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In a physics problem, I was expanding in a small parameter $x$, and arrived at the answer $$\cos^{-1}(1-x).$$ This function blows up if you take a Taylor expansion, because for small $x$, it looks like $\sqrt{x}$, which is singular at the origin. In particular you can't calculate a first order term, so it looks like the expansion fails.

However, if you switch your expansion parameter to $\sqrt{x}$, everything works out perfectly, and you can extract a Taylor series and everything. So the lesson was that nothing was actually singular; $x$ was just a bad parameter, and I should've been using $\sqrt{x}$ all along.

This issue can appear any time one wants to expand in a dimensionless parameter. For example, in relativity, you might want to expand in $gh$, which is dimensionless when $c = 1$. But then you'll run into the exact same issue if you want to calculate a velocity, since for slow velocities $v = \sqrt{2gh}$.

When you have a singular function $f(x)$, how can you systematically tell if it actually has an expansion in $x^\alpha$ for some $\alpha < 1$?

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closed as off-topic by David Z Jul 15 '16 at 7:39

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  • $\begingroup$ You can think about the Taylor expansion of $\text{sgn}(y) \cos^{-1}(1 - y^2)$, which should be well defined, and then substitute $y = \sqrt{x}$. I am not sure there is a general procedure, but you can check whether a substitution $x \to y^{1/\alpha}$ yields a analytical function, then you can expand this in $y$ and get the expansion of the original functions in terms of $x^\alpha$ by reversing the substituion. $\endgroup$ – Sebastian Riese Feb 24 '16 at 17:14
  • $\begingroup$ Crossposted from math.stackexchange.com/q/1664869/11127 $\endgroup$ – Qmechanic Feb 24 '16 at 20:42
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    $\begingroup$ By expanding $$\frac{d\log(f)}{dx}$$ in a Laurent series, the exponent $\alpha$ then becomes the residue. $\endgroup$ – Count Iblis Jul 15 '16 at 5:59
  • $\begingroup$ @CountIblis Whoa, thanks! Does that method have a name? $\endgroup$ – knzhou Jul 15 '16 at 6:00
  • $\begingroup$ It is the dlog-Padé method when you have a series expansion for f. E.g. in field theory or in statistical mechanics you can often obtain a formal perturbative expansion in a coupling, which is often does not converge. You can then extract the singular behavior of the function you are expanding by taking the logarithmic derivative and then computing the Padé approximants . This allows you to approximately determine the poles and the resides which are the critical points and the critical exponents, respectively. $\endgroup$ – Count Iblis Jul 15 '16 at 6:18

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