# Yugoslavia team selection for IMO 1987 (functions) [closed]

$$f(x)=\frac{\sqrt{2+\sqrt{2}}\;x\;+\;\sqrt{2-\sqrt2}}{-\sqrt{2-\sqrt{2}}\, x\;+\;\sqrt{2+\sqrt2}}$$.

Find: $$\underbrace {f(f(...(f(x))...).}_{1987\;times}$$

• Have you tried computing $f\circ f$, $f\circ f\circ f$ to find a pattern? – clark Nov 6 '19 at 22:09
• I have, but I probably made some mistakes in calculation, so I posted this because somebody might notice something I don't see at the moment. I remember the calculation and can't get rid of it so it stops me from continuing. I'm sure the term under the root can't be reformulated. That was another option. – Praskovya2.718281828 Nov 6 '19 at 22:20
• I see. I think if could include some of your calculations and some of your thoughts the post will be much more well received. – clark Nov 6 '19 at 22:23
• Have you tried rewriting $f$ before composing the function you want? – Andrew Chin Nov 6 '19 at 22:25
• Agree. It's pretty late in my time zone, but I'll see again in the morning. Thank you for the feedback! – Praskovya2.718281828 Nov 6 '19 at 22:25

This is probably not the intended solution, but here's a way to look at that problem using complex numbers :

For $$z=a+ib \in \mathbb{C}$$, denote $$h_z(x) = \frac{\ \ ax+b}{-bx+a}$$. Notice that composition of such functions behave like product of complex numbers : for all $$z_1, z_2 \in \mathbb{C}$$, we have

$$h_{z_1} \circ h_{z_2} = h_{z_1 \times z_2}$$

Now denote

$$c = \frac{\sqrt{2+\sqrt{2}}}{2}, \qquad s = \frac{\sqrt{2-\sqrt{2}}}{2}, \qquad \omega = c+is$$

We can rewrite your function as $$f(x)=\frac{\ \ cx+s}{-sx+c}$$, so it means $$f=h_{\omega}$$. And because of the above relation, we get

$$\underbrace{f \circ f \circ \ldots \circ f}_{1987 \text{ times}} = h_{\omega^{1987}}$$

So we need to compute $$\omega^{1987}$$. To do that, first compute that $$\omega^2 = \frac{1+i}{\sqrt{2}} = e^{i\pi/4}$$. This means that $$\omega$$ is a square root of $$e^{i\pi/4}$$, so either $$\omega = e^{i\pi/8}$$ or $$\omega = -e^{i\pi/8}$$. But since $$c \ge 0$$, we finally get $$\omega = e^{i\pi/8}$$. This enables us to compute the powers of $$\omega$$ really easily (because $$\omega^{16} = e^{i\frac{16 \pi}{8}} = 1$$ and $$\omega^{-1} = e^{-i\pi/8} = \overline{\omega}$$). By writing $$1987 = 16 \times 124 + 3$$, we finally get

$$\omega^{1987} = \omega^{3} = \omega^4 \times \frac{1}{\omega} = i \overline{\omega} = s+ic$$

And to conclude

$$\underbrace{f \circ f \circ \ldots \circ f}_{1987 \text{ times}}(x) = h_{s+ic}(x) = \frac{sx + c}{-cx + s} = \frac{\sqrt{2-\sqrt{2}}x + \sqrt{2+\sqrt{2}}}{-\sqrt{2+\sqrt{2}}x + \sqrt{2-\sqrt{2}}}$$

Note that $$\sqrt {2+\sqrt 2}\sqrt {2-\sqrt 2}=\sqrt 2$$

Which simplifies the function to $$f(x) = \frac {(1+\sqrt 2 )x+1}{-x+(1+\sqrt 2)}$$ $$f(f(x))= \frac {1+x}{1-x}$$ $$f^{(4)}(x) = \frac {-1}{x}$$ $$f^{(8)}(x)=x \implies f^{(1984)}(x)=x$$

Thus $$f^{(1987)} = f^3(x) = \frac {x+(1+\sqrt 2)}{-(1+\sqrt 2)x+1}$$