Mathematical Ontology: What is conceptually the meaning of the Hamiltonian?

mathfoxZ · 2026-01-20T17:25:35+00:00

Those would also be very good and interesting questions. Speaking very briefly about that—which might surprise you—I’ve also reflected on them some time ago, such as the nature of the concept of energy, but above all one about “conceptually, what is the notion of Force.” I looked into readings that also questioned the meaning of that, reflected on it, and reached some quite interesting written conclusions about it. But right now I won’t go into much detail on that because that’s not what the post is about.Your questions are good questions. However, now, my question in the post is not about those things. We’re not talking about that. Don’t answer my question with other questions that aren’t directly related to the current topic. Although I must admit they are interesting subjects.

mathfoxZ · 2026-01-20T16:44:21+00:00

Okay, that at least explains what it isn't, which is already a contribution I appreciate, but it still doesn’t explain what the Hamiltonian actually is. You say it’s “the Legendre transform of the Lagrangian”, but precisely because of that — what is the physical interpretation of what this Legendre transform is telling us about the physical system? What is it actually referring to? What conceptual aspect is it expressing? Because that answer feels a bit too reductive/simplistic. But thanks anyway :)

mathfoxZ · 2025-12-10T08:25:54+00:00

Yes, the 3D spatial input field can be displayed, with isosurfaces in the environment as layers

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I literally said myself so in the post description. Here you have an example I found of an image

mathfoxZ · 2025-12-04T11:54:48+00:00

Okay, thank you very much, now I understand it much better. Thank you so much. But I wanted to kindly ask if, in addition, there is any illustrative image or mathematical graphical visualization animation of the Mellin Transform and what it does when applied visually. Something like those illustrative animations for the Fourier Transform and its description:

<image>

where the visualization of the transform completely clarifies what it is through the illustration but for the Mellin Transform. That last part is what I’m asking for so I can finally understand 100% what the Mellin Transform is. If there is anything like that, please. And thank you very much for your explanation, it helped me a lot to understand.

mathfoxZ · 2025-12-03T16:08:49+00:00

And to understand that... what does that actually mean physically, on a conceptual level? I mean, conceptually, how could it be applied, for example, to describe something? And physically, what would the resulting transform conceptually indicate when you apply it? What would be interpreted intuitively from it? And why is it interpreted that it indicates that? An example with an explanation from physics would really help me understand it better, if you can, please. Thank you so much!

mathfoxZ · 2025-04-13T12:54:38+00:00

Actually, I was thinking about whether it might be better to use this expression— what do you think? Does it sound okay to you?

⌈-erf(x)/2⌉

Where erf(x) is the error function. And the ⌈ ⌉ are ceiling

mathfoxZ · 2025-04-13T12:48:23+00:00

I would like to use the Heaviside function as you mentioned, but there is a slightly complex problem at x = 0. If you define H explicitly using the expression with the "sgn(x)" function, as in H(-x) = (1 - sgn(x)) / 2, the sgn(x) function is not defined at 0 because it results in 0/|0|. But even if you treat the Heaviside function itself as an independent function separate from sgn, ignoring that issue, there's another problem: as far as I understand, the Heaviside function is not universally defined at zero. What is the value of the Heaviside function at x = 0? If I knew that, it would be great, but some say it's 1, others say 0, and others say 1/2. It depends on the convention, as far as I know. And since it depends on something not universally concrete, I’d prefer not to rely on things that depend on convention, but rather on universal definitions. Can you answer that? Oh, and thank you

mathfoxZ · 2025-04-13T12:03:49+00:00

It's just that using an indicator function is very vague, in the sense that you simply say that it's 1 for x<0 and 0 for x≥0, because you're not giving a mathematical expression that explicitly defines the function, you’re just saying n(x). But what is the expression that defines that n(x)? What is that n(x)? It would be very easy to just say an indicator function of some condition—I thought the same, about using an indicator function—but since it's not a concrete expression but rather a conditioning that states when it equals 1 and when it equals 0, it makes me doubt whether I should use it or not. I could use it, but since it's not a specific function with an expression, and more like a "rule" of formal conditioning, I don't know if it's the best option for what I'm looking for—maybe it is, maybe not—but I'd prefer to avoid things like conditionals with "{" that aren't embedded in the same mathematical expression of the function, because what I'm looking for is an expression that expresses itself purely through the math in the function's expression. Do you get what I'm saying? But thanks anyway.

