A textbook of Computer Based Numerical and Statiscal Techniques part 14 ppsx

116
COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES
Example 14. Evaluate the following:
I. ∆
2
(cos 2x)
II. ∆
2
(3e
x
)
III. ∆ tan
–1
x
IV. ∆(x + cos x)
the interval of differencing being h.
Sol. I. We have ∆
2
(cos 2x)= (E – 1)
2
cos 2x because
∆
= E – 1.
=(E
2
– 2 E + 1) cos 2x
= E
2
cos 2x – 2E cos 2x + cos 2x
= cos (2x + 4h) – 2 cos (2x + 2h) + cos 2x
= cos (2x + 4h) – cos (2x + 2h) – cos (2x + 2h) + cos 2x

= 2 sin (2x + 3h) sin (– h) – 2 sin (2x + h) sin h
= – 2 sin h [sin (2x + 3h) – sin (2x + h)]
= – 2 sin h [2 cos (2x + 2h) sin h]
= – 4 sin
2
h cos (2x + 2h).
II. We have ∆ (3e
x
) = 3(
∆
e
x
) = 3 (e
x +h
– e
x
)
=3e
x
(e
h
–1)
∴
∆
2
(3e
x
)=∆ (∆3e
x
) = ∆{3e

x
(e
h
–1)}
=3 (e
h
–1) (∆e
x
) = 3(e
h
–1) (e
x+h
– e
x
)
=3 (e
h
–1) e
x
(e
h
–1) = 3e
x
(e
h
–1)
2
.
III. We have ∆ tan
–1

x = tan
–1
(x + h) – tan
–1
x
= tan
–1
()
()
1
xh x
xhx
+−
++
= tan
–1

2
1
h
xh x


++


.
IV. We have ∆ (x + cos x) =
∆
x +

∆
cos x
={(x + h) – x} + {cos (x + h) – cos x}
= h + 2 sin
2
2
xh+
sin
2
h

−


= h – 2 sin
2
h
x

+


sin
.
2
h
Example 15. Evaluate
∆
2
E

sin (x + h) +
()
()
2
sin x h
Esin x h
∆+
+
, where h being the interval of differencing.
Sol. To evaluate the given problem we use the operator property that is,
∆
= E – 1
Now
2
E
∆
sin (x + h) +
()
()
2
sin
sin
xh
Exh
∆+
+
=
()
2
1

E
E
−
sin (x + h) +
()()
()
2
1sin
sin 2
Exh
xh
−+
+
CALCULUS OF FINITE DIFFERENCES
117
= (E – 2 + E
–1
) sin (x + h) +
−+ +
+
2
(21)sin()
(2)
EE xh
sin x h
= [sin (x + 2h) – 2 sin (x + h) + sin x] +
sin( 3 ) 2 sin( 2 ) sin( )
sin( 2 )
xh xh xh
xh

+− ++ +

+

= 2 sin (x + h) [cos h – 1] +
2sin( 2 )[cos 1]
sin( 2 )
xh h
xh
+−
+
= 2 (cos h – 1) {sin (x + h) – 1}.
Example 16. Show that B(m + 1, n) = (–1)
m
∆
m




1
n
where m is a positive integer.
Sol. We know that
0
nx
e
∞
−
∫

dx =
1
n
.
Therefore,
0
mnx
e
∞
−
∆
∫
dx = ∆
m

1
n



or
0
mnx
e
∞
−
∆
∫
dx = ∆
m

1
n



,
where for
m
∆
e
–nx
, n is to be regarded variable and x is to be regarded as constant.
Now, ∆
m
e
–nx
= ∆
m–1
[e
–(n+1)x
– e
–nx
]
= ∆
m–1
e
–nx
(e
–x
– 1) = (e

–x
– 1) ∆
m–1
e
–nx
= (e
–x
– 1)
2
∆
m–2
e
–nx
=
= (e
–x
– 1)
m
e
–nx
Therefore,
()
0
1
nx x
ee
∞
−−
−
∫

m
dx = ∆
m

1
n



Put e
–x
= z, so that –e
–x
dx = dz or dx = – (1/z) dz.
Then,
0
1
∫
n
z
(z–1)
m
(–1/z) dz = ∆
m

