Agnes Scott College
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.

bottomLimitGraphForZeros4-20v2

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°.).

bottomLimitGraphForZeros4-720

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).