lecture4 - Lecture 4: Methods of Lecture 4: Methods of...

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ecture 4: Methods of Lecture 4: Methods of Interpolation: Direct Method of Interpolation ewton’s Divided Difference Polynomial Method Newton s Divided Difference Polynomial Method of Interpolation Lagrangian Interpolation Spline Interpolation Method Please read Chapter 19.3-4 in Advanced Engineering 1 Mathematics
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What is Interpolation ? iven (x find the value of ‘y’ at a Given (x 0 ,y 0 ), (x 1 ,y 1 ), …… (x n ,y n ), find the value of y at a value of ‘x’ that is not given. 2 Figure 1 Interpolation of discrete.
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Interpolants Polynomials are the most common choice of interpolants because they py are easy to: Evaluate Differentiate, and Integrate 3
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Direct Method Given ‘n+1’ data points (x 0 ,y 0 ), (x 1 ,y 1 ),…………. . (x n ,y n ), pass a polynomial of order ‘n’ through the data as given elow: below: . .......... .......... n x a x a a y where a 0 , a 1 ,………………. a n are real constants. 1 0 n Set up ‘n+1’ equations to find ‘n+1’ constants. To find the value ‘y’ at a given value of ‘x’, simply substitute the value of ‘x’ in the above polynomial. 4
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eneral Problem General Problem The upward velocity of a rocket is given as a function of time in Table 1. Find the velocity at t=16 seconds using the direct method for linear interpolation. Table 1 Velocity as a function of time.   s , t     m/s , t v 00 10 227.04 15 362.78 20 517.35 22.5 602.97 5 30 901.67 Figure 2 Velocity vs. time data for the rocket example
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inear Interpolation Example 1: Linear Interpolation   t a a t v 1 0 hoose points closest to and bracket the target point:     78 . 362 15 15 a a v   1 1 , y x y Choose points closest to, and bracket the target point: 1 0  35 . 517 20 20 1 0 a a v 0 0 , y x   x f 1 Solving the above two equations gives , 93 . 100 0 a 914 . 30 1 a x Figure 3 Linear interpolation. Hence . 20 15 , 914 . 30 93 . 100 t t t v 6   m/s 7 . 393 16 914 . 30 93 . 100 16 v
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xample 2: Quadratic Interpolation Quadratic Interpolation Example 2:  2 2 1 0 t a t a a t v 4 27 0 0 2 1 1 , y x   2 2 , y x y       04 . 227 10 10 10 2 1 0 a a a v   78 . 362 15 15 15 2 2 1 0 a a a v   x f       35 . 517 20 20 20 2 2 1 0 a a a v   0 0 , y x 2 x Figure 6 Quadratic interpolation. Solving the above three equations gives 5 2 33 7 766 7 05 . 12 0 a 733 . 17 1 a 3766 . 0 2 a
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Quadratic Interpolation (cont.) 450 500 550 517.35  20 10 , 3766 . 0 733 . 17 05 . 12 2 t t t t v 300 350 400 y s f range () fx desired     2 16 3766 . 0 16 733 . 17 05 . 12 16 v m/s 19 . 392 10 12 14 16 18 20 200 250 227.04 20 10 x s range x desired he absolute relative approximate error obtained between The absolute relative approximate error obtained between the results from the first and second order polynomial is 00 70 . 393 19 . 392 a 8 % 38410 . 0
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lecture4 - Lecture 4: Methods of Lecture 4: Methods of...

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