1
n




or (–1)
m
1
1
0
n
z
−
∫
(1 – z)
m
dz = ∆
m

1
n



or
1
1
0
n
z
−
∫
(1 – z)
(m + 1)–1
dz = (–1)

m
∆
m

1
n



or B (m + 1, n) = (–1)
m
∆
m

1
n



118
COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES
Example 17. Show that e
x
=

∆


2
E

e
x
.
∆
x
2x
Ee
e
; the interval of differencing being h.
Sol. Let f(x) = e
x
, then Ef (x) = f(x + h), therefore Ee
x
= e
x+h
.
Now,
f∆
(x)= f (x + h) – f (x)
∴

x
e
∆
= e
x+h
– e
x
= e
x

(e
h
– 1)
∴
2 x
e
∆
=
∆
(
∆
e
x
) =
∆
{e
x
(e
h
– 1)}

2 x
e
∆
= (e
h
– 1)
∆
e
x

= (e
h
– 1)
2
e
x
∴
2
E

∆


e
x
= (
∆
2
E
–1
) e
x
=
∆
2
(E
–1
e
x
) =

∆
2
(e
x–h
)
=
∆
2
(e
x
e
–h
) = e
–h
∆
2
e
x
= e
–h
(e
h
– 1)
2
e
x
.
∴
2
E


∆


e
x
2
x
x
Ee
e
∆
= e
–h
(e
h
– 1)
2
e
x
()
2
1
xh
hx
e
ee
+
−
= e

–h
e
x+h
= e
x
.
Example 18. Evaluate ∆
2

+

++

2
5x 12
x5x6
; the interval of differencing being unity.
Sol. We have ∆
2

2
512
56
x
xx

+


++


Therefore, ∆
2
2
512
56
x
xx

+


++

= ∆
2
()()
512
23
x
xx

+


++


= ∆
2



+=∆∆ +∆


++ + +



23 2 3
23 2 3xx x x
= ∆
11 11
23
32 43xx xx


−+ −


++ ++



= – 2 ∆
()() ()()
11
3
23 3 4xx xx



−∆

++ ++


= – 2
()()()()
11
34 23xx xx

−

++ ++


– 3
()()()()
11
45 34xx xx

−

++ ++


=
()()()()()()
46
234 345xxx xxx

+
+++ +++
=
()
()()()()
25 16
2345
x
xxxx
+
++++
.
CALCULUS OF FINITE DIFFERENCES
119
Example 19. Evaluate ∆
n
e
ax+b
; where the interval of differencing taken to be unity?
Sol. Given ∆
n
e
ax+b
; which shows that f(x) = e
ax+b
.
Now ∆f (x)= f(x + 1) – f(x)
∴
∆(e
a + bx

)= e
a(x + 1)+b
– e
ax + b
= e
ax + b
(e
a
– 1)
∴
∆
2
(e
a + bx
)= ∆ (
∆
e
a + bx
) = ∆ {e
ax + b
(e
a
–1)}
= (e
a
– 1) (∆ e
ax + b
)
= (e
a

– 1) e
ax + b
(e
a
–1)
= (e
a
–1)
2
e
ax + b
.
Proceeding in the same way, we get
∆
n

e
ax + b
= (e
a
–1)
n
e
ax + b
Example 20. With usual notations, prove that,
∆
n





1
x
= (–1)
n
.
() ( )
!
n
nh
x x h x nh
++
Sol.
1
n
x

∆


=
1
1
n
x
−

∆∆



=
1
11
n
xhx
−

∆−

+

=
1
()
n
h
xx h
−

−
∆

+

=
−

−∆ ∆

+


2
1
()
()
n
h
xx h
=
−


−∆ ∆ −


+


2
11
(1)
n
xxh
=
2
11 1 1
(1)
2
n
xhx x hxh

−

 
−∆ − − −
 

+++
 

=
2
21 1
(1)
2
n
xh x x h
−

−∆ − −

++

=
2
2
2
(1)
()(2)
n
h

xx h x h
−

−
−∆

++

=
−

−∆

++

2
22
2!
(1)
()(2)
n
h
xx h x h
=
3
33
!
(1)
()(2)(3)
n

h
xx hx hx h
−

3
−∆

++ +

.
.
.
.
.
=
!
(1)
( ) ( )
n
n
nh
xx h x nh
−
++
Example 21. Prove that:
(a)
µµ −δδ

