Not another composed function

Let $x= \tan \theta$ . Then $f(x) = \frac{2\tan\theta }{1 - \tan^2 \theta} = \tan 2\theta$ . So, $f^5(x) = f^5(\tan \theta) = \tan(2^5 \theta) = \tan (32 \theta )$ . Hence the equation becomes $\tan 32 \theta = \tan \theta$ . The general solution to this equation is $32\theta = \theta + n \pi$ , where $n$ is a positive integer. Solving gives $\theta= \frac{n\pi }{31}$ . Note that the function $\tan$ repeats by periodicity $\pi$ , so after putting $n= 1, 2, ..., 31$ , when we put $n= 32, 33, ...$ , and so on, we will get the same value of $\tan \theta$ . Hence the total number of distinct real solutions for $x$ is $31$ .

The key is to notice that $f(\tan y) = \dfrac{2\tan y}{1-\tan^2 y} = \tan 2y$ .

If $x$ satisfies $f^{(5)}(x) = x$ , then write $x = \tan y$ for a unique $y \in (-\tfrac{\pi}{2},\tfrac{\pi}{2})$ . Then, $\tan y = f^{(5)}(\tan y)$ $= f^{(4)}(\tan 2y)$ $= f^{(3)}(\tan 4y)$ $= f^{(2)}(\tan 8y)$ $= f^{(1)}(\tan 16y)$ $= \tan 32y$ .

Since the $\tan$ function has period $\pi$ and is one-to-one on any period, we must have $32y = y+\pi k$ , i.e. $y = \dfrac{\pi}{31}k$ for some integer $k$ .

Since we restricted $-\dfrac{\pi}{2} < y < \dfrac{\pi}{2}$ , we must have $-\dfrac{\pi}{2} < \dfrac{\pi}{31}k < \dfrac{\pi}{2}$ , i.e. $-15.5 < k < 15.5$ . Since $k$ is an integer, we have $31$ values of $k$ which gives $31$ values of $y$ which yield $31$ distinct solutions for $x = \tan y$ .

Let $x = tan(y)$ . Then $f(x) = \frac{2\tan y}{1-\tan^2 y} = \tan(2y)$ . Then, $f^5(x) = f^5(\tan y)$ = tan(32y). Thus, we are asked to find the number of distinct solutions to $\tan y = \tan(32y)$ . Since tangent is periodic with period $\pi$ , we have $32y = y + n\pi$ , for some integer $n$ . Thus, $y = \frac{n \pi}{31}$ . Since we only find the $\tan y$ values and not the actual $y$ values, we see clearly that there are only $\boxed{31}$ solutions.

Since the domain of the function is the set of all real numbers except $\pm1$ , there exists $\theta \in \big(-\frac{\pi}{2},\frac{\pi}{2}\big)\setminus\{-\frac{\pi}{4},\frac{\pi}{4}\}$ such that $x=\tan\theta.$ By the double-angle formula for $\tan2\theta,$ we therefore have $f(\tan\theta) = \frac{2\tan\theta}{1-\tan^2\theta} = \tan 2\theta.$ Applying the double-angle formula for $\tan4\theta,$ we have $f^{(2)}(\tan\theta) = f(f(\tan\theta)) = f(\tan 2\theta) =\frac{2\tan 2\theta}{1-\tan^2 2\theta} = \tan 4\theta.$ Continuing in the same manner, we finally obtain $f^{(5)}(\tan\theta) = \tan 32\theta.$ So the problem is equivalent to solve the equation: $\tan 32\theta = \tan \theta$ with a restriction that $\tan\theta \neq \pm 1.$ The general solution is $32\theta = \theta +n\pi$ or $\theta = \frac{n\pi}{31}$ where $n\in\mathbb{Z}.$ Since tangent function is periodic with period of $\pi,$ it follows that all distinct solutions are $x=\tan\frac{n\pi}{31}, \quad n=0,1,2,\dots,30.$ This creates $\boxed{31}$ distinct real solutions in all.

Define $\theta=\arctan x$ , so $x=\tan \theta$ for $\theta\in (-\frac{\pi}{2},\frac{\pi}{2})$ . Then $f(\tan \theta)=\tan(2\theta)$ , so if $f^{(5)}(x)=x$ , $\tan \theta=f^{(5)}(\tan \theta)=\tan (32\theta)$ But this can happen iff $31\theta=32\theta-\theta$ is a multiple of $\pi$ , the period of $\tan x$ . Since $\theta\in (-\frac{\pi}{2},\frac{\pi}{2})$ , the solutions are $\theta=-\frac{15\pi}{31},-\frac{14\pi}{31},\cdots \frac{15\pi}{31}$ Giving $\boxed{31}$ corresponding solutions for $x$

let x=tg(a), then f(tg(a))=tg(2a), f(f(tg(a)))=f(tg(2a))=tg(4a) So f^(5)(tg(a))=tg(32a)=tg(a) 0=tg(32a)-tg(a)=tg(31a)(1+tg(a)tg(32a)) 31a=k(pi) k=-15..15 number=31

x=tant f(x)=tan 2t hence f(5)(x)=x is same as tan 32t=tan t in various divisions of domain of t we have only 31 distinct solutions of x hence 31