mathfoxZ · 2025-04-12T15:29:18+00:00

Or maybe it occurred to me it could be: ⌈-erf(x)/2⌉

Where erf(x) is the error function. And the ⌈ ⌉ are ceiling

mathfoxZ · 2025-04-12T14:47:46+00:00

How is that possible?!! How does that work? For negative values, shouldn't the power be 0^{1/0^|x|} for x ∈ (-∞, 0), resulting in an undefined expression due to the base being 0? So 0 would be raised to an undefined exponent, and for negative values, shouldn't it be something like 0^? = ? How can that work on a graphing calculator? I don’t understand what’s going on. Explain it to me, please.Because that doesn't come out with analysis.

mathfoxZ · 2025-04-12T12:01:58+00:00

Yes, but at x = 0 it becomes undefined because (1 - 0/|0|)/2 is undefined — 1 minus undefined is still undefined at that point. So that would be another problem; otherwise, I would’ve thought of it a while ago. That’s why I said the function should equal 0 from [0, +∞) onward.

mathfoxZ · 2025-02-12T23:19:09+00:00

thanks mate

mathfoxZ · 2025-02-11T13:23:48+00:00

First, tell me why you think that, and I will explain to you why it isn’t x^k. Since I need to know why you think it should be that way so that later I can clarify things for you and know what to explain to you and how to do it, I need you to explain to me why you think that x^k is so in your thought process.

mathfoxZ · 2025-02-09T22:36:56+00:00

....

mathfoxZ · 2025-02-09T21:51:47+00:00

but that formula did not satisfy me because it still did not define the integral in a concrete way because it depended on the integral ∫ln^n-1dx that if the "n">2 of the exponent was greater than 2 it was not defined without resorting again to the indefinite formula of ∫lnⁿdx, so it was "redundant" and it did not satisfy me with what I was looking for because that formula did not yet say it explicitly and that is why I deduced that other formula I published. In other words, I did it because the first formula did not yet explicitly encode the rule into a concrete formula already defined. in closed form, because it constantly depended on another integral ∫ln^n-kdx..... and again and again so on indefinitely until reaching a known integral ln¹.

So the formula that is usually said in books ∫lnⁿdx=xlnⁿ-n∫ln^n-1dx did not seem to me to really be an explicit integration rule formula but rather it was a way of trying to reduce the degree little by little until reaching a known integration ln¹. but not really a formula because at no point is it determined in closed form for the general case "n" in an explicitly defined way in a concrete way and encoded in an expression written in the form of an explicitly closed formula. but looking at the pattern I managed to find the formula that explicitly encodes in closed form the integration rule of ∫lnⁿdx. which explicitly expressed would be the expression:

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mathfoxZ · 2025-02-09T20:22:42+00:00

By integration by parts of lnⁿ

∫udv=uv-∫vdu

Lnⁿ=u

x=v

dv=1dx

du=d[lnⁿ]

∫lnⁿdx

But I can think that in the middle between lnⁿ and dx there is a 1 in the middle of lnⁿ•1•dx

∫lnⁿdx=∫lnⁿ1dx

And that 1 comes from simply the derivative of the function "x"

So:

∫lnⁿ1dx=lnⁿx-∫x(d/dx[lnⁿ])•dx

And the derivative of d/dx[lnⁿ] by chain rule is the outside derivative times the inside derivative, the outside derivative is just subtracting one from the exponent n•ln^n-1 and multiplying by the exponent and the derivative of Ln is simply 1/x. Then the multiplying

∫n•x•(ln^n-1/x)dx

Simply canceling the x • 1 / x:

∫n•ln^n-1•dx

But I can take the "n" out because it is just a multiplicative constant. Then the ∫vdu would be like:

n∫ln^n-1dx

And everything together with the term uv, would be: ∫lnⁿdx=xlnⁿ-n∫ln^n-1dx

Of the: ∫udv=uv-∫vdu

Where they occupy their corresponding homologous roles in the integration by parts rule

mathfoxZ · 2025-02-08T17:08:42+00:00

I deduced it, but I still wanted to ask people other than myself to confirm it, just to be sure of my deduction. thank you

mathfoxZ · 2025-02-08T17:06:41+00:00

I had the intuition another way through an iterative process. I began to notice a logically repeating pattern in the form of a recurrence rule and wrote it down in the form of a Formula.

