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Seemingly simple proof I can't seem to get

  1. Mar 26, 2006 #1
    I have:

    Ae^(iax)+Be^(ibx)=Ce^(icx)

    I have to show a=b=c and A+B=C but I can't... i've tried some standard tricks like squaring both sides, taking the derivative, then playing around with the equations but I can't get anything to stick. Any ideas?
     
  2. jcsd
  3. Mar 26, 2006 #2

    Hurkyl

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    It has to be true for all x, right? What about your favorite value of x?
     
  4. Mar 26, 2006 #3
    Yes, sorry, for all x... and I'm afraid I don't understand what you mean by my favorite value of x (x=0?)
     
  5. Mar 27, 2006 #4

    matt grime

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    There surely have to be some more restrictions than this, otherwise A+B=0=C, a=b=anything, and c is anything else not equal to a
     
  6. Mar 27, 2006 #5
    Question word for word...

    "Suppose Ae^(iax)+Be^(ibx)=Ce^(icx), for some nonzero constants A, B, C, a, b, c, and for all x. Prove that a=b=c and A+B=C"

    Any ideas?
     
  7. Mar 27, 2006 #6
    The Euler Formula expansion is key.
     
  8. Mar 27, 2006 #7

    Hurkyl

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    No it's not. You just have to use the fact it's true for all x.

    The equation is true for x=0, right? What does the equation look like when x=0?
     
    Last edited: Mar 27, 2006
  9. Mar 27, 2006 #8
    Nevermind then... I was thinking if it had to be true for any given constant value of x that would be the way to go about it.
     
  10. Mar 27, 2006 #9

    Hurkyl

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    Actually, what do you mean by Euler's expansion? I was thinking e^(ix) = cos x + i sin x... though I don't remember if I've ever heard it named that.

    If so, then this does give you a shortcut if you know the relevant theorems... but the same shortcut works directly for the problem at hand!
     
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