µ=



−+
1
f(x) g(x) f(x) g(x)
f(x)
4
11
g(x)
g(x )g(x )
22
Here, interval of differencing being unity.
120
COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES
(b)
∇
2
= h
2
D
2
– h
3
D
3
+
7
12
h
4
D

4 –
(c)
∇
– ∆ = –
∇
∆
(d) 1 +



2
2
δ
=
+
22
1
δµ
(e)
µδ
=
1
2
()∆+∇
Sol.
(a) Here R.H.S. is =
1
() () () ()
4
11

()()
22
fx gx fx gx
gx gx
µµ −δδ
−+
Now numerator of R.H.S. is given by:
=
1
2
[E
1/2
+ E
–1/2
] f(x) .
1
2
(E
1/2
+ E
–1/2
) g(x) –
1
4
(E
1/2
– E
–1/2
) f(x) (E
1/2

– E
–1/2
)g(x)
=
1
4
[ f (x +
1
2
) + f(x –
1
2
)] [g(x +
1
2
) + g(x –
1
2
)] –
1
4
[ f (x +
1
2
) – f(x–
1
2
] [g(x +
1
2

)
– g(x –
1
2
)]
=
1
4
[f (x +
1
2
)g(x +
1
2
) + f(x +
1
2
) g(x –
1
2
) + f (x –
1
2
) g(x +
1
2
) f(x –
1
2
)g(x –

1
2
)
–
1
4
[f(x +
1
2
) g(x +
1
2
) – f (x +
1
2
) g(x –
1
2
) – f (x –
1
2
) g(x +
1
2
) + f(x –
1
2
) f(x–
1
2

)]
=
1
2
[f (x +
1
2
) g(x –
1
2
) + f (x –
1
2
) g(x +
1
2
) ]
Therefore right hand side is =
111 11
222 22
11
22
fx gx fx gx
gx gx


+−+−+






−+


=
()
()
()
()
1/2 1/2
11
1
22
11
22
22
fx fx
fx fx
EE
gx gx
gx gx
−


+−



+


+= =µ





+−




(b) We know E = e
hD
and
∇
= 1 – E
–1
, therefore
∇
2
= (1 – e
–hD
)
2
.
=
() () ()
2
234

1 1
2! 3! 4!
hD hD hD
hD




−− + − + −





=
() () ()


−+−+



2
234

2! 3! 4!
hD hD hD
hD
CALCULUS OF FINITE DIFFERENCES
121

= h
2
D
2
()




−− +





2
2
1
26
hD
hD
= h
2
D
2
() ()
 


 

+− + − − +

 


 
2
22
1 2
26 26
hD hD
hD hD
= h
2
D
2
()


−++ −




2
11
1
43
hD hD
= h

2
D
2


−+ −


22
7
1
12
hD h D
= h
2
D
2
– h
3
D
3
+
7
12
h
4
D
4
–
(c)

∇−∆
= (1 – E
–1
) – (E – 1) =
1E
E
−



– (E – 1) = (E – 1) (E
–1
– 1)
= – (E – 1) (1 – E
–1
) = – ∇∆
(d) L.H.S. =
2
1
2
x
y


δ

+






=
()
2
1/2 1/2
1
2
x
EE
y
−

−

+



=
1
2
1
2
EE
−


+−


+






y
x
=
1
2
(E + E
–1
)y
x
.
R.H.S. =
(
)
22
1
+δ µ
y
x
=
()()
{}
1/2
22

1/2 1/2 1/2 1/2
1
1.
4
x
EE EE y
−−

+− +


=
()
1/2
2
1
1
4
EE
−


−


+







y
x
=
1/2
22
2
4
EE
−

++



y
x
=
1
2
EE
−

+



y
x

Hence, L.H.S. = R.H.S.
(e)
x
y
µδ
=
µ
(E
1/2
– E
–1/2
)y
x
=
µ

22
hh
xx
yy
+−

−



=
µ

2

h
x
y
+




–
µ

2
h
x
y
−




=
1
2
(E
1/2
+ E
–1/2
)
2
h

x
y
+




–
1
2
(E
1/2
+ E
–1/2
)
2
h
x
y
−




=
1
2
(y
x+h
+ y

x
) –
1
2
(y
x
+ y
x–h
) =
1
2
(y
x+h
– y
x
) +
1
2
(y
x
– y
x–h
)
=
1
2

()
x
y

∆
+
1
2

()
x
y
∇
=
1
2

()
∆+∇
y
x
Hence,
µδ
=
1
2

()
∆+∇
.
122
COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES
Example 22. Evaluate: ∆
n

