Approximaths

sqrt(2)

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Let’s imagine we don’t have a calculating machine or computer and we want to calculate the value of the root of 2 using perturbation theory. Remember that we always insert epsilon so that the problem is exactly solvable for epsilon equals zero. Here’s how we might proceed:

\displaystyle x = \sqrt{2} \quad \text{Problem}
\displaystyle x^2 = 2
\displaystyle x^2 - 1 = 1
\displaystyle x^2 - \epsilon = 1 \quad \text{Insert }\epsilon
\displaystyle (a_0 + a_1 \epsilon + a_2 \epsilon^2 + ...)^2 - \epsilon = 1 \quad \text{Insert }x_{\epsilon}
\displaystyle a_0^2 + 2a_0 a_1 \epsilon + 2 a_0 a_2 \epsilon^2 + a_1^2 \epsilon^2 + ... - \epsilon = 1
\displaystyle a_0^2 + (2a_0 a_1 - 1) \epsilon + (2 a_0 a_2 + a_1^2) \epsilon^2 + ... = 1

In this case, we keep only the powers of epsilon up to order two. We want to perform a perturbative calculation of order two.

\displaystyle \implies a_0 = 1, a_1 = \frac{1}{2}, a_2 = -\frac{1}{8}, ...
\displaystyle x_{\epsilon} = 1 + \frac{1}{2} \epsilon - \frac{1}{8}\epsilon^2 + ...
\displaystyle x_{\epsilon = 1} = 1 + \frac{1}{2} - \frac{1}{8} + ...

Let us solve the same problem but this time we keep the powers of epsilon up to order three.

\displaystyle x = \sqrt{2} \quad \text{Problem}
\displaystyle x^2 = 2
\displaystyle x^2 - 1 = 1
\displaystyle x^2 - \epsilon = 1 \quad \text{Insert }\epsilon
\displaystyle (a_0 + a_1 \epsilon + a_2 \epsilon^2 + a_3 \epsilon^3 + ...)^2 - \epsilon = 1 \quad \text{Insert }x_{\epsilon}
\displaystyle a_0^2+a_1^2\epsilon^2+2a_0a_1\epsilon+2a_0a_2\epsilon^2+2a_0a_3\epsilon^3+2a_1a_2\epsilon^3 + ... - \epsilon = 1
\displaystyle a_0^2+(2a_0a_1 -1)\epsilon+(2a_0a_2 + a_1^2)\epsilon^2+(2a_0a_3+2a_1a_2)\epsilon^3 + ... = 1
\displaystyle \implies a_0 = 1, a_1 = \frac{1}{2}, a_2 = -\frac{1}{8}, a_3 = \frac{1}{16}
\displaystyle x_{\epsilon} = 1 + \frac{1}{2} \epsilon - \frac{1}{8}\epsilon^2 + \frac{1}{16}\epsilon^3 + ...
\displaystyle x_{\epsilon = 1} = 1 + \frac{1}{2} - \frac{1}{8} + \frac{1}{16}

This suggests the following (using the double factorial notation):

\displaystyle \sqrt{2} = 1 + \sum_{k=1}^\infty (-1)^{k+1} \frac{(2k-3)!!}{(2k)!!}
\displaystyle = 1 + \frac{1}{2} - \frac{1}{2\cdot4} + \frac{1\cdot3}{2\cdot4\cdot6} - \frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot8} + \cdots
\displaystyle = 1 + \frac{1}{2} - \frac{1}{8} + \frac{1}{16} - \frac{5}{128} + \frac{7}{256} + \cdots

We could use Euler’s summation to speed up the convergence of this series. We observe that perturbing the initial problem in this form:

\displaystyle x^2 - \epsilon = 1

is similar to finding the roots of the polynomial

\displaystyle x^2 - \epsilon -1 = 0
\displaystyle x^2 - (\epsilon +1) = 0

whose solution is (see figure below)

\displaystyle x= \sqrt{\epsilon+1}

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