Devil Maths

We see that $f(1) = \dfrac{1}{\sqrt{2}}$ , $f^2(1) = \dfrac{1}{\sqrt{3}}$ , , $f^3(1) = \dfrac{1}{\sqrt{4}}$ , $f^4(1) = \dfrac{1}{\sqrt{5}}$ and so on.

So it seems that $f^n(1) = \dfrac{1}{\sqrt{n+1}}$ In order to see if this holds for all positive integral values of $n$ , we will use induction. Our base case is $n=1$ , and our equation holds. Now we assume that $f^n(1)=\dfrac{1}{\sqrt{n+1}}$ holds for $n$ , so now we try for $n+1$ .

$\begin{aligned} f^{n+1}(1)&=f(f^n(1))\\ &=\dfrac{f^n(1)}{\sqrt{1+(f^n(1))^2}}\\ &=\dfrac{\dfrac{1}{\sqrt{n+1}}}{\sqrt{1+\left(\dfrac{1}{\sqrt{n+1}}\right)^2}}\\ &=\dfrac{\dfrac{1}{\sqrt{n+1}}}{\sqrt{1+\dfrac{1}{n+1}}}\\ &=\dfrac{\dfrac{1}{\sqrt{n+1}}}{\sqrt{\dfrac{n+2}{n+1}}}\\ &=\dfrac{\dfrac{1}{\sqrt{n+1}}}{\dfrac{\sqrt{n+2}}{\sqrt{n+1}}}\\ &=\dfrac{1}{\sqrt{n+2}} \end{aligned}$

Hence the assumption is true.

Now we just plug in $n=99$ . $\boxed{f^{99}(1)=\dfrac{1}{10}}$ .

My method is Similar to Ahmed's, but I will creating a general formula for $f^{(99)} (x)$ : $f (f(x)) = \frac{\frac{x}{\sqrt{1+x^2} }}{\sqrt{1 +\frac{x^2}{1+x^2}}} \times\frac{\sqrt{1+x^2}}{\sqrt{1+x^2}} = \frac{x}{\sqrt{1+2x^2}}$ . Now I want to prove by induction that: $f^n (x) = \frac{x}{\sqrt{1+nx^2}} \ \ ....$ $f^{n+1} (x) = \frac{\frac{x}{\sqrt{1+x^2} }}{\sqrt{1 +n \frac{x^2}{1+x^2}}} \times\frac{\sqrt{1+x^2}}{\sqrt{1+x^2}}$ $= \frac{x}{\sqrt{1+x^2 +nx^2}} = \frac{x}{\sqrt{1+(n+1)x^2}} \ as \ required$ $\therefore\ f^{99} (1) = \frac{1}{\sqrt{100}} = 0.1 \ (or \ -0.1)$