[sin (ax + b)]
Sol. We know
f∆
(x) = f(x + h) – f (x) therefore
∆
sin (ax + b) = sin [a (x + h)+b] – sin)(ax + b)
= 2 sin
2
ah
cos
2
h
ax b


++




= 2 sin
2
ah
sin
2
ah
ax b
+π

++



Therefore,
∆
2
sin (ax + b)=
∆
2sin sin
22
ah
ah
ax b
+π


++




= (2 sin
2
ah
) (2 sin
2
ah
) sin [ax + b +
2
ah +π
+

2
ah +π
]
=
2
2sin
2
ah



sin
2
2
ah
ax b
π
+

++




On continuing in the same manner, we get
∆
3
sin (ax + b) =
3
2sin

2
ah



sin
3( )
2
ah
ax b
π
+

++






∆
n
sin (ax + b) =
2sin
2
n
ah




sin
()
2
nah
ax b
π
+

++




Example 23. Show that: u
0
– u
1
+ u
2
– =
1
2
u
0
–
1
4

∆
u

0
+
1
8

∆
2
u
0
–
Sol. On taking left hand side = u
0
– u
1
+ u
2
– u
3
+
= u
0
– Eu
0
+ E
2
u
0
– E
3
u

0
+
= (1 – E + E
2
– E
3
+ )u
0
=
()
1
1 E


−−


u
0
=
1
1 E


+

u
0
=
1

11


++∆

u
0
=
1
2


+∆

u
0
=
1
2

1
1
2
−
∆

+


u

0
=
1
2

∆∆ ∆
−+−+


23
1
24 8
u
0
=
1
2
u
0
–
1
4

∆
u
0
–
1
8
∆

2
u
0
–
1
16

∆
3
u
0
+ (R.H.S.)
CALCULUS OF FINITE DIFFERENCES
123
Example 24. Prove that:
(1)
() ()
δ


fxgx
=
fµ
(x)
()
gx
δ
+
()
gx

µ
()
fx
δ
(2)
()
()
() () () ()

µδ−µδ
δ=



−+


fx gx fx fx gx
11
gx
gx g x
22
Interval of differencing is unity.
Sol.
(1) R.H.S. =
fµ
(x)
()
gx
δ

+
()
gx
µ
()
fx
δ
=
1/2 1/2
2
EE
−
+
f(x). (E
1/2
– E
–1/2
) g(x) +
1/2 1/2
2
EE
−
+
g(x). (E
1/2
– E
–1/2
) f(x)
=
1

2
[{f(x +
1
2
) + f(x –
1
2
)} {g (x +
1
2
) – g (x –
1
2
)} + {g (x +
1
2
)
+ g (x –
1
2
)} {f (x +
1
2
) – f (x –
1
2
)}]
=
1
2

[{f(x +
1
2
) g (x +
1
2
)– f (x+
1
2
) g (x –
1
2
) + f(x –
1
2
) g (x +
1
2
)
– f (x –
1
2
) g (x –
1
2
)} + {f(x +
1
2
) g (x +
1

2
) + f(x +
1
2
) g (x –
1
2
)
– f (x –
1
2
) g (x +
1
2
) – f (x –
1
2
) g (x –
1
2
)}]
=
1
4
f(x +
1
2
) g(x +
1
2

) – f (x –
1
2
) g (x –
1
2
)
= E
1/2
f (x) g(x) – E
–1/2
f(x) g(x) = (E
1/2
– E
–1/2
) f (x) g(x) =
fδ
(x) g(x).
(2) R.H.S. =
() () () ()
11
22
gx f x f x gx
gx gx
µδ−µδ

−+


Now first we solve the numerator of right hand side.

