I am new around here and was hoping you will be able to help me with the following. I have the equation: $x^3 - 3x^2 +(3-\epsilon ) x + \epsilon = sin(\frac{\pi}{2} x +\frac{\pi \epsilon}{2} ) $ and need to find asymptotic expansions to everyone of its roots.
My question is: how can I determine the asymptotic expansion of the root satisfying $x(\epsilon ) = ord(1)$ ?
Obviously, the case $x>>1 $ is not possible , so we are left with the cases $x=ord(1)$ or $ x<<1 $ . In the second case, it is legitimate to use Maclaurin series of $sin$ and to equate highest order terms. My problem is with the other possiblity, where I cannot use Maclaurin... I know $x=1$ solves this equation when $\epsilon=0$ , but how does this help ?
Hope you will be able to help me
Thanks a lot
(This is intended to be a comment but the LaTeX would not render there for some reason.)
There are two roots which approach $x = 1$ as $\epsilon \to 0^+$. If we call them $x_+$ and $x_-$, then the first few terms of their expansions are given by $$\begin{align}x_\pm &= 1 + \left(\frac{4-\pi^2 \pm 2i\sqrt{2\pi^2-4}}{\pi^2}\right)\epsilon \\&\qquad + \frac{1}{\pi^6}\sqrt{\frac{2}{\pi^2-2}}\left(2\sqrt{2\pi^2-4}\pm(\pi^2-4)i\right)\epsilon^2 \\&\qquad + O(\epsilon^3).\end{align}$$