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5. Interpolation 5.1 Definition of interpolation. 5.2 Formulas for Interpolation. 5.3 Formulas for Interpolation for unequal interval. 5.4 Applications of interpolation
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5.1 Definition of interpolation. Definition
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5.2 DIRECT METHOD Now we consider a polynomial of degree n, i.e. y = a 0 +a 1 x+a 2 x 2 +……..+a n x n, where a 0,a 1,a 2,….a n are constants, on which we suppose (n+1) points like (x 0,y 0 ),(x 1,y 1 ),….,(x n,y n ). (i) Here we have to find (n+1) constants i.e. a 0,a 1,a 2,…..a n, for that we construct (n+1) equations. (ii) Substitute the value of x for the corresponding value of y in the above polynomial.
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Table 1 t246810 m/second1014213050
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Linear Interpolation
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Quadratic Interpolation
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Error
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Cubic Interpolation Example: - The velocity of a train is given in table 1. Find the velocity at t = 9 second using cubic Interpolation method. Solution:- Let v(t) =a 0 +a 1 t+a 2 t 2 +a 3 t 3. v(4)= a 0 +4a 1 +16a 2 +64a 3, v(6)= a 0 +6a 1 +36a 2 +216a 3, v(8)= a 0 +8a 1 +64a 2 + 512a 3, v(10)= a 0 +10a 1 +100a 2 +1000a 3.
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Continued………
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Comparison table Table:-Comparison of different degree of the polynomials. Degree of polynomial 123 v(t=9) meter/sec4038.62538.0625 Absolute relative Approximate error ---------------3.55987051.4778325
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Newton forward Interpolation formula
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Example The velocity of a train is given in table 1. Find the velocity at t= 5 second using Newton forward interpolation formula. Solution:- Forward Difference Table
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Continued tv1 st difference2 nd difference3 rd difference4 th difference 210 4 7 9 20 3 2 11 4 14 621 830 1050
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Result u=1.5. Putting these values in the Newton forward interpolation formula we get v(5)=17.2109 m/sec.
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Newton Backward Interpolation formula
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Example The velocity of a train is given in table 1. Find the velocity at t= 9 second using Newton backward interpolation formula. Solution: - u=-0.5. Putting these values in the Newton backward interpolation formula we get v(9)=37.6719 m/sec.
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5.3 Newton’s Divided Difference method Table 2 t m/secon d 210 519 725 1152 1660
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Linear Interpolation
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Example
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Quadratic Interpolation
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Example
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Error
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Cubic Interpolation
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Example
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Error
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Comparison Table Table:-Comparison of different degree of the polynomials. Degree of polynomial 123 v(t=12) meter/sec53.655.888958.06639 Absolute relative Approximate error ---------------4.09545 3.75
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Lagrange’s Interpolation
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Linear Interpolation
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Example The velocity of a train is given in table 2. Find the velocity at t= 12 second using linear method for interpolation. Solution:- Here t 0 =11sec, t 1 =16sec and t=12 sec. Putting these values in the v(t)= L 0 (t)v(t 0 )+ L 1 (t)v(t 1 ) we get v(12)=53.6 m/sec.
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Quadratic Interpolation
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Example The velocity of a train is given in table 2. Find the velocity at t= 12 second using quadratic method for interpolation. Solution:- Here t 0 =7sec, t 1 =11sec, t 2 =16 sec and t=12 sec. Putting these values in the v(t)= L 0 (t)v(t 0 )+ L 1 (t)v(t 1 ) + L 2 (t)v(t 2 ), we get v(12)=55.8889 m/sec.
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Error
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Cubic Interpolation
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Example The velocity of a train is given in table 2. Find the velocity at t= 12 second using cubic method for interpolation. Solution:- Here t 0 =5sec, t 1 =7sec, t 2 =11sec, t 3 =16 sec and t=12 sec. Putting the values in the v(t)= L 0 (t)v(t 0 )+ L 1 (t)v(t 1 ) + L 2 (t)v(t 2 ) + L 3 (t)v(t 3 ), we get v(12)=58.0657 m/sec.
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Error
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Comparison table Table:-Comparison of different degree of the polynomials. Degree of polynomial123 v(t=12) meter/sec53.655.888958.0657 Absolute relative Approximate error ---------------4.09544653.75
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5.4 Applications of interpolation. Form of the function. To fill the gaps in a table. Computer graphics. Census.
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