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What is wrong with the two difference usages of a function? If you treat function as a system that produces an output given an input - does it matter if it is a

What is wrong with the two difference usages of a function? 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?

deleted 4y ago
[deleted]
ogogmad 4y ago
> 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.
- thrown_22 4y ago
  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.
  - johnbcoughlin 4y ago
    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...
    - thrown_22 4y ago
      >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.
- patrick451 4y ago
  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.
  - deleted 4y ago
    [deleted]
  - 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.

What is wrong with the two difference usages of a function? 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?

deleted 4y ago

[deleted]

ogogmad 4y ago

> 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.

thrown_22 4y ago
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.
- johnbcoughlin 4y ago
  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...
  - thrown_22 4y ago
    >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.
patrick451 4y ago
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.
- deleted 4y ago
  [deleted]
- 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.