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I am struggling to see the great revelation. I am not a mathematician, but in the discrete domain of integers: 1) you have two functions, essentially we look

I am struggling to see the great revelation. I am not a mathematician, but in the discrete domain of integers: 1) you have two functions, essentially we look for where the two 3d space surfaces intersect, if z is taken as the result for each equation. 2) the integers "are disctinct", so (0,0) and (1,1) are out, plus (2,2) (3,3) etc. Basically a whole linear diagonal in the instersection of both 3d spaces is excluded (why though, to what useful end?) 3) Starting points for the ranges is therefore (0,1) and (1,0) 4) 1^y is always 1, and x^0 is always 1 so there is a constant starting value of 1 both on both axis 5) but x^y will always be larger than y^x, for y>x and x^y > y^x (prove this by taking the first derivative to get rate of change. Do you use Laplace z domain for discrete, instead of s for continuous?) 6) and the converse to 5. So once you have found one solution, you know to stop looking, the two surfaces keep diverging from each other. Why resort to the continuous domain to solve a problem n the discrete, is this even a valid approach? eg Is there a formal proof that says integers strictly follow that same rules as the continuous domain, just as a subset? I'm interested. Does this come under group theory, a set with an applied operation? As I said I am not a mathematician, I am an electrical engineer so probably one of the worst abusers of pure math in a formal sense, but the more I think about this the more questions this raises in my thinking. Can someone point out errors in thinking?

hgsgm 4y ago
> I am struggling to see the great revelation. No one said there was one. OP said "interesting problem".
- airbreather 4y ago
  but interesting why?
deleted 4y ago
[deleted]
chx 4y ago
I am not a mathematician either, I always maintained the university made a mistake granting make a math teacher degree :) It was also long ago enough a lot of you weren't even alive and I haven't done any such work in a quarter century. Nonetheless... > but x^y will always be larger than y^x, for y>x and x^y > y^x (prove this by taking the first derivative to get rate of change. Do you use Laplace z domain for discrete, instead of s for continuous?) I am struggling to follow what are you saying here > Why resort to the continuous domain to solve a problem n the discrete, is this even a valid approach? I can't make heads or tails of this question. The proof says, correctly, for all (real) x != y, x < y solutions it is true that 0<x<e and e<y. He found this by investigating the derivative and establishing the monotonically increasing / decreasing nature of the function. Only after finding this out using does he go back to the original question: what integer x could be? Since he restricted x to be 0 < x < e , we only need to investigate the case of 1 and 2.
zmgsabst 4y ago
A subset won’t have more solutions than its containing set; but it may have fewer. Many problems are easier to solve in the reals (due to being complete), and you can then restrict that solution to your (sub)set of interest — in this case, the integers. You see the same thing with Pythagorean triples being simpler to solve by doing the math over the complex numbers and then restricting your answers.
Sniffnoy 4y ago
> 5) but x^y will always be larger than y^x, for y>x and x^y > y^x (prove this by taking the first derivative to get rate of change. Do you use Laplace z domain for discrete, instead of s for continuous?) This is false, though! Or, it's almost always true, but there are some exceptions. We of course have an exception at (2,4) and (4,2), where they're equal, and of course at (n,n), where they're obviously equal. And of course 0's and 1's will cause problems for you. But also, most interestingly, there's an exception at (2,3) and (3,2)! 2<3, and yet, 3^2 > 2^3. Any proof has to account for this! (There's a fair bit I could say about this exception, but perhaps I should just let you think about it instead. :) ) > Why resort to the continuous domain to solve a problem n the discrete Because oftentimes this is easier. (In a number of cases it's much easier.) Also... you did this? Like to the extent that your step (5) is valid, you seem to have "proven" it by using the first derivative. That's a continuous tool! I'm not sure what you're talking about with the "Laplace z domain". Or are you using "first derivative" to mean "first difference", or something? > is this even a valid approach? Yes, why wouldn't it be? In fact this is actually one of the big reasons for introducing larger number systems, that they let you prove things about the original smaller number system. The rational numbers let you prove things about the integers; the complex numbers let you prove things about the reals; the real numbers and p-padic numbers let you prove things about the rationals; etc. (The integers let you prove things about the whole numbers!) Proving statements about integers by means of complex numbers is a whole field in itself, i.e., analytic number theory. And in the case of Goodstein's theorem, one famously proves a statement about the whole numbers by passing to the ordinals...
- airbreather 4y ago
  "The Z-transform is a mathematical tool which is used to convert the difference equations in discrete time domain into the algebraic equations in z-domain." from https://www.tutorialspoint.com/differentiation-in-z-domain-property-of-z-transform https://www.tutorialspoint.com/differentiation-in-z-domain-p.... I was taught in engineering math that z-domain is the Laplacian discrete equivelent of the s domain, which is continuous and used by EE as well in analog. z^-1 (or z(-1) means last sample. It is commonly used in digital signal processing, FFT etc. s domain is used for (from memory) the operater e^-jw, which is for electrical engineers the transform for use with sine waves, such that impedance is 1/sC for capacitance and sL for inductance. z domain has useful properties like "The differentiation in z-domain property of Z-transform states that the multiplication by n in time domain corresponds to the differentiation in zdomain." I am sure I have made some technical mistakes in the above, but it is how I remember it and they don't impact my ability to apply it for my limited EE needs. As to the question about the valid approach, intuitively I see this, but wondering if there was a formal proof of some kind, or is it taken as given?
armeehn 4y ago
I'm trying to understand 5). If you're claiming that both (A) y > x, and (B) x^y > y^x hold, then x^y > y^x ("will always be larger") holds. You're satisfying your claim by assumption. Nothing new is deduced. However, if only A) needs to be satisfied: y = 3, x = 2 is a counterexample, as x^y = 2^3 = 8 < 9 = 3^2 = y^x. edit: looks like someone had the same thought as me as I was typing my reply!
- Ensorceled 4y ago
  Your comment goes to the heart of the matter "claim by assumption". It's hilarious that the FIRST two integers greater than one form a counter example to their "proof".
colgate_total 4y ago
> eg Is there a formal proof that says integers strictly follow that same rules as the continuous domain, just as a subset? I'm interested. It may have come as a surprise to a programmer, but in the domain of math, the integers are indeed a subset of a real numbers.
bheadmaster 4y ago
> Basically a whole linear diagonal in the instersection of both 3d spaces is excluded (why though, to what useful end?) Because if m=n, then n^m=m^n is trivially true.
- airbreather 4y ago
  Yes, m=n is a solution, or set of solutions, but why arbitrarily exclude some solutions and not others? There is no explanation as to why - eg maybe it could be if the equations represent a physical system and m=n implied the same physical space was occupied by two objects, for want of a better example. But if the equations represented some kind of abstract concept, why are not all solutions valid and of interest? It seems like me saying, find an equation that only yields all the primes that use any digit once - but why, to what end and value?
- Ensorceled 4y ago
  This is when I stopped reading the original comment.
Someone 4y ago
> but x^y will always be larger than y^x, for y>x and x^y > y^x That’s trivially true, as that last condition equals the claim: but x^y > y^x, for y>x and x^y > y^x ^^^^^^^^^ ^^^^^^^^^ Also, if you leave out the and x^y > y^x part, that isn’t generally true because, for example 2⁴ = 4² and 1² < 2¹. Another, minor, point: you write > Why resort to the continuous domain to solve a problem n the discrete and > essentially we look for where the two 3d space surfaces intersect Those two are in conflict with each other.

