A textbook of Computer Based Numerical and Statiscal Techniques part 20 pptx

176
COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES
Sol. The difference table for the given date is as follows:
xy y y y∇∇ ∇
23
34.8
3.6
4 8.4 2.5
6.1 0.5
514.5 3
9.1 0.5
623.6 3.5
12.6 0.5
7 36.2 4
16.6 0.5
852.8 4.5
21.1
9 73.9
To obtain first term, we use Newton’s forward interpolation formula,
Here, a = 3, h = 1, x = 1 ∴ u = –2
Hence we have
f(1) =
23
33 3 3
(1) (1)(2)
2! 3!
uu uu u
yuy y y
−−−
+∆ + ∆ + ∆
On putting the subsequent values, we get

f(1) =
()
()
2(3)
(2)(3)(4)
4.8 2 3.6 (2.5) (0.5) 3.1
26
−−
−−−
+− × + + =
Similarly, to obtain tenth term, we use Newton’s backward interpolation formula. So
a + nh = 9, h = 1, a + nh + uh = 10
∴ u = 1
⇒ f (10) =
23
99 9 9
(1) (1)(2)
2! 3!
uu uu u
yuy y y
+++
+∇ + ∇ + ∇
= 73.9 + 21.1 + 4.5 + 0.5 = 100
Example 4. Find the value of an annuity at
3
5
8
%,
given the following table:
11

44 5 5 6
22
172.2903 162.888 153.7245 145.3375 137.6483
Rate
Annuity Value
INTERPOLATION WITH EQUAL INTERVAL
177
Sol. Difference table,
23 4

4 172.2903
9.4014
1
4 162.888 0.237
2
9.1644 0.537
5 153.7245 0.774 0.6132
8.387 0.0762
1
5 145.3375 0.6978
2
7.6892
6 137.6483
Rate Annuity Values ∇∇ ∇ ∇
−
−
−
−−
−
x =

343 1
5,6,
88 2
an===
y =
()
23
( 1.25)( 0.25) ( 1.25)( 0.25)(0.75)
(6) 1.25 (6) (6) (6)
2! 3!
yy y y
−− −−
+− ∇ + ∇ + ∇
=
( 1.25)( 0.25)
137.6483 ( 1.25)( 7.6892) (0.6978)
2!
−−
+− − +
( 1.25)( 0.25)(0.75) ( 1.25)( 0.25)(0.75)(1.75)
( 0.0762) (0.6132)
3! 4!
−− −−
+−+

= 147.2251 Approx.
Example 5. In an examination, the number of candidates who obtained marks between certain limits
are as follows:
Marks
No of candidates

−−−−−0192039405960798099
. 41 62 65 50 17
Find no. of candidates who obtained fewer than 70 marks.
Sol. First, we form the difference table.
Marks less than x No of candidates y ∇∇ ∇ ∇
−
−
−
−
234
() . ()
19 41
62
39 103 3
65 18
59 168 15 0
50 18
79 218 33
17
99 235
178
COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES
Here, we have h = 20, a = 99
∴ u =
70 99
1.45
20
−
=−
Now on applying ‘Newton’s backward difference formula, we get

f(70) =
23 4
(1)(1)(2)(1)(2)(3)
() () () () ()
2! 3! 4!
uu uu u uu u u
fa ufa fa fa fa
++++++
+∇ + ∇ + ∇ + ∇
= 235 + (–1.45)(17) +
( 1.45)( 0.45) ( 1.45)( 0.45)(0.55)
( 33) ( 18)
26
−− −−
×− + ×−
= 235 – 24.65 – 10.76625 – 1.076625
= 235 – 36.492875
= 198.507 { 198
∴ Total no. of candidates who obtained fewer than 70 marks are 198.
Example 6. The area A of a circle of diameter d is given for the following values:
d
A
80 85 90 95 100
5026 5674 6362 7088 7854
Find A for 105.
Sol. First of all we form the difference table as follow:
dA
xfx
∇∇∇∇
−

234
() ()
80 5026
648
85 5674 40
688 2
90 6362 38 4
726 2
95 7088 40
766
100 7854
Here, h = 5, a = 100, x = 105
∴ u =
105 100
1
5
−
=
Now on applying Newton’s backward difference formula, we have
(105)f
=
23 4
(1)(1)(2)(1)(2)(3)
() () () () ()
2! 3! 4!
uu uu u uu u u
fa ufa fa fa fa
++++++
+∇ + ∇ + ∇ + ∇
INTERPOLATION WITH EQUAL INTERVAL

