1. Interpolation with unequal intervals
The disadvantage for the previous interpolation formulas is that, they are used only for equal intervals.
The following are the interpolation with unequal intervals.
(i) Lagrange’s formula for unequal intervals
(ii) Newton’s divided difference formula
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2. Lagranges Interpolation formula
If y = f(x) takes the values y0, y1, y2, …, yn corresponding to x0, x1, x2, …, xn, then
Which is known as Lagrange’s formula.
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3. Divided Differences
If (x0, y0), (x1, y1), …, (xn, yn) are given points, then the first divided differences for the argument x0, x1 is defined by [x0, x1] = (y1 – y0)/ (x1 – x0) Similarly [x1, x2] = (y2 – y1)/ (x2 – x1) [x2, x3] = (y3 – y2)/ (x3 – x2) … [xn-1, xn] = (yn – yn-1)/ (xn – xn-1).
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4. Divided differences
The second divided differences for x0, x1, x2 [x0, x1, x2]= ([x1, x2] - [x0, x1])/(x2-x0) The third divided differences x0, x1, x2 , x3 [x0, x1, x2, x3]= ([x1, x2, x3] - [x0, x1, x2])/(x3-x0) And so on … All the divided differences systematically set out in a table called divided difference table.
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5. Divided difference tablex
f(x)
1st DD
2nd DD
3rd DD
4th DD x0 y0
[x0, x1] x1 y1
[x0, x1,x2]
[x1, x2]
[x0, x1,x2, x3] x2 y2
[x1, x2,x3]
[x0, x1,x2, x3,x4]
[x2, x3]
[x1, x2,x3, x4] x3 y3
[x2, x3,x4]
[x3, x4] x4 y4
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6. Newton’s divided difference formula
If y = f(x) takes the values y0, y1, y2, …, yn corresponding to x0, x1, x2, …, xn, then
f(x) = y0+(x-x0)[x0, x1] + (x-x0)(x-x1)[x0, x1, x2]
+ …+(x-x0)(x-x1)…(x-xn-1)[x0, x1, x2,
…, xn].
Which is known as Newton’s general interpolation formula with divided differences.
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Dr.G.Suresh University

7. Problem#1 Given the values Evaluate f(9), using Lagranges formula
Newton’s divided difference formula
Ans : 810
X : 5 7 11 13 17
f(x):
150
392
1452
2366
5202
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Dr.G.Suresh University

8. Divided difference formulax
f(x)
1st DD
2nd DD
3rd DD
4th DD 5
150
121 7
392 24
265 1 11
1452 32
457 13
2366 42
709 17
5202
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Dr.G.Suresh University

9. Problem#2
Determine f(x) as a polynomial in x for the following data, using Newton’s divided difference formula Ans: f(x)=3x4-5x3+6x2-14x+5. x -4 -1 2 5
f(x)
1245 33 9
1335
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Dr.G.Suresh University

10. Divided difference tablex
f(x)
1st DD
2nd DD
3rd DD
4th DD -4
1245
-404 -1 33 94
-28
-14 5 10 3 2 13 9 88
442
1335
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Dr.G.Suresh University

11. Problem#3
The following table gives the viscosity of an oil as a function of temperature. Using Lagrange’s formula find viscosity of oil at a temperature of 140o. Ans : 7.03
TemperatureO:
110
130
160
190
Viscosity:
150
392
1452
2366
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Dr.G.Suresh University

12. Problem#4
Apply Newton’s divided difference formula, evaluate f(8) and f(15) . Ans : f(8) = 448, f(15) = 3150 x 4 5 7 10 11 13
f(x) 48
100
294
900
1210
2028
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Dr.G.Suresh University

13. Problem#5 Certain corresponding values of x and log10x are given below
Calculate log10310, using
Lagrange’s formula
Newton’s divided difference formula.
Ans : by both formulae.
x :
300
304
305
307
log10x :
2.4771
2.4829
2.4843
2.4871
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Dr.G.Suresh University

14. Inverse interpolation formula
The process of estimating the value of x for a value y is called the inverse interpolation.
By interchanging x and y in Lagrange’s formula, is called inverse interpolation formula.
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Dr.G.Suresh University

15. Problem#1
The following table gives the values of x and y : Calculate the value of x corresponding to y=12, using Lagranges technique. Ans : 3.55
X :
1.2
2.1
2.8
4.1
4.9
6.2
f(x):
4.2
6.8
9.8
13.4
15.5
19.6
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Dr.G.Suresh University

16. Problem#2
Apply Lagrange’s formula inversely to obtain a root of the equation f(x)=0, given that f(30)=-30, f(34)=-13, f(38)=3, and f(42)=18. Ans : 37.23
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