5 ms·

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

WoahNoun 4y ago
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.
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BeetleB 4y ago
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.
- patrick451 4y ago
  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.
  - effie 4y ago
    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.
  - BeetleB 4y ago
    > 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".

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.

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BeetleB 4y ago

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.

patrick451 4y ago
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.
- effie 4y ago
  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.
- BeetleB 4y ago
  > 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".