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".
Taylor expansion at infinity
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0How about saying "Laurent series" instead of Taylor series? – 2011-07-16
2 Answers
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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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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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3For latex, you've just to put your formulas between dollar symbols, i.e. 'dollar' x=1 'dollar'. – 2011-07-16