# Find the limit of $f(n)$ where $f(n)=(2-f(n+1))^2$

Mathematics Asked by Amit Zach on December 2, 2020

Let $$f:mathbb{N}to[0,2]$$ such that $$f(1)=2$$ and:

$$f(n)=(2-f(n+1))^2qquadforall ninmathbb{N}$$

Prove that the limit of $$f(n)$$ as $$ntoinfty$$ exists and show it is equal to $$1$$; namely:

$$lim_{ntoinfty}f(n)=1$$

I couldn’t figure out how to prove the limit exists. Usually when I encounter these recursion relations I try to prove that $$f(n)$$ is bounded and monotonic, but after substituting some values of $$n$$ I realized (mayble falsley) that $$f(n)$$ is not monotonic at all. I’d be happy to hear your thoughts.

• Assuming the limit exists, I think I know how to prove it is equal to $$1$$ – All I need to do is to solve the equation $$L=(2-L)^2$$. The solutions are $$L=4$$ (not possible since $$f(n)leq2$$) and $$L=1$$ (which is the solution).

Thanks

Edit: I did manage to show that if for some $$kinmathbb{N}, f(k+1)geq f(k+2)$$, then also $$f(k+1)geq f(k)$$. This is because:

$$f(k+1)geq f(k+2)iff2-sqrt{f(k)}geq2-sqrt{f(k+1)}iff f(k)leq f(k+1)$$

Similarly, if $$f(k+1)leq f(k+2)$$, then also $$f(k+1)leq f(k)$$. This means that for odd values of $$n$$, $$f(n)$$ is monotonically decreasing, and for even values of $$n$$, $$f(n)$$ is monotonically increasing. $$f(n)$$ is obviously bounded, therefore, both the sequences $$f(2n)$$ and $$f(2n+1)$$ have a limit. The only problem I have know, is that I have to show they approach the same limit. In order to prove that, I thought maybe I should solve the following equation for $$L$$:

$$f(n+2)=2-sqrt{2-sqrt{f(n)}}implies L=2-sqrt{2-sqrt{L}}$$

But I don’t think I can.

It is easy to see $$f(n+1)=2-sqrt{f(n)}$$ and hence $$|f(n+1)-1|=|1-sqrt{f(n)}|=frac{|f(n)-1|}{1+sqrt{f(n)}}.$$ It is not hard to see $$2-sqrt2le f(n)le 2$$ and so $$|f(n+1)-1|=frac{|f(n)-1|}{1+sqrt{f(n)}}lefrac{1}{3-sqrt2}|f(n)-1|.$$ Thus $$|f(n)-1|le frac{1}{(3-sqrt2)^{n-1}}|f(1)-1|.$$ Letting $$ntoinfty$$ gives $$lim_{ntoinfty}f(n)=1.$$

Correct answer by xpaul on December 2, 2020

hint

$$f(n)-1=(2-f(n+1))^2-1$$

$$=(1-f(n+1))(3-f(n+1))$$

and

$$frac 13 le frac{1}{3-f(n+1)}<1$$

Answered by hamam_Abdallah on December 2, 2020