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Interpolation functions
 

Interpolation functions

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Interpolation functions

Interpolation functions

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    Interpolation functions Interpolation functions Presentation Transcript

    • Interpolation produces a function that matches the given data exactly. The function then can be utilized to approximate the data values at intermediate points.
    • Interpolation may also be used to produce a smooth graph of a function for which values are known only at discrete points, either from measurements or calculations.
    • Given data points Obtain a function, P(x) P(x) goes through the data points Use P(x) To estimate values at intermediate points
    • Given data points: At x0 = 2, y0 = 3 and at x1 = 5, y1 = 8 Find the following: At x = 4, y = ?
    • P(x) should satisfy the following conditions: P(x = 2) = 3 and P(x = 5) = 8 P( x ) = 3 L0 ( x ) + 8 L1 ( x ) P(x) can satisfy the above conditions if at x = x0 = 2, L0(x) = 1 and L1(x) = 0 and at x = x1= 5, L0(x) = 0 and L1(x) = 1
    • At x = x0 = 2, L0(x) = 1 and L1(x) = 0 and at x = x1= 5, L0(x) = 0 and L1(x) = 1 The conditions can be satisfied if L0(x) and L1(x) are defined in the following way. x−5 L0 ( x ) = and 2−5 x − x1 L0 ( x ) = x0 − x1 x−2 L1 ( x ) = 5−2 x − x0 and L1 ( x ) = x1 − x0
    • P( x ) = 3 L0 ( x ) + 8 L1 ( x ) P( x ) = L0 ( x ) y0 + L1 ( x ) y1 Lagrange Interpolating Polynomial P( x ) = L0 ( x ) f ( x0 ) + L1 ( x ) f ( x1 )
    • P( x ) = L0 ( x ) f ( x0 ) + L1 ( x ) f ( x1 )  x−5  x−2 P( x ) =  ( 3 ) +  ( 8 )  2−5 5−2 5x − 1 P( x ) = 3 5× 4 −1 P( 4) = = 6.333 3
    • The Lagrange interpolating polynomial passing through three given points; (x0, y0), (x1, y1) and (x2, y2) is: P( x ) = L0 ( x ) y0 + L1 ( x ) y1 + L2 ( x ) y2 P( x ) = L0 ( x ) f ( x0 ) + L1 ( x ) f ( x1 ) + L2 ( x ) f ( x2 )
    • ( x − x1 )( x − x2 ) L0 ( x ) = ( x0 − x1 )( x0 − x2 ) At x0, L0(x) becomes 1. At all other given data points L0(x) is 0.
    • ( x − x0 )( x − x2 ) L1 ( x ) = ( x1 − x0 )( x1 − x2 ) At x1, L1(x) becomes 1. At all other given data points L1(x) is 0.
    • ( x − x0 )( x − x1 ) L2 ( x ) = ( x2 − x0 )( x2 − x1 ) At x2, L2(x) becomes 1. At all other given data points L2(x) is 0.
    • General form of the Lagrange Interpolating Polynomial P( x ) = L0 ( x ) y0 + L1 ( x ) y1 + L2 ( x ) y2 + ........... + Ln ( x ) yn P( x ) = L0 ( x ) f ( x0 ) + L1 ( x ) f ( x1 ) + L2 ( x ) f ( x2 ) + ........... + Ln ( x ) f ( xn )
    • ( x − x0 )( x − x1 )...( x − xk −1 )( x − xk +1 )...( x − xn ) Lk ( x ) = ( xk − x0 )( xk − x1 )...( xk − xk −1 )( x k − x k +1 )...( xk − xn ) ( x − xi ) Lk ( x ) = ∏ i =0 ( x k − xi ) n i ≠k
    • Numerator of Lk ( x ) ( x − x0 ) ( x − x1 ) ( x − x2 ) ×LL × ( x − xk −1 ) ( x − xk +1 ) × LL × ( x − xn −1 ) ( x − xn )
    • Denominator of Lk ( x ) ( xk − x0 ) ( xk − x1 ) ( xk − x2 ) ×LL × ( xk − xk −1 ) ( xk − xk +1 ) × LL × ( xk − xn −1 ) ( xk − xn )
    • Find the Lagrange Interpolating Polynomial using the three given points. ( x0 , y0 ) = ( 2, 0.5) ( x1 , y1 ) = ( 2.5, 0.4) ( x2 , y2 ) = ( 4, 0.25)
    • ( x − x1 )( x − x2 ) L0 ( x ) = ( x0 − x1 )( x0 − x2 ) ( x − 2.5)( x − 4) L0 ( x ) = ( 2 − 2.5)( 2 − 4) = x − 6.5 x + 10 2
    • ( x − x0 )( x − x2 ) L1 ( x ) = ( x1 − x0 )( x1 − x2 ) ( x − 2)( x − 4) L1 ( x ) = ( 2.5 − 2)( 2.5 − 4) − x + 6x − 8 = 0.75 2
    • ( x − x0 )( x − x1 ) L2 ( x ) = ( x2 − x0 )( x2 − x1 ) ( x − 2)( x − 2.5) L2 ( x ) = ( 4 − 2)( 4 − 2.5) x − 4. 5 x + 5 = 3 2
    • P( x ) = L0 ( x ) f ( x0 ) + L1 ( x ) f ( x1 ) + L2 ( x ) f ( x2 )
    • ( ) P( x ) = x − 6.5 x + 10 ( 0.5 ) 2  − x + 6x − 8  ( 0.4 ) +   0.75   2  x − 4 .5 x + 5  ( 0.25 ) +   3   2
