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Consider these two quadratic equations,

$$\text{i)} \quad x^2+4x-5-\epsilon$$ $$\text{ii)} \quad \quad x^2+(4+\epsilon)x+4-\epsilon = 0$$

If we attempt to find an asymptotic approximation of the form $$x = x_0 + \epsilon x_1+...$$ for i) this works out fine, in ii) we get to an equation $-3 = 0$, which is rubbish.

From using the quadratic formula we find $$\text{i)} x = -2 \pm \sqrt{9+\epsilon}$$ $$\text{ii)} \quad x = \frac{-4-\epsilon \pm \sqrt{\epsilon}\sqrt{\epsilon+12}}{2}$$

The radical epsilon factor leads me to beleive that we should try an approximation of the form

$$x = x_0 + \sqrt{\epsilon} x_1+ \epsilon x_2 + ...$$

This works out fine. From this I would like to know if

Was this the correct way to deduce the new form of the asymptotic approximation? Is is there a way to "spot" that a standard asymptotic approximation is going to fail? For instance, above I did not have any idea that it was going to fail until it failed!

  • How did you get the formula ii)? when I tried solving it I got $x=(-4-\epsilon\pm\sqrt{\epsilon^2+12\epsilon+32})/2$ – MaudPieTheRocktorate May 27 '17 at 12:05
  • There is a typo in my question. sorry. –  May 27 '17 at 12:11
  • What is x0, what is x1, and how did you get -3 = 0? What you get depends on x0, x1. – TMM May 27 '17 at 12:14
  • They are just real numbers that need to be determined. I get -3 = 0 by substituting $x = x_0 + \epsilon x_1 +...$ into the quadratic equation ii) and evaluating the terms with coefficient $\epsilon$ –  May 27 '17 at 12:18
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    it's because $(x²+4x+4) = (x+2)^2$ has a double root at $-2$ – mercio May 27 '17 at 12:21
  • $b^2-4ac = (4+\epsilon)^2 - 4(4-\epsilon) = \epsilon(12 + \epsilon)$ unless I am being stupid again. –  May 27 '17 at 12:23
  • In general, a double root of a polynomial is very sensitive to perturbations of coefficients. – MaudPieTheRocktorate May 27 '17 at 12:23
  • I see! this is interesting. It has nothing to do with the alternating signs of the epsilon in the coefficients ? –  May 27 '17 at 12:24
  • Still you did not explain what x0 and x1 are. Based on the other comments I presume that x0 is the double root -2, but don't make us guess. – TMM May 27 '17 at 12:31
  • I said above that they are just real numbers. Perhaps the use of they is not great when many things are being discussed! If you are referring to their actual values then $x_0 = -2$ and on trying to find $x_1$ I find that -3=0 –  May 27 '17 at 12:33

1 Answers1

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Compute the differential of the equation.

  1. $(2x+4)dx-d\epsilon=0$. This shows that $\dfrac{dx}{d\epsilon}$ is finite a the roots.

  2. $(2x+4+\epsilon)dx+(x-1)d\epsilon=0$. At the double root $x=-2$, $\dfrac{dx}{d\epsilon}$ is infinite so that an "ordinary" approximation (entire function) cannot work.

  • When you say, compute the differential should 2. have $(2x + 4 + \epsilon) dx + (x-1)d \epsilon$? –  May 27 '17 at 12:43
  • Or have you evaluated it at $\epsilon =0$ because we are interested in the roots of the unperturbed equation? either way, how did you know to check the differential of the equations to show this? The answer to this question interests me most! –  May 27 '17 at 12:48
  • @HMPARTICLE Yep, typo. Fixed. An you indeed evaluate at $\epsilon=0$. The idea is to chack if there is a Taylor development. –  May 27 '17 at 13:10
  • Is this what you mean by a Taylor development?$$x = x_0 + \frac{dx}{d\epsilon}(x_0) \epsilon + ...$$ I have never heard the term before, so I am assuming it is the same as a Taylor series maybe? It makes sense to think of it this way I think. –  May 27 '17 at 13:13