Let x=tan a Therefore f(x)=2tan a/1-(tan a)^2 = tan 2a Similarly f^5(x)=tan 32a therefore as f^5(x)=x tan 32x = tan x 32x = n(180) + x x = n(180)/31 On solving this equation we get 31 different values of tan a = x

We can substitute $\theta = \arctan x$ , with $-\frac12\pi < \theta < \frac12\pi$ . Note that this substitution is one-to-one, so any solution we find for $\theta$ is a solution for $x$ and vice versa. We get

$f(x) = f(\tan\theta) = \frac{2\tan\theta}{1-\tan^2\theta} = \tan 2\theta.$

Therefore, through repeated use, we easily find

$f^{(5)}(x) = \tan(32\theta),$

Thus the equation to solve translates to

$\tan(32\theta) = \tan \theta.$

This equation holds if and only if

$32\theta = \theta + k\pi$

for some integer $k$ , which solves to

$\theta = \frac{k\pi}{31}.$

Considering the restraints on $\theta$ , the only valid solutions are for $-15 \leq k \leq 15$ , of which there are $\boxed{31}$

In general, $f^{(n)}(x) = x$ has $2^n-1$ solutions.

Let $x = \tan \theta$ , then $f(x) = f(\tan \theta) = \dfrac{2\tan \theta}{1 - (\tan \theta)^2}$ . This is the double-angle formula for tangent , so $f(\tan \theta) = \tan 2\theta$ . So $f^{(5)}(\tan \theta) = \tan 2(2(2(2(2\theta)))) = \tan 32\theta$ .

So $\tan 32\theta = \tan \theta$ . We can find $\tan 31\theta = \dfrac{\tan 32\theta - \tan \theta}{1 + \tan 32\theta \cdot \tan \theta} = 0$ because $\tan 32\theta - \tan \theta = 0$ . Note that $\tan x = 0$ has solutions $x = \pi k$ for all integer $k$ , so $\theta = \dfrac{\pi}{31}k$ .

Now, if $n \not\in [0,30]$ , let $m$ be the unique integer where $m \in [0,30]$ and $m \equiv n \pmod{31}$ . Then $\tan \dfrac{\pi}{31}m = \tan \dfrac{\pi}{31}n$ , since $\dfrac{\pi}{31}n = \dfrac{\pi}{31}m + \pi k$ for some integer $k$ and $\tan$ has period $\pi$ . So both result on the same value of $x$ , so we can discard $n$ .

So there are $31$ values of $n$ . Now suppose $n_1, n_2 \in [0,30]$ and $\tan \dfrac{\pi}{31}n_1 = \tan \dfrac{\pi}{31}n_2$ . Then,

$\dfrac{\sin \frac{\pi n_1}{31}}{\cos \frac{\pi n_1}{31}} = \dfrac{\sin \frac{\pi n_2}{31}}{\cos \frac{\pi n_2}{31}}$

$\sin \dfrac{\pi n_1}{31} \cos \dfrac{\pi n_2}{31} = \cos \dfrac{\pi n_1}{31} \sin \dfrac{\pi n_2}{31}$

$\sin \dfrac{\pi n_1}{31} \cos \dfrac{\pi n_2}{31} - \cos \dfrac{\pi n_1}{31} \sin \dfrac{\pi n_2}{31} = 0$

By the difference of angles formula for sine , we get $\sin \left( \dfrac{\pi n_1}{31} - \dfrac{\pi n_2}{31} \right) = 0$ , or $\sin \dfrac{\pi}{31} (n_1 - n_2) = 0$ . So $\dfrac{\pi}{31} (n_1 - n_2) = \pi k$ for some integer $k$ , so $n_1 - n_2 = 31k$ . But $n_1 - n_2 \le 30 - 0 = 30 < 31$ and $n_1 - n_2 \ge 0 - 30 = -30 > -31$ , so $k \ge 1$ and $k \le 1$ has no solutions. So $k = 0$ and so $n_1 = n_2$ . This means all $31$ values of $n$ gives distinct values of their tangents (all $31$ values of $\tan \dfrac{\pi}{31} n$ are distinct). Each of these values can be the value of $x$ (the only possibility that it cannot be the value of $x$ is if $\tan \dfrac{\pi}{31} n$ is undefined, but that requires $\dfrac{\pi}{31} n = \dfrac{\pi}{2} (1 + 2k)$ for some integer $k$ which means $n$ has a denominator of $2$ , impossible as $n$ is integer), so there are $\boxed{31}$ values of $x$ .

