Question

Lagrange Polynomials You just watched some poor kid let go of her balloon in a 8-m high mall food court. The four data points at right represent the height of a balloon (H, in meters) as a function of time (t, in seconds). (a) Create the 3rd-order Lagrange polynomial that exactly passes through all H(t)2 data, and use it to estimate (extrapolate) the balloon height at time 4 seconds. Show your work!! (b) Create the 3rd-order Lagrange polynomial to interpolate the height of the balloon at time 2.5 seconds (before hitting the ceiling) (c) Create a 1st-order Lagrange polynomial to interpolate the height of the balloon at time 2.5 seconds. This is basically doing a LINEAR SPLINE to interpolate h at t 2.5 sec. (d) Make a plot by-hand of the data (on the paper you're handing in), with the three interpolated points above super-imposed, and then comment on concerns using the Lagrange polynomials above for answering (a) (c). Do your results above make "common-sense"? Do you "trust" them, or are these just "bogus mathematical outputs"? 7

Answer 1

(a) 3^rd order Lagrange polynomial H(t) = (t - t_1) (t- t_2) (t - t_3)/ (t_0 - t_1) (t_0 - t_2) (t_0 - t_3) H(t_0) + ....+ (t - t_0) (t- t_1) (t - t_2)/ (t_3 - t_0) (t_3 - t_1) (t_3 - t_2) H(t_3) = (t - 1) (t - 2) (t - 3)/ (-1) (-2) (-3) (0) + (t - 0) (t - 2) (t - 3)/ (1) (-1) (-2) (1) + (t - 0) (t - 1) (t - 3)/ (2) (1) (-1) (7) + (t - 0) (t - 1) (t - 2)/ (3) (2) (1) 8 = + 1/2 t (t-2) (t-3) - 7/2 t (t-1) (t-3) + 4/3 t (t-1) (t-2) = t/2 (t^2 - 5t + 6) - 7/2 t (t^2 - 4t + 3) + 4/3 t (t^2 - 3t + 2) = (1/2 - 7/2 + 4/3) t^3 + (-5/2 + 14 - 4) t^2 + (3 - 21/2 + 8/3) t H(t) = -5/3 t^3 + 15/2 t^2 - 29/5 t Therefore H (t = 4) = -5/3 (4)^3 + 15/2 (4)^2 - 29/6 (4)