I have asked a related question here How to find asymptotic expansions of all real roots of $x \tan(x)/\epsilon=1?$, however, when I discussed with my adviser today, he argued the solution is flawed.

The question is

Find expansions of all the real roots of $$x\tan(x)=\epsilon?$$ (You have to consider the first root separately)

and the expansion in How to find asymptotic expansions of all real roots of $x \tan(x)/\epsilon=1?$ is as follows

$ x\tan(x)=\epsilon^{2\alpha}\left((x_0+\epsilon x_1+\epsilon^2x_2+\cdots)^2+\frac{1}{3}\epsilon^{2\alpha}(x_0+\epsilon x_1+\epsilon^2x_2+\cdots)^4+O(\epsilon^{4\alpha})\right) $

However, my adviser pointed out the steps are unknown, so he argued it should be

$ x\tan(x)=({\epsilon^\alpha}x_0+{\epsilon^\beta} x_1+{\epsilon^\gamma}x_2+\cdots)^2+\frac{1}{3}({\epsilon^\alpha}x_0+{\epsilon^\beta} x_1+{\epsilon^\gamma}x_2+\cdots)^4+O(\epsilon^{4\alpha}) $

He argued that in the first form the cases like $x_0+\epsilon x_0+\epsilon^{1/2}x_1+\epsilon x_1 \cdots$ is not considered.

Then I am really confused now. How to determine $\alpha, \beta, \gamma$ from the

$ x\tan(x)=({\epsilon^\alpha}x_0+{\epsilon^\beta} x_1+{\epsilon^\gamma}x_2+\cdots)^2+\frac{1}{3}({\epsilon^\alpha}x_0+{\epsilon^\beta} x_1+{\epsilon^\gamma}x_2+\cdots)^4+O(\epsilon^{4\alpha})=\epsilon ?$

It seems there is a lot of possibilities of $\alpha, \beta, \gamma$. Does that mean there is several possible asymptotic expansions?


$$x\tan(x)=(\epsilon^\alpha x_0+\epsilon^\beta x_1+\epsilon^\gamma x_2+\cdots)^2+\frac{1}{3}(\epsilon^\alpha x_0+\epsilon^\beta x_1+\epsilon^\gamma x_2+\cdots)^4+O(\epsilon^\kappa)=\epsilon$$ and you know that $\alpha<\beta<\gamma<\kappa$ and that $\alpha$ should be chosen to make the largest term on the LHS be $O(\epsilon)$.

Expanding gives (taking only some terms that are bigger than others) $$x\tan(x)=\epsilon^{2\alpha}x_0^2+2\epsilon^{\alpha+\beta }x_0x_1+\epsilon^{2\beta}x_1^2+\frac{1}{3}\epsilon^{4\alpha}x_0^4=\epsilon.$$

We can choose to set $2\alpha=1$ or $\alpha+\beta=1$ or $2\beta=1$, but whichever we choose, we require that all the other powers of $\epsilon$ are higher. We can't choose $2\beta=1$ because, since $\alpha<\beta$, $2\alpha$ would be less than 1, and hence $\epsilon^{2\alpha}>\epsilon^{2\beta}$ and there would be nothing to balance the $\epsilon^{2\alpha}$ term with. Choosing $\alpha+\beta=1$ means that $2\alpha<1$, so choosing $\alpha+\beta=1$ will not work either. So choose $2\alpha=1$.

Let $2\alpha=1$, then $$x\tan(x)=x_0^2+2\epsilon^{\beta-1/2}x_0x_1+\epsilon^{2\beta-1}x_1^2+\frac{1}{3}\epsilon x_0^4=1$$ and we know that $\beta>1/2$, so taking $O(1)$ terms gives $$ x_0^2=1.$$ Then we are left with $$2\epsilon^{\beta-1/2}x_0x_1+\epsilon^{2\beta-1}x_1^2+\frac{1}{3}\epsilon x_0^4=0$$ an we first see that $\epsilon^{2\beta-1}=\left(\epsilon^{\beta-1/2}\right)^{2}$ (and $2\beta-1>0$), so the first and second terms can't balance. Since the first term is larger than the second, we must balance the first and third terms. So choose$\beta-1/2=1\Rightarrow \beta=3/2\ $ to balance with the $x_0^4$ terms. We get $$2\epsilon x_0x_1+\epsilon^{2}x_1^2+\frac{1}{3}\epsilon x_0^4=0$$ and so the next equation is $$2x_0x_1+\frac{1}{3}x_0^4=0.$$

So the expansion is $x=\epsilon^{1/2}x_0+\epsilon^{3/2}x_1+O(\epsilon^\gamma)$, with $\gamma>3/2.$

You should be able to continue this onwards as long as you like, but expanding out the $x^2$ and $x^4$ terms gets tricky. If you assume that the difference between each successive power of $\epsilon$ is the same, you can now write the whole expansion. I'm not sure if this is guaranteed, but I think it might be.


I do not know if this is what you can be looking for as another possible approach; so, please forgive me if I am off-topic.

Assuming that $\epsilon$ is small, the solutions of $x\tan(x)=\epsilon$ are close to $x=k \pi$ (I do not consider the case of $k=0$). So, around this point, we can build Taylor series and, limiting to the second order, we should have, as an approximation, $$x\tan(x) \approx k\pi(x-k\pi)+(x-k\pi)^2$$ and so, the solution for the $k^{th}$ is given by $$x_k=\frac{1}{2} \left(\sqrt{\pi ^2 k^2+4 \epsilon }+\pi k\right)\approx k\pi+\frac{\epsilon}{k\pi}$$ which seems to be quite accurate.

For more accuracy, you could go to the next order to get $$x\tan(x) \approx k\pi(x-k\pi)+(x-k\pi)^2+\frac {k\pi}3 (x-k\pi)^3$$ which will require solving a cubic polynomial.


If you start with $$x=k\pi+\sum_{i=1}^n a_i \epsilon^i$$ you could find that $$a_1=\frac{1}{\pi k}$$ $$a_2=-\frac{1}{\pi ^3 k^3}$$ $$a_3=\frac{6-\pi ^2 k^2}{3 \pi ^5 k^5}$$ $$a_4=\frac{4 \pi ^2 k^2-15}{3 \pi ^7 k^7}$$ which seem to be very accurate.

  • $\begingroup$ Thanks! Is it possible that there exists several different asymptotic expansions for the same problem? $\endgroup$
    – Sherry
    Apr 18 '15 at 5:33
  • $\begingroup$ Big question ! By the way, I added some stuff to my answer. $\endgroup$ Apr 18 '15 at 5:54

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