Questions
Is the below approach and proof correct? If I take $x \to 1- \epsilon$ and $n $ is of order $ \frac{1}{\epsilon^s}$ where $s \geq 2$ (so that the error $\to 0$)
Then as $$ \sum_{r=1}^\infty \mu(r) x^r = f(x) = g(x) + O(x^{n+1})$$ And $$g(x) = \sum_{k=0}^\infty \frac{(\ln(1- \epsilon))^k }{\sum_{r=1}^n (r^k)(k!)}$$
How accurately does solve the problem? If it doesn't where in my reasoning do I go wrong? Also is it possible to transform the series back to:
$$g(x) = \sum_{k=0}^\infty \frac{(\ln(1- \epsilon))^k }{\sum_{r=1}^n (r^k)(k!)} = \sum_{r=1}^\infty \mu(r) x^r + O(x^{n+1}) $$
Functional equation
I was recently thinking of the functional equation:
$$ f(x) + f(x^2) + f(x^3) + \dots = x $$
This has the solution is:
$$ f(x) = \sum_{r=1}^\infty \mu(r) x^r $$
where $\mu(r)$ is the mobius function. Another way of solving this is. Let $x \to 0 \implies f(0) = 0$.
Differentiating the functional equation:
$$ f'(x) + 2x f'(x^2) + 3x^2 f'(x^2) + \dots = 1 $$
Setting $x \to 0 \implies f'(x) = 1$
Differentiating again:
$$ f''(x) + 2 f'(x^2) + 2xf''(x^2) + \dots = 0$$
Setting $x \to 0 \implies f''(0) + 2f'(0) = 0 \implies f''(0) = -2 \implies \mu(2) = -1 $
We can do so recursively as many times as we want.
Auxiliary Functional equation
Rather solving the above functional equation let us try to solve:
$$ \sum_{r=1}^n g(x^r) = g(x) + g(x^2) + g(x^3) + \dots + g(x^n) = x$$
This will match the solution of $f(x)$ upto $n$ terms:
$$ g(x) = \sum_{r=1}^n \mu(r) x^r + O(x^{n+1}) $$
Auxiliary Functional equation R.H.S
Let us make use of:
$$ x =e^{\ln x} = 1 + \ln x + \frac{(\ln (x))^2}{2!} + \frac{(\ln (x))^3}{3!} + \dots$$
Thus,
$$ \sum_{r=1}^n g(x^r) = \sum_{r=1}^\infty \frac{(\ln (x))^r}{r!} = x $$
However, in this basis $g(x)$ is obviously:
$$ g(x) = \frac{1}{n} + \frac{ 2 \ln (x)}{n(n+1)} + \frac{6 (\ln x)^2}{n(n+1)(2n+1))(2!)} + \dots $$
To convince yourself this the case take:
$$ g(x^r) = \frac{1}{n} + \frac{ 2 r \ln (x)}{n(n+1)} + \frac{6 r^2 (\ln x)^2}{n(n+1)(2n+1))(2!)} + \dots $$
And then summing over $r$ from $1$ to $n$ completes the proof.
Let $c_n(m) = 1_{m \in [1,n]}$ and $\mu_n(m)$ its Dirichlet convolution inverse and $g_n(x) = \sum_{m=1}^\infty \mu_n(m) x^m$ satisfying $\sum_{r=1}^n g_n(x^r) = x$.
Let $\zeta_n(s) = \sum_{m=1}^n m^{-s}$ then $$\sum_{m=1}^\infty \mu_n(m) m^{-s} = \frac{1}{\zeta_n(s)}$$
If $g_n(e^t)$ was analytic at $t=0$ then $g_n(x) = \sum_{k=0}^\infty a_n(k) (\ln x)^k$ and $$\sum_{k=0}^\infty a_n(k)(\ln x)^k \sum_{r=1}^n r^k = \sum_{r=1}^n g(x^r) = x= \sum_{k=0}^\infty \frac{1}{k!}(\ln x)^k \implies a_n(k) = \frac{ \frac{1}{k!}}{\sum_{r=1}^n r^k}$$
To estimate the $a_n(k)$ you'd need the Faulhaber formula.
But $g_n(e^t)$ isn't analytic and it is $\sum_{k=0}^\infty a_n(k)t^k$ plus a series over the zeros of $\zeta_n(s)$ (see this).
There is another method to investigate the $\mu_n(m)$. $$\sum_{m=1}^\infty \mu_n(m) m^{-s} = \frac{1}{\zeta_n(s)} = \sum_{L=0}^\infty (1-\zeta_n(s))^L= \sum_{L=0}^\infty (-1)^L (\sum_{v=2}^n v^{-s})^L \\= \sum_{L=0}^\infty(-1)^L \sum_{m=1}^\infty m^{-s} \sum_{u \in \mathbb{Z}^L, u_l \in [2,n], m = \sum_l u_l} 1 = \sum_{m=1}^\infty m^{-s} \sum_{L, u \in \mathbb{Z}^L, u_l \in [2,n], m = \sum_l u_l} (-1)^L$$
Let $$\frac{1}{2-\zeta(s)} = \sum_{L=0}^\infty (\zeta(s)-1)^L = \sum_{m=1}^\infty m^{-s} \sum_{L, u \in \mathbb{Z}^L, u_l \in [2,\infty), m = \sum_l u_l} 1= \sum_{m=1}^\infty b(m) m^{-s}$$ its abscissa of convergence is $\sigma\approx 1.729$ the positive real number such that $\zeta(\sigma)=2$.
For $\Re(s) \ge \sigma+\epsilon$ then $|\zeta(s)| \le \zeta(\sigma+\epsilon)$ so $|\zeta(s)-2| \ge 2 - |\zeta(s)|\ge 2-\zeta(\sigma+\epsilon)$ whence $b(m)m^{-\sigma-\epsilon} \le \frac{1}{2-\zeta(\sigma+\epsilon)}$. In particular $$|\mu_n(m)| \le b(m) \le \frac{m^2}{2-\zeta(2)}$$
With $g_\infty(x) = \lim_{n \to \infty} g_n(x) = \sum_{m=1}^\infty \mu(m) x^m$ then
$$|g_n(x)-g_\infty(x)| = |\sum_{m=n+1}^\infty (\mu_n(m)-\mu(m)) x^m| \le \sum_{m=n+1}^\infty (\frac{m^2}{2-\zeta(2)}+1) |x|^m$$