Are there case where does make sense to speak about the "Taylor expansion of a function ad infinity"?
By inversion, sending $x \to \frac{1}{x}$ one could exchange $0\leftrightarrows\infty$; then if the values of the derivatives of a function are finite at infinity I was wondering if it is possible to give some sense to $(x-\infty)^n$ in order to define the "Taylor expansion of a function ad infinity".

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    $\begingroup$ Yes, but it's more proper to think of them as "asymptotic expansions". For instance, you can have the expansion $\arctan(x)=\frac{\pi}{2}-\frac1{x}+\frac1{3x^3}-\cdots$ $\endgroup$ Jul 16 '11 at 11:56
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    $\begingroup$ Yes, but as per your suggestion (and the comment by J.M.) you don't try to use powers of $(x-\infty)$. You use powers of $1/x$ instead. The term is local parameter. The concept is more general, but on the Riemann sphere it amounts to a simple function that takes the value zero at the point of interest, and distinquishes the nearby points in a 'nice' way. At infinity $1/x$ is a perfectly good local parameter. At a finite point $x-x_0$ gets the job done. $\endgroup$ Jul 16 '11 at 12:06
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    $\begingroup$ One should note that the "Taylor series at infinity" can only make sense if the function and its derivatives converges suitably fast at infinity. For a function whose limit at infinity doesn't exist (such as the sine function), it would not make sense to speak of its asymptotic expansion. $\endgroup$ Jul 16 '11 at 12:58
  • $\begingroup$ How about saying "Laurent series" instead of Taylor series? $\endgroup$
    – GEdgar
    Jul 16 '11 at 15:58

While studying complex analysis concepts, we come across Laurent series at infinity.
There, if after substituting $z=\frac{1}{x}$, the function $f$ has Taylor expansion $c_0+c_1 z+c_2 z^2 + \dots$ for $z$ near $0$, then, substituting back $x=\frac{1}{z}$, the series $c_0+c_1 x^{-1} + c_2 x^{-2}+\dots$ could be called as Taylor series at $x_0=\infty$

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    $\begingroup$ For latex, you've just to put your formulas between dollar symbols, i.e. 'dollar' x=1 'dollar'. $\endgroup$ Jul 16 '11 at 20:39

Think of Taylor expansion as an approximation formula, with main term $f(x_0)$ and $\epsilon = x-x_0$ being a small parameter. When expanding around $x_0 = \infty$, $x-x_0$ is no longer small, but $x^{-1}$ is.


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