lagrange-newton interpolation yashoverdhan

Interpolation The word interpolation denotes the methodThe word interpolation denotes the method of computing the value of the function

f( ) for an gi en al e of heny = f(x) for any given value of x when (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 ( ) p y pfunction or polynomials ϕ(x) which has the same values as f(x) have forthe same values as f(x) have for x0, x1, x2….,xn. Compiled & Prepared by: Y

ASHOVERDHAN VYAS; Sir P

adampat Singhania University

, Udaipur; y

[email protected]

Compiled & Prepared by: YASHOVERDHAN VYAS; S

ir Padampat S

inghania University, U

daipur; [email protected]

Interpolation with unequal intervalsInterpolation with unequal intervalsIf x0, x1,…,xn are unequally spaced then, , , q y pwe use the following two interpolation formulae :formulae :

1. Lagrange’s Interpolation formula 2 N t ’ Di id d diff f l 2. Newton’s Divided difference formula.One advantage of these formulas is that they can also be used in case of equal intervals but the formulae for equal intervals qcannot be used in case of unequal intervals. Compiled & Prepared by: Y



, Udaipur; y

[email protected]

i) Lagranges formula for unequali) Lagranges formula for unequal intervals: If f( ) t k th lIf y = f(x) takes the values y0, y1, y2,….,yn corresponding to x = x0, x1, x2,…,xn then


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)x(f)xx)...(xx()xx()x(f 0n21 −−−

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

)x(f 0n02010 −−−

)xx) (xx()xx()xx( −−−−)x(f

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

1n1312101

n320

−−−−+

0 1 3 n2

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

− − − −+

2 0 2 1 2 3 2 n(x x ) (x x ) (x x )...(x x )....

− − − −+

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

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

1210

1n210 −

−−−−−−−−

+ )xx)...(xx()xx()xx( 1nn2n1n0n −


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is known as the Lagrange's interpolationis known as the Lagrange's interpolation formula. Ex.1. Find u5 by Lagrange’s Method if

1 19 49 181 u0 =1, u3 = 19, u4 = 49, u6 = 181.Solution:

x 0 3 4 6u 1 19 49 181

Solution:

At x = 5, u is given by u 1 19 49 181


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(5 3) (5-4)(5-6)u ( 1)−=u ( 1)

(0-3)(0-4)(0-6)=

(5 0)(5 4)(5 6)(5-0)(5-4)(5-6) (19) (3 0)(3 4)(3 6)

+− − −

( )( )( )

(5-0)(5-3)(5-6) ( 49)+ ( 49) (4 0)(4 3)(4 6)

+− − −

(5-0)(5-3)(5-4)(5-0)(5-3)(5-4) (181)(6 0)(6 3)(6 4)

+− − −

101=Compiled & Prepared by: Y



, Udaipur; y

[email protected]

Example-2 Prove that Lagrange’s formula for polynomial approximation Pn (x) of (n+1) pointspoints (xi, yi = f(xi)) i = 0, 1, 2, …, n can be expressed in the form:


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2

2

( ) 1( ) 1

L L nn

n

P x x x xf 2

0 0 0 02

( ) 10( ) 1 =

L L

L L

n

nf x x x xf x x x x1 1 1 1 0( ) 1 =L L

L L L L L L L

f x x x x

2( ) 1 L L nn n n nf x x x x

And also verify this result for n = 2. Apartial solution of Ex 2 is as follows :A partial solution of Ex-2 is as follows : Compiled & Prepared by: Y



, Udaipur; y

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Ex 3 If y(1) = 3 y(3) = 9 y(4) =30Ex.3. If y(1) = -3, y(3) = 9, y(4) =30, y(6) = 132 then find the Lagranges i l i l i l h k hinterpolating polynomial that takes the same values as y at the given points.y g pGiven:

x 1 3 4 6x 1 3 4 6y -3 9 30 132

Solution:


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

= −− − −

( 1)( 4)( 6) (x 1)(x 4)(x 6) .9(3 1)(3 4)(3 6)

− − −+

− − −

(3 1)(3 4)(3 6)(x 1)(x 3)(x 6) .30− − −

+ .30(4 1)(4 3)(4 6)

+− − −

(x 1)(x 3)(x 4)− − − (x 1)(x 3)(x 4) .132(6 1)(6 3)(6 4)

+− − −

3 2 = x3 - 3x2 + 5x - 6 Compiled & Prepared by: Y



, Udaipur; y

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Try thisTry this Q.1 Using Lagranges interpolation formula find a polynomial which passes through the points (0, -12), (1, 0) (3, 6). g p ( , ), ( , ) ( , )(4, 12) Answer: x3 7x2 +18x 12Answer: x - 7x +18x - 12


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To calculate the error from (7), the function y(x) and its (n+1) thderivative must be known. The point of interpolation x’ is used in place of x to determine the error at x’.


