Recent content by xkellyy

yj is the y co-ordinate of the data point...lj is the coefficient of x^n, where n is the varying degrees of the polynomial (in this example, 0-20 because there are 21 data points) a(n) was just my way of explaining lj.

so to find the nth coefficient for the x^n-1 term a(n) = p(n)/(x-x1)(x-x2)(x-x3)... x1, x2 terms are the x values of the data points provided from ascending order excluding the x(n) point if that makes sense but what I'm saying is, i get zero or 1/0 (for P(0) = 1) point for all the...

Homework Statement Find the polynomial p(x) of degree 20 satisfying: p(-10) =p(-9) = p(-8) = ...=p(-1) = 0 p(0) = 1 p(1) = p(2) = p(3) = ...p(10) = 0 Homework Equations L(x) := \sum_{j=0}^{k} y_j \ell_j(x) The Attempt at a Solution i tried using the formula above: a =...

im not sure if my chain rule is correct to begin..i found this in my textbook though on calculating partial derivatives with changes in variables: dh/du = dh/dg1 . dg1/du + dh/dg2 . dg2/du so i followed that instead... and got this since h (u,v) = 2(u^2 - v^2)^5 + 4(u^2-v^2)(uv) +...

Homework Statement Suppose f(x,y) = 2x^5 + 4xy + 2y^3 g1(u,v) = u^2 - v^2 g2(u,v) = uv h(u,v) = f(g1(u,v), g2(u,v)) Use chain rule to calculate: dh/du (1,-1) and dh/dv (1,-1) Homework Equations The Attempt at a Solution i let h (u,v) = 2(u^2 - v^2)^5 + 4(u^2-v^2)(uv) +...