# Dirac delta, math of implication?

So what we have so far is that any and all subsets are implied by a set. If there exist a set, then all the subsets within it are implied to exist also. This includes the elements of a set. The elements of a set are implied by the existence of a set.

One of the most natural things to do with sets is to count the number of elements it has. This can be done in order to give a relative size to subsets. So the number of elements works as a measure of a set.

You can count an element as a member of a set only if the element is included in the set. If we label the elements with arbitrary coordinates, we can describe this mathematically as,

$$$\int_{\rm{A}} {{\rm{\delta (x - x}}_0 ){\rm{dx}}} \, = \,1$$$

The integral is 1 if $$${\rm{x}}_{\rm{o}} \in \,\,{\rm{A}}$$$. In other words, if $$${\rm{A}} \to {\rm{x}}_0$$$ is true, then the integral is 1. But if $$${\rm{A}} \to {\rm{x}}_0$$$ is not true, then the integral is 0. The integral is 1 or 0 as the implication is true or false. The integrand is the dirac delta "function".

Now it seems to me that the above is all true no matter what the size of the region A is. We can let A become smaller and smaller and shrink down to the size of another point, x, and the above is still true. This procedure would be the same as taking the derivative of an integral, which is the same as the integrand, $$${{\rm{\delta (x - x}}_0 )}$$$. If $$$x \to {\rm{x}}_0$$$ is true then $$${\rm{\delta (x - x}}_0 ) \ne 0$$$. And if $$$x \to {\rm{x}}_0$$$ is not true, then $$${\rm{\delta (x - x}}_0 ) = 0$$$. The delta function is non-zero or zero as $$$x \to {\rm{x}}_0$$$ is true or false.

Using coordinates to label facts allow us to employ the dirac delta function to mathematically represent the implication between those facts. Does this much sound right? Thanks.

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EnumaElish
Homework Helper
Using coordinates to label facts allow us to employ the dirac delta function to mathematically represent the implication between those facts. Does this much sound right? Thanks.
Do you have an example?

Do you have an example?
The closest example I can think of is that of Ariel Caticha. In

http://arxiv.org/abs/quant-ph/9804012

Here he derives the path integral formulation of quantum theory from the integration of simple "setups". Each setup is described as a source of some event at x1 being detected as an event at x2. Later he reveals that these setups are actually the wavefunction and uses the delta function to describe the most primative setups (or wavefunctions).

But to have one event result in another event is just different terminology for implication. So it seems Caticha is using the delta function to represent implication. Caticha is careful not to draw too close an analogy with logic. But I think the generalization is clear.

Caticha then goes on to show how these simple setups can be combined to from a path integral. But I think this effort is unnecessary. I believe I can show that the Feynman path integral easily falls out of nothing more than the definition of the dirac delta function. Seriously, it's easy! And I'd be happy to show it to you. But I'd like to make sure of my assumptions first.

So I'm trying to get a better foundation for material implication being represented by the dirac delta function. And I think I have a start. Is it possible that the dirac delta function might be the simplest form of measure in measure theory from which all other types of measure can be derived? After all it does seem to say when you are allowed to count an element in a set. Or perhaps NO discernable causal relationship between facts is the simplest relationship. Then the delta function as a gaussian distribution for totally random processes would represent there being no detectable cause between two events.

I'm not theorizing, yet. Just trying to get information. Thanks.

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EnumaElish
Homework Helper
When I asked for an example, I meant something of the form, "let A = [0,1], let x0 = 1, then integral = 1 over A; alternatively if A = [0,1) then integral = 0 over A," and somehow relate this to your phrase "using coordinates to label facts."

You also seemed to be using the set notation and limit notation interchangeably ("x in A" equivalent to "x --> A"); is that a practical shorthand, or does it have theoretical significance in the context?

