Use Everett’s formula to determine k 0.25... Find the value of log 337.5 by using Laplace Everett’s formula... CHAPTER 5Interpolation with Unequal Interval 5.1 INTRODUCTION The interpola
Trang 1Difference table is:
−
−
−
308
896
By Everett’s formula,
(1.25 (0.25) 0.75) ( )( ) ( ) (1.75 0.75 ( 0.25))( ) ( )
= 4064
Hence f(30) = 4064.
Example 3 Apply Laplace Everett’s formula to find the value of log 2375 from the data given below: x
x
log 1.3222 1.3424 1.3617 1.3802 1.3979 1.4150
Sol Here h = 1
We take origin at 23
Now difference table is given by
−
−
−
log
0.0202
0.0171
Trang 2Here h = 1
∴ u = 23.75 23 0.75
1
x a h
w = 1 – 0.75 = 0.25
From Laplace Everett formula, we have
( 1 () 1) 2 ( 2)( 1) ( 1)( 2) 4
( 1) ( 1) 2 ( 2)( 1) ( 1)( 2) 4
= 0.75 1.3802 (1.75)(0.75)( 0.25) ( 0.0008) (1.75 2.75 0.75)( )( )( 0.25)( 1.25) (0.0002)
+ 0.25(1.25)( 0.75( 0.0008) (2.25 (1.25)(0.25)( 0.75)( 1.75))
= 1.035419 + 0.340455
= 1.375874
log 2375 = log (23.75 × 100) = log 23.75 + log 100
⇒ log 2375 = 1.375872 + 2 = 3.375872
Example 4 Find the value of e –x when x = 1.748 from the following data:
x
x
e−
0.1790 0.1773 0.1755 0.1738 0.1720 0.1703
Sol Here h = 0.01, take origin as 1.74.
The difference table for the given data is as:
x
−
−
−
−
−
−
1.72 0.1790
0.0017
0.0017 1.77 0.1703
Trang 31.748 1.74
0.8 0.01
w = 0.2
( )( )( )(2.8 1.8 0.8 0.2)( 1.2 (0.0004)) (0.8) 0.8(0.1738) (0.8)(1.8)( 0.2)( 0.00017)
120
( 0.8 0.2 1.2)( )( ) ( ) ( )( )( )(1.2 2.2 0.2 0.8)( 1.8) ( )
= 0.13904 + 0.0000816 + 0.000003225 + 0.0351 – 0.0000032 + 0.000002534
= 0.174224
Example 5 Prove that if third differences are assumed to be constant
of y 11 and y 16 if y 0 = 3010, y 5 = 2710, y 10 = 2285, y 15 = 1860, y 20 = 1560, y 25 =1510, y 30 = 1835.
Sol 1
11 10
0.2, 5
x = − =
2
16 15
0.2
5
x = − =
−
−
−
−
−
−
300
325
Using given formula,
x
11
1.2 (0.2)( 0.8) 0.8)(0.64 1
= 2196
Trang 4( )( ) ( )( )( )( ) ( )( ) ( )( ) ( )
16
= 1786
Example 6 Find the compound interest on the sum of Rs 10,000/- at 7% for the period 16 years
if,
n
(1.07) 1.40255 1.96715 2.75903 3.86968 5.42743 7.61236 Sol The difference table can be formed as:
−
−
−
−
0.5646
2.18493
3 30 7.61236
Here, h = 5
5
x a h
∴ w = 1− = −u 1 0.2 0.8=
On applying Laplace Everett formula, we have
0.2(0.2 1)(0.2 1) 0.2(0.2 2)(0.2 1)(0.2 1)(0.2 2)
0.8 2.75903 0.8(0.8 1)(0.8 1 0.8(0.8 2)(0.8 1)(0.8 1)(0.8 2)
= 0.776593 + 2.189027 = 2.96595 (Approx.)
