Using Gauss backward interpolation formula, find the population for the year 1936.. Given that Year Population in thousands Sol... Find the value of cos 51°421 by Gauss’s backward formul
Trang 1Example 5 If f(x) is a polynomial of degree four find the value of f(5.8) using Gauss’s backward formula from the following data:
f(4) = 270, f(5) = 648, ∆f(5) = 682, ∆3 f(4) = 132.
Sol Given ∆f(5) 682=
⇒ f (6) – f (5) = 682
⇒ f (7) – 3f (6) + 3f (5) – f (4) = 132
⇒ f (7) = 3 × 1330 – 3 × 648 + 270 + 132
f (7) = 2448
Now form difference table as:
x f x( ) ∆f x( ) ∆2f x( ) ∆3f x( )
378
1118
Take a = 6, h = 1 a + hu = 5.8
∴ u= −0.2
From Gauss backward formula
u u u
f − = f + ∆ − +u f + ∆ f − + + − ∆ f −
=1330 ( 0.2) 682+ − × +( 0.2)(0.8)− 2 ×436+( 0.2)(0.8)( 1.2)− 6 − ×132
= 1330 – 136.4 – 34.88 + 4.224
= 1162.944
Hence f (5.8) = 1162.944.
Trang 2Example 6 Using Gauss backward interpolation formula, find the population for the year 1936 Given that
Year Population in thousands
Sol Here h = 10 Take origin at 1941 to evaluate population in 1936
1936 1941 5
0.5
x a
u
h
Difference table for given data is as:
−
−
−
−
−
3
13
Gauss backward formula is
( ) (0) ( 1) ( 1) 2 ( 1) ( 1) ( 1) 3 ( 2)
4!
u+ u+ u u−
5
( 2)( 1) ( 1)( 2)
( 3) 5!
f
= 39 + ( 0.5) 12 ( 0.5)(0.5) 1 ( 0.5)(0.5)( 1.5) ( 4)
= 39 – 6.0 – 0.125 – 0.25
= 32.625 thousands
Hence, the population in 1936 is 32625 thousand
Trang 3PROBLEM SET 4.4
1 Given that 12500 = 111.803399, 12510 = 111.848111, 12520 = 111.892806, 12530
= 111.937483 Using Gauss’s backward formula show that 12516 = 111.8749301
2 Find the value of cos 51°421 by Gauss’s backward formula from the following data:
x x
[Ans 0.61981013]
3 The population of a town in the years are as follows:
Year Population in thousands
Find the population of the town in 1946 by applying Gauss’s backward formula
[Ans 22898]
4 Interpolate by means of Gauss’s backward formula, the population of a town KOSIKALAN for the year 1974, given that:
Year Population in thousands
[Ans 32.345 thousands appros]
5 Use Gauss interpolation formula to find y41 from the following data:
y30 = 3678.2, y35 = 2995.1, y40 = 2400.1, y45 = 1876.2, y50 = 1416.3
[Ans y41 = 2290.1]
6 Use Gauss’s backward formula to find the value of y when x = 3.75, given the following
table:
x
x
y
[Ans 19.704]
Trang 4Stirling’s Formula
Example 1 Evaluate sin (0.197) from the table given:
x
sin x
Sol The difference table is given by
0.01974
0.01952
−
−
−
−
h
−
= 0.197 0.19
0.02
− = 0.35 From Stirling formula, we have
f(u) = f(0) + ( )0 ( )1
2
u∆f + ∆ −f
+ 2
2!
u
( )
2f 1
3!
u+ u u−
3 ( )1 3 ( )2
2
+ ( 1) (2 1)
3!
u+ u u−
( )
4f 2
0.35 0.0196 0.01968
+
(0.35 1 0.35 0.35 1) ( ) ( )
6
× −0.000022 + (0.35)2 ( )2
0.35 1
0.00002 24
×
= 0.18886 + 0.0068741 – 0.0000049 + 0.0000005 – 0.00000009
= 0.19573 (Approx.)
Example 2 Find the value of e x = at x = 0.644 using by Stirling’s formula The following data given below:
x
x
y=e
1.840431 1.858928 1.877610 1.896481 1.915541 1.934792 1.954237
Trang 5Sol For the given data difference table is as:
x
−
−
−
−
3 0.61 1.840431
0.018497
0.019445
3 0.67 1.954237
0.01
x a h
By stirling formula, we get,
3 3
+
2
2
(0.4) (1.4)(0.4)( 0.6)
− +
(0.4) (0.4) 1 24
+ 0.000002 (1.4)(2.4)(0.4)( 0.6)( 1.6) 0.000006 0.000001
2 (0.4 2)(1.4)(0.4)( 0.6)( 0.4 2)
( 0.000007) 720
= 1.896481 + 0.0075862 + 0.00001512–0.000000112 – 0.0000000112 + 0.000000026 – 0.0000002
= 1.904081 (Approx.)