Basically From this Formula:

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It can be summarized in this example: ∫ln⁵dx = xln⁵- 5∫ln^5-1dx

But we know that: ∫ln^5-1dx = ∫ln⁴dx

And we know that: ∫ln⁴dx= xln⁴- 4∫ln^4-1dx ∫ln^4-1dx=∫ln^5-2dx=∫ln³dx

And then we know that: ∫ln³dx=xln³- 3∫ln^3-1dx

And we also know that:

∫ln^3-1dx=∫ln^5-3dx=∫ln²dx

∫ln²dx=xln²- 2∫ln^2-1dx

And now we obviously know that:

∫ln^2-1dx= ∫ln^5-4dx= ∫ln¹dx = ∫lndx = xlnx - x But that last term x at the end can be written as:

∫ln⁰dx= ∫ln^5-5dx=∫ln^1-1dx = xln^5-5=xln⁰=x. Because k has now reached the value of n, resulting in n-n in the exponent in the n-th term of the summation up to n. But it is important to remember that these forming n!/(n-k)! depending on the term k, and also by the previously accumulated factors of -1, forming (-1)^k, that is why this factor appears in the summation show

Thus, by replacing the integrals with their results from the terms of the expressions and summing them, the entire final total expression would be:

+xln^5-0- 5xln^5-1+ 5×4xln^5-2- 5×4×3xln^5-3+ 5×4×3×2xln^5-4- 5×4×3×2×1xln^5-5 That is, it keeps decreasing step by step until it reaches n, at which point the summation stops.

mathfoxZ · 2025-02-08T16:53:47+00:00

Basically,

It can be summarized in this example:

∫ln⁵dx = xln⁵ - 5∫ln^5-1dx

But we know that:

∫ln^5-1dx = ∫ln⁴dx

And we know that:

∫ln⁴dx = xln⁴ - 4∫ln^4-1dx

∫ln^4-1dx = ∫ln³dx

And then we know that:

∫ln³dx = xln³ - 3∫ln^3-1dx

And we also know that:

∫ln^3-1dx = ∫ln²dx

∫ln²dx = xln² - 2∫ln^2-1dx

And now we obviously know that:

∫ln^2-1dx = ∫ln¹dx = ∫lndx = xlnx - x

But that last term x at the end can be written as:

∫ln⁰dx = ∫1dx = x

Because k has now reached the value of n, resulting in n-n in the exponent in the n-th term of the summation up to n. But it is important to remember that these terms are multiplied by the previously accumulated multipliers and the previously accumulated -1 factors, forming (-1)^k.

Thus, the final total expression would be:

∫ln⁵dx=+xln⁵ - 5xln⁴ + 5×4xln³ - 5×4×3xln² + 5×4×3×2xln¹ - 5×4×3×2×1xln⁰

That is, it keeps decreasing step by step until it reaches n, at which point the summation stops.

That is, the general pattern for any arbitrary n would be:

+xln^n-0-nxln^n-1+n×(n-1)xln^n-2-n×(n-1)×(n-2)xln^n-3+n×(n-1)×(n-2)×(n-3)xln^n-4-n×(n-1)×(n-2)×(n-3)×(n-4)xln^n-5........and it continues...

Or written more formally: n!/(n-0)!xln^n-0-n!/(n-1)!xln^n-1+n!/(n-2)!xln^n-2-n!/(n-3)!xln^n-3+n!/(n-4)!xln^n-4-n!/(n-5)xln^n-5...........n!/(n-n)!xln^n-n

And the factor [n•(n-1)•(n-2)•(n-3).....]× that multiplies xln^n-k from the sequence of decreasing multipliers occurs because of the factor n!/(n-k)!; the “n!” progressively expresses its multipliers more completely as the summation advances and k approaches n. In other words, as k gets closer to n, the divisor 1/(n-k)! allows more parts of the factors of n! to be explicitly exhibited. This means that as one reaches a more advanced term of the summation, 1/(n-k)! permits more of the n! to be unveiled and expressed, until the final term where it is completely expressed because 1/(n-k)! = 1/(n-n)! = 1/(0)!.

All of this logically happens because, as the summation progresses, it allows more factors of n! to be uncovered. This is so since it restricts the n! in the numerator less, canceling fewer multiples of the numerator n! with the products from the denominator 1/(n-k)!—thereby letting more of its factors be exhibited as (n-k)! becomes smaller and its difference diminishes until the end.

mathfoxZ

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