I don't know the skill level of everyone on this forum, but this is a relatively easy olympiad problem for you to try:
Find all real valued functions f such that:
[f(x)+f(y)][f(u)+f(v)]=f(xu-yv)+f(xv+yu).
Find all real valued functions f such that:
[f(x)+f(y)][f(u)+f(v)]=f(xu-yv)+f(xv+yu).
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jerry811215
Re: Functional Equation
2007-02-06 17:44isomorphism
Re: Functional Equation
2007-02-06 19:00Usually, with functional equations, it is nice to play around, conjecture, and then prove.
isomorphism
Re: Functional Equation
2007-02-26 19:22First of all, letting x=y=u=v=0, we get that 4f(0)^2=2f(0). This means that f(0)=0 or f(0)=1/2.
If f(0)=1/2 then letting x=y=0 shows trivially that f(x) is constant and equals 1/2 for all x.
Now see what you can derive under the condition that f(0)=0.
isomorphism
Re: Functional Equation
2007-02-26 19:23(Well, there were 2 warm-up problems beforehand.)
aa963854116
Re: Functional Equation
2007-03-07 18:54i. f(x.) = 0
ii. f(x) = 1/2
iii. f(x) = x^2
jerry811215
Re: Functional Equation
2007-03-07 19:33So, everybody good luck!!
p.s. I think i will participate in the test in tainan , since the test won't be held in chia-yi!!
jerry811215
Re: Functional Equation
2007-03-07 19:37My math still need to be improved !!!
aa963854116
Re: Functional Equation
2007-03-08 19:57I think I just had a great fortune on the test last year.
(ps. The questions were also easier last year.)
isomorphism
Re: Functional Equation
2007-03-11 10:29About the AIME, last year's test indeed was easier. I'm aiming for a 14+ this year (even though a 7 will probably get me into USAMO).