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Month: June 2026

Illustration of a singular problem

Illustration of a singular problem:

$\displaystyle x^2-2x +1 = 0 \quad \text{Problem}$

$\displaystyle \epsilon x^2-2x +1 = 0 \quad \text{Insert }\epsilon$

The solutions to the perturbed problem are (see figure below):

$\displaystyle x_1(\epsilon) = \frac{1+ \sqrt{1-\epsilon}}{\epsilon}$

$\displaystyle x_2(\epsilon) = \frac{1- \sqrt{1-\epsilon}}{\epsilon}$

While one of the roots approaches $\frac{1}{2}$ as $\epsilon \to 0$ , the other goes to infinity. This is a manifestation of a singular behaviour. The problems above illustrate the importance of setting $\epsilon$ to a right position.

A solution of the regular problem $x^2-2x +\epsilon = 0$ is $x_1(\epsilon) = 1 - \sqrt{1- \epsilon}$ . We would like to calculate the corresponding perturbative series.
Using:

$\displaystyle (a+b)^n = \sum_{k=0}^{\infty} {n\choose k} a^{n-k}b^{k}$

$\displaystyle = a^n + n a^{n-1}b + \frac{n(n-1)}{2} a^{n-2}b^{2}+...+ \frac{n(n-1)\cdots(n-k+1)}{k!} a^{n-k}b^k +...$

We would like to find the perturbative series corresponding to:

$\displaystyle \sqrt{1-\epsilon} = (1-\epsilon)^{1/2}$

with $a=1$ , $b = -\epsilon$ , $n = \frac{1}{2}$ therefore:

$\displaystyle \sqrt{1-\epsilon} = 1 - \frac{1}{2}\epsilon - \frac{1}{8}\epsilon^2 - ...$

$\displaystyle 1- \sqrt{1-\epsilon} = \frac{1}{2}\epsilon + \frac{1}{8}\epsilon^2 + ...$

Here are the first series and their corresponding graphs:

$\displaystyle s_2 = \frac{1}{2} \epsilon + \frac{1}{8} \epsilon^2$

$\displaystyle s_3 =\frac{1}{2} \epsilon + \frac{1}{8} \epsilon^2 + \frac{1}{16} \epsilon^3$

$\displaystyle s_4 = \frac{1}{2} \epsilon + \frac{1}{8} \epsilon^2 + \frac{1}{16} \epsilon^3 + \frac{5}{128} \epsilon^4$

June 4, 2026