Let me respond with an example of my own. On the unit interval U = [0,1], I can ascribe a "scaled" Dirac measure $\bar\delta_x$ = δx(x)k to each x in U, where k is the scaling constant (assuming such a constant exists -- I am waving my hands). Specifically, let that measure be $\bar\delta_x$(x) = f(x), (e.g., f(x) = 1) and $\bar\delta_x$(y) = 0 for all y $\ne$ x.

There is a separate δx and a separate $\bar\delta_x$ for each x in U; let δ(y) = Σxδx(y) and $\bar\delta(y)=\sum_x\bar\delta_x(y)$. Then $\bar\delta$(y) = f(y) for all y in U.

Let F be the antiderivative of f (e.g. F(x) = x). Then $\int_U \bar\delta(x)dx = \int_U f(x)dx = \int_U dF(x)$ is the F-measure of the set U.

So in that "scaled delta" sense, I can define an arbitrary measure F in terms of $\bar\delta$ through f (the derivative of F), which is the scaled delta. But honestly, my example is a little like fitting a square peg into a round hole, because there is no scaling constant k (a real number) that would satisfy $\bar\delta$(x) = δ(x)k = $\infty$k = f(x) < $\infty$; that is, there is no real number k that could produce a finite measure out of the infinite Dirac delta. If you see a conceptual way to get you around this, then you will have answered your own question.

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When I asked for an example, I meant something of the form, "let A = [0,1], let x0 = 1, then integral = 1 over A; alternatively if A = [0,1) then integral = 0 over A," and somehow relate this to your phrase "using coordinates to label facts."
Oh! I thought you meant some particular types of facts being labeled with coordinates. But perhaps I'm overstating the necessity of needing facts to label. In the generalization of these concepts, we probably lose track of facts we are labeling and just notice how field values change as the coordinates change. I think in terms of mathematics we can restrict our concerns to elements at coordinates and ask what number describes how two different elements are related. Since the number of elements in a set give a relative probability, the relation between any two elements is most likely in terms of the probability of going from one to the other. If we start with a totally random process that leads from one to any other, then we are looking at a gausian probability distribution. Or does anyone else have any other ideas?

EnumaElish said:
You also seemed to be using the set notation and limit notation interchangeably ("x in A" equivalent to "x --> A"); is that a practical shorthand, or does it have theoretical significance in the context?
I'm using $$${\rm{A}} \to {\rm{B}}$$$ to symbolize that A implies B. This is fairly common notation, but there is other notation as well.

EnumaElish said:
Let me respond with an example of my own. On the unit interval U = [0,1], I can ascribe a "scaled" Dirac measure $\bar\delta_x$ = δx(x)k to each x in U, where k is the scaling constant (assuming such a constant exists -- I am waving my hands). Specifically, let that measure be $\bar\delta_x$(x) = f(x), (e.g., f(x) = 1) and $\bar\delta_x$(y) = 0 for all y $\ne$ x.
I'm not sure why you use a "scaled" dirac delta. As I understand it, the integral=1 when integrated in ANY interval that contains x, no matter how small that interval is. This is because the dirac delta has a shrinking parameter that renders it zero when the argument is not zero. Since that parameter shrinks to zero, any interval will contain the function being integrated.

EnumaElish said:
There is a separate δx and a separate $\bar\delta_x$ for each x in U; let δ(y) = Σxδx(y) and $\bar\delta(y)=\sum_x\bar\delta_x(y)$. Then $\bar\delta$(y) = f(y) for all y in U.

Let F be the antiderivative of f (e.g. F(x) = x). Then $\int_U \bar\delta(x)dx = \int_U f(x)dx = \int_U dF(x)$ is the F-measure of the set U.
The dirac delta function is the derivative of the unit step funciton, u(x), which equals zero for x<0 and equals 1 for x>0. This makes sense, since the slope of the unit step is 0 when x<0, is infinite at x=0, is 0 again when x>0, just like the delta function. So how does this help your diffinition of measure again?

