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I have trouble to find roots for the following problem:

$$\epsilon^{-1}x^3=\frac{e^x-e^{-x}}{e^x+e^{-x}}$$

I think this is a singular perturbation problem for $\epsilon$ showing up in the highest degree term. So I did the following:

  1. Move $\epsilon$ to the other side, and let it be $0$ at first. Then we get triple repeated roots $x=0,0,0$.
  2. Let $$x=0+\sum_{i=1}^\infty a_i\epsilon^i$$
    and then plug in to find $a_i$

However, it seems I can only get one $x$. How to find all three roots?

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For small $x$ the RHS $\approx x$ so $\epsilon^{-1} x^3 \approx x$ so the roots are $\approx \pm \epsilon^{1/2}$ and $0$.

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  • $\begingroup$ Isn't that small $x$, not small $\epsilon$? What guarantees that all the roots are small when $\epsilon$ is small? A common situation in singular perturbation theory is that the roots "destroyed" when $\epsilon=0$ come back in "from infinity". $\endgroup$ – Ian Feb 1 '18 at 3:21
  • $\begingroup$ @Ian Thanks, fixed. $\endgroup$ – Keith McClary Feb 1 '18 at 3:22
  • $\begingroup$ @Ian Just drawing a graph of the RHS and LHS curves. $\endgroup$ – Keith McClary Feb 1 '18 at 3:26
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In the same spirit as Keith McClary's answer, consider $$\epsilon^{-1}x^3=\frac{e^x-e^{-x}}{e^x+e^{-x}}=\tanh(x)=x-\frac{x^3}{3}+O\left(x^5\right)$$ making the non trivial root to be $$x=\pm\frac{1}{\sqrt{\frac{1}{\epsilon }+\frac{1}{3}}}$$

For illustration purposes, let $\epsilon=10^{-k}$ $$\left( \begin{array}{ccc} k & \text{approximation} & \text{exact} \\\ 0 & 0.866025 & 0.893395 \\ 1 & 0.311086 & 0.311273 \\ 2 & 0.0998337 & 0.0998344 \\ 3 & 0.0316175 & 0.0316175 \\ 4 & 0.00999983 & 0.00999983 \end{array} \right)$$

Edit

If you look for more accuracy, consider the series $$\tanh(x)= \sum^{\infty}_{n=1} \frac{B_{2n} 4^n \left(4^n-1\right)}{(2n)!} x^{2n-1} $$ write $$\epsilon=\frac{x^3}{\tanh(x)}$$ and use series reversion to get $$x=\sqrt \epsilon \left( 1-\frac{1 }{6}\epsilon+\frac{13 }{120}\epsilon ^2-\frac{1423 }{15120}\epsilon ^3+\frac{18901 }{201600}\epsilon ^4-\frac{807797 }{7983360}\epsilon ^5+O\left(\epsilon ^6\right)\right)$$

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