# A further question on asymptotic expansions of all real roots of xtan(x)=ϵ

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?

## 2 Answers

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

Edit

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.

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