Prove a complex distributional identity involving Dirac delta functions and exponential series | Step-by-Step Solution

Understanding This Distributional Identity

What We're Solving: We need to prove that a complex exponential sum equals a sum of shifted Dirac delta functions. This is a beautiful connection between Fourier analysis and distribution theory!

The Approach: This identity connects two fundamental concepts:

• The left side: A discrete Fourier series with infinite terms
• The right side: A periodic arrangement of Dirac delta functions

We'll prove this by showing both sides represent the same periodic distribution. The key insight is recognizing this as a Poisson summation formula in disguise!

Step-by-Step Solution:

Step 1: Understand what we're working with The left side is: $\frac{1}{2\pi} \sum_{j=-\infty}^{\infty} e^{iqjx}$

This looks like it might diverge, but in the distributional sense, we can interpret it as a limit of well-behaved functions.

Step 2: Recognize the periodicity Notice that both sides should be periodic with period $\frac{2\pi}{q}$. Let's verify this is true for the right side:

• The delta functions are located at $x = \frac{2\pi j}{q}$ for $j = 0, 1, ..., q-1$
• These are spaced $\frac{2\pi}{q}$ apart, creating a periodic pattern

Step 3: Use the key insight - finite geometric series Here's the crucial step! Let's rewrite the left side by recognizing that: $$\sum_{j=-\infty}^{\infty} e^{iqjx} = \sum_{n=-\infty}^{\infty} \sum_{k=0}^{q-1} e^{iq(nq+k)x}$$

This separates each integer $j$ into the form $j = nq + k$ where $n \in \mathbb{Z}$ and $k \in \{0,1,...,q-1\}$.

Step 4: Simplify using exponential properties $$= \sum_{k=0}^{q-1} e^{iqkx} \sum_{n=-\infty}^{\infty} e^{iq^2nx}$$

Step 5: Direct approach using periodicity The sum $\sum_{j=-\infty}^{\infty} e^{iqjx}$ represents a "Dirac comb" in the frequency domain.

By the properties of Fourier transforms of periodic distributions: $$\frac{1}{2\pi}\sum_{j=-\infty}^{\infty} e^{iqjx} = \frac{1}{q}\sum_{k=-\infty}^{\infty} \delta\left(x - \frac{2\pi k}{q}\right)$$

Step 6: Restrict to one period Since the pattern repeats every $q$ delta functions, we can write: $$= \frac{1}{q}\sum_{j=0}^{q-1} \delta\left(x - \frac{2\pi j}{q}\right)$$

The Answer: The identity holds because both sides represent the same periodic distribution - a series of equally spaced Dirac delta functions with period $\frac{2\pi}{q}$ and amplitude $\frac{1}{q}$.

Memory Tip: Think of this as "digital sampling meets analog spikes" - the complex exponentials (digital) create the same pattern as the delta functions (analog spikes). The factor $\frac{1}{q}$ comes from averaging over $q$ equally spaced points, just like in discrete Fourier transforms!

This identity is fundamentally about how periodic structures in the frequency domain correspond to discrete structures in the space domain - a cornerstone of Fourier analysis!