In my home I am trying to get the thesis for a specialization for my example* from a statement published related to functions that satisfy certain properties.
I've considered the function* $$f(x)=-\sum_{n=2}^\infty\frac{\mu(n)}{n}x^n,\tag{1}$$ where $\mu(n)$ denotes the Möbius function and our function $(1)$ is defined as a formal series over the real numbers of the unit interval. Using a CAS I think that my example satisfy (each of those) some of such properties, and with this purpose I wondered next question about one of main properties that I need.
Question. Is it possible to prove that $$f(x)=-\sum_{n=2}^\infty\frac{\mu(n)}{n}x^n$$ is bijective over $[0,1]$? Can you provide a good and explicit estimation (in terms of algebraic functions or elementary transcendental functions, see this Wikipedia $u(x)$ and $l(x)$) of $$l(x)\leq f^{-1}(x)\leq u(x)\tag{2}$$ over the unit interval? Many thanks.
*I know that my example of function $(1)$ was in the literature. If there was literature about if previous function is bijective and how to estimate its inverse $f^{-1}(x)$ over the unit interval, then answer this question as a reference request for such literature and I try to find and study such statements.
Remarkably $f$ is bijective only over $[0, 1-\delta]$ for a very small but positive value of $\delta$ (about $10^{-19}$), not over the entire unit interval $[0,1]$; indeed its derivative $$ f'(x) = - \sum_{n=2}^\infty \mu(n) x^n, $$ which one might expect to approach $1 - \frac1{\zeta(0)} = 3$ as $x \to 1$, instead changes sign infinitely often in $(1-\delta, 1)$, and becomes unbounded as $x \to 1$. This happens due to an expansion of $f'(x)$ that contains a term $$ -\frac{\Gamma(\rho)}{\zeta'(\rho)} (-\log x)^{-\rho} $$ for each nontrivial zero $\rho$ of the Riemann zeta function; as $x \to 1-$, the factor $(-\log x)^{-\rho}$ oscillates with amplitude approaching $\infty$, but the factor $\Gamma(\rho)$ is tiny due to the exponential decay of $\Gamma(s)$ of bounded real part and large $\left|{\rm Im}(s)\right|$, so $x$ must be extremely close to $1$ for the oscillation to overwhelm the term $1 - \frac1{\zeta(0)} = 3$. The first nontrivial pair of zeros $\frac12 \pm i \gamma_1$ already has $\gamma_1 > 14.13$ and $\left|\Gamma(\rho)\right| \sim 5.7 \cdot 10^{-10}$.
Combining the Dirichlet series $$ \frac1{\zeta(s)} = \sum_{n=1}^\infty \mu(n) n^{-s} \quad({\rm Re}(s)>1) $$ with the contour integral $$ \frac1{2\pi i} \int_{2-i\infty}^{2+i\infty} \Gamma(s) z^{-s} \, ds = \exp(-z), $$ we find that $$ f'(x) = x - \frac1{2\pi i} \int_{2-i\infty}^{2+i\infty} \Gamma(s) z^{-s} \frac{ds}{\zeta(s)} $$ where $x = \exp(-z)$.
We now move the contour to the left, which can be justified as in the complex-analytic proof of the Prime Number Theorem. The expected $1/\zeta(-1) = -2$ appears as the residue at the pole $s=0$; further poles at $s = -1, -2, -3, -4, \ldots$ (simple at odd $s$, double at even $s$ because of the trivial zeta zeros) contribute additional terms that are $O(\left|\log x\right|) = O(1-x)$ as $x \to 1-$. But there are also the nontrivial zeta zeros $\rho = \beta + i\gamma$, which contribute residues that grow as $x \to 1-$, oscillating with amplitude proportional to $(1-x)^{-\beta}$ (which is $(1-x)^{-1/2}$ under the Riemann hypothesis). But $x$ must get really close to $1$ for these oscillations to be visible because of the factor $\Gamma(\rho)$. The location of the first zero of $f'(x)$ depends on the phase of this oscillation, not just its magnitude; if I computed correctly, this happens for $x = 1 - \delta$ with $\delta \approx 1.6960 \cdot 10^{-19}$.