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My previous answer gives you a hint at analytic solution, but one may ask how to find the result in closed form.

To do so, one has to find family of flows $\phi_C(t)$ of the function $g(x)$ from the condition $\phi_C(t+1)=g(\phi_C(t))$. Then evaluate $\phi_C(1/2)$ and finally substitute $x$ instead of $C$.

For example, for the problem $f^{[2]}(x)=x^2$ the flow equation will be $\phi_C(t+1)=\phi_C(t)^2$ Taking logarithm with $C$ base one arrives at $\Delta\log_C(\phi_C(t))=\log_C\phi_C(t)$. Substituting $p(t)=\log_C\phi_C(t)$ we get a first-order linear difference equation $\Delta p(t)=p(t)$. Since $2^t$ is the only non-trivial function known to satisfy this equation, we get $\log_C\phi_C(t)=2^t$, and so $\phi_C(t)=C^{2^t}$. Evaluating at $1/2$ and substituting $x$ for $C$ we get $f(x)=x^{\sqrt{2}}$.

Of course, in many cases the equation may be non-linear and difficult to solve.

In general, any iterative equation of order $n$ (that is involving $f^{[n]}(x)$) is reducible to difference equation of order $n-1$ over flows. Such equation is expeted to have no more than $n-1$-dimentional space of solutions, except maybe some trivial cases so the original equation is also in non-trivial cases experted to have no more than $n-1$ solutions. Since your problem is iterative equation of order $2$, the solution is eqpected unique if exists, except some trivial cases such as $f^{[2]}(x)=x$ and $f^{[2]}(x)=-x$ where any 1-periodic and 1-antiperiodic functions (respectively) will satisfy as a flow.

My previous answer gives you a hint at analytic solution, but one may ask how to find the result in closed form.

To do so, one has to find family of flows $\phi_C(t)$ of the function $g(x)$ from the condition $\phi_C(t+1)=g(\phi_C(t))$. Then evaluate $\phi_C(1/2)$ and finally substitute $x$ instead of $C$.

For example, for the problem $f^{[2]}(x)=x^2$ the flow equation will be $\phi_C(t+1)=\phi_C(t)^2$ Taking logarithm with $C$ base one arrives at $\Delta\log_C(\phi_C(t))=\log_C\phi_C(t)$. Substituting $p(t)=\log_C\phi_C(t)$ we get a first-order linear difference equation $\Delta p(t)=p(t)$. Since $2^t$ is the only non-trivial function known to satisfy this equation, we get $\log_C\phi_C(t)=2^t$, and so $\phi_C(t)=C^{2^t}$. Evaluating at $1/2$ and substituting $x$ for $C$ we get $f(x)=x^{\sqrt{2}}$.

Of course, in many cases the equation may be non-linear and difficult to solve.

In general, any iterative equation of order $n$ (that is involving $f^{[n]}(x)$) is reducible to difference equation of order $n-1$ over flows. Such equation is expected to have no more than $n-1$-dimensional space of solutions, except maybe some trivial cases so the original equation is also in non-trivial cases expected to have no more than $n-1$ solutions. Since your problem is iterative equation of order $2$, the solution is expected unique if exists, except some trivial cases such as $f^{[2]}(x)=x$ and $f^{[2]}(x)=-x$ where any 1-periodic and 1-antiperiodic functions (respectively) will satisfy as a flow.

My previous answer gives you a hint at analytic solution, but one may ask how to find the result in closed form.

To do so, one has to find family of flows $\phi_C(t)$ of the function $g(x)$ from the condition $\phi_C(t+1)=g(\phi_C(t))$. Then evaluate $\phi_C(1/2)$ and finally substitute $x$ instead of $C$.

For example, for the problem $f^{[2]}(x)=x^2$ the flow equation will be $\phi_C(t+1)=\phi_C(x)^2$ Taking logarithm with $C$ base one arrives at $\Delta\log_C(\phi_C(t))=\log_C\phi_C(x)$. Substituting $p(t)=\log_C\phi_C(t)$ we get a first-order linear difference equation $\Delta p(t)=p(t)$. Since $2^t$ is the only non-trivial function known to satisfy this equation, we get $\log_C\phi_C(t)=2^t$, and so $\phi_C(t)=C^{2^t}$. Evaluating at $1/2$ and substituting $x$ for $C$ we get $f(x)=x^{\sqrt{2}}$.

Of course, in many cases the equation may be non-linear and difficult to solve.

In general, any iterative equation of order $n$ (that is involving $f^{[n]}(x)$) is reducible to difference equation of order $n-1$ over flows. Such equation is expeted to have no more than $n-1$-dimentional space of solutions, except maybe some trivial cases so the original equation is also in non-trivial cases experted to have no more than $n-1$ solutions. Since your problem is iterative equation of order $2$, the solution is eqpected unique if exists, except some trivial cases such as $f^{[2]}(x)=x$ and $f^{[2]}(x)=-x$ where any 1-periodic and 1-antiperiodic functions (respectively) will satisfy as a flow.

