A New Optimization Approach for Grey Model GM(1,1) - Advanced Research on Computer Education, Simulation and Modeling - page 119

(1)

dx

Firstly, calculate the integral of the white differential equation + ax

(1)

= b ,

dt

(1)

dx

k k

k
k

∫ ∫ ∫

i.e., dt + ax

(1)

dt = b

⇒ (1)

xk xk () −

(1)

( − 1) + a x t

(1)

= b ,

dt

k − 1 k − 1 − 1

then

( )

k
k

k

xk d
a

(0)

()

1

(1)

= ∫ =

(1)

∫

−− ak ( )

( )

zk () x t ge + g dt

= = .

xk xk

(1)

() −

(1)

( − 1) + d +

1 2

− 1 k − 1

a

(0) −− ak ( 1)

xk ( )

()

− e

= = e

a

⇒

(0) (0)

xk −−

(0) ak ( 1)

Due to ax = ln ( k − 1) − ln x ( ) k , () =

xk

(0)

e

− ak

k − 1

⎛ xk

(0)

( − ) ⎞

(0) ak ( − 1) (0)

gx ke = () = x k ()

Then ,

⎜ ⎟

xk

(0)

()

⎝ ⎠

k − 1

⎛

(0)

⎞

(0)

xk ( ) − xk ( ) −

( )

k

(0)

xk

(0)

()

xk ( − )

⎜ ⎟ (0) (0)

ge

a

xk () xk ()

⎝ ⎠

=

d = =

( ) ( )

k − 2

a (0)

xk

(0)

() xk

(0)

( −− 1) xk

(0)

()

e − 1 x k

xk

( − 1)

− 1

(0)

()

( )

k

(0)

xk ( − )

(0)

xk ()

(1)

Thus

zk () = + . (1.2)

( ) ( )

(0) (0) k − 2

ln xk ( ) − ln xk ( − 1)

xk

(0)

() xk

(0)

( −− 1) xk

(0)

()

Theorem 1. Taking (1.2) as the background value, grey model GM (1, 1) has the
white exponential superposition.

xk −

(0) ak

, then it suffices to prove a

Proof:

Let () = = a .Let

n

n

ce

− a

xk

(0)

( )

()

− e

−− ak ( 1)

∑ =

( )

ce −

ak

∑ =

(0) −− an ( ) a

D = xk () 1 − e , and = = e .

a (0) − ak

e − 1 xk e k = 2

k = 2

( )

k

xk

(0)

( ) −

(0)

xk
xk

()

Therefore zk =

(1) ()

+

( ) ( )

k − 2

(0) (0)

ln( ( )) − ln( xk ( − 1))

xk

(0)

() xk

(0)

( −− 1) xk

(0)

()

(0)

xk ( ) −

xk
xk
xk

(0)

()
()

(0)

xk ( ) −

k − 2

ce

−

c
ae

ak

cxk
a

(0) ()

(0)

⎛ xk ( ) − ⎞ xk

(0)

()

= = + = − .

⎛

(0)

⎞

⎜ ⎟

a a

− − 1 e − 1

xk

(0)

()

⎛ xk

xk

(0)

( ) 1
()

− ⎞

ln

⎝ ⎠

⎜ ⎟ −

⎜ ⎟

(0)

( ) −

⎝ ⎠

(0)

⎝ ⎠