A textbook of Computer Based Numerical and Statiscal Techniques part 15 doc

126
COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES
Similarly, ∆
n
1
x



=
()()() ()
()()()
−−− −
++ +
1 2 3
1 2
n
xx x x n
=
()
()()()
1!
12
n
n
xx x x n
−
++ +
(2) Similarly, ∆ (ab
cx
)= a ∆ b

cx
= a{b
c(x + 1)
– b
cx
}
= a{b
cx
b
c
– b
cx
} = a(b
c
–1)b
cx
.
2
()
cx
ab
∆
=
{}
() (1)
cx
ccx
ab a b b
∆∆ = ∆ −
=

2
(1) (1)
ccxccx
ab b ab b
−∆ = −
Proceeding in the same manner, we get
()
ncx
ab
∆
= a(b
c
– 1)
n
b
cx
.
Example 29. If p, q, r and s be the successive entries corresponding to equidistant arguments
in a table, show that when third differences are taken into account, the entry corresponding to the argument
half way between the arguments of q and r is
+
1
AB
24
, where A is the arithmetic mean of q, r and B is
the arithmetic mean of 3q – 2p – s and 3r – 2s – p.
Sol. On taking h being the interval of differencing the difference table is as:
∆∆ ∆
−
+−+

−−+−
+−+
−
+
23
2
33
22
3
xx x x
xu u u u
ap
qp
ah q r qp
rq s r qp
ahr srq
sr
ahs
The argument half way between the arguments of q and r is
1
2
(a + h + a + 2h) i.e.,
3
2
ah+
.
Hence, the required entry is given by,
u
a+(3/2)h
= E

3/2
u
a
= (1 + ∆)
3/2
u
a
=


+∆+ ∆+ − ∆




23
3311 3111
1
2222! 2223!
a
u
,
(Higher order differences being neglected).
Therefore u
a+(3/2)h
=
23
33 1
28 16
aa a a

uu u u
+∆ +∆ − ∆
=
33 1
()(2) (33)
28 16
pqp rqp srqp+−+−+− −+−
CALCULUS OF FINITE DIFFERENCES
127
=
331 333 33 1
1
2 8 16 2 4 16 8 16 16
pqrs

−++ + −− + + −


=
1991
16 16 16 16
pqrs−++−
=
1111
()
16 16 2 16
pqr s

−++ +−



=
11
() ( )
216
qr qrps++ +−−
(1)
Again A = arithmetic mean of q and r =
1
()
2
qr+
B = Arithmetic mean of 3q – 2p – s and 3r – 2s – p is
=
[]
()
13
32 32
22
q ps r sp qrsp
−−+−−= +−−
.
∴
1
24
AB+
=
1
().
216

qr
qrsp
+
++−−
Substituting this value in (1), we get u
a+(3/2)h
= A +
1
24
B.
Example 30. Given u
0
, u
1
, u
2
, u
3
, u
4
and u
5
. Assuming that, fifth order differences to be constant.
Show that:
−+ −
=+
1
2
2
1 25(c b) 3(a c)

uc
2 256
. where a = u
0
+ u
5
, b = u
1
+ u
4
, c = u
2
+ u
3
Sol. L.H.S.
1
2
2
u
= E
5/2
u
0
= (1 + ∆)
5/2
u
0
=
25
0

55 55555
.1 1234
5
22 22222
1
22! 5!
u

 
−−−−−
 

 
+∆+ ∆+ + ∆



=
23 4 5
00 0 0 0 0
515 5 5 3
2 8 16 128 256
uu u u u u+∆ + ∆ + ∆ − ∆ + ∆
=
010 210 3210
515 5
( ) ( 2 ) ( 3 3 )
28 16
u uu uuu uuuu+−+ −++ −+−+
543210

3
( 5 10 10 5 )
256
uu u uuu+−+−+−
128
COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES
=
05 14 23
32575
()()()
256 256 128
uu uu uu+− ++ +
=
32575
256 256 128
abc−+
=
325111
256 256 2 128
ab c

−++


=
3( ) 25( )
2 256
cac cb−+ −
+
(R.H.S.)

