sqrt(2) bis

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Let us revisit the problem of computing the value of $\sqrt{2}$ , but this time we slightly modify our perturbative approach to improve convergence.

Previously, we expanded around $x=1$ , which is relatively far from the true value. Instead, we now choose a better starting point.

$\displaystyle x = \sqrt{2} \quad \text{Problem}$

We observe that:

$\displaystyle \left(\frac{3}{2}\right)^2 = \frac{9}{4} = 2.25 \quad \text{so it is close to } 2$

This suggests writing:

$\displaystyle x = \frac{3}{2} + \delta$

The correction $\delta$ is expected to be small, which improves the convergence of the perturbative expansion.

We now insert this into the equation:

$\displaystyle x^2 = 2$

$\displaystyle \left(\frac{3}{2} + \delta \right)^2 = 2$

$\displaystyle \frac{9}{4} + 3\delta + \delta^2 = 2$

Rearranging, we obtain:

$\displaystyle 3\delta + \delta^2 = -\frac{1}{4}$

We now introduce a perturbation parameter $\epsilon$ :

$\displaystyle 3\delta + \delta^2 = -\frac{1}{4}\varepsilon$

We seek a solution of the form:

$\displaystyle \delta = b_1 \varepsilon + b_2 \varepsilon^2 + b_3 \varepsilon^3 + \cdots$

Substituting into the equation and expanding:

$\displaystyle 3b_1\varepsilon + 3b_2\varepsilon^2 + b_1^2\varepsilon^2 + \cdots = -\frac{1}{4}\varepsilon$

Matching powers of $\varepsilon$ :

$\displaystyle \varepsilon^1: \quad 3b_1 = -\frac{1}{4} \Rightarrow b_1 = -\frac{1}{12}$

$\displaystyle \varepsilon^2: \quad 3b_2 + b_1^2 = 0 \Rightarrow b_2 = -\frac{1}{432}$

Thus:

$\displaystyle \delta = -\frac{1}{12}\varepsilon - \frac{1}{432}\varepsilon^2 + \cdots$

We finally obtain:

$\displaystyle x = \frac{3}{2} - \frac{1}{12}\varepsilon - \frac{1}{432}\varepsilon^2 + \cdots$

Setting $\varepsilon = 1$ :

$\displaystyle x \approx \frac{3}{2} - \frac{1}{12} - \frac{1}{432} = 1.41435$

which is already a very good approximation of $\sqrt{2} \approx 1.41421$ .

This example illustrates a key idea in perturbation theory:

The choice of the unperturbed problem strongly affects the convergence of the series.

By expanding around a value closer to the true solution, we obtain a much more efficient approximation with fewer terms.

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