The answer is on the affirmative.
Consider the sequence
\begin{eqnarray}
\alpha_0 &=& 1,\\
\alpha_1 &=& 10005,\\
\alpha_2 &=& 1000505,\\
\alpha_3 &=& 10005050005,\\
\alpha_4 &=& 1000505000500000005,\\
\dots & &,
\end{eqnarray}
where, for $n>1$, $\alpha_{n+1}$ is obtained from $\alpha_n$ by appending a sequence of $2^{n-1}-1$ zeros and then a $5$:
$$\alpha_{n+1} = \alpha_n \cdot 10^{2^{n-1}}+5.$$
Let us first show, by induction, that
$$f\left(\alpha_n^2\right) = \alpha_n.\tag{1}\label{eq1}$$
Suppose that \eqref{eq1} holds true for a given $n$. Then
\begin{eqnarray}
f\left(\alpha_{n+1}^2\right) &=& f\left(\left(\alpha_n \cdot 10^{2^{n-1}}+5\right)^2\right)=\\
&=& f\left(\alpha_n^2\cdot 10^{2^n}+\alpha_n\cdot 10^{2^{n-1}+1}+25\right).
\end{eqnarray}
Note that
- By induction and self-similarity of $f$, $$f\left(\alpha^2_n \cdot 10^{2^n}\right) = \alpha_n \cdot 10^{2^{n-1}};$$
- For $n\geq 3$, the addition of the second term $\alpha_n\cdot 10^{2^{n-1}+1}$ does not modify any digits of the first term and has $0$'s in every even position;
- The first and last term of the sum never interfere.
As a consequence
\begin{eqnarray}
f\left(\alpha_{n+1}^2\right) &=& f\left(\alpha_n^2\cdot 10^{2^n} + 25\right) =\\
&=&\alpha_n \cdot 10^{2^{n-1}} + 5=\\
&=& \alpha_{n+1}.
\end{eqnarray}
Consider now the sequence
\begin{eqnarray}
\beta_0 &=& 1,\\
\beta_1 &=& 1.005,\\
\beta_2 &=& 1.00505,\\
\vdots && \vdots\\
\beta_n &=& \alpha_n\cdot 10^{-2^n-2}.
\end{eqnarray}
The sequence $(\beta_n)$ is monotonic and upper bounded, and thus convergent in $\Bbb R$, and so is the sequence $(\beta_n^2)$.
Let
$$(\beta_n^2) \to \xi.$$
Clearly $\xi$ is a non-terminating decimal. Thus, as shown in the answer to this question, $f(x)$ is continuous in $\xi$, and so is
$$ h(x) = f(x) - \sqrt x.$$
We therefore must have
$$\left(h\left(\beta^2_n\right)\right)\to h(\xi).$$
Since, by self-similarity of $h$, for each $n$
$$h\left(\beta_n^2\right) = 0,$$
it must be
$$h(\xi) = 0,$$
that is
$$f(\xi) = \sqrt \xi.$$
A little update
- The same reasoning applies to the sequences $(5.0005, 5.000505, 5.0005050005,\dots)$, and $(6.0005, 6.000505, 6.0005050005,\dots)$.
- In the above mentioned sequences any digit $5$ can be replaced by a $0$, obtaining thus a dense set of intersection points on the right neighborhoods of $1$, $5$, and $6$.
Best Answer
Obviously, it can be outputed by Turing Machine in real time. Thus under the Hartmanis-Stearns conjecture, it is a transcendental number.