179
=
40 2 1 2 3 1 2 3 4
7854 1 766 2 4
26 24
× ×× ×××
+× + + ×+ ×
= 7854 + 766 + 46
= 8666
Which is the required area for the given diameter of circle.
Example 7. The probability integral
2
1
t
2
0
2
Pe
−
=
∫
π
π
dt has the following values:
x
P
1.00 1.05 1.10 1.15 1.20 1.25
0.682689 0.706282 0.728668 0.749856 0.769861 0.788700
Calculate P for x = 1.235
Sol. First we form the difference table

xfx ∇∇ ∇∇ ∇
−
−
−
−
−
2345
()
1.00 0.682689
0.023593
1.05 0.706282 0.001207
0.022386 0.000009
1.10 0.728668 0.001198 0.000006
0.021188 0.000015 0.000004
1.15 0.749856 0.001183 0.000002
0.020005 0.000017
1.20 0.769861 0.001166
0.018839
1
.25 0.788700
Here, h = 0.05 a = 1.20
∴ u =
1.235 1.20
0.3
0.05
−
=−
()fx
=
23

(1) (1)(2)
() () () ()
2! 3!
uu uu u
fa ufa fa fa
+++
+∇ + ∇ + ∇ +
45
( 1)( 2)( 3) ( 1)( 2)( 3)( 4)
() ()
4! 5!
uu u u uu u u u
fa fa
+++ ++++
∇+ ∇
=
( 0.3)(0.7) ( 0.3)(0.7)(1.7)
0.788700 ( 0.3)(0.018839) ( 0.001166) (0.000017)
26
−−
+− + − +
()
0.3 (0.7)(1.7)(2.7)
( 0.3)(0.7)(1.7)(2.7)(3.7)
(0.000002) (0.000004)
24 120
−
−
++
180

COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES
= 0.788700 – 0.0056517 + 0.0012243 – 0.0000010115 – 0.00000008032
= 0.7888225488 – 0.00566189532
= 0.78316065356
Example 8: Calculate the value of tan 48°15′ from the following table:
45 46 47 48 49 50
1.00000 1.03553 1.07237 1.11061 1.15037 1.19175
x
tan x
°°°°°°°
Sol. Given that a + nh = 50
h = 1
a + nh + uh = 48°15′ = 48.25°
∴ 50 + u = 48.25
⇒ u = –1.75
The difference table for given data is as follows:
xyyyyyy°∇∇∇∇∇
°
°
°
−
°−
°
°
5 5 52 53 54 55
10 10 10 10 10 10
45 100000
3553
46 103553 131
3684 9

47 107237 140 3
3824 12 5
48 111061 152 2
3976 10
49 115037 162
4138
50 119175
()ya nh+
=
23
(1) (1)(2)
2! 3!
anh anh anh anh
uu uu u
yuy y y
++ + +
+++
+∇ + ∇ + ∇ +
45
(1)(2)(3) (1)(2)(3)(4)
4! 5!
anh anh
uu u u uu u u u
yy
++
+++ ++++
∇+ ∇
5
48.25
10

y
=
()
1.75)( 0.75
( 1.75)( 0.75)(0.25)
119175 ( 1.75) 4138 162 10
26
−−
−−
+− × + × + ×
( 1.75)( 0.75)(0.25)(1.25) ( 1.75)( 0.75)(0.25)(1.25)(2.25)
(2) (5)
24 120
−− −−
+×−+ ×−
5
48.25
10
y
= 112040.2867
⇒
48.25
y
= tan 48°15′ = 1.120402867.
INTERPOLATION WITH EQUAL INTERVAL
181
PROBLEM SET 4.2
1. The population of a town is as follows:
1921 1931 1941 1951 1961 1971
( )202429364651

Year
Population in lakhs
Estimate the increase in population during the period 1955 to 1961
[Ans. 621036.8 lakhs.]
2. From the following table find the value of tan 17
°
.
04 8 12 16 20 24
tan 0 0.0699 0.1405 0.2126 0.2867 0.3640 0.4402
θ
θ
[Ans. 0.3057]
3. From the given table find the value of log 5875
40 45 50 55 60 65
log 1.60206 1.65321 1.69897 1.74036 1.77815 1.81291
x
x
[Ans. 3.7690058]
4. From the following table, find y when x = 1.84 and 2.4
1.7 1.8 1.9 2.0 2.1 2.2 2.3
5.474 6.050 6.686 7.389 8.166 9.025 9.974
x
x
e
[Ans. 6.36, 11.02]
5. From the following table of half yearly premium for policies maturing at different ages,
estimate the premium for policy maturing at the age of 63:
45 50 55 60 65
( .) 114.84 96.16 83.32 74.48 68.48
Age