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21


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× 2× 2


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× 5× 5

× 7


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ii) i i iff (∆ )ii) Divided differences (∆d)1 0y yf ( ) [ ]−

∆ ∆ 1 0d 0 d 0 0 1

1 0

y yf (x ) y [x , x ]x x

∆ = ∆ = =−

2 1d 1 2 1

y yy [x ,x ]x x

−∆ = =

−

2 1x x

n n 1y yy [x x ]−−∆ = =d n 1 n 1 n

n n 1

y [x ,x ]x x− −

−

∆ = =−


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S d Di id d DiffSecond Divided Difference2 2 d 1 d 0y yf (x ) y ∆ − ∆

∆ ∆ d 1 d 0d 0 d 0

2 0

f (x ) yx x

∆ = ∆ =−

][]x,x[]x,x[ 0112 −

]x,x,x[xx

],[],[210

02

0112 =−

=


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ly 2 d 2 d 1d 1

y y|| y ∆ − ∆∆ =

3 1

3 2 2 1

x x[x ,x ] [x , x ]

−−3 2 2 1

3 1

[x ,x ] [x , x ]x x

=−

1 2 3[x , x , x ]=i il l 3 b d fi d∆similarly 3

d 0y ,.... can be defined∆


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Divided Difference Table x y ∆d y ∆d

2y ∆d3y ∆d

4yx0 y00 y0

∆dy0

x y ∆ 2y

1 0

1 0

y yx x

−=

−d 1 d 0y y∆ − ∆

=x1 y1 ∆d y0

∆dy1 ∆d3y0

∆ 2 ∆ 4

2 1

2 1

y yx x

−=

−

2 0x x=

−

d 2 d 1y y∆ − ∆

2 2d 1 d 0

3 0

y yx x

∆ − ∆=

−x2 y2 ∆d

2y1 ∆d4y0

∆dy2 ∆d3y1

3 2

3 2

y yx x

−=

−

d 2 d 1

3 1

y yx x

=−

y y∆ ∆

2 2d 2 d 1

4 1

y yx x

∆ − ∆=

−

x3 y3 ∆d2y2

∆dy34 3

4 3

y yx x

−=

−

d 3 d 2

4 2

y yx x

∆ − ∆=

−

x4 y4

4 3


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In general m 1 m 1

m d k 1 d ky yy− −

+∆ −∆∆ d d

d kk m k

yx x+

∆ =−k m k+

Provided all the involvedProvided all the involved quantities exist. Compiled & Prepared by: Y



, Udaipur; y

[email protected]

Newton's divided difference interpolation formulainterpolation formula

0 0 d 0y f(x) y (x x ) y= = + − ∆ 0 0 d 0

20 1 d 0(x x ) (x x ) y+ − − ∆

30 1 2 d 0(x x ) (x x )(x x ) y+ − − − ∆

n0 1 n 1 d 0... (x x ) (x x ) ...(x x ) y−+ + − − − ∆


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Ex.1 Given the valuesEx.1 Given the values x : 5 7 11 13 17f( ) 150 392 1452 2366 5202f(x) : 150 392 1452 2366 5202

Evaluate f(9) usingEvaluate f(9) using (i) Lagrange's formula (ii) N ' di id d diff f l(ii) Newton's divided difference formula.


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i) Lagranges formulai) Lagranges formulaf (9)

(9 7) (9 11) (9 13) (9 17) (150)(5 7) (5 11) (5 13) (5 17)

− − − −=

− − − −(5 7) (5 11) (5 13) (5 17)− − − −(9 5) (9 11) (9 13) (9 17) (392)− − − −

+ (392)(7 5) (7 11) (7 13) (7 17)

+− − − −

(9 5) (9 7) (9 13) (9 17)(9 5) (9 7) (9 13) (9 17) (1452)(11 5) (11 7) (11 13) (11 17)

− − − −+

− − − −(11 5) (11 7) (11 13) (11 17)Compiled & Prepared by: Y



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(9 5) (9 7) (9 11) (9 17)(9 5) (9 7) (9 11) (9 17) (2366)(13 5)(13 7)(13 11)(13 17)

− − − −+

− − − −( ) ( )( )( )(9 5) (9 7) (9 11) (9 13) (5202)− − − −

+ ( )(17 5)(17 7)(17 11)(17 13)− − − −

∴ f (9) = 810 ∴ f (9) 810


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ii) Newton Divided Difference formula)x y ∆d ∆d

