Interpolation

The word interpolation denotes the method of computing the value of the function y =

f(x) for any given value of x when a set (x0, y0), (x1, y1),…(xn, yn) are given.

Note:

Since in most of the cases the exact form of the function is not known. In such cases the

function f(x) is replaced by a simpler function (x) which has the same values as f(x) for

x0, x1, x2….,xn.

....y!3

)2u()1u(uy

!2

)1u(uyuy)x( 0

30

200

0n y

!n

)1nu...()2u()1u(u

is called the Newton Gregory forward difference formula

Note :

1. Newton forward interpolation is used to interpolate the values of y near the beginning

of a set of tabular values.

2. y0 may be taken as any point of the table but the formula contains those values of y

which come after the value chosen as y0.

The table gives the distances in nautical miles of the visible horizon for the given heights

in feet above the earths surface.

x = height 100 150 200 250 300 350 400

y = distance 10.63 13.03 15.04 16.81 18.42 19.90 21.27

Find the values of y when i) x = 120, ii) y =218

x y 2 3 4 5 6

100 10.632.40

150 13.03 -0.392.01 0.15

200 15.04 -0.24 -0.071.77 0.08 0.02

250 16.81 -0.16 -0.05 0.021.61 0.03 0.04

300 18.42 -0.13 -0.011.48 0.02

350 19.90 -0.111.37

400 21.27

Choose x0 = 100

i) 4.050

100120u,120x

)39.0(!2

)14.0()4.0()40.2(!14.063.10)120(f

)15.0(!3

)24.0()14.0()4.0(

)07.0(!4

)34.0()24.0()14.0()4.0(

)02.0(!5

)44.0()34.0()24.0()14.0()4.0(

649.11)02.0(!6

)54.0()44.0()34.0()24.0()14.0()4.0(

ii) Let x = 218, x0 = 200, 36.050

18

50

200218u

)16.0(2

)64.0(36.0)77.1(36.004.15)218(f

...)03.0(6

)64.1()64.0(36.0

= 15.7

2. Find the value of f(1.85)

x y y 2y 3y 4y 5y 6y1.7 5.474

0.5751.8 6.049 0.062

0.637 0.0041.9 6.686 0.066 0.004

0.703 0.008 -0.0042.0 7.389 0.074 0 0.004

0.777 0.008 02.1 8.166 0.082 0

0.859 0.0082.2 9.025 0.090

0.94923 9.974

5.01.0

8.185.1

h

xxu85.1x,8.1xChoose 0

0

)066.0(2

)5.0()5.0()637.0()5.0(049.6)85.1(f

)008.0(6

)5.1()5.0)(5.0(

0.00050.0008-0.31856.049

= 6.359

4. Given sin 45o = 0.7071, sin 50o = 0.7660, sin 55o =0.8192, sin 60o = 0.8660. Find sin

48o.

x y 2 3

45 0.7071

0.589

50 0.7660 -0.0057

0.0532 0.0007

55 0.8192 -0.0064

0.0468

60 0.8660

6.0h

xxu5h;45x,48x 0

0

)0589.0()6.0(7071.048sin o

)0057.0(2

)4.0)(6.0(

7431.0)0007.0(6

)4.1()4.0()6.0(

From the following data find the number of students who have obtained 45 marks. Also

find the number of students who have scored between 41 and 45 marks.

Marks 0 - 40 41 - 50 51 - 60 61 -70 71 - 80

No. of students 31 42 51 35 31

x y 2 3 4

40 31

42

50 73 9

51 -25

60 124 -16 37

35 12

70 159 -4

31

80 190

2

9)5.0()5.0()42()5.0(31)45(f

!3

)25()5.1()5.0()5.0(

488672.47)37(!4

)5.2()5.1()5.0()5.0(

f(45) - f(40) = 70 = Number of students who have scored between 41 and 45.

Find the interpolating polynomial for the following data:

f(0) = 1, f(1) = 0, f(2) = 1, f(3) = 10. Hence evaluate f(0.5)

