25 September 2014

Solving tan(x) = x

[UPDATE: 25/09/2014 with the iterative method] This kind of transcendental equation is often encountered in physics. Undergraduate students are usually shown (or asked to draw) the graphical solution:

The numerical solutions are easily found by an iterative method using a scientific calculator (see below), but how far can one go with only pen and paper?

Expansion

Aside from the trivial solution $$x_0 = 0$$, one clearly has $$x_k \simeq \frac{(2k +1) \pi}{2}$$ ($$k \geq 1$$), so we can write: $x_k = \frac{(2k +1) \pi}{2} - \varepsilon _k, \quad \mathrm{with} \quad \varepsilon _k < 1$ One would like to do an expansion in $$\varepsilon _k$$, but of course this will not work for the tangent around its divergence points. We can however use the cotangent, since $$\tan (x_k) = x_k \Rightarrow \cot (x_k) = 1/x_k$$. Using standard substitution formulas for the sine and cosine yields: $\cot \left [ \frac{(2k +1) \pi}{2} - \varepsilon _k \right ] = \tan(\varepsilon _k) \simeq \varepsilon _k \simeq \frac{2}{(2k + 1) \pi}$ where in the last equality we neglected $$\varepsilon _k$$ in the denominator. One can include it for a more rigorous treatment. Finally, we have: $x_k \simeq \frac{(2k +1) \pi}{2} - \frac{2}{(2k + 1) \pi}, \quad \mathrm{for} \quad k \geq 1 \, ,$ giving for the first three solutions 4.5002, 7.7267, and 10.9046, to be compared with the "exact" values 4.4934…, 7.7253…, and 10.9041…. The quality of the approximation increases with the order $$k$$, since $$\varepsilon _k$$ decreases (the intersections are closer and closer to the vertical asymptotes).

Iteration

Let us rewrite the initial equation by applying the arctangent to both members:
$x = \arctan (x) \tag{1}$
For the $$k$$-th solution, the initial estimate is: $$x^0_k = \frac{(2k +1) \pi}{2}$$. Let us plug it in the right-hand side of Eq. (1) to obtain the first order estimate $$x^1_k$$ and then iterate. Note that the arctangent is a multi-valued function, and the standard implementation reduces it to the first branch (the one going through the origin). We are looking for the solution sitting on the $$k$$-th branch, so we need to add $$k \pi$$ each time:
$x^{i+1}_k = \arctan (x^i_k) + k \pi \tag{2}$
For the first non-trivial solution ($$k = 1$$), the sequence is: 4.71239, 4.50328, 4.49387, 4.49343, 4.49341,... with the second iteration already reaching an excellent precision!

18 September 2014

Characterizing mixtures of gold nanoparticles

Our paper has just been published in Nanoscale as an Accepted Manuscript !

13 September 2014

False false friends

Whoever had to evolve between two related languages is familiar with the concept of false friends.
I would argue that there is a less visible category of terms (or more precisely, of relations between terms) namely similar words that one feels are false friends but that actually have a similar meaning in the two languages: these are false false friends (FFF).
This is a subjective relation, being a false perception of one speaker.
For me, when going from French (or Romanian) to English, the FFF are mainly terms of Latin origin for which I am tempted to substitute Saxon words or other Latin terms, but which have no immediate equivalent in Romance languages:
• salary becomes wages
• merits deserves
• evidentobvious
• hypothesis assumption
etc.