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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?

- Thread starter mathlete
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- #1

- 148

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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

Hurkyl

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

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Yes, sorry, for all x... and I'm afraid I don't understand what you mean by my favorite value of x (x=0?)Hurkyl said:It has to be true forallx, right? What about your favorite value of x?

- #4

matt grime

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Question word for word...matt grime said:

"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?

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The Euler Formula expansion is key.

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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?I'm afraid I don't understand what you mean by my favorite value of x (x=0?)

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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 said:No it's not. You just have to use the fact it's true for allx.

The equation is true for x=0, right? What does the equation look like when x=0?

- #9

Hurkyl

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If so, then this does give you a shortcut

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