x y 2 3

0 1

-1

1 0 2

1 6

2 1 8

9

3 10

x1

0xu

)2(!2

)1x(x)1(x1)x(f

1x2x6!3

)2x()1x(x 23

6. Find the interpolating polynomial for the following data:

x: 0 1 2 3 4

f(x) : 3 6 11 18 27

x y 2 3 4

0 33

1 6 25 0

2 11 2 07 0

3 18 29

4 27

x1

0xu

)2(2

)1x(x)3(x3)x(f

)0(!x

)1x(x

2xx23

Newton Gregory Backward Interpolation formula

....y!3

)2u()1u(uy

!2

)1u(uyuyy n

3n

2nn

h

xxuwhere n

2. The values of tan x are given for values of x in the following table. Estimate tan (0.26)

x 0.10 0.15 0.20 0.25 0.30

y 0.1003 0.1511 0.2027 0.2553 0.3093

x y 2 3 4

0.10 0.10030.0508

0.15 0.1511 0.00080.0516 0.0002

0.20 0.2027 0.0010 0.00020.0526 0.0004

0.25 0.2553 0.00140.0540

0.30 0.3093

8.005.0

3.026.0u

)054.0)(8.0(3093.0)26.0(f

)0014.0()2.0(2

)8.0(

2659.0)0004.0(6

)2.1()2.0()8.0(

3. The deflection d measured at various distances x from one end of a cantilever is given

by the following table. Find d when x = 0.95

x d 2 3 4 5

0 00.0347

0.2 0.0347 0.04790.0826 -0.0318

0.4 0.1173 0.0161 0.00030.0987 -0.0321 -0.0003

0.6 0.2160 -0.016 00.0827 -0.032

0.8 0.2987 -0.04810.0346

1.0 0.3333

95.0xwhen3308.0d25.02.0

195.0u

4. The area y of circles for different diameters x are given below:

x : 80 85 90 95 100

y : 5026 5674 6362 7088 7854

Calculate area when x = 98

x y y 2y 3y 4y80 5026

64885 5674 40

688 -290 6362 38 4

726 295 7088 40

766100 7854

Answer:

4.0n

xxu n

y = 7542

Find the interpolating polynomial which approximates the following data.

x 0 1 2 3 4

y -5 -10 -9 4 35

x y 2 3 4

0 -5-5

1 -10 61 6

2 -9 12 013 6

3 4 1831

4 35

1

4xu

!2

18)3x()4x()31()4x(35)x(f

!3

)6()2x()3x()4x(

5-6x2x x 23

Central difference interpolation formula

X y 2 3 4 5 6

x-3 y-3

y-3

x-2 y-2 2y-3

y-2 3y-3

x-1 y-1 2y-2 4y-3

y-1 3y-2 5y-3

x0 y0 2y-1 4y-2 6y-3

y-0 3y-1 5y-2

x1 y1 2y0 4y-1

y1 3y0

x2 y2 2y1

y2

x3 y3

a) Stirlings formula

2

yy

!3

)1u(uy

2

u

2

yyuyy 1

32

32

12

210

0

....2

yy

!5

)4u()1u(uy

!4

)1u(u 25

3522

24

22

h

xxuwhere 0

b) Bessel's formula

130

21

2

00 y!3

)1p(p)2

1p(

2

yy

!2

)1p(pypyy

....2

yy

!4

)2p()1p(p)1p( 14

24

h

xxpwhere 0

1. Apply sterling's formula to find f(14.2) from the following table:

x y 2 3 4

10 0.24

0.041

12 0.281 -0.004

0.037 0.001

14 0.318 -0.003 0

0.034 0.001

16 0.352 -0.002

0.032

18 0.384

1.02

142.14

h

xxu 0

}003.0{2

)1.0(

2

034.0037.01.0318.0y

2

2

001.0001.0

6

)1)1.0(()1.0( 2

= 0.3215

2.Apply stirling's formula to find a polynomial f(x) of degree 4 that approximates the

following data and hence find f(2.5)

x f(x) 2 3 4

1 1

-2

2 -1 4

2 -8

3 1 -4 18

-2 8

4 -1 4

2

5 1

3x1

3xu

2

88}1)3x{(

!3

)3x()4(

2

)3x(

2

)2(2)3x(1)x(f 2

2

161)3x(!4

)3x( 22

93x168x100x24x23

1 234

375.0)5.2(f

3. Using Bessels formula find f(12.5) for the data:

x f(x) 2 3

5 12

1

10 13 0

1 1

15 14 1

2

20 16

5.05

105.12

h

xxu 0

0)5.0(2

)5.0()5.0()1()5.0(13)5.12(f

=13.4375

4. Using Bessel's formula find 3rd degree polynomial that approximates the following

data:

f(0) = 2, f(1) = 3, f(2) = 8, f(3)=23

x y 2 3

0 2

1

1 3 4

5 6

2 8 10

15

3 23

1x1

1xu

6!3

)2x()1x()2

11x(

2

104

!2

)2x()1x(5)1x(3)x(f

2xxx 23

Interpolation with unequal intervals

Newton backward and forward interpolation is applicable only when x0, x1,…,xn-1 are

equally spaced.