Example 31. Given:
u
0
+ u
8
= 1.9243, u
1
+ u
7
= 1.9590, u
2
+ u
6
= 1.9823, u
3
+ u
5
= 1.9956. Find u
4
.
Sol. Since 8 entries are given, therefore we have ∆
8
u
0
= 0
i.e. (E – 1)
8
u
0
= 0

i.e. (E
8
–
8
C
1
E
7
+
8
C
2
E
6
–
8
C
3
E
5
+
8
C
4
E
4
–
8
C
5

E
3
+
8
C
6
E
2
–
8
C
7
E
1
+ 1)u
0
= 0
i.e. (E
8
– 8E
7
+ 28E
6
– 56E
5
+ 70E
4
– 56E
3
+ 28E

2
– 8E + 1)u
0
= 0
i.e. u
8
– 8u
7
+ 28u
6
– 56u
5
+ 70u
4
– 56u
3
+ 28u
2
– 8u
1
+ u
0
= 0
i.e. (u
8
+ u
0
) – 8(u
7
+ u

1
) + 28(u
6
+ u
2
) – 56(u
5
+ u
3
) + 70u
4
= 0
On putting the given values, we get
1.9243 – 8(1.9590) + 28(1.9823) – 56 (1.9956) + 70u
4
= 0
or –69.9969 + 70u
4
= 0
or u
4
= 0.9999557.
Example 32. Sum the following series 1
3
+ 2
3
+ 3
3
+ + n
3

using the calculus of finite
differences.
Sol. Let 1
3
= u
0
, 2
3
= u
1
, 3
3
= u
2
, , u
3
= u
n–1
. Therefore sum is given by
S = u
0
+ u
1
+ u
2
+ + u
n–1
= (1 + E + E
2
+ E

3
+ + E
n–1
)u
0
=
1
1
n
E
E

−


−

u
0
=
()
11
n

+∆ −

∆


u

0
=
1
∆

() ()()
23
112
1 1
2! 3!
n
nn nn n
n
−−−

+∆+ ∆+ ∆+ +∆−


u
0
=
()
()
()
2
00
12
1
2! 3!
nn n

nn
nu u
−−
−
+∆+ ∆
+
We know ∆u
0
= u
1
– u
0
= 2
3
– 1
3
= 7.
∆
2
u
0
= u
2
– 2u
1
+ u
0
= 3
3
–2(2)

3
+ 1
3
= 12.
Similarly we have obtained
∆
3
u
0
= 6 and
∆
4
u
0,
∆
5
u
0
, are all zero as

u
r
=

r
3

is a polynomial
of third degree.
CALCULUS OF FINITE DIFFERENCES

129
∴
S = n +
()
1
2!
nn
−
(7) +
()()
12
6
nn n
−−
12 +
()()()
123
24
nn n n
−−−
(6)
=
2
4
n
(n
2
+ 2n + 1) =
()
2

1
2
nn

+


Example 33. Prove that:
∞∞
==

∆∆
=+−+−


∑∑
2
2x
x
x0 x0
11
uu1
2424
u
0
.
Sol. Taking right hand side of the given expression
=
2
0

11
1
2424
x
x
u
∞
=

∆∆
+−+−


∑
u
0
=
1
2
(u
0
+ u
1
+ u
2
+ u
3
+ ) +
1
4

1
1
2
−
∆

+


u
0
=
1
2
(u
0
+ Eu
0
++ E
2
u
0
+ E
3
u
0
+ ) +
1
4

1
1
2
−
∆

+


u
0
=
1
2
(1 + E + E
2
+ E
3
+ )u
0
+
1
4

1
1
2
−
∆


+


u
0
=
1
2
(1 – E)
–1
u
0


+

1
2
(2 +
∆
)
–1
u
0
=
1
2
(1 – E)
–1

u
0


+

1
2
(1 +
∆
)
–1
u
0
=
1
2
[(1 – E)
–1
+ (1 + E)
–1
]u
0
=
1
2
. 2 [1 + E
2
+ E
4

+ E
6
+ ]u
0
= u
0
+ u
2
+ u
4
+ u
6
+
=
2
0
x
x
u
∞
=
∑
= L.H.S.
Example 34. Given that u
0
= 3, u
1
= 12, u
2
= 81, u

3
= 200, u
4
= 100, u
5
= 8. Find the value of
∆
5
u
0
.
Sol. We know
∆
= E – 1, therefore,
5
0
u∆
=

(E –1)
5
u
0
= (E
5
– 5E
4
+ 10E
3
– 10E

2
+ 5E – 1)u
0
=

u
5
– 5u
4
+ 10u
3
– 10u
2
+ 5u
1
– u
0
= 8 – 500 + 2000 – 810 + 60 – 3
= 755.
130
COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES
PROBLEM SET 3.1
1. Form the forward difference table for given set of data:
X:10203040
Y: 1.1 2.0 4.4 7.9
2. Construct the difference table for the given data and hence evaluate
3
(2)
f∆
.

