Use Gauss’s forward formula to find a polynomial of degree four which takes thefollowing values of the function fx: Sol.. Use Gauss’s forward formula to find y30 for the following data..
Trang 1Take the mean of equation (1) and (2)
f u = + + − + ∆f + − ∆ + ∆ −
( 1) 2) ( 1) ( 1) ( 1)( 2)
( 1)
+
( 3) ( 1) ( 2) ( 2) ( 1) ( 1)( 2)
u+ u u− u− − ∆ − + + ∆ − +
1 ( 1)
− −
5
1 ( 1) ( 2) ( 1 2) ( 1) ( 2)
( 1) ( 1)( 2)
( 2)
f
This formula is very useful when 1
2
u= and gives best result when 1 3
4< <u 4
4.4.5 Laplace-Everett’s Formula
Gausss forward formula is given by
( 2)( 1) ( 1)( _ 2) ( 1) ( 1)( 2)( 3)
We know
(0) (1) (0) ( 1) (0) ( 1) ( 2) ( 1) ( 2)
∆ − = ∆ − − ∆ − Therefore, using this in equation (1), we get
f u = f +u f −f + − ∆ f − + + − ∆ f − ∆ f −
(1 ) (0) (1) ( 1) 2 ( 1) ( 1) ( 1) 2 ( 1)
+( 1) ( 1)( 2) 4 ( 2) ( 2)( 1) ( 1)( 2) 4 ( 2)
Trang 2(1 ) (0) (1) ( 1)(2 ) 2 ( 1) ( 1) ( 1) 2 (0)
( 1)( 1) ( 2)( 1) ( 1)( 2)
Substitute 1 – u = w in second part of equation (2)
( 1) ( 1) ( 2)( 1) ( 1)( 2)
This is called Laplace-Everett’s formula It gives better estimate value when 1
2
u>
Example 1 From the following table, find the value of e 1.17 using Gauss forward formula:
x
x
e = f x
( ) 2.7183 2.8577 3.0042 3.1582 3.3201 3.4903 3.6693 Sol The difference table is as given:
1 2.7183
0.1394
0.179 1.30 3.6693
Trang 3Now, let taking origin at 1.15, here h = 0.05
Then, 1.17 1.15 0.4
0.05
On applying Gauss’s forward interpolation, we have
( ) (0) (0) ( 1) 2 ( 1) ( 1) ( 1) 3 ( 1) ( 1) ( 1)( 2) 4 ( 2)
f u = f + ∆u f + − ∆ f − + + − ∆ f − + + − − ∆ f −
5 ( 2)( 1) ( 1)( 2)
( 2) 5!
f
(0.4)( 0.6) (1.4)(0.4)( 0.6)
(1.4)(0.4)( 0.6)( 1.6)(2.4)
120
(0.4) 3.1582 0.06476 0.000948 0.0000224 0.0000010752
(0.4) 3.22199 (Approx.)
Example 2 Given that
x
x
log 1.39794 1.47712 1.54407 1.60206 1.65321
log 3.7 = ?
Sol
2
1
0
1
2
25 1.39794
0.07918
0.05115
45 1.65321
x
x
x
x
x
−
−
−
−
x = a + hu, x = 37, a = 35, h = 5
37 35
0.4
x a− = − =
Trang 42 3 4
1.54407 0.4 0.05798 (0.4)(0.4 1) ( 0.00896) (0.4)(0.4 1)(0.4 1) (0.00212)
(0.4)(0.4 1)(0.4 1)(0.4 2)
( 0.00115) 24
( ) 1.54407 0.023192 0.0010752 0.00011872 0.00002576
Since log 3.7 = log3.7 10 log37
⇒ log 3.7 = log 37 – log 10 = 1.56819272 – 1 = 0.56819272 Ans
Example 3 From the following table find y when x = 1.45.
x
Sol
−
−
−
0.112
1.008
u u u
f u = f + ∆u f + − ∆ f − + + − ∆ f − + + − − ∆ f −
u=1.45 1.40.2− =0.25
(0.25)( 0.75) (0.25)( 0.75)(1.25)
= 0.047875
Trang 5Example 4 Use Gauss’s forward formula to find a polynomial of degree four which takes the
following values of the function f(x):
Sol Taking center at 3 i.e., x0 = 3 and h = 1
h
−
Now, for the given data difference table becomes:
−
−
−
−
−
−
2
2
Gauss forward formula is
f u = f + ∆u f + − ∆ f − + + − ∆ f − + + − − ∆ f −
=1 2 6 2 2 14 24 4 3 12 2 104 32 2 4 28 3 142 2 308 80
Example 5 Use Gauss’s forward formula to find y30 for the following data.
