Lagrange Polynomial InterpolationDate: 08/08/97 at 14:08:29 From: Chun Wang Subject: Lagrange polynomial interpolation Dear Dr. Math: I looked all over on the internet for the textbook DEFINITION of Lagrange ploynomial interpolation, but I couldn't find it. Would you be so kind as to give me the definition and a brief example of how it works? A link will also do. I am only interested in the basics. Thanks. I really appreciate it. Sincerely, Chun Date: 08/14/97 at 13:13:04 From: Doctor Rob Subject: Re: Lagrange polynomial interpolation Lagrange interpolation is a method of finding a polynomial y = f(x) which passes through a specified set of n points (x(i),y(i)), 1 <= i <= n, in the plane. There is only one condition on the points, and that is that they should all have different x-coordinates, that is, x(i) = x(j) if and only if i = j. The polynomial is defined as follows. For 1 <= j <= n, let p(j,x) = PRODUCT (x - x(i)). i<>j This is zero at every one of the n points except x(j), where it is nonzero. Then set n y = f(x) = SUM y(j)*p(j,x)/p(j,x(j)). j=1 This polynomial has the property that f(x(i)) = y(i) for every i. It is nominally a polynomial of degree n-1. As an example, let's try it on two points (x(1),y(1)) and (x(2),y(2)). Then p(1,x) = x - x(2), and p(2,x) = x - x(1). Now y = f(x) = y(1)*[x-x(2)]/[x(1)-x(2)] + y(2)*[x-x(1)]/[x(2)-x(1)] = x*[y(1)-y(2)]/[x(1)-x(2)] + [x(1)*y(2)-x(2)*y(1)]/[x(1)-x(2)] y - y(2) = x*[y(1)-y(2)]/[x(1)-x(2)] + [x(2)*y(2)-x(2)*y(1)]/[x(1)-x(2)] = x*[y(1)-y(2)]/[x(1)-x(2)] - x(2)*[y(1)-y(2)]/[x(1)-x(2)][y-y(2)]/ [y(1)-y(2)] = [x-x(2)]/[x(1)-x(2)] This is the equation of the line through the two points. Here is a numerical example with three points (-2,5), (0,1), and (3,7). Then p(1,x) = (x-0)*(x-3) = x^2 - 3*x, p(2,x) = (x+2)*(x-3) = x^2 - x - 6, p(3,x) = (x+2)*(x-0) = x^2 + 2*x. Now y = f(x) = 5*(x^2-3*x)/10 + 1*(x^2-x-6)/(-6) + 7*(x^2+2*x)/15 = (4/5)*x^2 - (2/5)*x + 1 = (4*x^2 - 2*x + 5)/5 Let's check it: f(-2) = (16 + 4 + 5)/5 = 5, f(0) = (0 + 0 + 5)/5 = 1, f(3) = (36 - 6 + 5)/5 = 7. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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