Premium in Rs
[Ans. 70.585152]
6. The values of annuities are given for the following ages. Find the value of annuity at the
age of
1
27
2
.
Age
Annuity
25 26 27 28 29
16.195 15.919 15.630 15.326 15.006
[Ans. 15.47996]
7. Show that Newton’s Gregory interpolation formula can be written in the form as
23 4
00 0 0 0

x
uuxuxauxabuxabcu
=+∆−∆+∆− ∆+
where a =
111
1 ( 1), = 1 ( +1), = 1 ( 1)
234
xb xc x−+ − −+
etc.
182
COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES
4.4 CENTRAL DIFFERENCE FORMULAE
As earlier we study formulae for leading terms and differences. These formulae are fundamental

and are applicable to nearly all cases of interpolation, but they do not converge as rapidly as
central difference formulae. The main advantage of central difference formulae is that they give
more accurate result than other method of interpolation. Their disadvantages lies in complicated
calculations and tedious expression, which are rather difficult to remember. These formulae are
used for interpolation near the middle of a argument values. In this category we use the following
formulae:
4.4.1 Gauss Forward Difference Formula
We know Newton’s Gregory forward difference formula is given by
()fa hu+
=
23
(1) (1)(2)
() () () ()
2! 3!
uu uu u
fa ufa fa fa
−−−
+∆ + ∆ + ∆
+
4
( 1)( 2)( 3)
()
4!
uu u u
fa
−−−
∆+
(1)
Substitute a = 0, h = 1 in (1), we get

()fu
=
23 4
(1)(1)(2)(1)(2)(3)
(0) (0) (0) (0) (0)
2! 3! 4!
uu uu u uu u u
fuf f f f
−−−−−−
+∆ + ∆ + ∆ + ∆ +
(2)
Now obtain the values of
234
(0), (0), (0)
fff∆∆∆
To get these values,
3
(1)
f∆−
22
(0) ( 1)
ff=∆ −∆ −
⇒
2
(0)
f∆
32
(1) (1)
ff=∆ − +∆ −
Also,

4
(1)
f∆−
33
(0) ( 1)
ff=∆ −∆ −
⇒
3
(0)
f∆
43
(1) (1)
ff=∆ − +∆ −
5
(1)
f∆−
44
(0) ( 1)
ff=∆ −∆ −
⇒
4
(0)
f∆
54
(1) (1)
ff=∆ − +∆ −
6
(1)
f∆−
55

(0) ( 1)
ff=∆ −∆ −
⇒
5
(0)
f∆
65
(1) (1)
ff=∆ − +∆ −
and so on.
Substituting these values in equation (2)
()fu
=
32 43
(1) (1)(2)
(0) (0) ( 1) ( 1) [ ( 1) ( 1)]
2! 3!
uu uu u
fuf f f f f
−−−

+∆ + ∆−+∆−+ ∆−+∆−

54 65
( 1)( 2)( 3) ( 1)( 2)( 3)( 4)
(1) (1) (1) (1)
4! 5!
uu u u uu u u u
ff ff
−−− −−−−

 
+ ∆ −+∆ − + ∆ −+∆ − +
 
()fu
=
{}
23
(1) (1) (2)
(0) (0) (1) 1 (1)
22!3
uu uu u
fuf f f
−−−
+∆ + ∆ − + + ∆ −
+
{} {}
45
( 1)( 2) ( 3) ( 1)( 2)( 3) ( 4)
1 ( 1) 1 ( 1)
6 4 24 5
uu u u uu u u u
ff
−− − −−− −
+∆−+ +∆−
6
( 1)( 2)( 3)( 4)
( 1)
120
uuuuu
f

−−−−
+∆−+
INTERPOLATION WITH EQUAL INTERVAL
183
()fu
=
23 4
(1) (1)(1) (1)(1)(2)
(0) (0) (1) (1) (1)
2! 3! 4!
uu u uu u uu u
fuf f f f
−+−+−−
+∆ + ∆ − + ∆ − + ∆ −
5
(1)(1)(2)(3)
( 1)
5!
uuu u u
f
+−−−
+∆−+
(3)
But ∆
5
f (–2) = ∆
4
f (–1) – ∆
4
f (–2)

⇒∆
4
f (–1) = ∆
4
f (–2) + ∆
5
f (–2)
and ∆
6
f (–2) = ∆
5
f (–1) + ∆
5
f (–2)
⇒∆
5
f (–1) = ∆
5
f (–2) + ∆
6
f (–2)
The equation (3) becomes

()fu
=
23 4
(1) (1)(1) (1)(1)(2)
(0) (0) (1) (1) (2)
2! 3! 4!
uu u uu u uu u

fuf f f f
−+−+−−
+∆ + ∆ − + ∆ − + ∆ −
55
( 1) ( 1)( 2) ( 1) ( 1)( 2)( 3)
(2) (2)
4! 5!
uuu u uuu u u
ff
+−− +−−−
+∆−+ ∆−
+
6
(1)(1)(2)(3)
(2)
5!
uuu u u
f
+−−−
∆−
()fu∴
=
23 4
(1) (1)(1) (1)(1)(2)
(0) (0) ( 1) ( 1) ( 2)
2! 3! 4!
uu u uu u uu u
fuf f f f
−+−+−−
+∆ + ∆ − + ∆ − + ∆ −

56
( 2)( 1) ( 1)( 2) ( 1) ( 1)( 2)( 3)
( 2) ( 2)
5! 5!
uuuuu uuuuu
ff
++ −− + −−−
+∆−+∆−+
This formula is known as Gauss forward difference formula.
This formula is applicable when u lies between
1
0and
2
4.4.2 Gauss Backward Difference Formula
This formula is also solved by using Newton’s forward difference formula.
Now, we know Newton’s formula for forward interpolation is
23 4
(1)(1)(2)(1)(2)(3)
( ) () () () () ()
2! 3! 4!
uu uu u uu u u
fa hu fa ufa fa fa fa
−−−−−−
+= +∆+ ∆ + ∆ + ∆ +
(1)
Put a = 0, and h = 1, in equation (1), we get
23 4
( 1) ( 1)( 2) ( 1)( 2)( 3)
( ) (0) (0) (0) (0) (0)
2! 3! 4!

uu uu u uu u u
fu f uf f f f
−−−−−−
=+∆+ ∆ + ∆ + ∆ +
(2)
Now
2
22 3
33 4
44 5
(0) ( 1) ( 1)
(0) ( 1) ( 1)
(0) ( 1) ( 1)
(0) ( 1) ( 1) and so on.
ff f
ff f
ff f
ff f
∆=∆−+∆−
∆=∆−+∆−
∆=∆−+∆−
∆=∆−+∆−
On substituting these values in (2), we get

223 34
(1) (1)(2)
() (0) (1) (1) (1) (1) (1) (1)
2! 3!
uu uu u
fufuff ff ff

−−−

= +∆−+∆−+ ∆−+∆−+ ∆−+∆−

45
( 1)( 2)( 3)
( 1) ( 1)
4!
uu u u
ff
−−−

+∆−+∆−+

184
COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES
∴
23
(1) (1) (2)
() (0) (1) (1)1 (1)1
22 3
uuu u
fu f uf u f f
−− −
 
= +∆−+∆−+ + ∆−+
 
 
+
4

( 1)( 2) ( 3)
(1)1
34
uu u u
f
−− −

∆− + +


=
23
(1) (1)(1)
(0) ( 1) ( 1) ( 1)
2! 3!
uuuu
fuf uf f
++−
+∆− + ∆ − + ∆ −
+
4
(1)(1)(2)
( 1)
4!
uuu u
f
+−−
∆−+
(3)
Again

334
(1) (2) (2)
fff∆−=∆−+∆−
445
(1) (2) (2)
fff∆−=∆−+∆−
556
(1) (2) (2)
fff∆−=∆−+∆−
and so on.
Therefore, equation (3) becomes
234
(1) (1)(1)
() (0) (1) (1) (2) (2)
2! 3!
uu uuu
fu f uf f f f
++−

=+∆−+ ∆−+ ∆−+∆−

+
45 56
( 1) ( 1)( 2) ( 1) ( 1)( 2)( 3)
( 2) ( 2) ( 2) ( 2)
4! 5!
uuu u uuu u u
ff ff
+−− +−−−
 

∆ − +∆ − + ∆ − +∆ − +
 
23 4
( 1) ( 1) ( 1) ( 1) ( 1) ( 2)
() (0) (1) (1) (2) 1 (2)
2! 3! 3! 4
uu uuu uuu u
fu f uf f f f
++−+−−

= +∆−+ ∆−+ ∆−+ + ∆−


5
(1)(1)(2) (3)
1 ( 2)
4! 5
uuu u u
f
+−− −

++∆−+


23
(1) (1)(1) (2)(1)(1)
() (0) (1) (1) (2)
2! 3! 4!
uu uuu u uuu
fu f uf f f

++−++−
=+∆−+ ∆−+ ∆−+
45
(2)(1)(1)(2)
(2) (2)
5!
uuuuu
ff
++ −−
∆−+ ∆−+
This is known as Gauss Backward difference formula and useful when u lies between
1
and 0
2
−
.
4.4.3 Stirling’s Formula
This is another central difference formula and useful when
11 1
||
22 2
uor u<−<<
. It gives best
estimation when
11
.
44
u−<
This formula is obtained by taken mean of Gauss forward and Gauss
backward difference formula.

