Define functions f,g,h on Z x Z x Z as follows:?

Question

Define functions f,g,h on Z x Z x Z as follows:?

Answer 1

i've got no longer afflicted to be taught....is the unit step function H(x) seen "consumer-friendly?" i'm fairly specific it has a power series expansion that converges nicely sufficient. i'm going to apply the convention that H(0) = 0. yet besides, forgive the complexity... F(x,y,z) = x*{H(x-y) + H(x-z) - H(y-z) - H(z-y) } + y*{H(y-z) + H(y-x) - H(x-z) - H(z-x) } + z*{H(z-y) + H(z-x) - H(y-x) - H(x-y) } interior the (perchance) trivial case that x = y = z , F(x,y,z) could return 0, perfect seeing that NONE are correct. EDIT, ok i presumed-approximately yet in any different case this could behave badly, will behave badly....forgive much extra complexity. F(x,y,z) = x*H{ H(x-y) + H(x-z) - H(y-z) - H(z-y) } + y*H{ H(y-z) + H(y-x) - H(x-z) - H(x-z) } + z*H{ H(z-y) + H(z-x) - H(y-x) - H(x-y) } optimistically that association of H()'s will return one million provided that the respective variable is the terrific, on no account -one million

Answer 2

The question seems to say that any Pythagorean triplet can be obtained this way uniquely. I can show that not all Pythagorean triplets can be so obtained. (I haven't addressed the uniqueness question.) Here's my proof.



Let P(a,b) be the Pythagorean triplet



x = a^2 - b^2

y = 2ab

z = a^2 + b^2.



Then it is easy to show that



f(P(a,b)) = P(2a+b,a)

g(P(a,b)) = P(2a-b,a)

h(P(a,b)) = P(2b+a,b).



Forget the P, just think of f,g,h as the linear transformations



f(a,b) = (2a+b,a), etc.



Now the original triplet (5,4,3) is the (a,b) pair (2,1). It is not hard to show that any (k,1) for k>=1 can be obtained from f,g,h. But I claim the (a,b) pair (3,2), that is, the triplet (5,12,13) cannot be obtained from (2,1) by applying f,g,h.



Prove by induction that if (a,b) is in the range of f,g,h starting from (2,1) then a>=2, a>=b>=1. Assume this condition. Then apply f or g or h, and use the formulas above to derive the conclusion for the new pair f or g or h of (a,b).



Now suppose (3,2) is in the range of f,g,h from (2,1). If the last function is f then for some (a,b)



2a+b=c and

a=2.



Therefore b=1 or 2 by the above inductively proven statement. Similarly for g. Now suppose h was last to be applied. Then for some (a,b)



2b+a=c and

b=2.



So c=a+4. So c cannot be 3 since a cannot be -1.



Added later: I now see how to prove uniqueness for a pair (x,y) in the range of f,g,h starting with (2,1). Look at the three formulas:



(x,y) = f(a,b)



(x,y) = g(a,b)



(x,y) = h(a,b).



Using the above equations, one can solve for a and b in each case. Recall that a must be >= b and both must be positive. This is what will tell us which of the three conditions apply.



If it's g, then b= 2y-x. Otherwise in the other two cases a or b is x-2y. So if 2y-x>=0 then the previous function must have been g. If 2y-x<0 it was f or h.



Now assume 2y-x<0. In the f case a=y and in the h case b=y. So if y>=x-2y it must have been f. If y


So an (x,y) pair which is made up of f,g,h from (2,1) can be traced back uniquely to (2,1), peeling off each application of f or g or h by using the above criteria.



So the application of f,g,h is unique.



One final point: Pythagorean triples like 9,12,15 which are not

P(a,b) for any a,b are also not in the range of f,g,h applied to (2,1).

Answer 3

it is sufficient to prove that each of the functions f,g and h produce pythagorean triplets for any given pythagorean triplet, that is f(x,y,z) is an element of P where (x,y,z) is an element of P, P being the set of pythagorean triplets



it is trivial to prove the above; for example, consider h(x,y,z)

it is given that y^2+z^2 = x^2



now (2x-y+2z)^2+(2x-2y+z)^2 = 8x^2+5y^2+5z^2-12xy-8yz+12xz = (3x-2y+2z)^2



in fact, we can find many more such functions which produce pythagorean triplets; suppose



(ax+by+cz)^2+(mx+ny+pz)^2 = (rx+sy+tz)^2; then it follows that (a^2+m^2=r^2), etc; solving the set of resultant equations, we get all the functions that produce the set P



it is also obvious that any linear combination of them produces pythagorean triplets; there is no need for a separate proof since functions f, g, h are also in P

Answer 4

Okay, so your example doesn't satisfy the constraint x


My first step is to try and convince myself that all solutions even exist first.



My second step is to prove existence of solutions.



My last step is to prove uniqueness if uniqueness actually holds.



I will continue to think about this and add to my posting. It is a very hard problem, and the most luck you will get are likely suggestions to your problem.



In order to give me a better idea as to your problem--what are it's orignis? Is this for math research? Or is it a homework problem from a book? If so, what book is asking you this to help me guide what types of mathematics are used.

Answer 5

I will note that the first two respondents are totally misguided. Respondent #1 can't fathom that x^2 = y^2 + z^2 could be a Pythagorean triple. Respondent #2 has proven nothing - f, g, and h may propagate Pythagorean triples, but that's not the question. The question is to prove that any PPT that can be created in this fashion is created in a UNIQUE way. Meaning you'd have to show that something like:



f(g(g(h(f(g(f(5,4,3)))))))

can never coincide with

g(f(g(f(h(g(f(5,4,3)))))))



or any other such composition (not linear combination, fyi, that's the wrong terminology).



EDIT: Here is what I have so far. It's basically a complete argument, and I hope it's all helpful.



Basically, here's what happens. This is a linear map on ZxZxZ, and so we simply consider the subgroup of GL(3,Z) generated by f, g, and h.



Now, alot of that might be unknown to people. I refer you all to:

http://en.wikipedia.org/wiki/Linear_map

http://en.wikipedia.org/wiki/Generating_set_of_a_group

http://en.wikipedia.org/wiki/General_linear_group



Basically, we think of this as multiplication of matrices, rather than composition of functions. However, by arguments presented in this paper:

http://links.jstor.org/sici?sici=0002-9947(196311)109%3A2%3C313%3ANUFD%3E2.0.CO%3B2-2

we can determine that this is a noncommutative unique factorization domain (that is our goal).



I realize alot of this is way over people's heads, and that the paper I'm citing is inaccessible. But I read it and it seems to provide the algebraic machinery to decide this problem.