# Can a Gaussian distribution be represented as a sum of Dirac Deltas?

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tworitdash
We know that Dirac Delta is not a function. However, I just talk about the numerical version of it that we use every day. We can simply represent the Dirac delta function as a limiting case of Gaussian distribution when the width of the distribution ##\sigma->0##.

$$\delta(x - \mu) = lim_{\sigma -> 0} \frac{1}{\sqrt{2\pi \sigma^2}} e^{\frac{-(x - \mu)^2}{2\sigma^2}}$$

Is it possible to also say the reverse with a weighted sum of Dirac Deltas to construct a Gaussian spectrum?

$$\frac{1}{\sqrt{2\pi \sigma^2}} e^{\frac{-(x - mu)^2}{2\sigma^2}} = \sum_{i} w_i \delta(x - i)$$

Where, somehow the weights ##w_i## constitute how it is distributed (##\sigma##). If yes, how do we decide these weights?

Staff Emeritus
Any function can be represented as a sum of Dirac delta functions:

Let ##f(x)## be an arbitrary function of ##x##. Then you can represent it as:

##\int f(y) \delta(x-y) dy##

So that's a weighted sum (well, integral) of delta functions.

BvU
Staff Emeritus
Any function can be represented as a sum of Dirac delta functions:

Let ##f(x)## be an arbitrary function of ##x##. Then you can represent it as:

##\int f(y) \delta(x-y) dy##

So that's a weighted sum (well, integral) of delta functions.

If you really want a discrete sum, instead of an integral, then it can't be done for most functions. But I guess for some purposes, you can approximate a function by delta functions: Pick a small positive x increment ##\Delta x## and define ##\tilde{f}(x, \Delta x)## by:

##\tilde{f}(x, \Delta x) = \sum_j f(j \Delta x) \delta(x- j\Delta x) \Delta x##

where ##\Delta x## is some small real number. This approximation works in an integration sense: For any other smooth function ##g(x)##, we have:

##lim_{\Delta x \Rightarrow 0} \int \tilde{f}(x, \Delta x) g(x) dx = \int f(x) g(x) dx##

tworitdash