Larry Riddle, Agnes Scott College

### Taylor series for $$f_\theta (r)$$ near r = 0 (for side part of tree)

Recall that $$f_\theta (r) = x_m - x_k$$ where xk is the x-coordinate of the branch tip for the path Rk(LR) and xm is the x-coordinate of the branch tip for the path Rm(LR), where k is the smallest integer such that kθ ≥ 90° and m is the smallest integer such that mθ ≥ 450°. Note that k ≥ 1 and m ≥ 3. In fact, m ≥ k+2. For indeed, if m = k+1, then $m\theta = (k+1)\theta = (k-1)\theta + 2\theta < 180^\circ + 2\theta < 90^\circ + 360^\circ = 450^\circ$ which is a contradiction. Also kθ < 270° since $k\theta ≥ 270^\circ \Rightarrow (k-1)\theta = k\theta - \theta ≥ 270^\circ - \theta > 270^\circ - 180^\circ = 90^\circ$ which is again a contradiction.

Let $$\alpha = e^{\theta i}$$. The branch tip for Rk(LR) occurs at the point corresponding to the complex number $z_k = \left(i+ir\alpha^{-1} + ir^2\alpha^{-2} + ir^3\alpha^{-3} \ldots + ir^k\alpha^{-k} \right)+ \left(ir^{k+1}\alpha^{-k+1} + ir^{k+2}\alpha^{-k} + \ldots\right).$ Then \begin{align} x_k = -\text{Im}(z_k) &= 1 + r\sin(\theta) + r^2\sin(2\theta) + r^3\sin(3\theta) + \ldots + r^k\sin(k\theta) \\ \\ &\quad\quad+ r^{k+1}\sin((k-1)\theta) + r^{k+2}\sin(k\theta) + \ldots. \end{align} The value of xm is obtained in the same way to get \begin{align} x_m = \text{Im}(z_k) &= 1 + r\sin(\theta) + r^2\sin(2\theta) + r^3\sin(3\theta) + \ldots + r^k\sin(k\theta) \\ \\ &\quad\quad+ r^{k+1}\sin((k+1)\theta) + r^{k+2}\sin((k+2)\theta)\\ \\ &\quad\quad+ \dots + r^m\sin(m\theta) + r^{m+1}\sin((m-1)\theta) + r^{m+2}\sin(m\theta) + \ldots. \end{align} Subtracting the two expressions for the x-coordinates shows that \begin{align} f_\theta (r) &= r^{k+1}\left(\sin((k+1)\theta) - \sin((k-1)\theta)\right) + r^{k+2}\left(\sin((k+2)\theta) - \sin(k\theta)\right) + \ldots \\ \\ &= 2\sin(\theta)\cos(k\theta)r^{k+1} + \left(\sin((k+2)\theta) - \sin(k\theta)\right)r^{k+2} + \ldots. \end{align}

Suppose kθ > 90°. Since 90° < kθ < 270° we have $$\cos(k\theta) < 0$$ and $$\sin(\theta) > 0$$, and so the coefficient of rk+1 will be negative, and in particular, non-zero. Therefore $$f_\theta (r)$$ will behave like the function $${2\sin(\theta)\cos(k\theta)}r^{k+1}$$ when r is close to 0.

If kθ = 90°, then the rk+1 term in the formula for $$f_\theta (r)$$ will cancel. Therefore the behavior of $$f_\theta (r)$$ will be determined by the rk+2 term when r is small. Hence $$f_\theta (r)$$ will behave like the function $\left(\sin((k+2)\theta) - \sin(k\theta)\right)r^{k+2} = \left(\cos(2\theta) - 1\right) r^{k+2} .$ Again the coefficient will be negative. It is important to note that because m ≥ k+2, no term involving a power of r with exponent greater than or equal to m+1 will affect the rk+1 or rk+2 terms.

So in either case, when r is close to 0, the function $$f_\theta (r)$$ will behave like Arn with A < 0. Thus the graph of $$f_\theta (r)$$ will be decreasing and concave down for small values of r.

#### Examples

Asking about the behavior of a function near r = 0 is really asking about the Taylor series for the function centered at r = 0. A computer algebra system like Maple or Mathematica can calculate a Taylor series up to a given order.

Example 1: θ = 80° (so k = 2 and m = 6, with kθ > 90°)

$f_\theta (r) = -1.85083r^3 - 0.98481r^4 - 0.34202r^5 + \ldots$

Example 2: θ = 45° (so k = 2 and m = 10, with kθ = 90°)

$f_\theta (r) = -r^4 - \sqrt{2}r^5 - 2r^6 - \ldots$