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Lecture20101118_2

# Lecture20101118_2 - 0.2 0.4 1.00000 1.22140 1.49182 i D D 2...

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Numerical Analysis II Dr Abigail Wacher Nov 18, 2010 1 Example 1 Try with K 1, 1 , 0, 1 , 1, 2 , 2, 4 i x fx K 1 , K 2 , K 1 , K 3 ,.., 0 K 1 1 1 0 1 0 2 2 2 1 1 2 3 4 4 2 1 2 0 p 3 = 0 C 0 , 1 w 1 C 0 , 1 , 2 2 C 0 ,.., 3 3 = 1 C 0 C 1 C 1 2 C 1 C 0 \$ C 1 K 1 = 1 C 1 2 C 1 2 2 3 1 2 = 1 2 1 2 2 C 1 2 1 2 C 1 = 1.375 Example 2 K 1 , K 2 , K 1 , K 3 ,.., 0 0 1.0000000 1 0.2 1.0200668 0.100334 2 0.3 1.0453385 0.2527170 0.5079433 3 0.5 1.1276360 0.4224375 0.5290683 0.0422500 3 = 1 C 0.100334 C 0.5079433 K 0.2 C 0.0422500 K 0.2 K 0.3 This data was produced with = cosh Recall K 3 = 4 f 4 c 4! , e 0 , 3 % 9 \$ 10 4 \$ cosh 0.5 24 Newton form is for data not necessarily equally spaced. If we have equally spaced data points we can use the Newton forward difference.

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Numerical Analysis II Dr Abigail Wacher Nov 18, 2010 2 Theorem (Interpolating polynomial for equally spaced points) Let x = 0 C sh , then interpolant is ps = f 0 C s D 0 C ss K 1 2! D 2 0 C ... C K 1 ... K n C 1 ! D 0 [[ j = fx , D = C 1 K ]] Examples 1) Approximate 0.05 using the Newton forward difference.
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Unformatted text preview: 0.2 0.4 1.00000 1.22140 1.49182 i D D 2 1.00000 0.22140 0.04902 1.22140 0.27042 1.49182 = C D C D 2 2 K 1 = 1.00000 C 0.22140 C 0.02451 K 1 for = 0.05 = K h , since = 0.2, = 0.05 K 0.2 = 0.25 0.05 z p 0.25 = 1 C 0.25 0.22140 K 0.25 0.75 0.02451 = 1.05075 5 dp 2) On online supplementary notes...
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Lecture20101118_2 - 0.2 0.4 1.00000 1.22140 1.49182 i D D 2...

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