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# B f x x 4 x 3 x 2 x 1 approximate f 0 and f 0

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b. f ( x ) = x 4 - x 3 + x 2 - x + 1; approximate f (0) and f 0 (0). solution: Using the S 0 from Exercise 4b, we have S 0 (0) = 1 . 06641, and S 0 0 (0) = - 1 . 0625. The actual answers are f (0) = 1 and f 0 (0) = - 1. So the absolute error of approximation for f 0 (0) is 0.06641, and the absolute approximation error for f 0 (0) is 0.0625. 19. Suppose that f ( x ) is a polynomial of degree three. Show that f ( x ) is its own clamped cubic spline, but that it cannot be its own free cubic spline. solution: Given any set of endpoints [ a, b ] and any set of intermediate points x 1 , . . . , x n - 1 , the conditions for a clamped cubic spline are satisfied by f , (i.e. f matches the values of f and f matches the derivatives of f at all the required points. Since the clamped cubic spline interpolation is unique, it must be f . On the other hand, a condition for the free cubic spline interpolant is that the second derivative be zero at the endpoints. If f were its own free cubic spline on some interval, say [ a, b ], then we would need f 00 ( a ) = f 00 ( b ) = 0. However, the second derivative of any cubic polynomial is a polynomial of degree 1, and from the Fundamental Theorem of Algebra (or more precisely one of its corollaries) this degree one polynomial can only be zero at two distinct points a and b if it is the zero polynomial. This implies in particular that the coefficients of both the x 3 term and the x 2 term in f are zero and this contradicts the assumption that f was a cubic polynomial. Consequently, f cannot be its own free cubic spline.
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• Fall '10
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• Polynomials, Polynomial interpolation, Spline interpolation, benjamin johnson, cubic spline

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