quadratic equ. problem no.4
No I dont think I understand the second step in your proof.
You say f(x) - x = 0
Then you say f(f(x)) - x = 0 ???
How
The idea is functions obey "axiom of substitution",(AOS)
that is if $\displaystyle x=y \Rightarrow f(x) = f(y)$
So if f(x) = x for some x,
Then by AOS, f(f(x)) = f(x) = x for those x
in fact, he is right? didn't you notice it? he already omitted something there..
if i'll just follow the argument, it would be like this..
$\displaystyle f(x) - x = 0 \implies f(x) = x $
now, from your AOS, $\displaystyle f(f(x)) = f(x) = x $
therefore, $\displaystyle f(f(x)) - x = 0 $
hehe, you call it ""axiom of substitution",(AOS)".. first time to see/hear it.. Ü i simply say "function is well-defined"..
Yes, they are the same things. But if I tell people that it is well-defined, they think it is trivial. Somehow "well-defined" is often confused with "well, defined"
But if I say it obeys axiom of substitution, it looks like functions follow some kind of law. Makes it look more important
P.S:LOL, I recently learned that term from Terence Tao's Book on Real Analysis. He builds the number system and introduces this as a basic axiom . That funny name stuck me as something important, so i use it repeatedly to demonstrate wisdom
Isomorphism,
oic.. i have not seen his book yet.. anyhow, thanks for that info..
afeasfaerw23231233,
if you say that [2] equals [1], why don't you apply your result to [b]?
i don't know if this is the proper proof, but i'll write it anyway..
substitute $\displaystyle \alpha$ to $\displaystyle f(f(x)) - x $ and show that it is equal to 0..
here it is..
substitution: $\displaystyle f(f(\alpha)) - \alpha $
since $\displaystyle \alpha$ is a root of $\displaystyle f(x) - x = 0 $, we have $\displaystyle f(\alpha) - \alpha = 0 \implies f(\alpha) = \alpha $.
thus, $\displaystyle f(f(\alpha)) - \alpha = f(\alpha) - \alpha = 0$.
therefore $\displaystyle \alpha $ is a root of $\displaystyle f(f(x)) - x = 0 $.. (similarly, $\displaystyle \beta$ is a root.)
for [b], this is a hint.. $\displaystyle f(x + y) \neq f(x) + f(y) $..
for example, $\displaystyle f(x) = a^x $ and $\displaystyle f(1 + 2) = a^{1+2} = a^3 = a^1a^2 \neq a^1 + a^2 = f(1) + f(2) $
got it? so, your [b] should be $\displaystyle f(f(x)) - x = f(x^2 - 3x + 2) - x = (x^2 - 3x + 2)^2 - 3(x^2 - 3x + 2) + 2 - x $
just expand it and simplify..
Since we already know roots of f(x)=x are also roots of f(f(x)) = x,
Solve f(x) = x, i.e. x^2 - 3x + 2 = x. That is $\displaystyle x^2 - 4x + 2 = 0 \Rightarrow (x-2)^2 = 2 \Rightarrow x = 2 \pm \sqrt 2$.
f(f(x)) - x = (x^2 - 3x + 2)^2 - 3(x^2 - 3x + 2) + 2 - x
And it is a quartic equation. But we already know two roots of it by (i), so divide the quartic by $\displaystyle x^2 - 4x + 2$, you will get a quadratic quotient. Solve that quadratic to get the remaining 2 roots
Hey I did not get this... Is it some slick solution with tremendous insight???
Thanks