Maximum of $G=1+k\cos \theta+k^2\cos(2\theta)+k^3\cos(3\theta)+\cdots,$

Question

Given $$G=1+k\cos \theta+k^2\cos(2\theta)+k^3\cos(3\theta)+\cdots,$$ what is the maximum value of $G$ ( where $k=\frac{1}{2}$)?

Now i have thinking about approaching this by using Euler equation $$\cos(\theta)+i\sin(\theta)=e^{i\theta},$$ what I am not getting is how do I take the series individually. Is this is a valid way of solving this or is there any other way to do this?

Answer 1

No need of complex numbers in the present case. Indeed, as $k = 1/2$ is positive and cosine is maximal when $\theta \in 2\pi\mathbb{Z}$ with $\cos\theta = 1$, the series equal $1 + k + k^2 + \ldots = \frac{1}{1-k} = 2$ at most.

Answer 2

Euler's formula is totally a valid way to solve it. The key insight is $$ k^n \cos(n\theta) = \operatorname{Re}[k^n (\cos(n\theta)+i\sin(n\theta))] = \operatorname{Re}[(k^n e^{in\theta})] = \operatorname{Re}[(ke^{i\theta})^n]. $$ (This is often called De Moivre's formula.) Then set $k = 1/2$ and apply the sum of a geometric series: $$ \sum_{n=0}^\infty\operatorname{Re}[(e^{i\theta}/2)^n] = \operatorname{Re}\left[\sum_{n=0}^\infty (e^{i\theta}/2)^n\right] = \operatorname{Re}\left[\frac{1}{1-e^{i\theta}/2}\right] = \frac{4-2\cos\theta}{5-4\cos\theta}. $$ That last equality comes from $1/z = \bar{z}/z^2$ and doing some algebra. From here it's not too hard to show the maximum value is $2$, occurring when $\theta = 0$.

Answer 3

Note that $$\begin{align} 2G \cos \theta &= \sum_{n=0}^\infty k^n \, 2 \cos \theta \cos n \theta \\ &= 2\cos \theta + \sum_{n=1}^\infty k^n (\cos (n+1)\theta + \cos (n-1)\theta) \\ &= 2 \cos \theta + \frac{1}{k} \sum_{n=1}^\infty k^{n+1} \cos (n+1)\theta + k \sum_{n=1}^\infty k^{n-1} \cos (n-1)\theta \\ &= 2 \cos \theta + \frac{G - 1 - k \cos \theta}{k} + k G. \tag{1} \end{align}$$ Solving for $G$ yields $$G = \frac{1 - k \cos \theta}{1 - 2k \cos \theta + k^2}. \tag{2}$$ Hence $$\frac{dG}{d\theta} = \frac{k(k^2 - 1) \sin \theta}{(1 - 2k \cos \theta + k^2)^2} \tag{3}$$ and the critical points occur when $\sin \theta = 0$. If $k \in (-\infty, -1) \cup (0,1)$, then $\theta = 2m \pi$ are the maxima and $\theta = (2m+1)\pi$ are the minima; if $k \in (-1,0) \cup (1,\infty)$ the situation is reversed.