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?

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    $\begingroup$ We could be a bit more helpful if you'd say what your CDF is... $\endgroup$ – J. M. is a poor mathematician Oct 8 '11 at 17:24
  • $\begingroup$ @J.M. The function in question is $(1-(1-x)^n)^n$ But wouldn't the derivative of all CDFs defined over 0 to 1 be 0 at 0? $\endgroup$ – testing out Oct 8 '11 at 17:32
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    $\begingroup$ @testing, that's a polynomial -- it can't have all zero derivatives. Its leading term is $x^{n^2}$, so the $n^2$th derivative at 0 is $n^2!$. Probably you're making some mistake while differentiating it. Could you elaborate on your reasoning? $\endgroup$ – Henning Makholm Oct 8 '11 at 17:42
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    $\begingroup$ It is crucial that you specified that $f$ was defined only for $f\in[0,1]$. That allows $f(0)=0, f'(0)>0$ without $f$ ever needing to be negative (which doesn't make sense for a cdf). If instead you had said that $f$ is defined on $\mathbb R$ but $f(x)=0$ when $x<0$, then you're right that it couldn't have any nonzero derivatives. However, then most likely it wouldn't be differentiable arbitrary many times at all, and so it wouldn't have any Taylor series at 0, not even the zero one. $\endgroup$ – Henning Makholm Oct 8 '11 at 17:54
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    $\begingroup$ For your particular function, the first $n$ derivatives are going to be 0, because $1-(1-x)^n$ has zero constant term, and so $(1-(1-x)^n)^n = x^n p(x)^n$ for some polynomial $p$. $\endgroup$ – Henning Makholm Oct 8 '11 at 17:57

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 $$f(x) = \cases{0 & \text{when }x=0 \\ e^{-1/x^2} & \text{otherwise}}$$

  • $\begingroup$ You intended $\exp(-x^{-2})$ I presume? $\endgroup$ – J. M. is a poor mathematician Oct 8 '11 at 17:28
  • $\begingroup$ @J.M., yes. I copied from a one-sided example without thinking enough. $\endgroup$ – Henning Makholm Oct 8 '11 at 17:30

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