Newton Forward Difference Interpolation Method
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This document describes the Newton forward difference interpolation method. The method uses forward difference quotients to interpolate values of a function between given data points. It provides an example of using the method to interpolate the value of a function F(x) at x=0.36 using data points from x=0.2 to x=0.6. The interpolation formula involves summing terms containing forward difference quotients up to the fourth order difference. When calculated, the interpolated value of F(0.36) is determined to be approximately 0.3047248.
Newton Forward Difference Interpolation Method
- 3. Newton Forward 2 0 0 0 0 3 0 ( 1) ( ) ( ) ( ) ( ) 2 ( 1)( 2) ( ) 6 p p f x ph f x p f x f x p p p f x 0 0 2 0 3 0 ( ) First Value ( ) First Value ( ) First Value x x p h f x f x f x
- 4. x 0.2 0.3 0.4 0.5 0.6 F(x) 0.2304 0.2788 0.3222 0.3617 0.3979 X F(x) ∆F(x) ∆ 𝟐 F(x) ∆ 𝟑 F(x) ∆ 𝟒 F(x) 0.2 0.2304 0.0484 -0.005 0.0011 -0.0005 0.3 0.2788 0.0434 -0.0039 0.0006 0.4 0.3222 0.0395 -0.0033 0.5 0.3617 0.0362 0.6 0.3979 F(0.36) It’s Mean X=0.36
- 5. 1.6(1.6 1) (0.2 (1.6)(0.1)) 0.2304 (1.6)(0.0484) ( 0.005) 2 1.6(1.6 1)(1.6 2) 1.6(1.6 1)(1.6 2)(1.6 3) (0.011) ( 0.0005) 6 24 f 0 0 2 0 3 0 4 0 0.36 0.2 1.6 0.1 ( ) 0.0484 ( ) 0.005 ( ) 0.0011 ( ) 0.0005 x x p h f x f x f x f x