ISI interview

Similar solution to @Chris Lewis's

The infinite sum can be rewritten as:

$\begin{aligned} S & = \sum_{k=1}^\infty \frac {\sin k\theta}{2^{k-1}} & \small \color{#3D99F6} \text{By Euler's formula: }e^{i\theta} = \cos \theta + i \sin \theta \\ & = \Im \left(\sum_{k=1}^\infty \frac {e^{ik\theta}}{2^{k-1}}\right) & \small \color{#3D99F6} \text{where }\Im(z) \text{ is the imaginary part of complex number }z. \\ & = \Im \left(e^{i\theta}\color{#3D99F6} \sum_{k=1}^\infty \frac {e^{ik\theta}}{2^k}\right) & \small \color{#3D99F6} \text{Infinite sum of geometric progression of }\frac {e^{i\theta}}2 \\ & = \Im \left(e^{i\theta} \times \color{#3D99F6} \frac 1{1-\frac {e^{i\theta}}2} \right) & \small \color{#3D99F6} \text{since } \left| \frac {e^{i\theta}}2 \right| \le 1 \\ & = \Im \left(\frac {2e^{i\theta}}{2-e^{i\theta}} \right) & \small \color{#3D99F6} \text{By Euler's formula} \\ & = \Im \left(\frac {2(\cos \theta + i\sin \theta)}{2-\cos \theta - i\sin \theta} \right) \end{aligned}$

$\begin{aligned} \ \ & = \Im \left(\frac {2(\cos \theta + i\sin \theta)\color{#3D99F6}(2-\cos \theta + i\sin \theta)}{(2-\cos \theta - i\sin \theta) \color{#3D99F6}(2-\cos \theta + i\sin \theta)} \right) \\ & = \Im \left(\frac {2(\cos \theta(2+\cos \theta) - \sin^2 \theta + i {\color{#3D99F6}(2\sin \theta - \sin \theta \cos \theta + \sin \theta \cos \theta)})}{\color{#3D99F6}(2-\cos \theta )^2 + \sin^2 \theta} \right) \\ & = \color{#3D99F6} \frac {4\sin \theta}{5-4\cos \theta} \end{aligned}$

Therefore, $p(q-r) = 4(5-4) = \boxed 4$ .

Reference: Euler's formula

Write $S=\sin \theta + \frac{\sin 2\theta}{2} + \frac{\sin 3\theta}{2^2} + \cdots$

Since the numerators involve sines of multiple angles and the denominators involve a geometric progression (ie powers of $2$ ), this suggests using Euler's formula .

We have $S=\operatorname{Im} \left( e^{i\theta}+\frac{e^{2i\theta}}{2}+\frac{e^{3i\theta}}{2^2}+\cdots \right)$

This is a bit neater in the following form: $\frac{S}{2}=\operatorname{Im} \left(1+ \frac{e^{i\theta}}{2}+\frac{e^{2i\theta}}{2^2}+\frac{e^{3i\theta}}{2^3}+\cdots \right)$ . Note that we can add the real number $1$ without affecting the value of the imaginary part; but it helps recognise the pattern of a geometric series sum. We can now simplify to

$\frac{S}{2} = \operatorname{Im} \left(\frac{1}{1-\frac{1}{2}e^{i\theta}}\right)$

This is the key step; from here it's just a case of clearing the imaginary component of the denominator and tidying up the trig; this leaves

$S=\frac{4\sin\theta}{5-4\cos\theta}$

and gives the answer $\boxed4$ .

$\sum _{i=1}^{\infty } \frac{\sin (i\,\theta)}{2^{i-1}}$

$\sum _{i=1}^{\infty } i 2^{-i} \mathbb{e}^{-\mathbb{i}\, \theta i}-i 2^{-i} \mathbb{e}^{\mathbb{i}\,i\,\theta}$

The separation of terms is justified as there are no sign changes

$(\sum _{i=1}^{\infty } i 2^{-i} \mathbb{e}^{-\mathbb{i} \theta i})+(\sum _{i=1}^{\infty } -i 2^{-i} e^{\mathbb{i} \theta i})$

$\frac{\mathit{i}}{-1+2 \mathbb{e}^{\mathit{i} \theta }} + \frac{\mathit{i}\mathbb{e}^{\mathit{i} \theta }}{-2+\mathbb{e}^{\mathit{i} \theta }}$

$\frac{2 \mathit{i} \left(-1+\mathbb{e}^{2 \mathit{i} \theta }\right)}{-5 \mathbb{e}^{\mathit{i} \theta }+2 \mathbb{e}^{2 \mathit{i} \theta }+2}$

$\frac{4 \sin (\theta )}{5-4 \cos (\theta )}$