Page 73 - IJOCTA-15-4

Analysis and analytical solution of incommensurate fuzzy fractional nabla difference systems...
Theorem 6. Let µ, η, ζ, ψ ∈ R F , and assume that Proof. Since q is a m. i. c. function, from Equa-
µ ⊖ η = ζ is H-differenceable. Then: tion (17), we have
 −1
d−q (w)
(1) ζ ⊕ η = η ⊕ ζ = µ;  L d−a , w ≤ q(d),
r := q(µ(w)) =
(2) η ⊖ µ is not H-differenceable; q −1 (w)−d
 R , w ≥ q(d),
(3) µ ⊖ η ⊕ ψ = µ ⊕ ψ ⊖ η. b−d
(21)
and Supp(q(µ)) = [q(a), q(b)].
Proof. The proof follows directly from Equations
(11), (14), and the definition of R F . □ Then, from Equation (8), we get:
−1 −1
q (w) − d q (w) − d
f 2 (r) = f 2 R = ,
b − d b − d
The multiplication is defined as:
for w ≥ d and

 d − q −1 (w) d − q −1 (w)
ed−w
 L d(e−c)+e(d−a) , w ≤ e + d, f 1 (r) = f 1 L = ,
µ LR (w) ⊗ η LR (w) = d − a d − a
 R w−ed , w ≥ e + d,
d(h−e)+e(b−d)
for w ≤ d. By rearranging the equations from the
˜
˜
∼ (d(e − c) + e(d − a))f 1 , (ed, (d(h − e) + e(b − d))f 2 ) previous step, we can rewrite them equivalently
∼ (de, ef 1 + dg 1 , ef 2 + dg 2 ). as:
(15)
w = q(d + (b − d)f 2 (r)), q(w) ≥ q(d),
and
Let g : R → R be a function. It can be ex-
tended to a fuzzy function g : R F → R F by the w = q(d − (d − a)f 1 (r)), q(w) ≤ q(d).
extension theorem: On the other hand, by the definition of f 1 and
ˆ
ˆ
(g(µ))(w) = sup {µ(x), 0}. (16) f 2 , we should have:
x:g(x)=w
ˆ
w = q(d) + (q(b) − q(d))f 2 (r), q(w) ≥ q(d),
Fortunately, Equation (16) can be simplified
and
for one-to-one functions. In this case, (q(µ))(w)
ˆ
is equal to w = q(d) − (q(d) − q(a))f 1 (r), q(w) ≤ q(d).
( By comparing these equations, we can con-
µ(q −1 (w)), if q −1 (w)exists & q −1 (w) ∈ Supp(µ),
clude that:
0, oth.
ˆ
q(d) + (q(b) − q(d))f 2 (r) = q(d + (b − d)f 2 (r)),
(17)
It can be easy to verify that for f(x) = λx, the and
Definition Equations (9) and (17) are consistent. ˆ
q(d) − (q(d) − q(a))f 1 (r) = q(d − (d − a)f 1 (r)).
Most activation functions in neural networks
□
(NN) are m. i. c. functions such as the Sigmoid
function, rectified linear unit and hyperbolic tan- Corollary 1. Let q be invertible m. i. c. function
gent functions. To find a symmetric form of Equa- and µ ∼ (d, f 1 , f 2 ) where f 1 , f 2 ∈ U. Also, assume
tion (17), m. i. c. functions play an important a ≤ d ≤ b such that [a, b] = Supp(µ). Then,
role in achieving some interesting simplifications. q(µ) ∼ (q(d), f 1 , f 2 ) (22)
ˆ ˆ
Activation functions change the shape of the in-
where
put fuzzy number according to the following the-
ˆ
f 1 (r) = q(d) − q(d − f 1 (r)), (23)
orems.
and
ˆ
Theorem 7. Let q be an invertible m. i. c. func- f 2 (r) = q(d + f 2 (r)) − q(d). (24)
tion and µ ∼ (d, (b−a)f 1 , (d−a)f 2 ) where f 1 , f 2 ∈ 2.2. Nabla fractional differences and sum
S, a ≤ d ≤ b such that [a, b] = Supp(µ). Then,
There are different definitions for nabla fractional
ˆ
ˆ
q(µ) ∼ (q(d), (q(d) − q(a))f 1 , (q(b) − q(d))f 2 ),
differences and sums. We follow the definitions
(18) 21
in.
where
Definition 3. Let x : N a → R, α > 0 and
q(d) − q(d − (d − a)f 1 (r))
ˆ
f 1 (r) = , (19) N = ceil(α). Then, the fractional nabla sum is
q(d) − q(a)
defined as
and t
q(d + (b − d)f 2 (r)) − q(d) ∇ −α 1 X Γ(t − s + α)
ˆ
f 2 (r) = . (20) x(t) = Γ(α) Γ(t − s + 1) x(s),
q(b) − q(d) s=a
615