hgsgm 4y ago

> I am struggling to see the great revelation. No one said there was one. OP said "interesting problem".

airbreather 4y ago
but interesting why?

deleted 4y ago

[deleted]

chx 4y ago

I am not a mathematician either, I always maintained the university made a mistake granting make a math teacher degree :) It was also long ago enough a lot of you weren't even alive and I haven't done any such work in a quarter century. Nonetheless... > but x^y will always be larger than y^x, for y>x and x^y > y^x (prove this by taking the first derivative to get rate of change. Do you use Laplace z domain for discrete, instead of s for continuous?) I am struggling to follow what are you saying here > Why resort to the continuous domain to solve a problem n the discrete, is this even a valid approach? I can't make heads or tails of this question. The proof says, correctly, for all (real) x != y, x < y solutions it is true that 0<x<e and e<y. He found this by investigating the derivative and establishing the monotonically increasing / decreasing nature of the function. Only after finding this out using does he go back to the original question: what integer x could be? Since he restricted x to be 0 < x < e , we only need to investigate the case of 1 and 2.

zmgsabst 4y ago

A subset won’t have more solutions than its containing set; but it may have fewer. Many problems are easier to solve in the reals (due to being complete), and you can then restrict that solution to your (sub)set of interest — in this case, the integers. You see the same thing with Pythagorean triples being simpler to solve by doing the math over the complex numbers and then restricting your answers.

