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first of all, I searched that question, could not find any.
If by any chance you find, I ask you, "Dont close thread cus of duplicate, because I want to understand myself".
I have this question: With induction, proof the following this: $fᵤ(x) = f(· · ·(f(x)))$
When:
$$f(x)=-\frac{x}{1+x}$$
and $x≠-1$ and I get:
$f_n(x) = f(\cdots(f(x))\cdots)$ which equals ( = ) $n$

I know how to do induction. but, my problem is understand what I am supposed to do here.
Can I get any tip? not an answer, only a tip.
I tried doing something like that:
$f((f(x)) $= ( put the X of $f(x)$ in here).
got: $f((f(x)) = 1/x$
after it, tried again, for:
$f(f(f(x)))$ = and when X, put the x of $f(f(x))$, got something else.

tried doing base induction, $n=0$ ( thats the examples I got before of $f(f(x))$ and $f(f(f(x)))$

but something really is weird in this question.. Will be happy to get a tip. thanks!

Jean Marie
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    The point of closing questions that are duplicates is not to prevent you from understanding for yourself. It's just that, if the answers to an existing question X are not enough for you, the question you should be asking is not the same question X again, but rather "I did not understand this part of the answers to question X. How does it work?" – Misha Lavrov Nov 02 '21 at 20:41
  • @MishaLavrov I understand what you say, but anyway, this question doesnt have it here, its a pretty much unique one ( you may search and see ) - if you find, tell me what you wrote in the search, I wrote many types of fn(x) induction and other similliar stuff. Anyway, I had a question lately, searched all here, found many explanations, but nothing I could understand, in no part... ( the induction I found was not according to how I learn )... I couldn't ask specifically.. thats why I prefer in my own post to ask my own questions according ot my own experience. –  Nov 02 '21 at 20:47

1 Answers1

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Your calculation of $f(f(x))$ is not correct. $$f(x)=\frac {-x}{1+x}$$ we get $$f(f(x))=\frac {\frac {x}{1+x}}{1-\frac {x}{1+x}}=x$$

The pattern should be clear, so you can take it from here.

Mohammad Riazi-Kermani
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  • wait what? how is it possible? I will calculate it once again. The pattern, as I understand is that I should get, if n=2, then I get instead of x, I will get 2, that is what I understand. but there is the x also, I mean, its or x or n that I can delete, but I still stay with x Or n ( one of them ). How am I supposed to get rid of it? anyway, I will try it for a minute and update if I am still stuck, thanks. –  Nov 02 '21 at 21:30
  • The pattern is $f(x),x,f(x),x,.....$ – Mohammad Riazi-Kermani Nov 02 '21 at 22:20
  • Hi, thanks alot! I will try it in a few more time, I wanna see if I manage to do it by myself from this point on. thanks again. –  Nov 02 '21 at 22:31
  • Good, try it and you will get it. – Mohammad Riazi-Kermani Nov 02 '21 at 22:32
  • Okay, you are right, the pattern is 1,3,5,7 - x 2,4,6,8 - the f(x) I still dont understand the fn(x). what does n mean? if n=1: f1(x) is f(x)? if n=2 f2(x) is f(f(x))? if n=3 f3(x) is f(f(fx))))? or its something else? –  Nov 03 '21 at 11:19
  • All I understand, if fn(x) = n, it means thats its the 1,3,5,7 part. But if its fn+1(x) then it willb e the 2,4,6,8 part... but thats it, nothing more nothing less... –  Nov 03 '21 at 11:22
  • Okay, I managed to do it I think. I got to a solution like that ( I dont know if its a solution ), showed that if n=k, and I realized n is even, so f2k(x) = x, f2k+1(x) = the function. and also in f2k+2(x) = x ( even (, and f2k+3(x) = the function. Is that the point? mentioning I got help from a teacher, but still, the question is not very clear... –  Nov 03 '21 at 19:19
  • You have the right idea about odds and even iteration. Now you prove it by induction. – Mohammad Riazi-Kermani Nov 03 '21 at 21:19
  • What do you mean? there aint any way to prove by induction. I did it, I mean. I showed that ( what I said above: ) "howed that if n=k, and I realized n is even, so f2k(x) = x, f2k+1(x) = the function. and also in f2k+2(x) = x ( even (, and f2k+3(x) = the function" –  Nov 03 '21 at 22:40