4.6 Lagrange interpolation. In exercise 3.1 we considered the problem of finding a polynomial p(t) = x1 + x2t + + xnt n1 (4) with specified values p(t1) = y1, p(t2) = y2, , . . . , p(tn) = yn. (5) The polynomial p is called the interpolating polynomial through the points (t1, y1), . . . , (tn, yn). Its coeffi- cients can be computed by solving the set of linear equations 1 t1 t n2 1 t n1 1 1 t2 t n2 2 t n1 2 . . . . . . . . . . . . 1 tn1 t n2 n1 t n1 n1 1 tn t n2 n t n1 n x1 x2 . . . xn1 xn = y1 y2 . . . yn1 yn . (6) The coefficient matrix is called a Vandermonde matrix. As we have seen in the lecture, a Vandermonde matrix is nonsingular if the points ti are distinct (ti 6= tj for i 6= j). As a consequence, the interpolating polynomial is unique: if the points ti are distinct, then there exists