Complex Numbers - Infinite Series 3

Submitted by roamingfree on Sun, 06/27/2021 - 18:34
skills
You need to be able to move between trigonometric and exponential representations
Question
Let $T$ be the infinite series defined by $$ T = \sin(\theta) - \frac{1}{2}\cos(2\theta) + \frac{1}{4}\sin(3\theta) - \frac{1}{8}\cos(4\theta) + \dots $$ Determine the integer coefficients such that the following is true $$ T = \frac{A\sin \theta + B\cos 2\theta + C}{D + E\cos 2\theta} $$
answer --- press button to toggle display
$$T = \frac{20\sin \theta -8\cos 2\theta + 2}{17 - 8\cos 2\theta}$$
solution --- press button to toggle display

Let us first split the series into two sub-series, $T = T^S + T^C$ $$ \begin{align} T^S &= \sin\theta + \frac{1}{4}\sin 3\theta + \frac{1}{16}\sin 5\theta +\dots\\ T^C &= -\frac{1}{2}\cos 2\theta - \frac{1}{8}\cos 4\theta - \frac{1}{32}\cos 6\theta +\dots \end{align} $$

If we now express $T^S$ in terms of exponentials, we can start the process of rewriting it as a rational function. $$T^S = \frac{1}{2i}(e^{i\theta} - e^{-i\theta}) + \frac{1}{8i}(e^{3i\theta} - e^{-3i\theta}) + \dots$$

We can now split this series into two sub-series, $T^S = T^{S-} + T^{S+}$ and treat them as geometric series. $$ \begin{align} T^{S+} &= \frac{1}{2i}e^{i\theta} + \frac{1}{8i}e^{3i\theta} + \frac{1}{32i}e^{5i\theta} \\ &= \frac{\frac{1}{2i}e^{i\theta}}{1 - \frac{1}{4}e^{2i\theta}} \\ T^{S-} &= \frac{-1}{2i}e^{-i\theta} + \frac{-1}{8i}e^{-3i\theta} + \frac{-1}{32i}e^{-5i\theta} \\ &= \frac{\frac{-1}{2i}e^{-i\theta}}{1 - \frac{1}{4}e^{-2i\theta}} \\ \end{align} $$

We can now add these two rational functions together, giving $$ \begin{align} T^S &= T^{S+} + T^{S-} \\ &= \frac{\frac{5}{8i}e^{i\theta} - \frac{5}{8i}e^{-i\theta}}{(1- \frac{1}{4}e^{2i\theta})(1- \frac{1}{4}e^{-2i\theta})} \\ &= \frac{\frac{5}{4}\sin\theta}{\frac{17}{16} - \frac{1}{2}\cos 2\theta} \\ &= \frac{20\sin\theta}{17 - 8 \cos 2\theta} \end{align} $$

Performing an identical process on $T^C$ yields $$ T^C = \frac{-8\cos 2\theta + 2}{17 - 8\cos 2\theta} $$

Finally, we get $$ T = \frac{20\sin \theta -8\cos 2\theta + 2}{17 - 8\cos 2\theta} $$

quick check

With such an involved question, a quick check is advisable, in this case, I plotted the series, truncated after the sixth term, against the final answer, the curves are virtually identical, suggesting our final answer is correct.