{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

curve_fitting - C v Port PE 5 Curve Fitting and...

Info icon This preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: C v Port: PE. 5;: Curve Fitting and Interpolation 105 Polynomial Approximation to a Function 100 —~ + 80 Suppose y = f (21:) where f (at) is an unknown function. However, Suppose we have N + 1 pairs of values (56],, yk), k = O, 2, ..., N. An approximation to f(:1:) is the N m order polynomial, pN(:c) that passes through the N + 1 points. This can be very useful. For example derivatives or integrals of f can be approximated by the corresponding derivatives or integrals of pN. Also, pN is an interpolating function for f. One obvious way to determine the required N + 1 coefficients, (2, for pN is to write the N + 1 equations: N . Z 3:20,- : yk, k = 0, 1, ..., N i=0 This is equivalent to the matrix equation, ch = y There is another way to determine an approximating (interpolating) polynomial that does not require solution of a matrix equation. To introduce it, suppose we seek the polynomial that passes through just two points ($0,110), (331,3;1). It is easy to show that the polynomial is given by: p1(w)=<x_$l)yo+($-$O)yi $0 " 17,1 331 — C’30 p is the linear combination of two order—1 polynomials L and can be written as: 101 (113) = L1,0($)yo + L1,1($)yi 106 L17! LP?— T he polynomials LN,k(m) are called Lagrange Polynomials. The polynomial rep- resentation can be extended to the case of N + 1 points as: N PN(€E) = Z LN,kf($k) k=0 The Lagrange Polynomials, L N,k(x) are polynomials of order N and have the following properties: 1 forj=k LN’k(xj) : { 0 forjaék The polynomials that have these properties are: N x _ :1:- LN,k($) = ll 1 j=o,#k wk * 539‘ 107 H L: ,€379)3!'31.§§§£é3‘E§jEE:¥__":%;_Q§§)_,_WW.-_ L. k W: W: _ 043*“, J; Q); 3 6,:Etél: If},,.frf_?(1:1)walfl w W _ 21’1“”3(373????3:iafl‘é’éfli} I V .W. . w-» ? . W V . , 108 CW”. «mm. .. ,7-“WW-7-V~~.W_~_F-_..._M.._. mm“.-- Mame-1'- IL;— Numerical Differentiation Numerical Differentiation is used when: 1. A functional form is so complicated that it is more convenient to do numerical integration, 2. when we have a table of values of [rich f (3:1)] and we wish to find df/drr for some given value(s) of m. Examples of situations for which derivatives are needed include: . . . . . . _ _ i“; _ Q? 1. Quantities given in terms of derivatives. 1) — dt, u — 635' 2. Mathematical procedures requiring derivatives: 0 A function y = f(ac) is to be approximated by 3) 2 fix) and f contains constants to be determined which minimize the error in t2he fit of the function to N points at xi. Error : xiv [ f (331) — f (332)] 0 Finding the roots of y = f (as) In other words, find the values of a: such that f(:c) : 0. Two principal methods for obtaining numerical estimates of f’ ($3) when we have a set (table) of pairs of values lay, f(:rz) E fi] , i = 1, 2, ..., N are: 1. Develop relatively simple formulae that provide estimates of the deriva— tive in terms of values of fi and mi, 2. Determine an analytic function g(:c) Which is a good approximation to f (x) and differentiate g(a:) analytically. We will consider the first method here. The second is in the category of functional estimation or approximation. 109 ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern