 [a, b] and f (m+1) (a) 6= 0, then the intermediate
point of the mean-value formula (23) satisfies the relation:
cb − a
1
=
b→a b − a
m+2
lim

(26)

Proof. From (23) and (25) we can written
Zb

2s+1
f (x)Ps,m
(x)w(x)dx

f (m) (cb )
=
m!

Zb

2s+1
xm Ps,m
(x)w(x)dx

a

a

m (m)

= (−1) f

Z

(cb )

b

Vm (x)dx.

(27)

a

From relations (25) and (27) we obtain
Z

b

f

(m)

(x)Vm (x)dx = f

(m)

(cb )

Z

b

Vm (x)dx.

a

a

We consider the functions
Z b
Z b
(m)
(m)
Vm (x)dx,
f (x)Vm (x)dx − f (a)
F (b) =
a

a

G(b) = (b − a)m+2 .

Since

k−1 
X
k−1
F (b) =
f (m+k−1−ν) (b)Vm−ν (b) − f (m) (a)Vm−k+1 (b),
ν
ν=0
m−1
X m 


(m+1)
F
(b) =
f (2m−ν) (b)Vm−ν (b) + f (m) (b) − f (m) (a) V0 (b),
ν
(k)

ν=0

by using successive l ′ Hospital rule, we have
F (b)
f (m+1) (a)
=
V0 (a),
b→a G(b)
(m + 2)!
lim

but

30

(28)

A. M. Acu - About an intermediate point property in some...

f

(m)

(cb )

Zb

Vm (x)dx − f

(m)

(a)

F (b)
a
= lim
b→a G(b)
b→a
(b − a)m+2
cb − a
V0 (a) (m+1)
f
(a) · lim
.
=
b→a b − a
(m + 1)!
lim

Zb

Vm (x)dx

a

(29)

From (28) and (29) it follows that the intermediate point cb from the
generalized mean-value formula of N. Cior˘anescu verifies the property (26).
References
[1] A.M. Acu, An intermediate point property in the quadrature formulas
of Gauss-Jacobi type , Advances in Applied Mathematical Analysis, Vol. 1,
Nr.1, 2006.
[2] A.M. Acu, An intermediate point property in the quadrature formulas
, Advances in Applied Mathematical Analysis, (to appear)
[3] A.M. Acu, Monosplines and inequalities for the remaider term of
quadrature formulas, General Mathematics, Vol. 15, No. 1, 2007, 81-92.
[4] D.Acu, P. Dicu, About some properties of the intermediate point in
certain mean-value formulas, General Mathematics, Vol. 10, No. 1-2, 2002,
51-64.
[5] P. Cerone and S.S.Dragomir, Trapezoidal-type Rules from an Inequalities Point of View,Handbook of Analytic-Computational Methods in Applied
Mathematics, Editor: G. Anastassiou, CRC Press, New York, (2000),65-134.
[6] N. Cior˘anescu, La g´en´eralisation de la premi´ere formule de la moyenne,
L ′ einsegment Math., Gen´eve, 37 (1938), 292 - 302.
[7] P. Dicu, An intermediate point property in some of classical generalized
formulas of quadrature, General Mathematics, Vol. 12, No. 1, 2004, 61-70.
[8] P. Dicu, An intermediate point property in the quadrature formulas of
type Gauss , General Mathematics, Vol. 13, No. 4, 2005, 73-80.
[9] D.D. Stancu, G. Coman, P.Blaga, Analiza numerica si Teoria aproximarii, Vol.II, Presa Universitara Clujeana, Cluj-Napoca, 2002.
[10] G. Szeg¨o , Orthogonal Polynomials, American Mathematical Society,
Colloquium Publication, Volume 23, 1939.
31

A. M. Acu - About an intermediate point property in some...
Author:
Ana Maria Acu
Department of Mathematics
University ”Lucian Blaga” of Sibiu
Str. Dr. I. Rat.iu, No. 5-7
550012 - Sibiu, Romania
email:acuana77@yahoo.com

32