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As a physics student I see the Taylor series being (ab)used very often, mostly for the expansion of $\exp(x)$. The usual line of though is that the error (remainder) term of a high order is likely negligible with a very small deviation $\Delta x$ from the expansion point $x_0$. I've been wondering about the distribution of the error that this might produce and how much control we have over it.

Let's assume I'm expanding a function $f \in C^{(n+1)}(x_0, x)$ at point $x_0$ and I want to approximate $f(x)$ with devation $\Delta x = x - x_0$, $x \gt x_0$. Let's also assume $f(x)$ has any other necessary properties for any formulas to come. The Taylor expansion will have the form $$f(x) = T_n(x_0, x) + R_n(x_0, x)$$ where $T_n$ is the Taylor series of order $n$ and $R_n$ the remainder (which I refer to as error).

From my calculus classes I remember that the Taylor expansion only guarantees that the error will be 0 at the expansion point, but the further away I deviate from it, the worse it is likely to get. I say "likely", because the Lagrange or Cauchy form of the remainder [1] $$R_n(x, x0) \frac{1}{(n+1)!}f^{(n+1)}(\xi_L)\Delta x^{n+1} = \frac{1}{n!}f^{(n+1)}(\xi_C)\Delta x^{n}(x-\xi_C)$$ depend on $\xi_L, \xi_C \in (x_0, x)$, so unless $f^{(n+1)}$ or at least its upper bound on (x_0, x) is known, one cannot be sure about the error. Nevertheless, for a very high order $n$ the term will likely be very small for $\Delta x \lt 1$ and possibly shoot through the roof for $\Delta x \gt 1$.

The integral form of the remainder [1] $$R_n(x_0, x) = \int\limits_{x_0}^x \frac{f^{(n+1)}(t)}{n!}(x-t)^n\mathrm{d}t$$ also suggests that the error depends highly on the behavior of $f^{(n+1)}$. I'm not sure how to interpret the fact that this could represent convolution for $x_0 = 0$ as discussed in this question. Nevertheless, it suggests that for a very high order $n$ and deviation $\Delta x \le 1 $ only the values of $f^{(n+1)}$ around $x_0$ will significantly contribute to the error term, because $(x - t)^n$ will be close to 0 within much of the interval $(x_0, x)$ and will quickly rise towards 1 only around $x_0$.

So I'm wondering whether there are some alternative function expansions (preferably as a power series) that offer more control over the error. For instance, if the requirement that error must be 0 at the expansion point was be relaxed, some error could be, perhaps, re-distributed from the deviation bound towards the expansion point. Ideally, I'd like to specify an error distribution function and find an expansion that would satisfy it at least as an upper bound. The distribution function could be inferred from e.g. the error estimation of $x$ if it represents experimental data. The error distribution is very important with experimental data in general so I'm also wondering whether there are better ways to manipulate it within the mathematical calculations which Taylor expansion doesn't make easy.

References: [1] http://en.wikipedia.org/wiki/Taylor%27s_theorem#Explicit_formulae_for_the_remainder

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  • $\begingroup$ Being a physicist, I agree with you statement related to the abuse of Taylor series. You could be interested by Pade approximants : for the same number of terms, they are always much better than Taylor but the result can be difficult to manipulate. There is also a way to create best rational approximations over a given range. $\endgroup$ – Claude Leibovici May 23 '15 at 15:24
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The chebycheff-polynomials

http://en.wikipedia.org/wiki/Chebyshev_polynomials

are useful for approximations with the property that the error spreads over the whole interval.

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Have you considered

  1. Pade's Approximation: https://en.wikipedia.org/wiki/Pad%C3%A9_approximant
  2. Other versions of rational polynomial approximation: http://mathfaculty.fullerton.edu/mathews/n2003/RationalApproxMod.html
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