# MTH603 Handouts Lec 20

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```					Numerical Analysis –MTH603                                                                        VU

Shift operator, E
Let y = f (x) be a function of x, and let x takes the consecutive values x, x + h, x + 2h, etc.
We then define an operator having the property
E f ( x ) = f ( x + h)
Thus, when E operates on f (x), the result is the next value of the function. Here, E is
called the shift operator. If we apply the operator E twice on f (x), we get
E 2 f ( x) = E[ E f ( x)]
= E[ f ( x + h)] = f ( x + 2h)
Thus, in general, if we apply the operator ‘E’ n times on f (x), we get
E n f ( x) = f ( x + nh)
OR
E n yx = yx + nh
Ey0 = y1 ,   E 2 y0 = y2 ,        E 4 y0 = y4 , … ,                  E 2 y2 = y 4
-1
The inverse operator E is defined as
E −1 f ( x) = f ( x − h)
Similarly
E − n f ( x) = f ( x − nh)
Average Operator, µ ;
it is defined as
1               h              h 
µ f ( x) =  f  x +  + f  x −  
2       2         2 
1
=      y x + ( h / 2) + y x −( h / 2) 
2                                 
Differential Operator, D
it is defined as
d                    
Df ( x) =     f ( x) = f ′( x) 
dx                   
2                
d
D f ( x) = 2 f ( x) = f ′′( x) 
2

dx                  


Important Results Using { ∆, ∇, δ , E , µ }

∆y x = y x + h − y x = Ey x − y x
= ( E − 1) y x
⇒ ∆ = E −1
Also
∇y x = y x − y x − h = y x − E −1 yx
= (1 − E −1 ) y x

Numerical Analysis –MTH603                                                                     VU

E −1
⇒ ∇ = 1 − E −1 =
E
And
δ yx = yx + ( h / 2) − yx −( h / 2)
= E1/ 2 yx − E −1/ 2 y x
= ( E1/ 2 − E −1/ 2 ) y x
δ = E1/ 2 − E −1/ 2
The definition of µ and E similarly yields

1
µ yx =  yx + ( h / 2) + yx −( h / 2) 
2                              
1
= ( E1/ 2 + E −1/ 2 ) y x
2
1
⇒ µ = ( E1/ 2 + E −1/ 2 )
2
We know that
Ey x = y x + h = f ( x + h)
h2
Ey x = f ( x) + hf ′( x) +         f ′′( x) +
2!
h2 2
D f ( x) +
= f ( x) + hDf ( x) +
2!
 hD h 2 D 2     
= 1 +    +    +  f ( x) = e hD y x
    1!   2!     
Thus
hD = log E
Example:
Prove that
hD = log(1 + ∆)
= − log(1 − ∇)
= sinh −1 ( µδ )
Solution:
Using the standard relations we have
hD = log E
= log(1 + ∆)
= − log E −1
= − log(1 − ∇)

Numerical Analysis –MTH603                                                                           VU

1
µδ = ( E1/ 2 + E −1/ 2 )( E1/ 2 − E −1/ 2 )
2
1
= ( E − E −1 )
2
1 hD − hD
= (e − e )
2
= sinh(hD)
⇒ hD = sinh −1 µδ
Example
Prove that
2
 δ2 
1)   1 + δ µ = 1 + 
2       2

   2 
δ
2)        E1/ 2 = µ +
2
δ   2
3)     ∆=              + δ 1 + (δ 2 / 4)
2
∆E −1 ∆
4)      µδ =       +
2     2
∆+∇
5)       µδ =
2
Solution
From the definition, we have:
1                                 1
(1)        µδ = ( E1/ 2 + E −1/ 2 )( E1/ 2 − E −1/ 2 ) = ( E − E −1 )
2                                 2
1                      1
∴ 1 + µ 2δ 2 = 1 + ( E 2 − 2 + E −2 ) = ( E + E −1 ) 2
4                      4
δ 2
1                      1
1+     = 1 + ( E1/ 2 − E −1/ 2 ) 2 = ( E + E −1 ) 2
2        2                      2

(2)
µ + (δ / 2)
1
= ( E1/ 2 + E −1/ 2 + E1/ 2 − E −1/ 2 ) = E1/ 2
2

(3)

Numerical Analysis –MTH603                                                                                              VU

1 1/ 2
(E          − E −1/ 2 ) 1 +     ( E − E −1/ 2 )
2

(E          − E −1/ 2 )
2            1/ 2
1/ 2
δ   2
4
+ δ 1 + (δ 2   / 4) =                               +
2                                         2                                             1

E − 2 + E −1 1 1/ 2
=               + ( E − E −1/ 2 )( E1/ 2 + E −1/ 2 )
2       2

E − 2 + E −1 E − E −1
=               +
2          2

= E −1 = ∆
(4)
1                                                    1
µδ = ( E1/ 2 + E −1/ 2 )( E1/ 2 − E −1/ 2 ) = ( E − E −1 )
2                                                    2
1                 ∆ 1
= (1 + ∆ − E −1 ) = + (1 − E −1 )
2                 2 2
∆ 1  E −1  ∆ ∆
= +             = +
2 2  E  2 2E
(5)
1
µδ = ( E1/ 2 + E −1/ 2 )( E1/ 2 − E −1/ 2 )
2
1
= ( E − E −1 )
2
1                1
= (1 + ∆ − 1 + ∇) = (∆ + ∇)
2                2