Seemingly simple proof I can't seem to get

[mathlete]

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?

 

[Hurkyl]

It has to be true for all x, right? What about your favorite value of x?

 

[mathlete]

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?)

 

[matt grime]

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

 

[mathlete]

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?

 

[Stephan Hoyer]

The Euler Formula expansion is key.

 

[Hurkyl]

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?

 

[Stephan Hoyer]

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.

 

[Hurkyl]

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!