My previous answer gives you a hint at analytic solution, but one may ask how to find the result in closed form.

To do so, one has to find family of flows $\phi_C(t)$ of the function $g(x)$ from the condition $\phi_C(t+1)=g(\phi_C(t))$. Then evaluate $\phi_C(1/2)$ and finally substitute $x$ instead of $C$.

For example, for the problem $f^{[2]}(x)=x^2$ the flow equation will be $\phi_C(t+1)=\phi_C(t)^2$ Taking logarithm with $C$ base one arrives at $\Delta\log_C(\phi_C(t))=\log_C\phi_C(t)$. Substituting $p(t)=\log_C\phi_C(t)$ we get a first-order linear difference equation $\Delta p(t)=p(t)$. Since $2^t$ is the only non-trivial function known to satisfy this equation, we get $\log_C\phi_C(t)=2^t$, and so $\phi_C(t)=C^{2^t}$. Evaluating at $1/2$ and substituting $x$ for $C$ we get $f(x)=x^{\sqrt{2}}$.

Of course, in many cases the equation may be non-linear and difficult to solve.

In general, any iterative equation of order $n$ (that is involving $f^{[n]}(x)$) is reducible to difference equation of order $n-1$ over flows. Such equation is expeted to have no more than $n-1$-dimentional space of solutions, except maybe some trivial cases so the original equation is also in non-trivial cases experted to have no more than $n-1$ solutions. Since your problem is iterative equation of order $2$, the solution is eqpected unique if exists, except some trivial cases such as $f^{[2]}(x)=x$ and $f^{[2]}(x)=-x$ where any 1-periodic and 1-antiperiodic functions (respectively) will satisfy as a flow.

My previous answer gives you a hint at analytic solution, but one may ask how to find the result in closed form.

To do so, one has to find family of flows $\phi_C(t)$ of the function $g(x)$ from the condition $\phi_C(t+1)=g(\phi_C(t))$. Then evaluate $\phi_C(1/2)$ and finally substitute $x$ instead of $C$.

For example, for the problem $f^{[2]}(x)=x^2$ the flow equation will be $\phi_C(t+1)=\phi_C(x)^2$ Taking logarithm with $C$ base one arrives at $\Delta\log_C(\phi_C(t+1))=\log_C\phi_C(x)$. Substituting $p(t)=\log_C\phi_C(t+1)$ we get a first-order linear difference equation $\Delta p(t)=p(t)$. Since $2^t$ is the only non-trivial function known to satisfy this equation, we get $\log_C\phi_C(t)=2^t$, and so $\phi_C(t)=C^{2^t}$. Evaluating at $1/2$ and substituting $x$ for $C$ we get $f(x)=x^{\sqrt{2}}$.

Of course, in many cases the equation may be non-linear and difficult to solve.

In general, any iterative equation of order $n$ (that is involving $f^{[n]}(x)$) is reducible to difference equation of order $n-1$ over flows. Such equation is expeted to have no more than $n-1$ independent solutions, except maybe some trivial cases so the original equation is also in non-trivial cases experted to have no more than $n-1$ solutions. Since your problem is iterative equation of order $2$, the solution is eqpected unique if exists, except some trivial cases such as $f^{[2]}(x)=x$ and $f^{[2]}(x)=-x$ where any 1-periodic and 1-antiperiodic respectively functions will satisfy as flow.

My previous answer gives you a hint at analytic solution, but one may ask how to find the result in closed form.

To do so, one has to find family of flows $\phi_C(t)$ of the function $g(x)$ from the condition $\phi_C(t+1)=g(\phi_C(t))$. Then evaluate $\phi_C(1/2)$ and finally substitute $x$ instead of $C$.

For example, for the problem $f^{[2]}(x)=x^2$ the flow equation will be $\phi_C(t+1)=\phi_C(x)^2$ Taking logarithm with $C$ base one arrives at $\Delta\log_C(\phi_C(t))=\log_C\phi_C(x)$. Substituting $p(t)=\log_C\phi_C(t)$ we get a first-order linear difference equation $\Delta p(t)=p(t)$. Since $2^t$ is the only non-trivial function known to satisfy this equation, we get $\log_C\phi_C(t)=2^t$, and so $\phi_C(t)=C^{2^t}$. Evaluating at $1/2$ and substituting $x$ for $C$ we get $f(x)=x^{\sqrt{2}}$.

Of course, in many cases the equation may be non-linear and difficult to solve.

In general, any iterative equation of order $n$ (that is involving $f^{[n]}(x)$) is reducible to difference equation of order $n-1$ over flows. Such equation is expeted to have no more than $n-1$-dimentional space of solutions, except maybe some trivial cases so the original equation is also in non-trivial cases experted to have no more than $n-1$ solutions. Since your problem is iterative equation of order $2$, the solution is eqpected unique if exists, except some trivial cases such as $f^{[2]}(x)=x$ and $f^{[2]}(x)=-x$ where any 1-periodic and 1-antiperiodic functions (respectively) will satisfy as a flow.