- #36
mtayab1994
- 584
- 0
micromass said:OK, and fof=... ??
f(x^2)=f^-1of(-x^2)
micromass said:OK, and fof=... ??
mtayab1994 said:f(x^2)=f^-1of(-x^2)
micromass said:No. How did you get this??
What is f(f(y))?? What is f(f(x^2+f(y)))? How would you simplify f(f(x^2+f(y)))=f(y-x^2)??
micromass said:Take a look at what you did in post 20.
mtayab1994 said:why is it not true?
mtayab1994 said:yea alright since we chose x=0 we got f(f(y))=0 so therefore for every x in ℝ: f(f(x))=x that's what let's us say that: f(k^2+f(f(0))=-k^2+f(0)
micromass said:Oh OK. But that's something different from what you wrote there!
You have now basically that
[tex]f(k^2)=-k^2+f(0)[/tex]
mtayab1994 said:yea i had a typo on my paper i was writing the stuff fast so what's wrong with that?
mtayab1994 said:yea and for x<0:
[tex]f(0)=f(k^{2}f(f(-k^{2}))[/tex] which then equals
micromass said:Typo here??
yea i fixed it before you posted and btw the proof is as follows:
[tex]f(x)=-x+c[tex] for c+f(0)
indeed: [tex]f(x^{2}+f(y))=x^{2}-f(y)+c[/tex]
as: [tex]f(y)=-y+c[/tex] then [tex]f(x^{2}+f(y))=-x^{2}+y-c+c[/tex]
and finally [tex]f(x^{2}+f(y))=y-x^{2}[/tex] for every x in ℝ