Seemingly simple proof I can't seem to get

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In summary, the equation Ae^(iax)+Be^(ibx)=Ce^(icx) is true for all x, but to solve for a, b, and c, you need to use the Euler Formula expansion.
  • #1
mathlete
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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?
 
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  • #2
It has to be true for all x, right? What about your favorite value of x?
 
  • #3
Hurkyl said:
It has to be true for all x, right? What about your favorite value of x?
Yes, sorry, for all x... and I'm afraid I don't understand what you mean by my favorite value of x (x=0?)
 
  • #4
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
 
  • #5
matt grime said:
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
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?
 
  • #6
The Euler Formula expansion is key.
 
  • #7
No it's not. You just have to use the fact it's true for all x.

I'm afraid I don't understand what you mean by my favorite value of x (x=0?)
The equation is true for x=0, right? What does the equation look like when x=0?
 
Last edited:
  • #8
Hurkyl said:
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?
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.
 
  • #9
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!
 

FAQ: Seemingly simple proof I can't seem to get

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