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Comment by ogogmad | Hacker News Reader

> If you treat function as a system that produces an output given an input - does it matter if it is a graph or a probability distribution?

Because that's wrong. For example, take the Dirac delta function (point mass). You can only integrate it, not evaluate it.

GCR was saying that it's OK to use the same notation for both kinds of function. Mathematicians use separate notations, which is ugly and awkward. Engineers and physicists conflate these objects in their terminology, which is misleading. The right answer is to use the same notation but explain there's a difference.

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Now I'm thinking what the type of the Dirac delta function is.

I'm leaning to something like: (X (R -> R) ((R -> R) -> (R -> R))) -> R where we are given a function (R -> R) on which it acts and a linear operation (R -> R) -> (R -> R), like integration, which acts on it and gives us a single real number as the answer.

This is basically correct, the theory of distributions can be presented in terms of linear functionals. A functional is a map that takes a function and returns a scalar. I'm not sure how to encode linearity in your type system though...

>I'm not sure how to encode linearity in your type system though...

You don't really, the same way that a regular function R->R has no specific type information that tells you ti's linear or not, e.g. both x and sin(x^2) have the same type.

What do you mean you can't evaluate it? It evaluates to zero everywhere but the origin.

The rigorous definition of dirac delta is a function that doesn't take points as inputs. It's a function that takes in sets and then integrates over that set to product a value.

This is what is often taught in engineering/physics, but is not correct. One physics professor tried to explain it by pointing out that it only has meaning under an integral sign (not sure it's correct).

The other obvious problem is its value at the origin. Infinity as a value is not something mathematicians like - at least not the way it is used here. And integrating it to get 1 makes no sense mathematically merely by setting its value to infinity.

These concerns never really made much sense to me. 1/x also has a singularity, but as far as I'm aware, it's a perfectly normal function. So why is it suddenly a problem that the dirac delta is zero almost everywhere? I recall that in my analysis class, they wanted to define the dirac delta as the limit of a sequence of functions. That never seemed very different from defining a matrix exponential as the limit as n -> infinity of a taylor series.

Dirac delta being zero almost everywhere is sometimes assumed (when describing infinitely short impulse) but often it is a superfluous assumption, that is not needed for the use of delta. This is because delta is a distribution, not a function. It does not need to be ascribed values.

For example, delta can be put as initial condition for psi function in the time-dependent Schroedinger equation to search for Green's function. Evolution of this initial condition in time is a regular oscillatory function that looks nothing like 0 or almost 0 almost everywhere for any $t>0$. So if it isn't 0 almost everywhere for t>0, why should we impose that at t=0.

> 1/x also has a singularity, but as far as I'm aware, it's a perfectly normal function.

No one integrates it over 0, but they do the delta function.

> I recall that in my analysis class, they wanted to define the dirac delta as the limit of a sequence of functions.

In my engineering courses, they often tried to make it a limit of the "rectangle" function. Consider a function that is 0 everywhere except between [-delta, delta], where its value is whatever is needed to make the area under this rectangle 1. Then let delta approach 0.

This definition still fails using the standard limits/analysis that is taught (convergence problems, etc). In mathematics, the delta function needs the theory of distributions to "work".

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