WORKED OUT PROBLEMS
The problems in this section are solved using Newton's divided
difference formula and Lagrange's formula. Since By Sheperd's Zig-Zag rule
any aritrary path from function values to its highest divided difference
to compute the value of f(x) in all these examples first fuction
value and its higher divided differences are been used to compute f(x).
1. Find f(2) for the data f(0) = 1, f(1) = 3
and f(3) = 55.
Solution :
By Newton's divided difference formula :
Divided difference table
|
xi
|
fi
|
|
|
|
0
|
1
|
|
|
|
|
|
2
|
|
|
1
|
3
|
|
8
|
|
|
|
26
|
|
|
3
|
55
|
|
|
Now Newton's divided difference formula is
f(x) = f [x0] + (x - x0) f [x0,
x1] + (x - x0) (x - x1) f [x0,
x1, x2]
f(2) = 1 + (2 - 0) 2 + (2 - 0)(2 - 1) 8
= 21
By Lagrange's formula :
| |
(x - x1) (x - x2)
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(x - x0)(x - x1)
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|
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f(x) =
|
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f0+ . . . +
|
|
f2
|
| |
(x0 - x1) (x0 - x2)
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(x2 - x0)(x2 - x1)
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|
| |
(2 - 1)(2 - 3)
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(2 - 0)(2 - 3)
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(2 - 0)(2 - 1)
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|
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f(2) =
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1+
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3 +
|
|
55
|
| |
(0 - 1) (0 - 3)
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(1 - 0)(1 - 3)
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(3 - 0)(3 - 1)
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|
f(2) = 21
2. Find
f(3) for
|
x
|
0
|
1
|
2
|
4
|
5
|
6
|
|
f
|
1
|
14
|
15
|
5
|
6
|
19
|
Solution :
By Newton's divided difference formula :
Divided difference table
|
xi
|
fi
|
|
|
|
|
|
|
0
|
1
|
|
|
|
|
|
|
|
|
13
|
|
|
|
|
|
1
|
14
|
|
-6
|
|
|
|
|
|
|
1
|
|
1
|
|
|
|
2
|
15
|
|
-2
|
|
0
|
|
|
|
|
-5
|
|
1
|
|
0
|
|
4
|
5
|
|
2
|
|
0
|
|
|
|
|
1
|
|
1
|
|
|
|
5
|
6
|
|
6
|
|
|
|
| |
|
13
|
|
|
|
|
|
6
|
19
|
|
|
|
|
|
Now Newton's divided difference formula is
f(x) = f [x0] + (x - x0) f [x0,
x1] + (x - x0) (x - x1) f [x0,
x1, x2] + (x - x0) (x - x1)
(x - x2)f [x0, x1, x2,
x3]
+ (x - x0) (x - x1) (x - x2)(x
- x3)f [x0, x1, x2, x3,
x4]
+ (x - x0) (x - x1) (x - x2)(x
- x3)(x - x4)f [x0, x1,
x2, x3, x4, x5]
f(3) = 1 + (3 - 0) 13 + (3 - 0)(3 - 1) -6 + (3 - 0) (3 - 1) (3 - 2)
1
= 10
By Lagrange's formula :
| |
(3 - 1)(3 - 2)(3 - 4)(3 - 5)(3 - 6)
|
|
(3 - 0)(3 - 2)(3 - 4)(3 - 5)(3 - 6)
|
|
|
f(3) =
|
|
1+
|
|
14+ |
| |
(0 - 1)(0 - 2)(0 - 4)(0 - 5)(0 - 6)
|
|
(1 - 0)(1 - 2)(1 - 4)(1 - 5)(1 - 6)
|
|
|
(3 - 0)(3 - 1)(3 - 4)(3 - 5)(3 - 6)
|
|
(3 - 0)(3 - 1)(3 - 2)(3 - 5)(3 - 6)
|
|
|
15+
|
|
5+ |
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(2 - 0)(2 - 1)(2 - 4)(2 - 5)(2 - 6)
|
|
(4 - 0)(4 - 1)(4 - 2)(4 - 5)(4 - 6)
|
|
|
(3 - 0)(3 - 1)(3 - 2)(3 - 4)(3 - 6)
|
|
(3 - 0)(3 - 1)(3 - 2)(3 - 4)(3 - 5)
|
|
|
6+
|
|
19 |
|
(5 - 0)(5 - 1)(5 - 2)(5 - 4)(5 - 6)
|
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(6 - 0)(6 - 1)(6 - 2)(6 - 4)(6 - 5)
|
|
f(2) = 10
3. Find
f(0.25) for
|
x
|
0.1
|
0.2
|
0.3
|
0.4
|
0.5
|
|
f
|
9.9833
|
4.9667
|
3.2836
|
2.4339
|
1.9177
|
Solution :
By Newton's divided difference formula :
Divided difference table
|
xi
|
fi
|
|
|
|
|
|
0.1
|
9.9833
|
|
|
|
|
|
|
|
-50.166
|
|
|
|
|
0.2
|
4.9667
|
|
166.675
|
|
|
|
|
|
-16.83
|
|
-416.68
|
|
|
0.3
|
3.2836
|
|
41.67
|
|
833.42
|
|
|
|
-8.497
|
|
-83.32
|
|
|
0.4
|
2.4339
|
|
16.675
|
|
|
|
|
|
-5.162
|
|
|
|
|
0.5
|
1.9177
|
|
|
|
|
Now Newton's divided difference formula is
f(x) = f [x0] + (x - x0) f [x0,
x1] + (x - x0) (x - x1) f [x0,
x1, x2] + (x - x0) (x - x1)
(x - x2)f [x0, x1, x2,
x3]
+ (x - x0) (x - x1) (x - x2)(x
- x3)f [x0, x1, x2, x3,
x4]
f(3) = 9.9833 + (0.25 - 0.1) -50.166 + (0.25 - 0.2)(0.25
- 0.3) 166.675 +
(0.25 - 0.1) (0.25 - 0.2) (0.25 - 0.3) -416.68 + (0.25
- 0.1) (0.25 - 0.2) (0.25 - 0.3)(0.25 - 0.4) 833.42
= 3.912
By Lagrange's formula :
f(0.25) =
| |
(.25 - .2)(.25 - .3)(.25 - .4)(.25 - .5)
|
|
(.25 - .1)(.25 - .3)(.25 - .4)(.25 - .5)
|
|
|
|
|
9.9833+
|
|
4.9667 + |
| |
(.1 - .2)(.1 - .3)(.1 - .4)(.1 - .5)
|
|
(.2 - .1)(.2 - .3)(.2 - .4)(.2 - .5)
|
|
|
(.25 - .1)(.25 - .2)(.25 - .4)(.25 - .5)
|
|
(.25 - .1)(.25 - .2)(.25 - .3)(.25 - .5)
|
|
|
3.2836+
|
|
2.4339 + |
|
(.3 - .1)(.3 - .2)(.3 - .4)(.3 - .5)
|
|
(.4 - .1)(.4 - .2)(.4 - .3)(.4 - .5)
|
|
|
(.25 - .1)(.25 - .2)(.25 - .3)(.25 - .4)
|
|
1.9177 |
|
(.5 - .1)(.5 - .2)(.5 - .3)(.5 - .4)
|
|
Work out with the Newton's
divided difference here
Work out with the Lagrange's
Interpolation formula here
|