Now we use two interpolation formulae for unequally spaced values of x.

i) Lagranges formula for unequal intervals:

If y = f(x) takes the values y0, y1, y2,….,yn corresponding to x = x0, x1, x2,…,xn then

)x(f)xx)...(xx()xx(

)xx)...(xx()xx()x(f 0

n02010

n21

)x(f)xx)...(xx()xx()xx(

)xx)...(xx()xx()xx(1

n1312101

n320

....)x(f)xx)...(xx()xx()xx(

)xx)...(xx()xx()xx(2

n2321202

n310

)x(f)xx)...(xx()xx()xx(

)xx)...(xx()xx()xx(n

1nn2n1n0n

1n210

is known as the lagrange's interpolation formula

ii) Divided differences ()

]x,x[xx

yyy)x(f 10

01

0100

]x,x[xx

yyy 12

12

121

]x,x[xx

yyy n1n

1nn

1nn1n

second divided difference

02

010

20

2

xx

yyy)x(f

]x,x,x[xx

]x,x[]x,x[210

02

0112

]x,x,x[xx

]x,x[]x,x[

xx

yyy|| 321

13

1223

13

121

2ly

similarly definedbecan,....y03

Newton's divided difference interpolation formula

03

21002

10000 y)xx)(xx()xx(y)xx()xx(y)xx(y)x(fy

0n

n10 y)xx...()xx()xx(...

is called the Newton's divided difference formula.

Note:

Lagrange's formula has the drawback that if another interpolation value were inserted,

then the interpolation coefficients need to be recalculated.

Inverse interpolation: Finding the value of y given the value of x is called interpolation

where as finding the value of x for a given y is called inverse interpolation.

Since Lagrange's formula is only a relation between x and y we can obtain the inverse

interpolation formula just by interchanging x and y.

0n02010

n21 x.)yy)...(yy()yy(

)yy)....(yy()yy(x

...x)yy...()yy()yy()yy(

)yy)...(yy()yy()yy()yy(1

n1312101

n3200

n1nn1n0n

1n10 x.)yy)...(yy()yy(

)yy)...(yy()yy(...

is the Lagranges formula for inverse interpolation

5. The following table gives the values of x and y

x : 1.2 2.1 2.8 4.1 4.9 6.2

y : 4.2 6.8 9.8 13.4 15.5 19.6

Find x when y = 12 using Lagranges inverse interpolation formula.

Using Langrages formula

05040302010

54321 x)yy()yy()yy()yy()yy(

)yy()yy()yy()yy()yy(x

445251505

43210 x)yy)...(yy()yy()yy(

)yy()yy()yy()yy()yy(....

055.0964.0419.3252.1234.0022.0

= 3.55

1. Given the values

x : 5 7 11 13 17

f(x) : 150 392 1452 2366 5202

Evaluate f(9) using (i) Lagrange's formula (ii) Newton's divided difference formula.

i) Lagranges formula

392.)177()137()117()57(

)179()139()119()59()150(

)175()135()115()75(

)179()139()119()79()9(f

)2366()1713()1113()713()513(

)179()119()79()59()1452(

)1711()1311()711()511(

)179()139()79()59(

810)5202()1317()1117()717()517(

)139()119()79()59(

f(9) = 810

ii)

5 150

121

7 392 24

265 1

11 1452 32 0

457 1

13 2366 42

709

17 5202

f(9) = 150 + 121 (9 - 5) + 24 (9 - 5) (9 - 7) + 1(9 - 5) (9 - 7) (9 - 11) = 810

2. Using I) Langranges interpolation and ii) divided difference formula. Find the value of

y when x = 10.

x : 5 6 9 11

y : 12 13 14 16

Lagranges formula

13)116)(96)(56(

)1110)(910)(510(12

)115()95()65(

)1110()910()610()10(fy

16)911)(611)(511(

)910)(610)(510(14

)119)(69)(59(

)1110)(610)(510(

3

44

Divided difference

x y 2 3

5 12

1

6 13

6

1

4

3/2

3

1

20

1

6

10/3

690

27

5116

1

15

2

9 14

15

2

5

3/2

12

2

11 16

20

1)910)(610)(510(

6

1).610)(510()510(12)10(f

3

44

3. If y(1) = -3, y(3) = 9, y(4) =30, y(6) = 132 find the lagranges interpolating polynomial

that takes the same values as y at the given points.

Given:

x 1 3 4 6

y -3 9 30 132

9.)63)(43)(13(

)6x)(4x)(1x()3(.

)61)(41)(31(

)6x()4x()3x()x(f

132.)46)(36)(16(

)4x)(3x)(1x(30.

)64)(34)(14(

)6x)(3x)(1x(

= x3 - 3x2 + 5x - 6

4. Find the interpolating polynomial using Newton divided difference formula for the

following data:

x 0 1 2 5

y 2 3 12 147

x y 2 3

0 2

1

1 3 4

9 1

2 12 9

45

5 147

F(x) = 2 + (x - 0)(1) + (x - 0) (x - 1) (4) + (x - 0) (x - 1) (x - 2) 1

= x3 + x2 - x + 2

Try this

5. Using Lagranges interpolation formula find a polynomial which passes through the

points (0, -12), (1, 0) (3, 6). (4, 12)

Answer: x3 - 7x2 +18x - 12