Gauss forward formula for interpolating central difference is,
23 4
(1) (1)(1) (1)(1)(2)
() (0) (0) (1) (1) (2)
2! 3! 4!
uu u uu u uu u
fu f u f f f f
−+−+−−
= +∆ + ∆−+ ∆−+ ∆−

56
( 2)( 1) ( 1)( 2) ( 1) ( 1)( 2)( 3)
(2) (2)
5! 5!
uuuuu uuuuu
ff
++ −− + −−−
+∆−+∆−+
(1)
Gauss Backward difference is,
23
(1) (1)(1) (2)(1)(1)
() (0) (1) (1) (2)
2! 3! 4!
uu uuu u uuu
fu f uf f f
++−++−
=+∆−+ ∆−+ ∆−+
INTERPOLATION WITH EQUAL INTERVAL
185

45
(2)(1)(1)(2)
( 2) ( 2)
5!
uuuuu
ff
++ −−
∆−+ ∆−+
(2)
Take mean of Equation (1) and (2)
[]
2
(0) ( 1)
(1)(1)
() (0) (1)
22! 2
uf f
uuu
fu f f
∆+∆−
−++

=+ +∆−


33
4
(1) (2)
(1)(1) (1)(1) (2)(2

(2)
3! 2 3! 2
ff
uuu uuu u u
f

∆−+∆−
+− +− −++

++∆−





5
( 2)( 1) ( 1)( 2)
(2)
5!
uuuuu
f
++ −−
+∆−+
33
2
2
(0) ( 1) ( 1) ( 2)
(1)(1)
() (0) (1)
22! 3! 2

ff f f
uuu
u
fu f u f

∆+∆− ∆−+∆−
+−

=+ +∆−+





2
45
( 1) ( 1) ( 2)( 1) ( 1)( 2)
( 2) ( 2)
3! 5!
uuu u uuu u
ff
+− ++−−
+∆−+ ∆−+
This is called Stirling’s formula.
4.4.4 Bessel’s Interpolation Formula
This is one of the another type of central difference formula and obtained by (1) shifting the origin
by 1 in Gauss backward difference and then (2) replacing u by (u – 1), (3) take mean of this
equation with Gauss forward formula.
Gauss backward difference formula is,
()

23 4
(1) (1)(1) (2)(1)(1)
() (0) (1) (1) (2) 2
2! 3! 4!
uu uuu u uuu
fu f uf f f f
++−++−
=+∆−+ ∆−+ ∆−+ ∆−
5
(2)(1)(1)(2)
(2)
5!
uuuuu
f
++ −−
++∆−+
Now shift the origin by one, we get
()
23 4
(1) (1)(1) (2)(1)(1)
( ) (1) (0) (0) ( 1) 1
2! 3! 4!
uu uuu u uuu
fu f uf f f f
++−++−
=+∆+ ∆ + ∆−+ ∆−
5
( 2)( 1) ( 1)( 2)
( 1)
5!

uuuuu
f
++ −−
+∆−+
On replacing u by (u – 1)
()
23 4
( 1) ( 1) ( 2) ( 1) ( 1)( 2)
() (1) ( 1) (0) (0) (1) 1
2! 3! 4!
uu uuu uuu u
fu f u f f f f
−−−+−−
=+−∆+ ∆ + ∆−+ ∆−
5
(1)((1)(2)(3)
(1)
5!
uuuuu
f
+− −−
+∆−+
…(1)
Gauss forward difference formula is,
23 4
(1) (1)(1) (1)(1)(2)
() (0) (0) (1) (1) (2)
2! 3! 4!
uu u uu u uu u
fu f uf f f f

−+−+−−
= +∆ + ∆−+ ∆−+ ∆−
56
(2)(1)(1)(2) (1)(1)(2)(3)
( 2) ( 2)
5! 5!
u u uu u u uu u u
ff
++ −− + −−−
+∆−+∆−+
…(2)