f(x) = (x + 3)(x + 1.5)(x - 2)(x - 5)/4
Consider generalized Rolle's theorem and show in detail the process to obtain μ(0) and μ(4).
Note:
Let data set x = {x_j}_(j)=0^(n) contains (n + 1) unique points in the interval [a, b] and f(x) in C^((n+1))[a, b]. Then, for each element of the interval, a generally unknown number μ(x) in (a, b) exists with f(x) = P_(n)(x) + R_(n)(μ(x)) where P_(n)(x) is the Lagrange interpolating polynomial and R_(n)(μ(x)) the remainder term given by
R_(n)(μ(x)) = (f^((n+1))(μ(x)))/((n + 1)!)prod_(k)=0^(n)(x - x_(k))
Extra Notes:
Proof Note first that if x = x_(k), for any k = 0, 1, ..., n, then f(x_(k)) = P(x_(k)), and choosing μ(x_(k)) arbitrarily in (a, b) yields Eq. (3.3).
If x ≠ x_(k), for all k = 0, 1, ..., n, define the function g for t in [a, b] by
g(t) = f(t) - P(t) - [f(x) - P(x)]((t - x_(0))(t - x_(1))⋯(t - x_(n)))/((x - x_(0))(x - x_(1))⋯(x - x_(n)))
= f(t) - P(t) - [f(x) - P(x)]prod_(i)=0^(n)((t - x_(j)))/((x - x_(i))).
Since f in C^(n+1)[a, b], and P in C^∞[a, b], it follows that g in C^(n+1)[a, b]. For t = x_(k), we have
g(x_(k)) = f(x_(k)) - P(x_(k)) - [f(x) - P(x)]prod_(i)=0^(n)((x_(k) - x_(i)))/((x - x_(i))) = 0 - [f(x) - P(x)] * 0 = 0.
Moreover,
g(x) = f(x) - P(x) - [f(x) - P(x)]prod_(i)=0^(n)((x - x_(i)))/((x - x_(i))) = f(x) - P(x) - [f(x) - P(x)] = 0.
Thus g in C^(n+1)[a, b], and g is zero at the n + 2 distinct numbers x, x_(0), x_(1), ..., x_(n). By Generalized Rolle's Theorem 1.10, there exists a number μ in (a, b) for which g^((n+1))(μ) = 0.
So
0 = g^((n+1))(μ) = f^((n+1))(μ) - P^((n+1))(μ) - [f(x) - P(x)](d^(m+1))/(dt^(n+1))[prod_(i)=0^(n)((t - x_(j)))/((x - x_(i)))]_(r) = μ.
However, P(x) is a polynomial of degree at most n, so the (n + 1)st derivative, P^((n+1))(x), is identically zero. Also, prod_(i)=0^(n)[(t - x_(i))/(x - x_(i))] is a polynomial of degree (n + 1), so
prod_(i)=0^(n)((t - x_(i)))/((x - x_(i))) = [(1)/(prod_(i)=0^(n)(x - x_(i))]t^(n+1) + (lower-degree terms in t), and
(d^(n+1))/(dr^(n+1))prod_(i)=0^(n)((t - x_(i)))/((x - x_(i))) = ((n + 1)!)/(prod_(i)=0^(n)(x - x_(i)).
Equation (3.4) now becomes
0 = f^((n+1))(μ) - 0 - [f(x) - P(x)]((n + 1)!)/(prod_(i)=0^(n)(x - x_(i)),
and, upon solving for f(x), we have
f(x) = P(x) + (f^((n+1))(μ))/(n + 1)!prod_(i)=0^(n)(x - x_(i))
Lagrange Interpolating Polynomial
Extra Notes:
Proof Note first that if x =, for any k = 0, 1,...,n, then f(x) = P(x), and choosing (x) arbitrarily in (, b) yields Eq. (3.3). If x ≠ x, for all k = 0, 1,..,n, define the function g for in [, b] by r-xorxr-x gr=fr)Pr [fxP(x] (x xo(xx) (x x)
