I am trying to wrap my head around the reason why the moment of approximation matters for the end result of my analysis.

As an example, let's take an equation for which we can still find the full solution without approximations: $$\tag{1} \frac{a^4}{x^2} + (a + b) x^2 -c = 0$$ where $a$, $b$ and $c$ are positive and real. I can solve this equation analytically for $x$ and find:

$$\tag{2} x=\pm \frac{\sqrt{\frac{\sqrt{-4 a^5-4 a^4 b+c^2}}{a+b}\pm\frac{c}{a+b}}}{\sqrt{2}}$$

If I now make an approximation to this function by assuming $a$ to be small and taking a first order Taylor expansion I find: $$\tag{3} x\approx \pm \frac{\sqrt{c} (a-2 b)}{2 b^{3/2}} \; \text{and} \; x\approx\pm 0$$

However, if I would already take the Taylor series for small $a$ for equation (1) then I find $$\tag{4} x\approx\pm \sqrt{\frac{c}{a+b}}$$ which is close to (3) but not quite the same (except for $a=0$).

My question is: what is the reason that the moment of approximation matters? Is this related to the fact that the $1/x^2$ term is lost with the second strategy or is something different at play? And what is normally the appropriate moment to do an approximation? As early as possible, as late as possible, or does this depend on the case?

  • $\begingroup$ Te leading order, both of your approximations are equal. Expanding different things in Taylor series will surely yield different intermediate steps, but both methods will eventually end up at the same asymptotic series in $a$, for example. $\endgroup$ – Antonio Vargas Oct 30 '13 at 14:52
  • $\begingroup$ So would it be appropriate then to take again a Taylor series of (4) because this function is no longer linear in $a$? Because doing that would indeed result in the same answer as (3) $\endgroup$ – Michiel Oct 30 '13 at 14:57
  • $\begingroup$ I guess it depends on your application. Often a power series in a small variable ($a$ in this case) will yield a decent approximation, but not always the best one. For example, robjohn's use of Newton's method in this answer yields an approximation which is much sharper than the usual series approximation. I wonder if the same approach could be applied to your problem. $\endgroup$ – Antonio Vargas Oct 30 '13 at 15:06
  • $\begingroup$ @AntonioVargas - I have tried the Newton's method approach for the full equation I am working on (the one here is just a minimal example to reproduce the behaviour) and it does indeed give a somewhat better approximation, although at the expense of having a significantly bigger equation. Thanks! $\endgroup$ – Michiel Nov 1 '13 at 6:19

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