Ch E 310 - Fall 10 - Lecture 17

# Ch E 310 - Fall 10 - Lecture 17 - Lecture 17 – Agenda •...

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Unformatted text preview: Lecture 17 – October 21, 2010 Agenda: • Polynomial Interpolation (last topic until Exam 2) • HW 4 (due 10/27) questions and help • If you haven’t found a partner yet, try the WebCT discussion board or class roster (email) • Remember: Exam 2 is on 10/28 (next Thursday) • Two data points: (1,6) and (3,9) : • Linear point-slope form: y – y 1 = [( y 2 – y 1 )/( x 2 – x 1 )]( x – x 1 ) • This gives: y – 6 = [(9 – 6)/(3 – 1)]( x – 1) • Rearranging/simplifying: y = 1.5 x + 4.5 Linear Interpolation • Plotting the interpolation function: • Provides us with a functional description of points residing between our two original data points • Extrapolated values: values outside the original data Linear Interpolation y = 1.5 x + 4.5 Polynomial Interpolation • We often need to estimate values between data points • Most commonly, this is done using a polynomial interpolation: • The above is the format MATLAB uses for polynomials • For n data points, a unique polynomial of order ( n-1 ) will pass through all of the points • Examples: a straight line will uniquely pass through two points, a parabola (quadratic) will uniquely pass through three points, etc. n n n n p x p x p x p x f 1 2 2 1 1 ... Polynomial Interpolation • Polynomial interpolation examples: Polynomial Interpolation • Determining polynomial coefficients • n data points needed to determine n coefficients • Example: fit density data using a parabola T (°C) r (kg/m 3 ) 300 0.616 400 0.525 500 0.457 200 250 300 350 400 450 500 550 600 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Polynomial Interpolation...
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Ch E 310 - Fall 10 - Lecture 17 - Lecture 17 – Agenda •...

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