SELECTED CHAPTERS FROM ALGEBRA I. R. Shafarevich Preface

8 I. R. Shafarevich(x;x 1 ) (x;x k;1 )(x;x k+1 ) (x;x n+1 ) is equal to F (x) . Putting F (x) =x ; x k x ; xF k k (x), we get(9) c k = y kF k (x k ) f k(x) = y kF k (x k ) F k(x):Passing on to the general interpolation problem with the table (7) weonlyhaveto notice that its solution is the sum of all polynomials f k (x) which correspond toall the simplest interpolation problems:f(x) =f 1 (x)+f 2 (x)++ f n+1 (x):Indeed, if we putx = x k then all the members on the right-hand side become 0,except f k (x k ), and since f k (x) is the solution of the k-th simplest interpolationproblem, we have f k (x k ) = y k . Finally, the degrees of f 1 (x), ... , f n+1 (x) arenot greater than n and the same holds for their sum. We can write the obtainedformula in the form(10) f(x) = y 1F 1 (x 1 ) F 1(x)+ y 2F 2 (x 2 ) F y n+12(x)++F n+1 (x n+1 ) F n+1(x)where F k (x) = F (x)x ; x k, F (x) =(x ; x 1 )(x ; x 2 ) (x ; x n+1 ).There is an unexpected identity which follows from the formula for the interpolationpolynomial. Consider the interpolation problem corresponding to thetablex j x 1 x 2 ... x n+1f(x) j x k 1x k 2... x k n+1where k is a positive integer not greater than n or k =0. On one hand it is evidentthat the polynomial f(x) =x k is the solution of this interpolation problem. Onthe other hand, we can write it down using formula (10) and we obtain thatx k = xk 1F 1 (x 1 ) F 1(x)+ xk 2F 2 (x 2 ) F x k n+12(x)++F n+1 (x n+1 ) F n+1(x)where F (x) =(x ; x 1 )(x ; x 2 ) (x ; x n+1 ) and F i (x) = F (x)x ; x i. The polynomialsF i (x) have degree n and the coecient ofx n is 1. If k