Trang 6Example 3 Employ Stirling’s formula to evaluate y 12.2 from the data given below
(y x = 1+ log 10 sin x).
5
x y
°
Sol For the given data difference table is as:
10
4093
3159
x
−
−
∴ u = 12.2 – 12 = 0.2
From Stirling formula
[ ] 2 0.2 (0.2){ 2 1} (0.2)2{ ( )0.2 2 1}
= 31788 – 714.9 – 6.14 – 1.648 + 0.0208
y12.2 = 32495
105y12.2 = 32495
∴ y12.5 = 0.32495 Ans
Trang 7Example 4 Apply Stirling’s formula to find a polynomial of degree three which takes the following values of x and y:
x
Sol Let u = 6
2
x−
Now, we construct the following difference table:
3
12
Stirling’s formula is
2 2
+
2
4 ( 1) ( 1)
( 2) 3!
u u u
f
u
y = +u + + × + − + +
3 7 3 2 2( 3 )
2u 2u 3 u u
3 2 3 3 2 7 2
3u 2u 2u 3u
3 2 3 3 2 17
3u 2u 6 u
= + + +
= 0.0833x3 – 1.125x2 + 8.9166x – 19.
Trang 8Example 5 Apply Stirling’s formula to find the value of f(1.22) from the following table which gives the value of
2
x x 2 0
1
2π
= ∫ dx at intervals of x = 0.5 from x = 0 to 2.
( )
x
f x
Sol Let the origin be at 1 and h = 0.5
∴ x = a + hu, u = 1.22 1.00 0.44
0.5
x a h
Applying Stirling’s formula
2
+
2
4 ( 1) ( 1)
( 2) 3!
u u u
f
∴ f(0.44) = f(0) + (0.44)
2 2
2 ∆f + ∆ − +f 2 ∆ f −
= f(0) + (0.22) [∆f(0) + ∆f(–1)] + 0.0968 ∆2 f(–1)
– 0.029568 [∆3 f(–1) + ∆3f(–2)] – 0.06505∆4 f(–2) .(1)
The difference table is as follows:
−
−
−
−
191
44
Trang 9f(0) and the differences are being multiplied by 103
∴ 103 f(0.44) ≈341 + 0.22 × (150 + 92) + 0.0968 × (–58)
– 0.029568 × (–17 + 10) – 0.006505 × 27
≈ 341 + 0.22 × 242 – 0.0968 × 58 + 0.029568 × 7 – 0.006505 × 27
≈ 341 + 53.24 – 5.6144 + 0.206276 – 0.175635
≈ 388.66
f(0.44) = 0.389 at x = 1.22 Ans
Example 6 Use Stirling’s formula to find y 28 given.
y 20 = 49225, y 25 = 48316, y 30 = 47236,
y 35 = 45926, y 40 = 44306
Sol Let the origin be at 30 and h = 5
a + hu = 28
The difference table is as follows:
909
1620
−
−
−
−
By Stirling’s formula,
2
( 0.4)
1080 1310
(.6)( 4)( 1.4) 3!
+ 59 80 ( ).4 2{( 4.4)2 1}
( 21)
− −
Hence, y28 = 47691.8256
Trang 10PROBLEM SET 4.5
1 Use Stirling’s formula to find y32 from the following table:
x
y
[Ans 13.062]
2 Use the following table to evaluate tan 16° by Stirling’s formula:
tan
θ°
[Ans 0.2866980499]
3 Use Stirling’s formula to find the value of f(1.22) from the following data:
[Ans 0.9391002]
4 Find f(0.41) using Stirling’s formula if,
x
f x
[Ans 0.15907168]
5 Use Stirling’s formula to find y35, data being:
y20 = 512, y30 = 439, y40 = 346, and y50 = 243,
[Ans 394.6875]
6 From the following table find the value of f(0.5437) by Stirling’s formula:
( ) 0.529244 0.537895 0.546464 0.554939 0.663323 0.571616 0.579816
x
f x
[Ans 0.558052]