    • P( x ) = 0.05 x − 0.425 x + 1.15 2 The three given points were taken from the function 1 f ( x) = x
    • 1 f ( 3) = = 0.333 3 An approximation can be obtained from the Lagrange Interpolating Polynomial as: P( 3) = 0.05( 3) − 0.425( 3) + 1.15 = 0.325 2
    • Newton’s Interpolating Polynomials Newton’s equation of a function that passes through two points ( x0 , y 0 ) and ( x1 , y1 ) is P( x ) = a 0 + a1 ( x − x0 )
    • P( x ) = a 0 + a1 ( x − x0 ) Set x = x 0 P ( x0 ) = y0 = a0 Set x = x1 P ( x1 ) = y1 = a0 + a1 ( x1 − x0 )
    • y1 − y 0 a1 = x1 − x0 Newton’s equation of a function that passes through three points ( x0 , y 0 ) ( x1 , y1 ) is and ( x2 , y2 )
    • P ( x ) = a0 + a1 ( x − x0 ) + a2 ( x − x0 ) ( x − x1 ) a 2 , set x = x 2 P ( x2 ) = a0 + a1 ( x2 − x0 ) + a2 ( x2 − x0 ) ( x2 − x1 ) To find
    • y2 − y1 y1 − y0 − x2 − x1 x1 − x0 a2 = x2 − x0
    • Newton’s equation of a function that passes through four points can be written by adding a fourth term . P ( x ) = a0 + a1 ( x − x0 ) + a2 ( x − x0 ) ( x − x1 ) + a3 ( x − x0 ) ( x − x1 ) ( x − x2 )
    • P ( x ) = a0 + a1 ( x − x0 ) + a2 ( x − x0 ) ( x − x1 ) + a3 ( x − x0 ) ( x − x1 ) ( x − x2 ) The fourth term will vanish at all three previous points and, therefore, leaving all three previous coefficients intact.
    • Divided differences and the coefficients The divided difference of a function, f with respect to xi is denoted as f [ xi ] It is called as zeroth divided difference and is simply the value of the function, f at xi f [ xi ] = f ( xi )
    • The divided difference of a function, f with respect to xi and i +1 called as the first divided difference, is denoted x f [ xi , xi +1 ] f [ xi , xi +1 ] f [ xi +1 ] − f [ xi ] = xi +1 − xi
    • The divided difference of a function, f with respect to xi , i +1 and i +2 called as the second divided difference, is denoted as x x f [ xi , xi +1 , xi + 2 ] f [ xi , xi +1 , xi +2 ] f [ xi +1 , xi +2 ] − f [ xi , xi +1 ] = xi + 2 − xi
    • The third divided difference with respect to , , xi + 2 and i + 3 i i +1 x x f [ xi , xi +1 , xi +2 , xi +3 ] x f [ xi +1 , xi + 2 , xi +3 ] − f [ xi , xi +1 , xi + 2 ] = xi +3 − xi
    • The coefficients of Newton’s interpolating polynomial are: a0 = f [ x0 ] a1 = f [ x0 , x1 ] a 2 = f [ x0 , x1 , x 2 ] a3 = f [ x0 , x1 , x 2 , x3 ] a 4 = f [ x0 , x1 , x 2 , x3 , x 4 ] and so on.
    • First divided differences Second divided differences Third divided differences
    • Example Find Newton’s interpolating polynomial to approximate a function whose 5 data points are given below. x f ( x) 2.0 0.85467 2.3 0.75682 2.6 0.43126 2.9 0.22364 3.2 0.08567
    • i 0 xi f [ xi ] 2.0 f [ xi −1 , xi ] f [ xi − 2 , xi −1 , xi ] f [ xi −3 , , xi ] f [ xi −4 , , xi ] 0.85467 -0.32617 1 2.3 0.75682 -1.26505 -1.08520 2 2.6 0.43126 2.13363 0.65522 -0.69207 3 2.9 0.22364 4 3.2 0.08567 -0.29808 0.38695 -0.45990 -2.02642
    • The 5 coefficients of the Newton’s interpolating polynomial are: a0 = f [ x0 ] = 0.85467 a1 = f [ x0 , x1 ] = −0.32617 a2 = f [ x0 , x1 , x2 ] = −1.26505 a3 = f [ x0 , x1 , x2 , x3 ] = 2.13363 a4 = f [ x0 , x1 , x2 , x3 , x4 ] = −2.02642
    • P ( x ) = a0 + a1 ( x − x0 ) + a2 ( x − x0 ) ( x − x1 ) + a3 ( x − x0 ) ( x − x1 ) ( x − x2 ) + a4 ( x − x0 ) ( x − x1 ) ( x − x2 ) ( x − x3 )
    • P ( x ) = 0.85467 − 0.32617 ( x − 2.0 ) -1.26505 ( x − 2.0 ) ( x − 2.3 ) + 2.13363 ( x − 2.0 ) ( x − 2.3) ( x − 2.6 ) −2.02642 ( x − 2.0 ) ( x − 2.3) ( x − 2.6 ) ( x − 2.9 ) P(x) can now be used to estimate the value of the function f(x) say at x = 2.8.
    • P ( 2.8 ) = 0.85467 − 0.32617 ( 2.8 − 2.0 ) -1.26505 ( 2.8 − 2.0 ) ( 2.8 − 2.3 ) + 2.13363 ( 2.8 − 2.0 ) ( 2.8 − 2.3 ) ( 2.8 − 2.6 ) −2.02642 ( 2.8 − 2.0 ) ( 2.8 − 2.3) ( 2.8 − 2.6 ) ( 2.8 − 2.9 ) f ( 2.8 ) ≈ P ( 2.8 ) = 0.275