I know it is possible to expand an expandable function for a real, and for infinite by setting $x=\dfrac1y$ and then expanding for $0$.

But my question is, how do we do if the evaluation of the new function and its derivatives is not possible ? I mean I find things like $\sqrt{\left(\dfrac1y\right)^2 - \dfrac1y +1}$, but I can't evaluate it at $0$ ... Wolfram|Alpha says it can be expanded and gives me the result which works perfectly for the rest of my problem.

  • $\begingroup$ Could you clarify ? There is an ambiguity between the title and your last sentence. Thanks. $\endgroup$ – Claude Leibovici Dec 6 '13 at 12:03
  • $\begingroup$ I have a fonction, i want to expand it at -inf, i know that it is possible, but i don't see how $\endgroup$ – user113865 Dec 6 '13 at 12:18
  • $\begingroup$ Post your function, please. $\endgroup$ – Claude Leibovici Dec 6 '13 at 12:20

If you want $\sqrt{x^2 - x +1}\;$ for $x\rightarrow \infty,\;$ you have $x > 0\;$ and write $$\sqrt{x^2 - x +1} = x \sqrt{1 - \frac{1}{x} +\frac{1}{x^2}}.$$ Now substitute $y=\frac{1}{x}$ and compute the series for $y\rightarrow 0$ $$\sqrt{1 - y + y^2} = 1-\frac{1}{2}y+\frac{3}{8}y^2 + \frac{3}{16}y^3 + O(y^4)$$ Reverse the substitution, multiply by $x$ and get for $x\rightarrow \infty$ $$\sqrt{x^2 - x +1} = x-\frac{1}{2}+\frac{3}{8}\frac{1}{x} + \frac{3}{16}\frac{1}{x^2} + O(\frac{1}{x^3})$$

  • $\begingroup$ You have the talent to tell what I wanted to mean. Thanks. $\endgroup$ – Claude Leibovici Dec 6 '13 at 12:19
  • $\begingroup$ Oh, i see clearly now !! Thanks man $\endgroup$ – user113865 Dec 6 '13 at 12:20

What does it mean to "expand a series at infinity?" Infinity isn't a point. Taylor series are only defined wherever the function is k-times differentiable, and $x=1/y$ is not differentiable at zero.

The key here is differentiability; the k-th order Taylor series are only defined on an interval on which the function is k-times differentiable. We know that differentiability implies continuity, so clearly $f$ has to be continuous wherever the Taylor series is defined. In your example of $f(y) = \sqrt{\frac{1}{y^2} + \frac{1}{y} + 1}$, $f$ isn't continuous at 0 because it isn't defined there, and hence the Taylor series can't be expanded there.

Taylor series give us approximations of $f$ in a given neighborhood of a point $a$, so does it make sense to "approximate" $f$ at a point at which it is not continuous?

  • $\begingroup$ I do not totally agree. If you develop y * f(y), you have a Taylor series. I suppose that this is what do CAS. $\endgroup$ – Claude Leibovici Dec 6 '13 at 12:11
  • $\begingroup$ @ClaudeLeibovici - in that case, that would create a differentiable function on a neighborhood of $0$. Pac - think of taking a sum "at infinity" as taking the limit, not evaluated at infinity $\endgroup$ – Lost Dec 6 '13 at 12:28

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