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| Author : | Topic: Maclaurin's expansion | Bottom |
| ferme Posts : 85 |
Suppose f(x) = exp(-1/x) for x > 0, f(x) = 0 otherwise. Then f is C^oo on R, and all MacLaurin polynomials of f are 0. So clearly another polynomial will do better in almost any kind of approximation, including the integral approximation. However, given any smooth f and any polynomial P of degree n that is not the MP of degree n, there exists b > 0 such that that the integral approximation of f over (0,b) with MP_n is better than that with P. |
| saucer admin Posts : 673 A Good Tautology is Hard to Find! ![]() |
- I will move this Thread to; ' Math themes ' - saucer |
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