Table Of ContentOn a representation of the inverse F -transform
q
Sabir Umarov1, Constantino Tsallis2,3
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0 1 Department of Mathematics, Tufts University, Medford, MA 02155, USA
0
2 Centro Brasileiro de Pesquisas Fisicas, Xavier Sigaud 150, 22290-180 Rio de Janeiro-RJ, Brazil
2
3 Santa Fe Institute, 1399 Hyde Park Road, Santa Fe, NM 87501, USA
n
a
J Abstract
8
A representation formula for the inverse q-Fourier transform is obtained in the class of
functions = , where = f =ae−βx2, a>0, β >0 .
] G 1≤q<3Gq Gq { q }
h
c S
e
m
InthispaperwefindarepresentationformulafortheinverseF -transforminaclassoffunctions
- q
t
a defined below. Fq-transform is a useful tool in the study of limit processes in nonextensive
G
st statistical mechanics (see [1] and references therein). Note that, in this theory, random states are
t. correlated in a special manner, and a knowledge on Fq-inverse of data is helpful in understanding
a
the nature of such correlations.
m
Throughout the paper we assume that 1 q < 3. The F -transform, called also q-Fourier
- ≤ q
d transform, of a nonnegative f(x) L (R1) is defined by the formula (see [2, 3])
1
n ∈
o
c F [f](ξ) = eixξ f(x)dx, (1)
[ q Zsuppf q ⊗q
1 where isthesymbolof theq-product,andex isaq-exponential. Thereaderisreferredfordetails
v ⊗q q
1 of q-algebra and q-functions to [4, 5, 6]. Fq coincides with the classic Fourier transform if q = 1. If
1 1 < q < 3 then F is a nonlinear mapping in L . The representation formula for the inverse, F 1,
q 1 q−
3
1 is defined in the class of function of the form Ae−q1Bξ2, since it uses a specific operator I defined in
. this class. The question on extension of this operator to wider classes is remaining a challenging
1
0 question.
8 The obvious equality eixξ f(x) = f(x)eixξ[f(x)]q−1, which holds for all x suppf, implies
0 q ⊗q q ∈
: the following lemma, which gives an expression for the q-Fourier transform without usage of the
v
q-product.
i
X
r Proposition 0.1 The q-Fourier transform can be written in the form
a
F [f](ξ) = f(x)eixξ[f(x)]q−1dx. (2)
q q
Zsuppf
Introduce the operator
F [f](ξ) = f(x)e ixξ[f(x)]q−1dx. (3)
q∗ −q
Zsuppf
1
Forarbitrarynonnegativef L (R)bothoperators,F andF arecorrectlydefined. Moreover,
∈ 1 q q∗
sup F [f](ξ) f and sup F [f](ξ) f . (4)
| q | ≤ k kL1 | q∗ | ≤ k kL1
ξ R1 ξ R1
∈ ∈
Introduce the set of functions
= f : f(x)= ae βx2, a> 0, β > 0 . (5)
Gq { q− }
Obviously, L for all q < 3. The set is fully identified by the triplet (q,a,β). We denote
q 1 q
G ⊂ G
= (q,a,β) :q < 3,a > 0,β > 0 .
R { }
For any function f we have f( x) = f(x), so f is symmetric about the origin. Moreover,
q
∈ G −
√β
it follows from the symmetry that F [f](ξ) = F [f](ξ). Further, a function f with a = ,
q q∗ ∈ Gq Cq
where
2 π/2(cos t)13−−qqdt = 2√πΓ(1−1q) , < q < 1,
√1 q 0 (3 q)√1 qΓ( 3−q ) −∞
− − − 2(1−q)
Cq = √√qπ2,1R0∞(1+y2)q−−11dy = √√qπΓ1(Γ2((3q−−q11))), q1=<1q,< 3. (6)
is called a q-Gaussian, and−isRdenoted by Gq(β;x). −Thusq,−t1he set of all q-Gaussians forms a subset
of .
q
G
The following statement was proved in [2].
Proposition 0.2 Let 1 q < 3. For the q-Fourier transform of a q-Gaussian, the following
≤
formula holds:
ξ2
F [G (β;x)](ξ) = e−4β2−qCq2(q−1) 3−2q. (7)
q q q
(cid:16) (cid:17)
Assume a sequence q is given by
k
2q k(q 1) 2
q = − − , < k < 1, (8)
k
2 k(q 1) −∞ q 1 −
− − −
for q > 1, and q = 1 for all k = 0, 1,..., if q = 1.
k
±
It follows from this proposition the following result.
Corollary 0.3 Let 1 q < 3 and k < 2 1. Then
≤ q 1 −
−
Fqk[Gqk(β;x)](ξ) = eq−kβ+k1+1ξ2, (9)
where qk+1 = 31+−qqkk and βk+1 = 8β2−q3k−Cqq2kk(qk−1).
Remark 0.4 It follows from (9) that
aC
F [ae βx2](ξ) = qk e Bξ2,
qk q−k √β −qk+1
where B = a2(qk−1)(3−qk).
