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How is that always proven?

  1. Aug 25, 2013 #1
    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?

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  2. jcsd
  3. Aug 25, 2013 #2

    haruspex

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    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.
     
  4. Aug 25, 2013 #3

    LCKurtz

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    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##.
     
  5. Aug 26, 2013 #4
    But didn't that use the restriction that x->0 and y->1? How did it prove it in general?
     
  6. Aug 26, 2013 #5
    I didn't see this reply at first.

    But if it holds for the limit, how does it prove it in general?
     
  7. Aug 26, 2013 #6

    D H

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    That [itex]\exp(-\rho x y) = 1/((1+\rho x)(1+\rho y))[/itex] 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 [itex]\exp(-\rho x y) = 1/((1+\rho x)(1+\rho y))[/itex] for all x,y>0. Finding any particular x,y that restricts rho to 0 suffices to show that rho must necessarily be zero.
     
  8. Aug 26, 2013 #7
    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.
     
  9. Aug 26, 2013 #8

    Ray Vickson

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    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.
     
  10. Aug 26, 2013 #9

    D H

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    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).
     
  11. Aug 26, 2013 #10
    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.
     
  12. Aug 26, 2013 #11

    D H

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    Do you understand the difference between "for all" and "there exists"?
     
  13. Aug 26, 2013 #12
    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?
     
  14. Aug 26, 2013 #13

    Ray Vickson

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    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!
     
  15. Aug 26, 2013 #14
    I suspect the answer lies closer to the first reply. That it's related to continuity.
     
  16. Aug 26, 2013 #15

    LCKurtz

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    @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.
     
  17. Aug 26, 2013 #16
    That makes perfect sense.

    Per the first reply, doesn't continuity play a role?
     
  18. Aug 26, 2013 #17

    LCKurtz

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    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.
     
  19. Aug 26, 2013 #18

    LCKurtz

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    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.
     
  20. Aug 26, 2013 #19

    D H

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    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.
     
  21. Aug 26, 2013 #20

    D H

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    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.
     
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