X:01234
Y: 1.0 1.5 2.2 3.1 4.6 [Ans. 0.4]
3. Find the value of E
2
x
2
when the values of x vary by a constant increment of 2.
[Ans. x
2
+ 8x + 16]
4. Evaluate E
n
e
x
when interval of differencing is [Ans. E
n
e
x
= e
x+nh
]
5. Evaluate ∆
3
(1 –x) (1–2x) (1–3x) ; the interval of differencing being unity.
[Ans.
3
() 36
fx∆=−
]
6. If f(x) = exp (ax), evaluate

()
n
fx
∆
[Ans.
(1)
nax ah nax
ee e
∆=−
]
7. Evaluate
2
3
x
E

∆


[Ans. 6x]
8. Find the value of
24
2
22
(1)
xx
aa
a

+

∆

−

[Ans.
2224
(1)
xx
aa a
++
]
9. Evaluate:
(a)
cot 2
x
∆
[Ans. –Cosec 2
x+1
]
(b) ∆+sin ( )ha bx [Ans.
++2sin cos ( )
22
bb
hhabx
]
(c)
tan ax∆
[Ans.
sin
cos cos ( 1)

a
ax a x
+
]
10. Prove that:
(a)
1/2
1
2
E =µ= δ
(b)
1/2 1/2 1
()
EE E
−−
δ+ =∆+∆
(c)
1/2 1/2
(1 ) (1 )
−−
δ=∇ −∇ =∆ +∆
(d)
1/2 1/2
EE
−
δ=∆ =∇
(e)
2
∇∆ = ∆∇ = δ
(f)

11 1
1
EE E
−− −
∇=∆ = ∆= −
11. Show that:
(a)
sin
cot( )
sin( ) sin( )
b
abx
abx abbx
−
∆+=
+++
(b)
(( )
sin( ) (2sin ) sin(
22
n
n
bnb
abx abx
+π
∆+= ++
(c)
()
cos( ) (2 sin ) cos(
22

n
n
bnb
abx abx
+π
∆+= ++
CALCULUS OF FINITE DIFFERENCES
131
12. What is the difference between
2
E
∆



u
x
and
2
2
x
x
u
Eu

∆



. If u

x
= x
3
and the interval of
differencing is unity. Find out the expression for both. [Ans. 6h
2
(3x – h),
23
3
66
(2)
xh h
xh
+
+
]
13. If f(x) = e
ax
, show that f(0) and its leading differences form a geometrical progression.
14. A third degree polynomial passes through the points (0, –1), (1, 1), (2, 1) and (3, 2). Find
the polynomial. [Ans.
32
1
(3166)
6
xx x−+−+
]
15. Prove that ∆ sin
–1
x =

22
[( 1) 1 1 ( 1) ]
xxxx
+−−−+
.
3.5 FUNDAMENTAL THEOREM ON DIFFERENCES OF POLYNOMIAL
Statement: If f(x) be the nth degree polynomial in x, then the n
th
difference of f(x) is constant and
∆
n+1
f(x) and all higher differences are zero when the values of the independent variables are at
equal interval.
Proof: Consider the polynomial f(x) = a
0
+ a
1
x + a
2
x
2
+ + a
n
x
n
(1)
Where n is a positive integer and a
0
, a
1

, a
2
, a
n
are constants.
We know ∆f(x)= f(x + h) – f(x).
On applying the operator ∆ on equation (1), we get
∆f(x)= ∆(a
0
+ a
1
x + a
2
x
2
+ + a
n
x
n
)
⇒ f(x + h) – f(x)= [a
0
+ a
1
(x + h) + a
2
(x + h)
2
+ + a
n

(x + h)
n
]
–[a
0
+ a
1
x + a
2
x
2
+ +a
n
x
n
]
22 33
12 3
[( ) ] [( ) ] [( ) ]
nn
n
ahaxhxaxhx axhx
⇒+ +−+ +−+ + +−
⇒ a
1
b + a
2
[
2
C

1
xh + h
2
] + a
3
[
1
2
323 23 1 2
12 1
] [ ]
n
nn nCx nn
nn
Cxh Cxh h a Cx h h Ch
−
−
++++ + +
221
12 3 1

nn
nn
bbxbx bx nahx
−−
−
⇒+ + + + +
(2)
where b
1

, b
2
, b
n–1
are constant coefficients.
According to equation (2), we have the first difference of equation (1) is again a polynomial
of degree n – 1.
From this we say that ∆f(x) is one degree less than the degree of original polynomial.
Again, on taking a difference of equation (2) i.e. second difference of equation (1), we get

2222
23 4
( ) ( 1)
n
n
fx C Cx Cx nn hax
−
∆=++++−
…(3)
This is a polynomial of degree n – 2.
Thus, on continuing this process up to nth difference we get a polynomial of degree zero.
Such that:
()
n
fx
∆
=
( 1)( 2) 1.
nnn
n

nn n h ax
−
−−
=
0
!
n
n
nhax
=
!
n
n
nha
132
COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES
Hence, we have nth difference of the polynomial is constant and so all higher differences are
each zero. i.e.
12
() () 0
nn
fx fx
++
∆=∆==
3.6 ESTIMATION OF ERROR BY DIFFERENCE TABLE
Let y
0
, y
1
, y

