# How is that always proven?

1. Aug 25, 2013

### cdux

Teacher wanted to prove that for that condition to be satisfied ρ must be zero, but the limits give the idea that it applies only when those limits are satisfied. By the way, x>0 and y>0. Is it a universal proof or only for those limits?

2. Aug 25, 2013

### haruspex

The given condition is that the equality holds for all x, y > 0. If the limits exist and the functions are continuous (e.g. limx → 0 ex = e0) then it must also hold in the limit.

3. Aug 25, 2013

### LCKurtz

I don't know what you mean by "universal proof". You have this proposed identity that looks really unlikely to be true unless $\rho=0$. To refute the identity, all you have to do is find something that implies that $\rho$ must be $0$. That argument surely works. You could undoubtedly find other arguments that would imply $\rho=0$.

4. Aug 26, 2013

### cdux

But didn't that use the restriction that x->0 and y->1? How did it prove it in general?

5. Aug 26, 2013

### cdux

I didn't see this reply at first.

But if it holds for the limit, how does it prove it in general?

6. Aug 26, 2013

### D H

Staff Emeritus
That $\exp(-\rho x y) = 1/((1+\rho x)(1+\rho y))$ is true for all x,y in the case that rho=0 is obvious. That this trivial solution for rho is the only solution in the case that one but not both of x or y is zero is also obvious.

And that is all that is needed. The claim is that $\exp(-\rho x y) = 1/((1+\rho x)(1+\rho y))$ for all x,y>0. Finding any particular x,y that restricts rho to 0 suffices to show that rho must necessarily be zero.

7. Aug 26, 2013

### cdux

I don't get this logic. If you find an x and a y that point to ρ = a then how does that prove ρ must always be a? There is something missing in the explanations in text. I'm sure it's possible that's correct but I don't see it when reading the texts here about it.

8. Aug 26, 2013

### Ray Vickson

You have an identity of the form $F(x,y,\rho) = 0$ that must hold for all x and y. For more-or-less arbitrary values of x and y it might not be easy to see what this implies about $\rho$, but by taking $x, y \to 0$ we can quite easily see that we must have $\rho = a$. Once we do have $\rho = a$ we can plug that back into F to check that, indeed, $F(x,y,a) \equiv 0,$ as desired.

9. Aug 26, 2013

### D H

Staff Emeritus
That is not what the conjecture is saying. To the contrary, for any given x,y>0 there are three real solutions (not necessarily distinct) of ρ such that eρxy=(1+ρx)(1+ρy).

10. Aug 26, 2013

### cdux

Hence it's insufficient for the needs of the exercise. It was explicitly asked to prove that "only when p=0" that expression can be true.

11. Aug 26, 2013

### D H

Staff Emeritus
Do you understand the difference between "for all" and "there exists"?

12. Aug 26, 2013

### cdux

And I'm not convinced it proves it for all. What am I missing?

I keep hearing "If I set x and y tending to those numbers ρ must be 0". Yeah, fine, that's obvious. Why does that restrict ρ to one number?

13. Aug 26, 2013

### Ray Vickson

That is NOT what it says. It says that something holds FOR ALL x,y. Of course if ρ ≠ 0 the equation F(x,y,ρ) = 0 has some (x,y) solutions; for any given value of ρ you will get a curve C(ρ) in x-y space so that for all points (x,y) in the curve, the equation is satisfied. However, that means the equation is satisfied on the curve, not on all of x-y space. If you change the value of ρ you will get a different curve. What the result is saying is that for the special value ρ = 0, and for only that value, the "curve" spreads out to become the whole of x-y space!

14. Aug 26, 2013

### cdux

I suspect the answer lies closer to the first reply. That it's related to continuity.

15. Aug 26, 2013

### LCKurtz

@cdux: You were given a proposed identity. That means it must be true for all x and y. $\rho$ is presumably a constant so it doesn't change. So if it is shown to be zero under any circumstance or argument, it is zero period. Here's a simpler example of the same idea:
$$(x + y)^2=x^2 + cxy+y^2$$Show that $c$ must b $2$. You could put, for example, $x=1,~y=1$ in both sides to see it. You could try lots of other values to see it too. Nothing but $c=2$ can work. Any other value of $c$ won't work when $x=1,~y=1$. Your "identity" is a little more complicated, but it's the same idea.

16. Aug 26, 2013

### cdux

That makes perfect sense.

Per the first reply, doesn't continuity play a role?

17. Aug 26, 2013

### LCKurtz

Yes, it plays a role in the particular argument that was given. But they could have used other arguments. Any logical argument at all that shows that $\rho=0$ would do. Just like in my simpler example, you could have used other values of x and y.

Your original problem could probably be worked by choosing particular x and y also. It's just that you would need a calculator to work it out, where the limit argument they gave is "simpler" that way.

18. Aug 26, 2013

### LCKurtz

Actually, why don't you try $x=1,~y=1$ in your problem. See that you can get $\rho=0$ out of that. It's easy.

19. Aug 26, 2013

### D H

Staff Emeritus
That's not quite right. One has to show that ρ=0 is a solution under all circumstances *and* that this is the only solution under all circumstances. The first condition is trivial to prove; just substitute ρ=0 and you get 1=1. The second condition is where the given proof comes into to play. If there is some circumstance where the only solution is the trivial solution ρ=0, then that's all one needs.

20. Aug 26, 2013

### D H

Staff Emeritus
How about ρ ≈ -1.47767 and ρ ≈ 2.51286? Both of are also solutions to eρxy=(1+ρx)(1+ρy) at x=y=1. The problem is that these non-trivial solutions for this specific x,y pair are not valid for all x,y>0. Use some other values of x and y and the equality no longer holds.