0
$\begingroup$

Suppose I have $\lambda = e^{i\mu}$, where $\mu = \mu(\epsilon)$ with $\epsilon$ a small parameter. Suppose $\mu$ can be expanded in series in $\epsilon$ as $\mu = \mu_{0} + \epsilon \mu_{1} + O(\epsilon)^{2}$, with $\mu_{0}=0$. In other words, for small $\epsilon$, $\mu$ tends to $0$.

Is it possible for $\lambda^3 =1 $ or $\lambda^{4} = 1$ when taking the exact value of $\mu$ for some small $\epsilon$? More generally, can we have $\lambda^{k} = 1$ for $k<5$ ?

1 Answers 1

1

$$1=\lambda^k=e^{ik\mu}\implies\mu=\frac{2n\pi}k$$

So for small enough $\epsilon>0$, since $\mu\to0$ but $\mu\ne0$, then $\lambda^k\ne1$.

  • 0
    But for $k = O(1/ \epsilon)$ with $n<$\lambda^{k} = 1$, as then $\mu$ will be small, right? @Simple Art – 2017-01-20
  • 0
    Then heuristically, an initial point $e^{i\mu}$ on the circle near $1$ will do $k$ "shifts" along the circle and eventually hit exactly $1$ for large enough $k$... (I know that this $k$ range is outside of the specified one in my initial question) @Simple Art – 2017-01-20