=
1/2 1/2
2
EE
−
+
g(x) (E
1/2
– E
–1/2
) f(x) –
1/2 1/2
2
EE
−
+
f(x) (E
1/2
– E
–1/2
) g(x)
=
1
2
[{g(x +
1
2
) + g(x –
1
2

)} {f(x +
1
2
) – f(x –
1
2
)} – {f(x +
1
2
)
+ f(x –
1
2
)} {g(x +
1
2
) – g(x –
1
2
)}]
=
111 11 11 1 1
[( )( ) ( )( ) ( )( ) ( ) ( )]
222 22 22 2 2
f x gx f x gx fx gx f x gx++++−−−+−− −
–
111 11 11 1 1
[( )( ) ( )( ) ( )( ) ( ) ( )]
222 22 22 2 2
fx gx fx gx fx gx f x gx++−+−+−+−− −

= f (x +
1
2
) g(x –
1
2
) – f(x –
1
2
) g(x +
1
2
)
124
COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES
Therefore right hand side as
=
11 11
()()()()
22 22
11
()()
22
fx gx fx gx
gx g x
+−−−+
−+
=
1
()

2
1
()
2
fx
gx
+
+
–
1
()
2
1
()
2
fx
gx
−
−
= E
1/2
()
()
fx
gx



– E
–1/2

()
()
fx
gx



= (E
1/2
– E
–1/2
)
()
()
fx
gx



=
()
()
fx
gx

δ


.
Example 25. Evaluate: (a)

x
2
(x 1)!
∆
+
; differencing 1. (b)
∆(log)
ax
ebx
.
Sol.
(a) Let f(x)= 2
x
, g(x) = (x + 1)!, therefore
1
() 2 2 2
xxx
fx
+
∆= −=
and
g∆
(x) = (x + 1 + 1)! – (x + 1)! = (x + 1) (x + 1)!
()
()
fx
gx

∆



=
∆− ∆
+
() () () ()
( )()
gx fx fx gx
gx hgx
=
(1)!.22.(1)(1)!
( 1 1)!( 1)!
xx
xxx
xx
+−++
++ +
(Because h = 1)
=
2 ( 1) (!(1 1)
2
(2)!(1)! (2)!
x
x
xx
x
xx x
+−−
=−
++ +
.

(b) Again let, f(x)= e
ax
, g(x) = log bx, therefore
∆f(x)=
()
(1)
ax h ax ax ah
eeee
+
−= −
∆g(x)=
+− = +log ( ) log log(1 )
h
bx h bx
x
We know that
()()fxgx∆
=
( ) () () ()fx h gx gx fx+∆ + ∆
Therefore
(log)
ax
ebx
∆
=
()
+

++ −



)
log 1 (log ) ( 1)
ax h
ax ah
h
ebxee
x
∆(e
ax
log bx)= e
ax
[e
ah
log
1
h
x

+


+ (e
ah
–1) log bx].
Example 26. Prove that, hD =– log (1 –
∇
) = sin h
–1
(µδ).

Sol. Because, E
–1
=1 –
∇
therefore,
hD = log E = – log (E
–1
) = – log (1 –
∇
)
CALCULUS OF FINITE DIFFERENCES
125
Also, µ =
1/2 1/2
1
()
2
EE
−
+
δ = E
1/2
– E
–1/2
Therefore, µδ =
−−
−= − =
1
11
()( )sin()

22
hD hD
EE e e hhD
hD = sin h
–1
(µδ).
Example 27. Show that ∆ log f(x) =

∆
+


f(x)
log 1
f(x)
Sol. L.H.S.
log ( )fx∆
=
log ( ) log ( )fx h fx+−
=
() ()
log log
() ()
fx h Efx
fx fx

+
=



=
(1 ) ( )
log
()
fx
fx

+∆


=
() ()
log
()
fx fx
fx

+∇


=
()
log 1
()
fx
fx

∆
+



Example 28. Evaluate (1)

∆


n
1
x
(2)
n
∆
(ab
cx
).
Sol.
(1) We have,
1
n
x

∆


= ∆n
–1
∆
1
x




.
Now, ∆
1
x



=
1
1x
+
–
1
x
=
()
()
1
1
xx
xx
−+
+
=
()
()
1
1xx

−
+
∆
2
1
x



= ∆∆
1
x



= ∆
()
()
1
1xx

−


+


= (–1) ∆
()
1

1xx



+


= (–1)
()()()
11
12 1xx xx


−

++ +


= (–1)
()
()()
2
12
xx
xx x
−+
++
=
()()
()()

−−
++
12
12xx x
∆
3
1
x



=
()()()
()()()
123
123xx x x
−−−
+++