Proof Help: Solving Griffith's Problem 9.15 for Ae^iax + Be^ibx = Ce^icx

[pt176900]

from Griffith's, problem 9.15: 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

I'm definitely confused on where to begin my manipulation. It seems quite reasonable to meet that the constants should be equal, and the amplitudes should sum up to C but I don't know how to get there mathematically.

Can someone give me a hint to get me started?

 

[dextercioby]

HINT:make x=0.

Daniel.

 

[pt176900]

if I make x = 0 then the relation a=b=c need not hold.

By this I mean that e^n*0 = 1 for all n, hence a, b, and c can be any real number.

So basically I'd trade proving a=b=c for A+B=C

 

[pt176900]

ok so wait... I can use x = 0 to prove A+B=C (I did it by contradiction ie assume A+B != C for all x. then insert x=0 and you get A+B=C - a contradiction).

then can I use the fact that A+B = C to prove that a=b=c?

 

[dextercioby]

Nope.That x=0 will probe that A+B=C.Now u'll have to make use of the independence of the e^{ix} and e^{x} for x\in \mathbb {R}.

Daniel.

 

[pt176900]

I'm sorry but I don't understand what you mean by the independence of e^ix and e^x.

 

[dextercioby]

Ae^{ix}+Be^{x}=0\Leftrightarrow A=B=0


Daniel.