ATTENTION! Anybody can solve this challenging problem?

ryan_kidz

Junior Member
Joined
Sep 11, 2005
Messages
89
Let f(x)= 1/1-x ans let F1(x)= x

Now supposed Fn+1(x)= (foFn)x for each n= 1,2,3...

Who can solve this problem?? pretty hard right?
 
I'm sorry, but your formatting is ambiguous. Do you mean the following?

. . . . .Let f(x) = 1/(1 - x).
. . . . .Let F<sub>1</sub>(x) = x.
. . . . .Let F<sub>n+1</sub>(x) = (f(F<sub>n</sub>(x))
. . . . .for each n = 1, 2, 3, 4,....

Then you ask "Who can solve this problem?" but I'm afraid I'm not clear on just what the "problem" is...?

Eliz.
 
Because F<SUB>1</SUB>(x)=x, F<SUB>2</SUB>(x)=f(F<SUB>1</SUB>(x))=f(x)=1/[1-x].
F<SUB>3</SUB>(x)=f(F<SUB>2</SUB>(x))=f[f(x)]=1/[1-f(x)] = [x-1]/x.
F<SUB>4</SUB>(x)=f(F<SUB>3</SUB>(x))=1/[1- F<SUB>3</SUB>(x)] = x.

Now you should see a pattern! DO YOU?
 
thnx for the correction stapel i dunno know how to change the letter to become small :lol:
and thnx for trying to answer pka. anybody has different answer??
 
Subscripts are formatted using the "<sub>" and "</sub>" tags. So "F<sub>1</sub>" yields "F<sub>1</sub>". Superscripts work the same way, using the "<sup>" and "</sup>" tags.

Eliz.
 
thnx for trying to answer pka. “trying? nonsense”
I did not try to answer, I DID ANSWER THE QUESTION!

For each n,
F<SUB>n</SUB>(x)=x if (n mod 3)=1;
F<SUB>n</SUB>(x)= 1/[1-x] if (n mod 3)=2;
F<SUB>n</SUB>(x)= [x-1]/x if (n mod 3)=0.
THAT IS THE ANSWER.
 
hehehe... sorry! just kidding pka! yeah.. thnx alot for ur answer u right!

u r so smart.. but can u find F2005 (2005)=

if u can, u r truly smart
 
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