2 ∆d3 ∆d

4

5 1505 150121

7 392 247 392 24265 1

11 1452 32 0457 1

13 2366 42709

17 5202Compiled & Prepared by: Y



, Udaipur; y

[email protected]

f(9) 1 0 121 (9 )f(9) = 150 + 121 (9 - 5) + 24 (9 - 5) (9 - 7) ( ) ( ) + 1(9 - 5) (9 - 7) (9 - 11) = 810 Ex 2 Using i) Lagrange’s interpolationEx.2. Using i) Lagrange s interpolation and ii) divided difference formula. Find h l f h 10the value of y when x = 10.

x : 5 6 9 11x : 5 6 9 11y : 12 13 14 16


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Lagranges formula: (10 6) (10 9) (10 11)y f (10) 12

(5 6) (5 9) (5 11)− − −

= = ×− − −

(5 6) (5 9) (5 11)− − −(10 5)(10 9)(10 11) 13− − −

+ × 13(6 5)(6 9)(6 11)

+ ×− − −

(10 5)(10 6)(10 11) (10 5)(10 6)(10 11) 14(9 5)(9 6)(9 11)

− − −+ ×

− − −

( )( )( )(10 5)(10 6)(10 9) 16(11 5)(11 6)(11 9)

− − −+ ×

(11 5)(11 6)(11 9)− − −Compiled & Prepared by: Y



, Udaipur; y

[email protected]

Divided difference

d d d


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f (10)1⎛ ⎞112 (10 5) (10 5)(10 6).6

⎛ ⎞= + − + − − −⎜ ⎟⎝ ⎠

1(10 5)(10 6)(10 9)

⎝ ⎠⎛ ⎞+ − − − ⎜ ⎟(10 5)(10 6)(10 9)

20+ ⎜ ⎟

⎝ ⎠44 344

=


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Ex 3 Find the interpolating polynomialEx.3. Find the interpolating polynomial using Newton divided difference formula for the following data:

x 0 1 2 5x 0 1 2 5y 2 3 12 147


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x y ∆d ∆d2 ∆d

3

0 21

1 3 49 19 1

2 12 94545

5 147


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

3 2 = x3 + x2-x + 2


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Ex.4. Find f(4) given f(0) = -4, f(2) = 2, f(3) =14 and f(6) = 158

The divided difference table is The divided difference table is x y ∆d ∆d

2 ∆d3

0 -43

2 2 312 1

3 14 948

6 158Compiled & Prepared by: Y



, Udaipur; y

[email protected]

∴ f(4)= -4 + (4 ) (3)+ (4) (2) (3)

+(4) (2) (1) (1) = 40 +(4) (2) (1) (1) = 40

Ex.5. Fit an interpolating polynomial forEx.5. Fit an interpolating polynomial for u4 = 48, u5 = 100, u6 = 180 , u8 = 448, u10 =900 and u11 = 1210


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x y ∆d ∆d2 ∆d

3 ∆d4

4 4852

5 100 1480 1

6 180 18 0134 1

8 448 23 0226 1226 1

10 900 28310310

11 1210Compiled & Prepared by: YASHOVERDHAN VYAS; S

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By applying Newton’s Divided diff f l tdifference formulae, we get

48 + ( 4) 52 + ( 4) ( 5) 14∴ y = 48 + (x -4) 52 + (x-4) (x-5) 14 + (x 4) (x 5) (x 6) = x3 x2 + (x-4) (x-5) (x -6) = x - x


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E l 6 Gi th f ll i t blExample 6. Given the following table, find f(x) as a polynomial in powers of ( 5)(x – 5)

0 2 3 4 7 9x: 0 2 3 4 7 9f(x): 4 26 58 112 466 922. Sol. The divided difference table is:


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By Newton’s divided difference formulaBy Newton’s divided difference formula, we get gf(x) = 4 + (x – 0)(11) + (x – 0)(x – 2)7 + ( 0)( 2)( 3) 1(x – 0)(x – 2)(x – 3) 1= x3 + 2x2 + 3x + 4 x 2x 3x 4In order to express it in power of (x – 5), we use synthetic division, as:


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∴ 2x2 + x3 + 3x + 4

= (x – 5)3 + 17(x – 5)2 + 98 (x – 5) + 194. Compiled & Prepared by: Y



, Udaipur; y

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Example 7. Calculate the approximateExample 7. Calculate the approximate

value of mode of a certain frequency

Curve y = f (x). The values of frequency

density f(x) for x = 8.9, 9.0 and 9.3 are

respectively equal to 0 30 0 35 and 0 25respectively equal to 0.30, 0.35 and 0.25.