18.4708 17.8144 17.1070 16.3432 15.5154 Sol Let us take the origin at x = 29
30 29 1
0.25
Trang 6Now, for the given data difference table is:
−
−
−
−
21 18.4708
0.6564
0.8278
37 15.5154
Putting these values in Gauss forward interpolation formula, we have
0.25
(0.25)( 0.750) 17.1070 (0.25) ( 0.7638) ( 0.0564)
2
+ (0.25)( 0.750)(1.25) ( 0.0076) (1.25)(0.25)( 0.750)( 1.75) ( 0.0022)
y0.25 = 17.1070 – 0.19095 + 0.0052875 + 0.00002968 – 0.00000375
y0.25 = 16.9216 Ans
PROBLEM SET 4.3
1 The values of e –x at x = 1.72 to x = 1.76 are given in the following table:
( )
0.17907 0.17728 0.17552 0.17377 0.17204
x
f x
Find the values of e–1.7425 using Gauss forward difference formula
[Ans 0.1750816846]
2 Apply Gauss’s forward formula to find the value of f(x) at x = 3.75 from the table:
( )
24.145 22.043 20.225 18.644 17.262 16.047
x
f x
[Ans 19.407426]
3 Apply Central difference formula to obtain f(32) Given that :
f(25) = 0.2707, f(30) = 0.3027, f(35) = 0.3386, f(40) = 0.3794. [Ans 0.316536]
4 Apply Gauss forward formula to find the value of U9, if
Trang 75 Apply Gauss forward formula to find a polynomial of degree three which takes the values
of y as given on next page:
x
3
6 x 2x 3x
6 Use Gauss’s forward formula to find the annuity value for 27 years from the following data:
10.3797 12.4622 14.0939 15.3725 16.3742 17.1591
Year
Annuity
[Ans 14.643]
7 Use Gauss’s forward formula to find the value of f(x), when 1,
2
x= given that:
( )
x
f x
−
GAUSS BACKWARD
Example 1 Given that
tan 1.1918 1.2349 1.2799 1.3270 1.3764
x
x
°
°
Using Gauss’s backward formula, find the value of tan 51° 42′
Sol Take the origin at 52° and given h = 1
h
Now using Gauss backward formula
f(u) = f(0) + u∆f (–1) + ( 1) 2 ( )
1 2!
f
+
( 1) ( 1) 3 ( ) ( 2)( 1) ( 1)
2
f
Trang 8Difference table for given data is:
50 1.1918
0.0431
0.0494
54 1.3764
From equation (1)
f(–0.3 °) = 1.2799 + (–0.3) (0.045) + ( 0.3 0.7)( ) ( 0.3 0.7 1.7) ( ) ( )
= 1.2799 – 0.0135 – 0.0002205 – 0.0000119
= 1.266167 (Approx.)
Example 2 Apply Gauss backward formula to find sin 45 ° from the following table
Sin
θ°
θ°
0.34202 0.502 0.64279 0.76604 0.86603 0.93969 0.98481 Sol Difference table for given data is:
20 0.34202
0.15998
70 0.93
sinθ°
−
−
−
−
−
0.04512
80 0.98481
−
Trang 9∴ u = x a
h
− = 45 40 10
− = 5
10 = 0.5 Now using Gauss backward formula
f(u) = f(0) + u∆f (–1) + ( 1)
2!
u+ u ∆2f( )−1 + ( 1) ( 1)
3!
u+ u u−
( )
3f 2
+ ( 2)( 1) ( 1)
4!
u+ u+ u u−
( )
4f 2
∆ − + ( 2)( 1) ( 1)( 2)
5!
u+ u+ u u− u−
( )
5f 2
f(0.5) = 0.64279 + 0.5 × 0.14079 + 0.5 1.5
2
×
× (–0.01754) + 0.5 1.5( 0.5)
6
× (0.00165)
+ 0.5 1.5( 0.5 2.5)( )
24
× (–0.00737)
= 0.64279 + 0.070395 – 0.0065775 – 0.000103125 + 0.00028789
= 0.706792 Ans
Example 3 Apply Gauss backward formula to find the value of (1.06) 19 if
(1 06) 10 = 1.79085, (1.06) 15 = 2.39656, (1 06) 20 = 3.20714, (1.06) 25 = 4.29187, (1.06) 30 = 5.74349
Sol The difference table is given by
10 1.79085
0.60571
1.45162
30 5.74349
h
−
= 19 20 5
−
= – 0.2 From Gauss backward formula
f(u) = f(0) + u∆f (–1) + ( 1)
2!
u+ u
( )
2f 1
∆ − + ( 1) ( 1)
3!
u+ u u−
( )
3f 2
+ ( 2)( 1) ( 1)
4!
u+ u+ u u−
∆4f( )−2 +
f(u) = 3.20714 – 0.2 × 0.81058 – 0.2 0.8( )
2 × 0.27415 – 0.2 0.8( )( 1.2)
6
0.06928 + 0.2 0.8( ) (−1.2 1.8) ( )
× 0.02346
Trang 10f(u) = 3.20714 – 0.162116 – 0.021932 + 0.002216 + 0.00033782
= 3.0256458 (Approx.)
Example 4 Using Gauss backward formula, Estimate the no of persons earning wages between Rs.
60 and Rs 70 from the following data:
( )
No of Persons in thousands
Sol Difference table for the given data is as:
−
−
−
−
120
50
∴ u = x a
h
− = 70 80 20
− = 10 20
− = – 0.5 From Gauss backward formula
4 ( 2)( 1) ( 1)
( 2) 4!
f
= 470 ( 0.5) ( 100) ( 0.5)(0.5) ( 30) ( 0.5)(0.5)( 1.5) ( 10)
+ ( 0.5)(0.5)( 1.5)(1.5) (20)
24
= 470 50 3.75 0.625 0.46875− + − +
= 423.59375
Hence No of Persons earning wages between Rs 60 to 70 is 423.59375 – 370 = 53.59375 or
54000 (Approx.)