Absolute value of the exponential sum $\sum_{n \leq x} e^{2 i \pi n \theta}$

Question

Let: $$S(x, \theta) = \sum_{n \leq x} e^{2 i \pi n \theta}.$$

We have $$|S(x, \theta)| = \frac{|e^{2 i \pi x \theta} - 1|}{|e^{2 i \pi \theta}-1|}.$$

When $x$ is a fixed integer, and $\theta \rightarrow 0$, we have naturally $|S(x,\theta)| \rightarrow x$.

But how can we describe more precisely this behaviour (see following curve)?

For $x$ fixed, it looks like a $\theta \mapsto x \frac{|\sin(\theta)|}\theta$ function.

With $x=100$:

With $x=1000$:

Answer 1

Notice that

$$e^{i\phi}-1=e^{i\phi/2}(e^{i\phi/2}-e^{-i\phi/2})=2 i e^{i\phi/2}\sin(\phi/2)$$

When you take the absolute value, the phase terms disappear. This means that you can write your expression as:

$$|S(x,\theta)|=\left|\frac{\sin(\pi x \theta)}{\sin(\pi\theta)}\right |$$

From this, you can get all of the limiting behaviour you noticed. If $\theta\ll 1$, then the denominator is approximately $\pi\theta$, and for $\theta\to 0$, you get $x$.