# 14 Aug 2007 KKKQ 3013 PENGIRAAN BERANGKA Week 6 – Interpolation & Curve Fitting 14 August 2007 8.00 am – 9.00 am.

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14 Aug 2007 KKKQ 3013 PENGIRAAN BERANGKA Week 6 – Interpolation & Curve Fitting 14 August 2007 8.00 am – 9.00 am

14 Aug 2007 Week 6 Page 2 Topics Introduction Newton Interpolation: Finite Divided Difference Lagrange Interpolation Spline Interpolation Polynomial Regression Multivariable Interpolation

14 Aug 2007 Week 6 Page 3 Tutorial Example 1 (adapted courtesy of ref. [1]) [1] KQ3013 lecture notes From table 1 below, estimate f(x) at x = 5.5 using the third order (cubic) Newton interpolating polynomial. As a start, use points x = 1, 4, 5 & 6.

14 Aug 2007 Week 6 Page 4 Tutorial Example 1 (adapted courtesy of ref. [1]) [1] KQ3013 lecture notes The third order Newton interpolating polynomial, eq. (3.5) i.e. n = 3:

14 Aug 2007 Week 6 Page 5 Tutorial Example 1 (adapted courtesy of ref. [1]) [1] KQ3013 lecture notes where from equation (3.6c)

14 Aug 2007 Week 6 Page 6 Tutorial Example 1 (adapted courtesy of ref. [1]) [1] KQ3013 lecture notes Note for b 3 :

14 Aug 2007 Week 6 Page 7 Tutorial Example 1 (adapted courtesy of ref. [1]) [1] KQ3013 lecture notes The first divided difference:

14 Aug 2007 Week 6 Page 8 Tutorial Example 1 (adapted courtesy of ref. [1]) [1] KQ3013 lecture notes Therefore, the second divided difference: and hence, the third divided difference:

14 Aug 2007 Week 6 Page 9 Tutorial Example 1 (adapted courtesy of ref. [1]) [1] KQ3013 lecture notes Summarising the divided difference calculation: Therefore, 3 rd order Newton polynomial: and ln 5.5 Now, repeat using points x = 4, 5, 6 & 7. Which group of points gives closer approximation to true value ?

14 Aug 2007 Week 6 Page 10 Tutorial Example 1 (adapted courtesy of ref. [1]) Summarising the divided difference calculation for next group of points: and ln 5.5 Exact or true value: ln 5.5 = 1.704748 !!

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