I've said previously that I can derive the Feynman path integral solely in terms of the dirac delta function. But there is some controversy about whether the Feynman path integral is mathematically well defined. They say that "measure" of differential paths in not well defined, or something like that. So this brings me to wonder if the dirac delta function is well defined or not. Does anyone have any insight into the dirac delta function being a well defined measure or distribution? Thanks.

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rbj
So this brings me to wonder if the dirac delta function is well defined or not. Does anyone have any insight into the dirac delta function being a well defined measure or distribution?
The dirac delta function is not strictly, as mathematicians routinely define "function", a function. Strictly speaking, what they teach us in Real Analysis (or a similarly titled course) is that the Lebesgue integrals of two different functions f(x) and g(x) that are equal "almost everywhere" must be equal. Thinking of the dirac delta as a regular function that is zero everywhere but at x=0 leads one to contradict the above since the integral of the dirac, f(x) is 1, yet the integral of g(x)=0 is zero, yet the two functions agree almost everywhere.

Mathematicians do not like how engineers (and some physicists) deal with the dirac delta (as if we are playing fast and lose with the facts).

The Dirac delta function is not strictly, as mathematicians routinely define "function", a function. Strictly speaking, what they teach us in Real Analysis (or a similarly titled course) is that the Lebesgue integrals of two different functions f(x) and g(x) that are equal "almost everywhere" must be equal. Thinking of the Dirac delta as a regular function that is zero everywhere but at x=0 leads one to contradict the above since the integral of the Dirac, f(x) is 1, yet the integral of g(x)=0 is zero, yet the two functions agree almost everywhere.
Does infinity only at one point still qualify as equal almost everywhere to a finite value only at that point? Or does infinity throw off that calculation since you would then be giving an infinite weight to the contribution for that point as opposed to a finite weight for that point.

rbj said:
Mathematicians do not like how engineers (and some physicists) deal with the Dirac delta (as if we are playing fast and lose with the facts).
I wonder if all concerns are well defined and valid when the zero approaching parameter of the delta function is some arbitrary value that's NOT zero, then wouldn't those concerns also be valid as the parameter APPROACHES zero? At what value of the parameter does one draw the line? If the concerns (I'm not sure what) are valid independent of the value of the parameter, then I would think it is still valid as the parameter approaches zero, since it is NOT technically zero.

EnumaElish
Homework Helper
I'm not sure why you use a "scaled" dirac delta. As I understand it, the integral=1 when integrated in ANY interval that contains x, no matter how small that interval is. This is because the dirac delta has a shrinking parameter that renders it zero when the argument is not zero. Since that parameter shrinks to zero, any interval will contain the function being integrated.
Dirac's delta "DD" is defined for an interval that contains a singular element x, for which DD is being defined (or associated with). In that respect, ANY superset of {x} has the same measure (=1), which makes DD not particularly useful to define a measure. I was using the scaled DD to get around that property -- or impose some structure onto the problem so that DD can be used to construct a useful measure that discriminates between a set and its numerous supersets.

Dirac's delta "DD" is defined for an interval that contains a singular element x, for which DD is being defined (or associated with). In that respect, ANY superset of {x} has the same measure (=1), which makes DD not particularly useful to define a measure. I was using the scaled DD to get around that property -- or impose some structure onto the problem so that DD can be used to construct a useful measure that discriminates between a set and its numerous supersets.
I think I understand what you're saying. But the Dirac delta would only be useful in deciding whether to count an element as a member of a set. The definition of the delta precludes any information about the set as a whole, since any size superset containing the element gives the same number. The measure of the set to which the element belongs would be accomplished by some count of ALL the elements in the set. I suppose that means there would have to be a Dirac delta function for each element and some integration of all the deltas in the region occupied by the set.

There is another thing to consider. The Dirac delta for each element assigns a equal weight to each element. But what if different element have different weights. Wouldn't that mean that you would then integrate the Dirac delta multiplied by some function that weighs each element?

EnumaElish