PROOF EXERCISE) The steps are similar to those in the Lagrange Interpolation The
ID: 3281782 • Letter: P
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PROOF EXERCISE) The steps are similar to those in the Lagrange Interpolation Theorem First show that the Theorem holds if r-To Next assume x is arbitrary, but r -ro. (Consider x as fixed.) Define c(r) - o)n+1 Define F(z)=f(z)_p(z) _ c(z) (z-ro)"+1 Show that F(k)(zo) = 0 , k = 0, 1, . . . ,n , and that Give a qualitative graph of F(z) . F(x) = 0 . . . 212 . Show F,(So)-0 for some between 20 and r. Graph F,(2) Show F"(6)-0 for some between xo and . Graph F"(z) etc . Show that Fn+(n)0 for somebetween zo and n-1 From this derive that fn+( . Show how Taylor's Theorem follows from this last step.QED! 213Explanation / Answer
Fix a point x [a, b]. If x is one of the interpolation points x0, . . . , xn, the result holds trivially. We therefore assume that x != xj 0 <= j<= n. We now let F(y) = f(y) Qn(y) w(y),
where is chosen as to guarantee that F(x) = 0, i.e., = f(x) Qn(x) w(x) . Since the interpolation points x0, . . . , xn and x are distinct, w(x) does not vanish and is well defined. We now note that since f C n+1[a, b] and since Qn and w are polynomials, then also F C n+1[a, b]. In addition, F vanishes at n + 2 points: x0, . . . , xn and x. According to Rolle’s theorem, F 0 has at least n + 1 distinct zeros in (a, b), F 00 has at least n distinct zeros in (a, b), and similarly, F (n+1) has at least one zero in (a, b), which we denote by n. We have 0 = F (n+1)(n) = f (n+1)(n) Q (n+1) n (n) (x)w (n+1)(n) (3.28) = f (n+1)(n) f(x) Qn(x) w(x) (n + 1)! Here, we used the fact that the leading term of w(x) is x n+1, which guarantees that its (n + 1)th derivative is w (n+1)(x) = (n + 1)! (3.29) we conclude with f(x) Qn(x) = 1 f^ (n+1)(n)w(x)/(n + 1)!.
In addition to the interpretation of the divided difference of order n as the coefficient of x n in some interpolation polynomial, it can also be characterized in another important way. Consider, e.g., the first-order divided difference f[x0, x1] = f(x1) f(x0) x1 x0 . Since the order of the points does not change the value of the divided difference, we can assume, without any loss of generality, that x0 < x1. If we assume, in addition, that f(x) is continuously differentiable in the interval [x0, x1], then this divided difference equals to the derivative of f(x) at an intermediate point, i.e., f[x0, x1] = f 0 (), (x0, x1). In other words, the first-order divided difference can be viewed as an approximation of the first derivative of f(x) in the interval. It is important to note that while this interpretation is based on additional smoothness requirements from f(x) (i.e. its being differentiable), the divided differences are well defined also for non-differentiable functions. This notion can be extended to divided differences of higher order as stated by the following lemma.
Let Qn(y) interpolate f(y) at x0, . . . , xn1, x. Then according to the construction of the Newton form of the interpolation polynomial, we know that Qn(y) = Qn1(y) + f[x0, . . . , xn1, x] nY1 j=0 (y xj ). Since Qn(y) interpolated f(y) at x, we have f(x) = Qn1(x) + f[x0, . . . , xn1, x] nY1 j=0 (x xj ). We know that the interpolation error is given by f(x) Qn1(x) = 1 n! f (n) (n1) nY1 j=0 (x xj )
f[x0, . . . , xn1, x] = f (n) ()/ n! .
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