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The divided difference table for given f d i i f llfrequency density is as follows:


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Applying Newton’s divided difference formulaApplying Newton s divided difference formula100 f(x)

50 3500⎛ ⎞= 30 + (x – 8.9)× 509

+ (x – 8.9) (x – 9) 350036

⎛ ⎞−⎜ ⎟⎝ ⎠

= – 97. 222x2 + 1745.833x – 1759.7217 ∴ f(x) = – 0 9722x2 + 17 45833x – 17 597217∴ f(x) 0.9722x2 + 17.45833x 17.597217f ' (x) = – 1.9444x + 17.45833 Putting f ' (x) 0 we getPutting f ' (x) = 0, we get


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17.45833x = 17.45833 8.97881.9444

=

Also, f '' (x) = – 1.9444 i.e., (–) ve ∴ f(x) is maximum at x = 8.9788( ) Hence mode is 8 9788Hence, mode is 8.9788.


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Example 7. Apply the Newton divided difference formula for inverse interpolation to find the real root of ( ) 0=f x , where ( )f

3( ) 2 5= − −f x x x by considering the values of ( )f x at x = 1.9, 2, 2.1, 2.2.

Solution: The values of f(x) at ( )x = 1.9, 2, 2.1, 2.2 are

1 941 1 000 0 061 1 248 respectively– 1.941, – 1.000, 0.061, 1.248, respectively.


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To find the root we change some notations To find the root, we change some notations as follows: x: – 1.941 – 1.000 0.061 1.248 ux: 1.9 2.0 2.1 2.2 So we have to find ux at x = 0. The divided difference table is:


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x ux ∆dux ∆d2ux ∆d

3ux x ux ∆dux ∆d ux ∆d ux

-1.941 1.9 0.1062699

1 000 2 0 0 0060035-1.000 2.0 -0.0060035 0.0942507 0.0004869

0.061 2.1 -0.0044505 0.0842459

1.248 2.21.248 2.2 Compiled & Prepared by: Y



, Udaipur; y

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Applying the Newton-divided difference Applying the Newton divided difference formula, u = 1 9 + (x + 1 941) × 0 1062699ux = 1.9 + (x + 1.941) × 0.1062699 + (x + 1.941)(x + 1)(– 0.0060035) + (x + 1.941)(x + 1)(x – 0.061) × 0.0004869.For x = 0 u0 = 1.9 + 0.2062698 – 0.0116527

– 0 0000576 = 2 0945595 0.0000576 2.0945595∴ The required root is 2.0945595.


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Try this. Question. Apply the Newton divided

difference formula for inverse interpolationdifference formula for inverse interpolation

to find the real root of ( ) 0=f x , where ( )f

3( ) 3= + −f x x x by considering the values

of ( )f x at x = 1.1, 1.2, 1.3, 1.4.

Ans. 1.2134 approx.


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Example 8 The following are the mean Example 8. The following are the mean temperatures (°F) on three days, 30 days apart round the periods of summer andapart round the periods of summer and winter. Estimate the approx. dates and values of max and min temperaturevalues of max. and min. temperature.

Summer Winter Day Date Temp. Date Temp.

0 15-Jun. 58.8 16-Dec. 40.730 15-July 63.4 15-Jan. 38.1 60 14-Aug 62 5 14-Feb 39 360 14-Aug. 62.5 14-Feb. 39.3


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Solution: The divided difference table forSolution: The divided difference table for summer is:

x f(x) ∆df(x) ∆d2f(x)

0 58.8 4.6

1 63.4 -2.7

0 9 -0.9 2 62.5

Compiled & Prepared by: Y



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∴ f(x) = 58 8 + (x 0)(4 6) + (x 0)(x 1)( 2 75) ∴ f(x) 58.8 + (x – 0)(4.6) + (x – 0)(x – 1)(– 2.75) = – 2.75 x2 + 7.35 x + 58.8 F i d i i f f( ) h For maximum and minimum of f(x), we have f ' (x) = 0

5 5 7 35 0 1 342 ⇒ – 5.5 x + 7.35 = 0 ⇒ x = 1.342 Again, f “ (x) = – 5.5 < 0 ∴ f(x) is maximum at x = 1.342 Since unit 1 = 30 days ∴ 1.342 = 30 × 1.342 = 40.26 days ∴ The maximum temperature was on p


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15 J + 40 d i 25 J l d 15 June + 40 days, i.e., on 25 July, and the value of the maximum temperature is [f( )] [f( )] 63 711°F i t l [f(x)]max. = [f(x)]1.342 = 63.711 F. approximately. …… N t T th i i t f th l ti Note: Try the remaining part of the solution yourself.


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LAGRANGE Interpolation Formula1. Read n, x2. for i =1 to (n) in steps of 1 do ( )3. Read xi, fi end for4 sum ← 04. sum ← 05. for i =1 to (n) in steps of 1 do6. prodfunc ← 17. for j =1 to (n) in steps of 1 do


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8. if ( j ≠ i ) then9. prodfunc ← prodfunc×(x − xj )/ (xi − xj )

endfor10.sum ← sum + (fi × prodfunc)

endforendfor11.Write x, sum12 STOP12.STOP


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lagrange-newton interpolation yashoverdhan

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