Sniffnoy 4y ago

> 5) but x^y will always be larger than y^x, for y>x and x^y > y^x (prove this by taking the first derivative to get rate of change. Do you use Laplace z domain for discrete, instead of s for continuous?) This is false, though! Or, it's almost always true, but there are some exceptions. We of course have an exception at (2,4) and (4,2), where they're equal, and of course at (n,n), where they're obviously equal. And of course 0's and 1's will cause problems for you. But also, most interestingly, there's an exception at (2,3) and (3,2)! 2<3, and yet, 3^2 > 2^3. Any proof has to account for this! (There's a fair bit I could say about this exception, but perhaps I should just let you think about it instead. :) ) > Why resort to the continuous domain to solve a problem n the discrete Because oftentimes this is easier. (In a number of cases it's much easier.) Also... you did this? Like to the extent that your step (5) is valid, you seem to have "proven" it by using the first derivative. That's a continuous tool! I'm not sure what you're talking about with the "Laplace z domain". Or are you using "first derivative" to mean "first difference", or something? > is this even a valid approach? Yes, why wouldn't it be? In fact this is actually one of the big reasons for introducing larger number systems, that they let you prove things about the original smaller number system. The rational numbers let you prove things about the integers; the complex numbers let you prove things about the reals; the real numbers and p-padic numbers let you prove things about the rationals; etc. (The integers let you prove things about the whole numbers!) Proving statements about integers by means of complex numbers is a whole field in itself, i.e., analytic number theory. And in the case of Goodstein's theorem, one famously proves a statement about the whole numbers by passing to the ordinals...

airbreather 4y ago
"The Z-transform is a mathematical tool which is used to convert the difference equations in discrete time domain into the algebraic equations in z-domain." from https://www.tutorialspoint.com/differentiation-in-z-domain-property-of-z-transform https://www.tutorialspoint.com/differentiation-in-z-domain-p.... I was taught in engineering math that z-domain is the Laplacian discrete equivelent of the s domain, which is continuous and used by EE as well in analog. z^-1 (or z(-1) means last sample. It is commonly used in digital signal processing, FFT etc. s domain is used for (from memory) the operater e^-jw, which is for electrical engineers the transform for use with sine waves, such that impedance is 1/sC for capacitance and sL for inductance. z domain has useful properties like "The differentiation in z-domain property of Z-transform states that the multiplication by n in time domain corresponds to the differentiation in zdomain." I am sure I have made some technical mistakes in the above, but it is how I remember it and they don't impact my ability to apply it for my limited EE needs. As to the question about the valid approach, intuitively I see this, but wondering if there was a formal proof of some kind, or is it taken as given?

armeehn 4y ago

I'm trying to understand 5). If you're claiming that both (A) y > x, and (B) x^y > y^x hold, then x^y > y^x ("will always be larger") holds. You're satisfying your claim by assumption. Nothing new is deduced. However, if only A) needs to be satisfied: y = 3, x = 2 is a counterexample, as x^y = 2^3 = 8 < 9 = 3^2 = y^x. edit: looks like someone had the same thought as me as I was typing my reply!

Ensorceled 4y ago
Your comment goes to the heart of the matter "claim by assumption". It's hilarious that the FIRST two integers greater than one form a counter example to their "proof".

colgate_total 4y ago

> eg Is there a formal proof that says integers strictly follow that same rules as the continuous domain, just as a subset? I'm interested. It may have come as a surprise to a programmer, but in the domain of math, the integers are indeed a subset of a real numbers.

bheadmaster 4y ago

> Basically a whole linear diagonal in the instersection of both 3d spaces is excluded (why though, to what useful end?) Because if m=n, then n^m=m^n is trivially true.

airbreather 4y ago
Yes, m=n is a solution, or set of solutions, but why arbitrarily exclude some solutions and not others? There is no explanation as to why - eg maybe it could be if the equations represent a physical system and m=n implied the same physical space was occupied by two objects, for want of a better example. But if the equations represented some kind of abstract concept, why are not all solutions valid and of interest? It seems like me saying, find an equation that only yields all the primes that use any digit once - but why, to what end and value?
Ensorceled 4y ago
This is when I stopped reading the original comment.

Someone 4y ago

> but x^y will always be larger than y^x, for y>x and x^y > y^x That’s trivially true, as that last condition equals the claim: but x^y > y^x, for y>x and x^y > y^x ^^^^^^^^^ ^^^^^^^^^ Also, if you leave out the and x^y > y^x part, that isn’t generally true because, for example 2⁴ = 4² and 1² < 2¹. Another, minor, point: you write > Why resort to the continuous domain to solve a problem n the discrete and > essentially we look for where the two 3d space surfaces intersect Those two are in conflict with each other.