Half root (corrected)

As Calvin Lin has pointed out, no one has currently gotten a rigorous solution. What I detail here is more of a guideline of a rigorous solution, which explains why there are several possibilities for the function.

Firstly, since the RHS can take any positive real value and this would be defined for any $f$ , we have $f$ surjective. Since $f$ is monotonically increasing, we have $f$ injective. Thus, $f$ bijective, so an inverse exists.

Now, we consider removing the monotonic condition and redefining $f$ as a bijective function in the positive reals satisfying $f(f(x))=\sqrt{x}$ . Note that by associativity, we have $f(f(f(x)))=f(\sqrt{x}) = \sqrt{f(x)}$ .

We determine that the only fixed point is $x=1$ .

Now let $g: \mathbb{R}^+ \to \mathbb{R}^+$ satisfy $g(x)=\sqrt{x}$ . We consider chains (iterations) of $g$ as $A=\{\ldots, g^{-3}(x), g^{-2}(x), g^{-1}(x), x, g(x), g^2(x), g^3(x), \ldots\}$ and $B=\{\ldots, g^{-3}(f(x)), g^{-2}(f(x)), g^{-1}(f(x)), f(x), g(f(x)), g^2(f(x)), g^3(f(x)), \ldots\}$ , where we choose $x$ such that all elements are distinct i.e. $x$ not a fixed point so $x \neq 1$ .

We notice that what $f$ does is create a relation from an element in $A$ to an element in $B$ and vice versa by the associative relation we found earlier. However, this pairing can occur arbitrarily. Thus, without this monotonic condition, we can have a lot of weird functions satisfying.

Now all that's left to show is that if $f$ is indeed monotonic increasing, then the arrows must be arranged such that we have $f(x)=x^{\frac{1}{\sqrt{2}}}$ .

After further working, I have found there are other solutions that don't give $f(2)=2^{\frac{1}{\sqrt{2}}}$ , so this question is flawed. Consider $h(h(x))=\frac{x}{2}$ for $h: \mathbb{R}^+ \to \mathbb{R}^+$ . Any solution to this is equivalent to any solution for $f(f(x))=\sqrt{x}$ because taking the log of the expression in $f$ gives the expression in $h$ . ( $f(x)=\exp(h(\log(x)))$ )

Thus, we want to find positive, monotonic increasing solutions to $h(h(x))=\frac{x}{2}$ . If we set $h(1)= \frac{1}{\sqrt{2}}$ and $h(\sqrt{2}) = 1$ , and force $h$ to be continuous increasing in the interval $[1,\sqrt{2}]$ , we can define $h$ on the rest of $\mathbb{R}^+$ accordingly, so there exist other solutions.

The above image shows one such example of $h$ I simulated on Geogebra. I fixed the curve between points $A(1,\frac{1}{\sqrt{2}})$ and $B(\sqrt{2},1)$ , and then used the trace-point feature and animated more points that should be on the curve. This function (after iterating it) remains monotonic increasing and satisfies $h(h(x))=\frac{x}{2}$ .

Therefore, this question is flawed.

First approch: If we assume a power function $f(x) = c \cdot x^\alpha$ it follows $\begin{aligned} f(f(x)) &= c (c x^\alpha)^\alpha = c^{1 + \alpha} \cdot x^{\alpha^2} = \sqrt{x} = x^{1/2} \\ \Rightarrow \quad \alpha &= \pm \frac{1}{\sqrt{2}}, \quad c = \pm 1 \end{aligned}$ We yield a positive, monotonic increasing function only if both signs are positve. Therefore, we get the result $f(2) = 2^{1/\sqrt{2}} = \sqrt[\sqrt{2}]{2} \approx 1.6325$

Second approch: Without making any assumptions about the function $f(x)$ , we know that $\sqrt{x} \leq f(x) \leq x, \quad x \geq 1$ since the assumptions $f(x) > x$ and $f(x) < \sqrt{x}$ can be contradicted: $\begin{aligned} f(x) > x \quad \Rightarrow & \quad \sqrt{x} = f(f(x)) > f(x) > x \\ f(x) < \sqrt{x} \quad \Rightarrow & \quad \sqrt{x} = f(f(x)) < \sqrt{x} \end{aligned}$ In particular, $f(1) = 1$ follows. We can also calculate the derivatives at $x = 1$ : $\begin{aligned} f'(f(x)) f'(x) &= \frac{1}{2} x^{- \frac{1}{2}} & \Rightarrow \quad f'(1) &= \frac{1}{\sqrt{2}} \\ f''(f(x))(f'(x))^2 + f'(f(x))f''(x) &= - \frac{1}{4} x^{- \frac{3}{2}} & \Rightarrow \quad f''(1) &= -\frac{1}{2(1 + \sqrt{2})} \\ f'''(f(x)) f'(x) (f'(x) + 1) + (f''(x))^2(2 f'(x) + 1) &= \frac{3}{8} x^{- \frac{5}{2}} & \Rightarrow \quad f'''(1) &= \frac{2\sqrt{2} - 1}{2(2 + \sqrt{2})} \end{aligned}$ Therefore, we can use the Taylor expansion $\begin{aligned} f(x) &\approx f(1) + f'(1) (x-1) + \frac{1}{2} f''(1) (x-1)^2 + \frac{1}{12} f'''(1) (x-1)^3 \\ \Rightarrow \quad f(2) &\approx 1.63 \end{aligned}$

A function f(f(x)) can also be written as (f o f) (x). A function is a special kind of relation where you have 1 input and it spits out 1 output. That's the definition of a function. Since you take the one function f after the other function f you do the same transformation 2 times in a row.

The question is what is the middle or what is f(x)?

When you start from X and end in Root X you have to change the exponents. So how do you go from exponent 1 to exponent 1/2? Remember you have to do 2 operations or transformations rather to go from X to root X or from 1 to 1/2.

I called the operation p and we have to do this operation 2 times in a row to get from exponent 1 (input) to exponent 1/2 (output):

So now we fill in x = 2 and you get the solution 1.63

|For this problem, I decided to find the distance between the equation $y=x$ and $y=\sqrt{x}$ . We know that all three lines have the same value at $x=1$ .

$\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

We plug in the value of $x$ which is $2$

$\sqrt{(2-2)^2+(\sqrt{2}-2)^2}$

We get the value as $0.59$ once rounded to the second decimal place.

Finally, I added $1$ to that value because we know that at $x=1$ all the points have the same value.

We end up with the answer $1.59$ , which is an accepted approximation according to the problem.