2
, y
n
be the exact values of a function y = f(x) corresponding to arguments
x
0
, x
1
, x
2
, , x
n
. Now to determine error in such a case and to correct the functional values,
let an error δ is made in entering the value of y
3
in the table so that erroneous value of y
3
is
y
3
+ δ.
2
00
11 0
2
22 1 0
2
33 2 1
2
44 3 2

2
55 4 3
4
66 5
2
xy y y
xy
xy y
xy y y
xy y y
xy y y
xy y y
y
xy y
∆∆
∆
∆∆
+δ ∆ +δ ∆ +δ
∆−δ ∆ −δ
∆∆+δ
∆
∆
From the above difference table we noted that:
1. The error in column y affects two entries in column ∆y, three entries in column ∆
2
y and
so on. i.e. the error spreads in triangular form.
2. The error increases with the order of differences.
3. The coefficients of δ’s are binomial coefficients with alternative signs +, –,
4. In various difference columns of the above table the algebraic sum of the errors is zero.,

5. The errors in the column ∆
i
y are given by the coefficients of the binomial expansion
(1 –δ)
i
.
6. In even differences columns of ∆
2
y, ∆
4
y, , the maximum error occurs in a horizontal line
in which incorrct value of y lies.
7. In odd difference columns of ∆
1
y, ∆
3
y, , the maximum error lies in the two middle
terms and the incorrect value of y lies between these two middle terms.
Example 1. Find the error and correct the wrong figure in the following functional values:
Sol.
1234567
2 5 10 18 26 37 50
x
y
CALCULUS OF FINITE DIFFERENCES
133
∆∆ ∆
−
−
23

12
3
25 2
51
310 3
83
418 0
83
526 3
11 1
637 2
13
750
xy y y y
Here the sum of all the third differences is zero and the adjacent values –3, 3 are equal in
magnitude. Also horizontal line between –3 and 3 points out the incorrect functional value 18.
Therefore coefficient of first middle term on expansion of (1 – p)
3
= –3
⇒ –3e = –3 ⇒ e = 1
∴ Correct functional value = 18 – 1 = 17.
Example 2. Find and correct by means of differences the error in the following table:
20736, 28561, 38416, 50625, 65540, 83521, 104976, 130321, 160000
Sol. For the given data we form the following difference table:
∆∆∆∆∆
−
−
2345
20736
7825

28561 2030
9855 324
38416 2354 28
12209 352 20
50625 2706 8
14915 360 40
65540 3066 48
17981 408 40
83521 3474 8
21455 416 20
104976 3890 28
25345 444
130321 4334
29679
160000
y y y yyy
→
←
134
COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES
From this table we have the third differences are quite iregular and the irregularity starts
around the horizontal line corresponding to the value y = 65540.
Since the algebraic sum of the fifth differences is 0, therefore –5ε = –20 ⇒ ε = 4.
Therefore the true value of y
5
= 65540 – 4 = 65536.
Example 3. Locate the error in following entries and correct it.
1.203, 1.424, 1.681, 1.992, 2.379, 2.848, 3.429, 4.136
Sol. Difference table for given data is as follows:
∆∆ ∆ ∆

−
−
3 3 32 33 34
10 10 10 10 10
1203
221
1424 36
257 18
1681 54 4
311 22
1992 76 16
387 6
2379 82 24
469 30
2848 112 16
581 14
3429 126
707
4136
yy y y y
Sum of all values in column of fourth difference is –0.004 which is very small as compared
to sum of values in other columns.
∴∆
4
y = 0
Errors in this column are e, –4e, 6e, –4e and e.
Term of Maximum value = 24 ⇒ 6e = 24 ⇒ e = 4.
Error lies in 2379.
Hence, required correct entry = 2379 – 4 = 2375.
Hence, correct value = 2.375

→
←
CALCULUS OF FINITE DIFFERENCES
135
Example 4. One number in the following is misprint. Correct it.
1 2 4 8 16 26 42 64 93.
Sol. Difference table for given data it as follows:
∆∆∆∆∆
−
−
−
−
−
2345
11
1
22 1
21
34 2 1
42 5
48 4 4
8210
516 2 6
10 4 10
626 6 4
16 0 5
742 6 1
22 1
864 7
29

993
xyyyyyy
In the above table, the fourth difference column have algebraic sum of all the values is 0.
The middle term of this difference column is 6.
∴ 6e = 6 or e = 1.
∴ Correct value is given by 16 – 1 = 15.
Example 5. Locate the error in the following table:
1234567891011
1.0000 1.5191 2.0736 2.6611 3.2816 3.9375 4.6363 5.3771 6.1776 7.0471 8.0000
x
y