1

I see that when $θ$ is a multiple of $2π$, $\frac {e^{in θ}} {1+ e^{i2nθ}}$ is constant and thus has limit $\frac 1 2$.

When $θ$ isn't a multiple of $2π$, however, it seems that the value $\frac {e^{in θ}} {1+ e^{i2nθ}}$ of oscillates in a way that the limit doesn't exist.

For certain special cases (i.e. when $e^{iθ}=-1$ or $i$ or $i^{0.3}$), this can be demonstrated by brute force. But I am struggling to prove it for the general case, especially since $θ$ isn't necessarily rational.

Pascal's Wager
  • 1,827

1 Answers1

2

Note that $e^{in\theta}\ne 0$. Therefore, $$ f_n(\theta)=\frac{e^{in\theta}}{1+e^{2in\theta}}=\frac{1}{e^{-in\theta}+e^{in\theta}}=\frac{1}{2}\frac{2}{e^{in\theta}+e^{-in\theta}}=\frac{1}{2}\frac{1}{\cos n\theta}. $$

Just a reminder that this result would be invalid if $1+e^{2in\theta}=0$, or equivalently, if $\cos n\theta=0$.

To explore the convergence of $f_n(\theta)$ for a fixed $\theta$, it suffices to explore the convergence of $g_n(\theta)=\cos n\theta$. In this case, we may generalize the answers in this thread.

Suppose $$ \lim_{n\to\infty}g_n(\theta)=g(\theta) $$ for some function $g$ of $\theta$. Then it is a must that $$ \lim_{n\to\infty}g_{n+1}(\theta)=g(\theta). $$ Now, note that $$ g_{n+1}(\theta)=\cos\left(n+1\right)\theta=\cos n\theta\cos\theta-\sin n\theta\sin\theta, $$ or equivalently, $$ \sin n\theta=\frac{g_n(\theta)\cos\theta-g_{n+1}(\theta)}{\sin\theta} $$ for $\sin\theta\ne 0$. In this case, the convergence of $g_n(\theta)$ implies the convergence of $h_n(\theta)=\sin n\theta$, i.e., $$ \lim_{n\to\infty}h_n(\theta)=h(\theta) $$ for some function $h$ of $\theta$.

Besides, note that $$ g_n^2(\theta)+h_n^2(\theta)=\cos^2n\theta+\sin^2n\theta=1. $$ It follows that \begin{equation} g^2(\theta)+h^2(\theta)=1.\tag{1} \end{equation}

Finally, if $\sin\theta=0$, we have $\theta=k\pi$ with $k\in\mathbb{Z}$. In this case, $g_n(\theta)=\pm 1$, whose convergence depends on the choice of $k$, which is obvious to figure out. If $\sin\theta\ne 0$, recall that $$ g_{n+1}(\theta)=g_n(\theta)\cos\theta-h_n(\theta)\sin\theta. $$ Taking the limit on both sides yields $$ g(\theta)=g(\theta)\cos\theta-h(\theta)\sin\theta, $$ or equivalently, \begin{equation} h(\theta)=-g(\theta)\frac{1-\cos\theta}{\sin{\theta}}.\tag{2} \end{equation} Besides, apply the same trick to \begin{align} h_{n+1}(\theta)=\sin\left(n+1\right)\theta&=\sin n\theta\cos\theta+\cos n\theta\sin\theta\\ &=h_n(\theta)\cos\theta+g_n(\theta)\sin\theta, \end{align} and we obtain $$ h(\theta)=h(\theta)\cos\theta+g(\theta)\sin\theta, $$ or equivalently, \begin{equation} g(\theta)=h(\theta)\frac{1-\cos\theta}{\sin\theta}.\tag{3} \end{equation} You may find an immediate contradiction between (1), (2), and (3).

In conclusion, the convergence of $g_n(\theta)$ could be discussed for $\theta=k\pi$ with $k\in\mathbb{Z}$. Otherwise, $g_n(\theta)$ always diverges.

hypernova
  • 6,072