The Question:

Derive the error term in Theorem 3.9. [Hint: Use the same method as in the Lagrange error derivation, Theorem 3.3, defining (1 - xo) ... (1 - Xn)- g (1) = f(t) - Hanti(t ) - (x - XO)2 ... (x - Xn)2 [f (x) - Han+1 (x) ] and using the fact that g'(t) has (2n + 2) distinct zeros in [a, b].]Theorem 3.9 If f E C' [a, b] and Xo, ... , Xn E [a, b] are distinct, the unique polynomial of least degree agreeing with f and f' at Xo, .. . , An is the Hermite polynomial of degree at most 2n + 1 given by n n Han +1 ( x ) = > f ( x; ) Hn. ; (x ) +> f' (x;) Hn.; (x), j=0 j=0 where, for Ln. (x) denoting the jth Lagrange coefficient polynomial of degree n, we have gave a general Hn,j (x) = [1 - 2(x - x;)L'" (x;)]L, (x) and Hnj (x) = (x -x;)L (x). omial in a letter ardt, to whom he Moreover, if f E C2n+2[a, b], then new results. His an interesting f (x) = H2n+1 (x) + (x - XO )? ... ( x - Xn) 2 f ( 2n + 2 ) (E (x ) ) , use of complex (2n + 2)! iques to solve a lem. for some (generally unknown) (x) in the interval (a, b).Proof First, recall that Ln, j (x; ) = Jo, ifi # j, 1, ifi = j. Hence, when i # j, Hnj (xi) = 0 and Hn,; (xi ) = 0, whereas, for each i, Hn.i (x/) = [1 - 2(x - x)L, (x)] . 1=1 and An.i (x; ) = (x1 - X1) . 12 = 0. As a consequence, Han + 1 ( x / ) = >f ( x ; ) . o+ f ( x ) . 1 + _ f ( x; ) . 0 = f(xi ). j=0 1=0 so Hen+1 agrees with f at Xo, X1, . . . , Xn. To show the agreement of Hint , with f' at the nodes, first note that Ln. ; (x) is a factor of H, (x), so H, (x;) = 0 when i # j. In addition, when i = j, we have Ln,i (x;) = 1, so Hi (x; ) = -2Ln. ( x;) . Ln, ( x; ) + [1 - 2(x; - x;) L'n.; ( x; ) 12Ln. (x;) In; (x;) = -2Ln. (x/) + 2Ln, (x;) = 0. Hence, H. (x;) = 0 for all i and j. Finally, An , ( x ; ) = Ly , ( x;) + ( x; - x; )2Ln.; (x:)Lin; (x;) = Ln, j ( x;)[Ln. j ( x; ) + 2(x; - x;)In, (x/)1, so H) ; (x;) = 0ifi # j and H, (x;) = 1. Combining these facts, we have Han+1 ( x / ) = _f ( x; ) . 0 + Ef ( x;) . 0+ f' (x) . 1 = f'(xi ). 1=0 Therefore, Han+ 1 agrees with f and Hint , with f' at Xo, X1, . . . , Xn. The uniqueness of this polynomial and the derivation of the error formula are considered in Exercise 11.Theorem 3.3 Suppose Xo, X1, .. . , Xn are distinct numbers in the interval [a, b] and f e Cat [a, b]. Then, for each x in [a, b], a number 5 (x) (generally unknown) between min{xo, X1, . . . , Xn), and the max (xo, X1, .. ., X }and hence in (a, b), exists with f( +1)(5 (x) ) ways that the f (x) = P(x)+ (n + 1)! (x - xo) (x - x1 ) . .. (x - Xn), (3.3) he Lagrange be expressed, but where P(x) is the interpolating polynomial given in Eq. (3.1). useful form and st closely agrees Proof Note first that if x = Xx, for any k = 0, 1, ... , n, then f(xk) = P(Xk), and rd Taylor choosing (xx) arbitrarily in (a, b) yields Eq. (3.3). or form.If x * XK, for all k = 0, 1, . . ., n, define the function g for t in [a, b] by 8 (t) = f (t) - P(t) - [f (x) - P(x ) ]- (t - xo) (t - x1 ) . .. (t - xn) ( x - xo ) ( x - x1 ) ... (x - Xn) n = f(t) - P(t) - [f(x) - P(x)III (t - xi ) 1=0 (x - xi ) Since f E Cut [a, b], and P E Coo [a, b], it follows that g E Cut [a, b]. For t = Xk, we have n 8 ( XK ) = f(XK) - P(XK) - [f(x) - P()III (Xk - Xi) -=0 - [f(x) - P(x)] . 0 = 0. 1=0 (x - xi ) Moreover, 8 (x) = f(x) - P(x) - If(x) - P()III ( x - Xi) = f (x) - P(x) - [f (x) - P(x)] = 0. i=0 (x - xi ) Thus, g E C"it [a, b], and g is zero at the n + 2 distinct numbers x, Xo, X1, . .. , Xn. By Generalized Rolle's Theorem 1.10, there exists a number & in (a, b) for which g("+1) (5) = 0. So, dn+1 n 0=8("+") (8) = (+1(5) - P(+1)(5) -If(x) - P(x)] (t - xi ) din+1 (3.4) =0 I=E However, P(x ) is a polynomial of degree at most n, so the (n + 1)st derivative, P(+1) (x), is identically zero. Also, II"_o[(t - x;)/(x - x;)] is a polynomial of degree (n + 1), so II - Xi ) "+ + (lower-degree terms in t), 1=0 (x - xi ) and dntl n (n + 1)! din+1 II (t - Xi ) (x - xi ) i=0 I17-o(x - xi) Equation (3.4) now becomes 0 = f("+"() -0-[f(x) - P(x)]; (n + 1)! II'-o(x - X;) and, upon solving for f (x), we have n f (x) = P(x) +: (n + 1)! I (x - x;). 1=0