Trang 5Example 7 The values of the ellipitc integral ( ) / 2( 2 )1/ 2
0
k m =π∫ 1 m sin− θ − dθ for certain equidistant values of m are given below Use Everett’s formula to determine k (0.25).
m
k m
( ) 1.659624 1.669850 1.680373 1.691208 1.702374 1.713889
Sol Here h = 0.02, take origin as 0.24
u = 0.25 0.24 0.5
0.02
x a h
w = 1 – u = 0.5
−
−
( ) 0.20 1.659624
0.010226
0.011515 0.30 1.713889
= ( )(0.5 1.691208) (0.5)(1.5)( 0.5)(0.000331) ( 1.5)( )( )( )( )(.5 5 1.5 2.5 0.000001)
( )(0.5 1.680073) ( )( )( )(.5 1.5 .5 0.000312) ( )( ).5 1.5 ( 5) 2.5( )( 1.5) (.000004)
= 0.845604 – 0.00002069 – 0.00000014 + 0.84018650 – 0.0000195 + 0.00000005
= 1.685750
Trang 6Example 8 Find the value of log 337.5 by using Laplace Everett’s formula Given that:
x
x
log 2.49136 2.50515 2.51851 2.53148 2.54407 2.55630
Sol Here h = 10, take origin as 340
337.5 340
0.25 10
x a
u
h
w = 1 – u = 1+ 0.25 = 1.25
For the given data difference table is as:
−
−
−
−
3 310 2.49136
0.01379
0.01223
2 360 2.55630
( ) ( ) ( 1) ( 1) 2 ( ) ( 2)( 1) ( 1)( 2) 4 ( )
( ) ( 1) ( 1) 2 ( ) ( 2)( 1) ( 1)( 2) 4 ( )
( 0.25) 2.54407 ( 1.25)( 0.25 0.75)( ) ( 0.00036) 1.25 2.53148
6
(0.25 1.25 2.25)( )( )( 0.00038)
6
+ − (3.25 2.25 1.25 0.25)( )( )( )( 0.75)(0.00001)
120
− +
= –0.6360175–0.0000140625 + 3.16435 – 0.00004453125 – 0.0000001428
= 2.528273 (Approx.)
Trang 7PROBLEM SET 4.7
1 Eliminate odd difference from the Gauss Forward formula to drive Everett’s formula:
u
y = −u f + δ =u − − S δ + + − S δ +
where u x x0
h
−
=
2 From the following table of values of x and y = e x , interpolate the value of y when x = 1.91:
5.4739 6.0496 6.6859 7.3891 8.1662 9.0250
x
x
y=e
[Ans 6.7531]
3 From the following present value annuity a n table:
n
x
a
11.4699 12.7834 13.7648 14.4982 15.0463
Find the present value of the annuity a31, a33
[Ans 13.9186, 14.2306]
4 Find the value of x1/3 when x = 51 to 54 from the data:
x
x1/3
3.4200 3.3569 3.6840 3.8030 3.9149 4.0207
[Ans 3.7084096, 3.7325079, 3.7563005, 3.7797956]
5 From the following data, find the value of f(31), f(32),
f(20) = 3010, f(25) = 3979, f(30) = 4771, f(35) = 5441,
f(40) = 6021, f(45) = 6532
[Ans 4913, 5052]
6 Apply Everett’s formula to find the value of f(26) and f(27) from the data given below:
x
f x
( ) 12.849 16.351 19.524 22.396 24.999 27.356
[Ans 20.121431, 20.707077]
7 The following table gives the values of e x for certain equidistant values of x:
1.840431 1.858928 1.877610 1.896481 1.915541 1.934792 1.954237
x
x
e
Find the value of e x when x = 0.644 using Everett’s formula.
