Interpolation. s(s 1)(s 2) Δ 3 f ! s(s 1) Δ 2 f 0 + 2! f(x s )=f 0 + sδf 0 +

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1 Interpolation For equally spaced sample points, the Newton-Gregory formula can be used effectively in two ways: (1) Use for derivation of well-known intepolation formulas. (2) Use directy for numerical computation. The formula can be written using forward differnces in the form: f(x s )=f 0 + sδf 0 + s(s 1) Δ 2 f 0 + 2! s(s 1)(s 2) Δ 3 f ! in which s =(x s x 0 )/h is the normalized distance from x 0 and h = x 1 x 0 is increment of the sample points. (1) For the purpose of derivations, the forward differences can be written as: f Δf Δ 2 f Δ 3 f Δ 4 f f 0 f 1 f 0 f 1 f 2 2f 1 + f 0 f 2 f 1 f 3 3f 2 +3f 1 f 0 f 2 f 3 2f 2 + f 1 f 4 4f 3 +6f 2 4f 1 + f 0 f 3 f 2 f 4 3f 3 +3f 2 f 1 f 3 f 4 2f 3 + f 2 f 5 4f 4 +6f 3 4f 2 + f 1 f 4 f 3 f 5 3f 4 +3f 3 f 2 f 4 f 5 2f 4 + f 3 f 5 f 4 f 5 (2) For the purpose of direct computation, the forward differences can be written in terms of the forward differences of one order lower, as: f Δf Δ 2 f Δ 3 f Δ 4 f f 0 f 1 f 0 f 1 Δf 1 Δf 0 f 2 f 1 Δ 2 f 1 Δ 2 f 0 f 2 Δf 2 Δf 1 Δ 3 f 1 Δ 3 f 0 f 3 f 2 Δ 2 f 2 Δ 2 f 1 f 3 Δf 3 Δf 2 Δ 3 f 2 Δ 3 f 1 f 4 f 3 Δ 2 f 3 Δ 2 f 2 f 4 Δf 4 Δf 3 f 5 f 4 f 5

2 Interpolation With Equally Spaced Sample Points Two-sample-point Lagrange Formula Two sample points: Abscissas: x 0 and x 1. Sample functional values: f 0 and f 1. Let s =(x x 0 )/h. In this case, let p = s =(x x 0 )/h. The Newton=Gregory formula has the form: f(x) =f 0 + sδf 0 = f 0 + s(f 1 f 0 ). Substitute p for s, the Lagrange Two-Point Interpolation formula is: f(x) =f 0 + p (f 1 f 0 )=(1 p) f 0 + pf 1. Three-sample-point Lagrange Formula Three sample points: Abscissas: x 1, x 0 and x 1. Sample functional values: f 1, f 0 and f 1. Let s =(x x 1 )/h. In this case, let p be the normalized distance from the center point x 0, i.e.,p =(x x 0 )/h. With some simplication, p =[(x x 1 )+(x 1 x 0 )]/h = s 1. The Newton=Gregory formula has the form: In terms of the sample values, f(x) =f 1 + sδf s(s 1)Δ2 f 1. f(x) =f 1 + s (f 0 f 1 )+ 1 2 s(s 1) (f 1 2f 0 + f 1 ). Substitute for s = p +1, the formula can be written in terms of p as f(x) =f 1 +(p +1)(f 0 f 1 )+ 1 2 (p +1)p (f 1 2f 0 + f 1 ). Now collect terms according to the sample values: f(x) =f 1 [1 (p +1)+ 1 2 (p2 + p)] + f 0 [(p +1) (p 2 + p)] + f 1 [ 1 2 (p2 + p)] Finally, the 3-point Lagrange Formula could be written as f(x) = 1 2 p(p 1) f 1 +(1 p 2 ) f p(p +1)f 1.

