ATTENTION! Anybody can solve this challenging problem?

[ryan_kidz]

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?

 

[stapel]

I'm sorry, but your formatting is ambiguous. Do you mean the following?

. . . . .Let f(x) = 1/(1 - x).
. . . . .Let F₁(x) = x.
. . . . .Let F_n+1(x) = (f(F_n(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.

 

[pka]

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?

 

[ryan_kidz]

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??

 

[stapel]

Subscripts are formatted using the "_{" and "}" tags. So "F₁" yields "F₁". Superscripts work the same way, using the "^{" and "}" tags.

Eliz.

 

[pka]

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.

 

[stapel]

[...gives a complete solution...]

thnx for trying to answer pka.

I did not try to answer, I DID ANSWER THE QUESTION!

Indeed.

It would be nice if people actually read the answers you give them... :roll:

Eliz.

 

[ryan_kidz]

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

 

[stapel]

can u find F2005 (2005)=

Seriously, dude: Read what he gave you. :roll:

Eliz.