[Ans 1.904082]
Trang 88 Apply Everett’s formula to find the values of e –x for x = 3.2, 3.4, 3.6, 3.8 if,
0.36788 0.13534 0.04979 0.01832 0.00674 0.00248
x
x
e−
[Ans 0.04087, 0.03354, 0.02749, 0.02248]
9 Use Everett’s formula to find the present value of the annuity of n = 36 from the table:
12.7834 13.7648 14.4982 15.0463 15.4558 15.7619
x
x
a
[Ans 14.620947]
10 Obtain the value of y 25, given that:
y20 = 2854, y24 = 3162, y28 = 3544, y32 = 3992 [Ans 3250.875]
GGG
Trang 9CHAPTER 5
Interpolation with Unequal Interval
5.1 INTRODUCTION
The interpolation formulae derived before for forward interpolation, Backward interpolation and central interpolation have the disadvantages of being applicable only to equally spaced argument values So it is required to develop interpolation formulae for unequally spaced argument values
of x Therefore, when the values of the argument are not at equally spaced then we use two such
formulae for interpolation
1 Lagrange’s Interpolation formula
2 Newton’s Divided difference formula
The main advantage of these formulas is, they can also be used in case of equal intervals but the formulae for equal intervals cannot be used in case of unequal intervals
5.2 LAGRANGE’S INTERPOLATION FORMULA
Let f(x0), f(x1) f(x n ) be (n + 1) entries of a function y = f(x), where f(x) is assumed to be a polynomial corresponding to the arguments x0, x1, x2, x0 So that
The polynomial f(x) may be written as
f(x) = A0 (x – x1) (x – x2) (x – x n ) + A1 (x – x1) (x – x2) (x – x n) +
+ A n (x – x1) (x – x2) (x – x n–1) (1)
where A0, A1, A n, are constants to be determined
Putting x = x0, x1, x2, , x n in (1) successively, we get
For x = x0 , f(x0) = A0 (x 0 – x1) (x 0 – x2) (x0 – x n)
( 0 1)( 0 02) ( 0 n)
f x
For x = x1 , f(x1) = A1 (x 1 – x0)(x 1 – x2) (x1 – x n)
( 1 0) ( 1 12) ( 0 n)
f x
Similarly,
For x = x n , f(x n ) = A n (x n – x0)(x n – x1) (x n – x n–1)
224
Trang 10A n = ( )
( 0)( 2) ( 1)
n
f x
Substituting the values of A0, A1, A n, in equation (1), we get
( 1)( 2 ) ( ) ( )0 ( ( 0)( )( 2) ( ) ( ) ) ( )1
+
( 00) ( 11) ( 11) ( )
n
n
f x
−
−
This is called Lagrange’s interpolation formula In equation (5), dividing both sides by
(x – x0) (x – x1) (x – x n), Lagrange’s formula may also to written as
( 0)( 1) ( ) ( 0 1)( 0 2) (0 0 )( 0)
( 1 0)( 1 12) ( 1 ) ( 1) ( 0)( 1) ( 1) ( )
n
Corollary Show that Lagrange’s formula can be put in the form
( ) 0( ) ( ))
( ) ( '
n
r n
r
x f x
P x
=
φ
=
∑
where ( )
0
n
r
x
=
φ = ∏(x – x r) and φ’(x r) = d { } ( )
x dx
( 0)(0 11) ( 11)( 11) ( )
0
n
r
=
( ) ( )( ) ( ( )− )( + ) ( )
=
∑
n
r
r
f x x
0
n
x xr
r∏ −
Therefore, φ( )x = (x x− 0)(x−x1)(x−x2) (x−x r−1) (x−x r) (x−x r+1) (x−x n)
∴ φ′(x) = (x−x1) (x−x2) ( x−x r) ( x−x n) (+ x−x0)(x−x2) ( x−x r) ( x x− n)
+ (x−x0)(x−x1) ( x−x r)(x x− r+1) ( x−x n) (+ x−x0)(x−x1) ( x−x r) ( x x− n−1)
⇒ φ′(x) = [φ′(x)] x = xr = (x r – x 0 ) (x r – x1) (x r – x r–1 ) (x r – x r+ 1 ) (x r – x n) (2) Hence from equation (1)
( ) ( )
( ) ( ) 0
( )
'
n
r n
r
x f x
P x
=
φ
=