I am sorry for such a basic question... but I want to try to do a Taylor expansion on my function, which is a CDF defined over 0-1. However, when I expand around 0, which is what I read is typical, then everything cancels out, because all the derivatives at 0 are going to be 0. Clearly, I am missing something obvious.... can anyone point it out for me? I have never done this before and am reading up on Internet sites, but none seem to explain this basic of a point. Am I misunderstanding the whole idea of a Taylor expansion? Are you only allowed to use some values for $a$ (where $a$ is the constant you are expanding around?
Really basic question about the Taylor expansion of a CDF
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$\begingroup$
probability-distributions
power-series
taylor-expansion
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0@Henning: That is so, so helpful, thank you! – 2011-10-08
1 Answers
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You have discovered that not all functions are equal to their Taylor expansions. That's just a fact of life.
Functions that are equal to their Taylor expansion are called analytic. In practice, most functions you can write down explicitly without resorting to case analysis are analytic, but arbitrary functions don't have to be.
A standard example of a non-analytic function is $f(x) = \cases{0 & \text{when }x=0 \\ e^{-1/x^2} & \text{otherwise}}$
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0@J.M., yes. I copied from a one-sided example without thinking enough. – 2011-10-08