Larry Riddle, Agnes Scott College

$$\lim_{r \rightarrow 1^{-}} f_\theta (r)$$ when θ is a zero of N(θ)

Here we have $$\theta = \frac{720^\circ}{p}$$ for some integer p ≥ 4 with N(θ) = 0. \begin{align} f_\theta (r) &= \sum_{n={k+1}}^{m}{r^n \cos(n \theta)} \\ \\ &+ \frac{r^{m+1} \cos((m-1)\theta) + r^{m+2} \cos(m \theta) - r^{k+1} \cos((k-1)\theta) - r^{k+2} \cos(k \theta)}{1-r^2} \\ \\ \end{align}

where $$k = \left\lceil \frac{180^\circ}{\theta}\right\rceil$$ and $$m = \left\lceil \frac{540^\circ}{\theta}\right\rceil$$. Then

$\lim_{r \rightarrow 1^{-}} \frac{r^{m+1} \cos((m-1)\theta) + r^{m+2} \cos(m \theta) - r^{k+1} \cos((k-1)\theta) - r^{k+2}\cos(k\theta)}{1-r^2} = \frac{N(\theta)}{0} = \frac{0}{0}.\\ \\$

Therefore we can use L'Hospital's rule to evaluate the limit.

\begin{align} \lim_{r \rightarrow 1^{-}} f_\theta (r) &= \sum_{n=k+1}^{m} \cos(n \theta) \\ \\ &\quad + \frac{(m+1)\cos((m-1)\theta) + (m+2)\cos(m \theta)}{-2} \\ \\ &\quad - \frac{(k+1)\cos((k-1)\theta) + (k+2)\cos(k\theta)}{-2}\\ \\ \end{align}

Let's first consider a special case when $$\theta = \frac{720^\circ}{4k} = \frac{180^\circ}{k}$$. Then kθ = 180° and mθ = 540° since m = 3k. In this case

\begin{align} \sum_{n=k+1}^{m} \cos(n \theta) &= \sum_{n=k+1}^{3k} \cos(n \theta) = \text{Re}\left(\sum_{n=k+1}^{3k} e^{n\theta i}\right) \\ \\ &= \text{Re}\left(\frac{e^{(k+1)\theta i} - e^{(3k+1)\theta i}}{1-e^{\theta i}}\right) \\ \\ &= \text{Re}\left( \frac{e^{k\theta i}e^{\theta i} - e^{3k\theta i}e^{\theta i}}{1-e^{\theta i}}\right) \\ \\ &= \text{Re}\left(\frac{-e^{\theta i} - (-1)e^{\theta i}}{1-e^{\theta i}}\right) = 0 \end{align} and \begin{align} &\frac{(m+1)\cos((m-1)\theta) + (m+2)\cos(m \theta) - (k+1)\cos((k-1)\theta) - (k+2)\cos(k\theta)}{-2} \\ \\ &= \frac{-(m+1) \cos(\theta) - (m+2) + (k+1) \cos(\theta) + (k+2) }{-2} \\ \\ &= \frac{-(3k+1) \cos(\theta) - (3k+2) + (k+1) \cos(\theta) + (k+2) }{-2} \\ \\ &= (1+\cos(\theta))k. \\ \\ \end{align}

So when $$\theta = \frac{180^\circ}{k}$$, then $$\lim_{r \rightarrow 1^{-}} f_\theta (r) = (1+\cos(\theta))k$$.

What can we say in general? Let \begin{align} g(p) &= \lim_{r \rightarrow 1^{-}} f_\theta (r) = \sum_{n=k+1}^{m} \cos(n \theta) \\ \\ &\quad + \frac{(m+1)\cos((m-1)\theta) + (m+2)\cos(m \theta)}{-2} \\ \\ &\quad - \frac{(k+1)\cos((k-1)\theta) + (k+2)\cos(k\theta)}{-2} \\ \\ \end{align} where $$\theta = \frac{720^\circ}{p}$$, $$k = \left\lceil\frac{180^\circ}{\theta}\right\rceil = \left\lceil\frac{p}{4}\right\rceil$$ and $$m = \left\lceil\frac{540^\circ}{\theta}\right\rceil = \left\lceil\frac{3p}{4}\right\rceil$$. Here is the graph of g(p) for p = 4, 5, ..., 20 (black dots). The red dots are at p = 4, 8, 12, 16, and 20, where θ is one of the special cases discussed above. The red graph is $$(1+\cos(\theta))k$$ for the corresponding theta values for p between 4 and 20.

We see that g(p) is positive for p ≥ 5. This behavior continues as shown in the graph below for p from 4 to 720 (representing the zeros between 1° and 180°.).

We see, therefore, that $$\lim_{r \rightarrow 1^{-}} f_\theta (r)$$ exists and is positive when $$\theta = \frac{720^\circ}{p}$$ is a zero of N(θ).