We will solve for the general formula $f^{(n)} (x) = x$

Recall the double angle formula $\tan (2A) = \frac {2 \tan (A) } { 1 - \tan^2 (A) }$

Let $x=\tan y \Rightarrow f(x) = \tan (2y)$

$f^{(2)} (x) = \tan (2^2 y)$

$f^{(3)} (x) = \tan (2^3 y)$

$\ldots$

$f^{(n)} (x) = \tan (2^n y)$

When $f^{(n)} (x) = x$ , $\tan (2^n \space y) = \tan (y)$

$\large \Rightarrow \frac {\sin (2^n y) }{ \cos (2^n y) } = \frac {\sin (y) }{ \cos (y) }$

$\large \Rightarrow \sin (y(2^n - 1)) = 0$

$\large \Rightarrow y = \frac {\pi \space M}{2^n - 1}$ , for $M=0,1,2,3, \ldots$

$\large \Rightarrow x = \tan (\frac {\pi \space M}{2^n - 1} )$ , for $M=0,1,2,3, \ldots , 2^n - 2$ , we stop at $M=2^n - 2$ because of the periodicity of $\large \tan (\frac {\pi \space n}{2^n - 1} )$

Hence, there's a total of $(2^n - 1)$ solution. In this case, $n=5$ . The answer is $2^5 - 1 = \boxed{31}$

Put $f(x) = \tan (\tan ^{ - 1} \frac{2x}{1 - x^2}) = \tan (2 \tan ^{ - 1} x)$

So, $f^{2} (x) = \tan ( 2^2 \tan ^{ - 1} x)$ Continuing, $f^{5} (x) = \tan ( 2^5 \tan ^{ - 1} x)$

Now $f^{5} (x) = x \Rightarrow \tan ( 2^5 \tan ^{ - 1} x) = x ...............(1)$

Applying $\tan ^{ - 1}$ to both sides & rearrange, $2^5 \tan ^{ - 1} x = n\pi +\tan ^{ - 1}x \rightarrow \tan ^{ - 1} x = \frac{n\pi}{31}$ Now this equation has infinite solutions but for only certain distinct solutions, (1) holds. As $n$ ranges over all integers, it has only 31 (0-30) values $\pmod {31}$ , hence $\tan x$ has only 31 correspondingly distinct values (since $\tan x = \tan (n\pi + x)$ ). So answer is 31 .

f(x)=2x/(1-x^2). f(f(x))=2[2x/(1-x^2)]/[1-(2x/(1-x^2))^2] =4x(1-x^2)/(1-6x^2+x^4). similarly,applying it for f(f(f(x))),f(f(f(f(x)))),f(f(f(f(f(x))))) step by step we will recieve an expression in the variable 'x' which when equated to 'x' as given in the condition will give a polynomial in 'x' which has 31 solutions.thus the answer is 31.

When looking to $f(x)$ , comes to mind the expression of $tan(2α)$ . Therefore, let $x = tan(α)$ :

$f(tan(α)) = \frac{2tan(α)}{1-tan^2(α)} = tan(2α)$ .

Now, $f^{(5)}(tan(α)) = tan(2^{5}α) = tan(32α)$ .

After that, the equation becomes $tan(32α) = tan(α)$ . Its general solution is given by:

$32α = α + nπ$

But that can be expressed as:

$α = \frac{nπ}{31}$

That expression gives us different solutions for $n = 1, 2, 3, ... 30, 31$ . After that, we have only duplicates because $tan(α)$ is periodical.

Thus, the number of distinct real solutions we are looking for is $31$ .

Hint: Put $x=tany$ , then you will reach to $tan32y=tany$ and find general solutions.

put x=tanθ

then f(tanθ) 2tanθ/1-tan^2θ=tan2θ

now

f^5(x) = f^5(tanθ) = tan(2^5θ) = tan(32θ).

then equation becomes tan32θ=tanθ then general solutions of this equation is given ny

The general solution is 32θ=θ+nπ or

θ=nπ/31 so n=1 2 3 ....31 give different solutions after that give the duplicate solutions due to periodicity of tanθ

so 31 solutions are there

Let $tan\theta = x$ , for some real value of y as x is real,

$f(x)=\frac{2x}{1-x^2} = \frac {2tan\theta}{1 - tan^2 \theta} = tan 2\theta$

Similarly,

$f^{(5)}(tan\theta)= tan 32\theta = x = tan \theta$ (given)

On solving $tan 32\theta=tan\theta$ gives value of $y = \frac{n\pi}{31}$

Therefore the number of distinct possible values of $tan\theta$ is 31 ie possible values of x

Therefore answer 31

Putting $x=\tan{\theta}$ the condition reduces to $\tan{32\theta}=\tan{\theta}$ . Since $\tan$ has a period of $\pi$ this is equivalent to $32\theta=\theta+n\pi$ and hence $x=\tan{(\frac{n\pi}{31})}$ .

The values $n=0,1,2,\dots,30$ give 31 distinct values for $x$