# Thread: quadratic equ. problem no.4 quadratic equ. problem no.4

1. ## quadratic equ. problem no.4 quadratic equ. problem no.4

quadratic equ. problem no.4

2. 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 $x=y \Rightarrow f(x) = f(y)$

So if f(x) = x for some x,
Then by AOS, f(f(x)) = f(x) = x for those x

3. Originally Posted by Isomorphism
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
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..

$f(x) - x = 0 \implies f(x) = x$

now, from your AOS, $f(f(x)) = f(x) = x$
therefore, $f(f(x)) - x = 0$

Originally Posted by Isomorphism
The idea is functions obey "axiom of substitution",(AOS)
that is if $x=y \Rightarrow f(x) = f(y)$

So if f(x) = x for some x,
Then by AOS, f(f(x)) = f(x) = x for those x
hehe, you call it ""axiom of substitution",(AOS)".. first time to see/hear it.. Ü i simply say "function is well-defined"..

4. Originally Posted by kalagota
in fact, he is right? didn't you notice it?

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

5. 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 $\alpha$ to $f(f(x)) - x$ and show that it is equal to 0..

here it is..
substitution: $f(f(\alpha)) - \alpha$

since $\alpha$ is a root of $f(x) - x = 0$, we have $f(\alpha) - \alpha = 0 \implies f(\alpha) = \alpha$.

thus, $f(f(\alpha)) - \alpha = f(\alpha) - \alpha = 0$.

therefore $\alpha$ is a root of $f(f(x)) - x = 0$.. (similarly, $\beta$ is a root.)

for [b], this is a hint.. $f(x + y) \neq f(x) + f(y)$..
for example, $f(x) = a^x$ and $f(1 + 2) = a^{1+2} = a^3 = a^1a^2 \neq a^1 + a^2 = f(1) + f(2)$

got it? so, your [b] should be $f(f(x)) - x = f(x^2 - 3x + 2) - x = (x^2 - 3x + 2)^2 - 3(x^2 - 3x + 2) + 2 - x$

just expand it and simplify..

6. 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 $x^2 - 4x + 2 = 0 \Rightarrow (x-2)^2 = 2 \Rightarrow x = 2 \pm \sqrt 2$.

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 $x^2 - 4x + 2$, you will get a quadratic quotient. Solve that quadratic to get the remaining 2 roots

Originally Posted by kalagota
for [b], this is a hint.. f(x + y) \neq f(x) + f(y)..
for example, [/tex]f(x) = a^x [/tex] and f(1 + 2) = a^{1+2} = a^3 = a^1a^2 \neq a^1 + a^2 = f(1) + f(2).
Hey I did not get this... Is it some slick solution with tremendous insight???
Thanks

7. Originally Posted by Isomorphism
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 $x^2 - 4x + 2 = 0 \Rightarrow (x-2)^2 = 2 \Rightarrow x = 2 \pm \sqrt 2$.

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 $x^2 - 4x + 2$, 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
which one you didn't get?

8. Originally Posted by kalagota
which one you didn't get?
What is the connection with $f(x) + f(y) \neq f(x+y)$ and the problem??

9. Originally Posted by Isomorphism
What is the connection with $f(x) + f(y) \neq f(x+y)$ and the problem??
oic.. in fact, nothing.. there is no connection..

i'm just wondering where the "2" go after composition..