3 Five-sample-point Lagrange Formula Five sample points: Abscissas: x 2, x 1, x 0, x 1 and x 2. Sample functional values: f 2, f 1, f 0, f 1 and f 2. Let s =(x x 2 )/h. In this case, let p be the normalized distance from the center point x 0, i.e.,p =(x x 0 )/h. With some simplication, p =[(x x 2 )+(x 2 x 0 )]/h = s 2. The Newton=Gregory formula has the form: f(x) =f 2 + sδf s(s 1)Δ2 f s(s 1)(s 2)Δ3 f s(s 1)(s 2)(s 3)Δ4 f 2. In terms of the sample values, f(x) =f 2 + s (f 1 f 2 )+ 1 2 s(s 1) (f 0 2f 1 + f 2 ) s(s 1)(s 2) (f 1 3f 0 +3f 1 f 2 ) s(s 1)(s 2)(s 3) (f 2 4f 1 +6f 0 4f 1 + f 2 ) Substitute for s = p +2, the above formula can be written in terms of p as f(x) =f 2 +(p +2)(f 1 f 2 )+ 1 2 (p + 2)(p +1)(f 0 2f 1 + f 2 ) (p + 2)(p +1)p (f 1 3f 0 +3f 1 f 2 ) (p + 2)(p +1)p(p 1) (f 2 4f 1 +6f 0 4f 1 + f 2 ) Now collect terms according to the sample values: f(x) =f 2 [1 (p +2)+ 1 2 (p + 2)(p +1) 1 6 (p + 2)(p +1)p + 1 (p + 2)(p +1)p(p 1)] 24 + f 1 [(p +2) (p + 2)(p +1)+ 1 2 (p + 2)(p +1)p 1 (p + 2)(p +1)p(p 1)] 6 + f 0 [ 1 2 (p + 2)(p +1) 1 2 (p + 2)(p +1)p 1 (p + 2)(p +1)p(p 1)] 4 + f 1 [ 1 6 (p + 2)(p +1)p 1 (p + 2)(p +1)p(p 1)] 6 + f 2 [ 1 (p + 2)(p +1)p(p 1)] 24 Finally, the 3-point Lagrange Formula could be written as f(x) = 1 24 (p2 1)p(p 2) f (p 1)p(p2 4) f (p2 1)(p 2 4) f (p +1)p(p2 4) f (p2 1)p(p +2)f 2

4 Example - Application of 3-point Lagrange Formula Given x 1 =0.9 f 1 = x 0 =1.4 f 0 = x 1 =1.9 f 1 = Solution: h = = =0.5. To interpolate for f(1.53), let p =( )/0.5 = The interpolation is then accomplished by f(1.53) 1 2 (0.26)(0.26 1)( ) + (1 (0.26)2 )( ) + 1 (0.26)( )( ) 2 Hence, the interpolated value is f(1.53) Since the values were obtained as sin x, the exact value for f(1.53)= But we need to do the interpolation because the tabulated values didn t include that for x =1.53.

5 Numerical Interpolation using Newton Gregory Formula If the number of sample points is large, it might be simpler to use the Newton Gregory formula directly. It is also quite simple to create a computer program to accomplish the task. Example - 3-point Forward Difference Table Given f Δf Δ 2 f Solution: h = = =0.5. To interpolate for f(1.53), let s =( )/0.5 = The application of the Newton-Gregory Formula will yield f(1.53) (1.26)( ) + 1 (1.26)(0.26)( ) = Example - 5-point Forward Difference Table Given f Δf Δ 2 f Δ 3 f Δ 4 f Solution: h = = = To interpolate for f(1.53), let s =( )/0.25=2.52. The application of the 4-th order Newton-Gregory Formula will yield f(1.53) (2.52)( ) (2.52)(1.52)( ) (2.52)(1.52)(0.52)( ) + 1 (2.52)(1.52)(0.52)( 0.48)( ) = The more refined interpolation yields a value of , very near the exact value of

6 Interpolation With Non-Equally Spaced Sample Points Three-sample-point Formula With three non-equally spaced abscissa, x 1, x 2 and x 3, and three function values f 1, f 2 and f 3, the interpolated value at x can be determined as f(x) =f 1 δ 1 (x)+f 2 δ 2 (x)+f 3 δ 3 (x). The requirements for δ i (x) is such that it must be equal to 1 when x = x i and it must be equal to 0 when x = j with j i. For this particular case, the interpolation functions could be written as δ 1 (x) = (x x 2)(x x 3 ) (x 1 x 2 )(x 1 x 3 ) δ 2 (x) = (x x 1)(x x 3 ) (x 2 x 1 )(x 2 x 3 ) δ 3 (x) = (x x 1)(x x 2 ) (x 3 x 1 )(x 3 x 2 ) The above functions satisfy the order of the polynomial as well, i.e., a parabola can be determined uniquely by 3 sample points. N-sample-point Formula It is clear that the interpolation formula was written as a sum of products. Therefore, the general case of the interpolation formula could be written as f(x) = N f i δ i (x). i=1 in which δ i (x) = N j=1,j i (x x j ) (x i x j ).