8β
2
Theorem 0.5 The operator F : is invertible.
qk Gqk → Gqk+1
Proof. With the operator F : we associate the mapping defined as
qk Gqk → Gqk+1 R → R
(qk,a,β) → (qk+1,A,B), where A = a√Cqβk and B = a2(qk−81β)(3−qk). Consider the system of equations
1+q
k
= Q,
3 q
k
−
aC
qk = A,
√β
a2(qk−1)(3 qk)
− = B
8β
with respect to q ,a,β assuming that Q,A and B are given. The first equation is autonomous and
k
has a unique solution q = (3Q 1)/(Q+1). If the condition k < 2 1 is fulfilled then the other
k − q 1−
two equations have a unique solution as well, namely −
1 1
A√3 qk 2−qk A2(qk−1)(3 qk) 2−qk
a = − , β = − . (10)
2Cqk√2B! 8Cq2k(qk−1)B !
It follows from (8) that Q and q are related as Q = q . Hence, the inverse mapping (F ) 1 :
k k+1 q −
exists and maps each element Ae Bξ2 to the element ae βx2 with a and β
Gk+1 → Gk −qk+1 ∈ Gk+1 q−k ∈ Gk
defined in (10).
Now we find a representation formula for the inverse operator F 1. Denote by T the mapping
q−
T : (a,β) (A,B), where A = aCqk and B = a2(qk−1)(3−qk), as indicated above. We have seen
→ √β 8β
that T is invertible and T 1 : (A,B) (a,β) with a and b in (10). Assume (a¯,β¯) = T 2(A,B) =
− −
→
T 1(T 1(A,B)) = T 1(a,β).Further,weintroducetheoperatorI : defined
− − − (qk+1,qk−1) Gqk+1 → Gqk−1
by the formula
I [Ae Bξ2]= a¯e β¯ξ2. (11)
(qk+1,qk−1) −qk+1 −qk−1
Consider the composition H = F I . By definition, it is clear that I :
qk q∗k−1 ◦ (qk+1,qk−1) (qk+1,qk−1)
. Since F : , we have H : . Let fˆ , that is
Gqk+1 → Gqk−1 q∗k−1 Gqk−1 → Gqk qk Gqk+1 → Gqk ∈ Gk+1
fˆ(ξ) = Ae Bξ2. Then, taking into account the fact that supp fˆ = R1 if q 1, one obtains an
q−k+1 ≥
explicit form of the operator H :
qk
H [fˆ(ξ)](x) = ∞ a¯e β¯ξ2 e ixξdξ = ∞ I [fˆ(ξ)] e ixξdξ. (12)
qk −qk−1 ⊗qk−1 −qk−1 (qk+1,qk−1) ⊗qk−1 −qk−1
Z−∞(cid:16) (cid:17) Z−∞
Theorem 0.6 1. Let f . Then H F [f]= f;
∈ Gqk qk ◦ qk
2. Let f . Then F H [f]= f.
∈ Gqk+1 qk ◦ qk
Proof. 1. We need to show validity of the equation F I F = , where is the
qk−1◦ (qk+1,qk−1)◦ qk J J
identity operator in . This equation is equivalent to T T 2 T = J (J is the identity operator
Gqk ◦ − ◦
in with fixed q), which is correct by construction.
R ′ ′
2. Now the equation F F I = ( is the identity operator in ) is
qk ◦ qk−1 ◦ (qk+1,qk−1) J J Gqk+1
equivalent to the identity T2 T 2 = J.
−
◦
3
Corollary 0.7 The operator H :G G is the inverse to the q Fourier transform: H =
qk qk+1 → qk k− qk
F 1.
q−k
Corollary 0.8 For q = 1 the inverse F 1 coincides with the classical inverse Fourier transform.
q−k
Proof. If q = 1, then by definition one has q = q = q = 1. We find a¯ and β¯ taking
k k 1 k+1
(A,B) = (1,1). It follows from relationships (10) that (a−,β) = T 1(1,1) = ( 1 , 1). Again using
− 2√π 4
(10) we obtain (a¯,β¯) = T 2(1,1) = T 1( 1 ,1) = ( 1 ,1). This means that I fˆ(ξ) = 1 fˆ(ξ).
− − 2√π 4 2π (1,1) 2π
Hence, the formula (12) takes the form
1
F 1[fˆ(ξ)](x) = ∞ fˆ(ξ)e ixξdx,
1− 2π −
Z−∞
recovering the classic formula for the inverse Fourier transform.
Summarizing, we have proved that, if fˆ(ξ) is a function in , where q with k < 2 1 is
Gqk+1 k q 1 −
defined in Eq. (8) for q [1,3), then −
∈
F 1[fˆ(ξ)](x) = ∞ I [fˆ(ξ)] e ixξdξ, (13)
q−k (qk+1,qk−1) ⊗qk−1 −qk−1
Z−∞
with the operator I given in (11). This might constitute a first step for finding a repre-
(qk+1,qk−1)
sentation of F 1[fˆ(ξ)](x) for generic fˆ(ξ) L (R), which would be of great usefulness.
q− ∈ 1
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