Lecture Notes Course à MA 190 Numerical Mathematics, First ...

polynomial Q. We next prove uniqueness:Let Q, R be two interpolating polynomials of degree < n Next setS(t) = P (t) − R(t).Thus S is a polynomial of degree < n such thatS(t i ) = 0, i = 1, . . . , n.Thus S is of degree < n having n distinct zeros. This is only possible, if S isidentically 0, establishing uniqueness.•6.1.3 Interpolation formula with remainderTheorem 6.1.3 Let I = [a, b] be a real interval, f be a function defined on Ihaving n continuous derivatives there and let t 1 < t 2 < . . . < t n be n distinctpoints in I. Let also Q be the polynomial of degree < n satisfyingThen we establishQ(t i ) = f(t i ), i = 1, 2, . . . , n.f(x) = Q(x) + K(x)P (x), P (x) =where ξ depends on x.n∏(x − t i ), K(x) = f n (ξ),n!Proof: If x coincides with one of the points t i , it is seen at once that thestatement is true. We treat next the case when x is different from all of thesen points. Let now x be fixed. Then we havei=1f(x) = Q(x) + K(x)P (x), K(x) =Next form the new function F , given byF (t) = f(t) − Q(t) − K(x)P (t).f(x) − Q(x).P (x)We verify that F is 0 at t 1 , . . . , t n , x and hence F has in total n + 1 zeros. ByRolle’s theorem F (n) has a zero at a point which we denote ξ. Differentiating ntimes with respect to t we arrive atF (n) (t) = f (n) (t) − K(x) · n!.Using the fact that F (n) (ξ) = 0 we finally getas claimed.•K(x) = f (n) (ξ),n!Remark 6.1.4 Only in special cases are the higher order derivatives availableas useful formulas, a fact which often causes difficulties in practical situations.42