# Möbius function and polynomials

Let $$\mu$$ be the Möbius function. It is well known that $$\sum_{n|k} \mu(n) = 0$$ for $$k>1$$. What could be said about the polynomials $$R_k = \sum_{n|k} \mu(n) x^n$$ for $$x \in [0,1]$$? There does not seam to an easy answer to that question. It could be a good idea to take the average. Let $$Q_k=1/k \sum_{n=1}^{k} \mu(n)[k/n]x^n$$, where $$[x]$$ denotes the largest integer smaller then $$x$$. We have $$(k+1)Q_{k+1} - k Q_k = R_{k+1}$$. Set $$Q_1=R_1=x$$. Then by induction $$Q_k=1/k \sum_{n=1}^{k} R_n$$, is indeed the average. Let $$p$$ be a prime number then $$R_p=x-x^p$$ and it is concave and positive on $$[0,1]$$.

The plot of $$Q_k$$ on $$[0,1]$$, for a finite number of values of $$k$$, suggests that $$Q_k$$ is positive and could be concave. Moreover for all $$c<1$$, $$Q_k=\sum_{n=1}^{+\infty} (\mu(n)[k/n]x^n)/k$$ converges uniformly on $$[0,c]$$ towards the function $$g(x)=\sum_{n=1}^{+\infty} (\mu(n)/n)x^n$$. Indeed, $$|\mu(n)[k/n](x^n)/k|, for all $$x \in [0,c]$$. As a consequence of Abel summation and the prime number theorem the function $$g$$ is continuous on $$[0,1]$$. The function $$g$$ and its derivative were already studied in the literature, for example Fröberg,Numerical studies of the Möbius power series.

According to the paper above, the function g is concave on $$[0,1-\delta]$$, and then it starts to oscillate infinitely often, where $$\delta = 10^{-18}$$. So the $$Q_k$$'s can not be concave on the whole [0,1], in general. I was wondering how good of approximation is $$Q_k$$ of $$g$$? I believe that $$\max_{x \in [0,1]}(|Q_k(x)-Q_{k+1}(x)|)=O(1/k^2)$$. However I do know how to prove it.

Here is an "explanation" why $$O(1/k^2)$$. Since the convergence of $$Q_k$$ towards $$g$$ is uniform on every compact of $$[0,1)$$, "all the points converge at approximately the same rate/speed". But $$|Q_k(1)-Q_{k+1}(1)|=1/(k(k+1))$$. Then for any point in the small enough neighborhood of 1, the same bound "should hold by continuity" of the polynomial function. By uniform convergence property on every compact, "this bound should propagate on the whole [0,1]". Of course this is not a proof. The statements in quotes in this paragraph are intended to convey a rough idea.

Let $$||f||=\max_{x\in[0,1]}|f(x)|$$. The function $$Q_k$$ appears to be "almost concave" on $$[0,1]$$. We could compare it to something concave via Legendre transform. This transform is defined as follows, for a function $$f$$ defined on $$[0,1]$$ let $$f^{\ast}(s)=\sup_{x\in[0,1]} ({sx-f(x)})$$ (cf. p.36 of Convex analysis and minimization algorithms. I, Grundlehren der Mathematischen Wissenschaften 305). Then by the properties of $$()^{\ast}$$, described on page 44 of that book, we can say that $$C_k:=-(-Q_k)^{\ast\ast}$$ is the smallest concave function majorizing $$Q_k$$ on $$[0,1]$$. So that we have $$C_k \geq Q_k$$ and $$C_k(0)=Q_k(0)=0$$.

Let me suggest the following. By comparing $$Q_k$$ and $$C_k$$, I hope that we could prove that $$||Q_k-Q_{k+1}||=O(1/k^2)$$. I believe that the following statements should hold:

1 $$C_k$$ converges uniformly on every compact of $$[0,1)$$ to $$-(-g)^{\ast\ast}$$ and $$||C_k-C_{k+1}||=O(1/k^2)$$, for all k large enough.

2 $$||Q_k-Q_{k+1}||=O(||C_k-C_{k+1}||)$$, for all $$k$$ large enough.

Question Do we have $$||Q_k-Q_{k+1}||=O(1/k^2)$$, for $$k$$ large enough?