Help with Mobius Inversion in Riemann's Zeta Function by Edwards (J to Prime Pi)

[SychoScribler]

Help with Mobius Inversion in "Riemann's Zeta Function" by Edwards (J to Prime Pi)

Can someone please add more detail or give references to help explain the lines of math in "Riemann's Zeta Function" by Edwards.

At the bottom of page 34 where it says "Very simply this inversion is effected by performing ...", the two lines of math that follow could do with a few more lines perhaps to make it clear enough for me to understand. I am already lost on the first line. One line or two before this might be all the help that I need.

I have been studying up on Mobius inversion, but with all my uni books and online references, I am having great difficulty in finding examples of the use of the inversion in the case of inverting a summation over the integers. For example in "Elementary Number Theory" by Rosen, I see the definition of the Mobius Function, some proofs of properties of the Mobius Function, a proof of the Mobius Inversion Formula, and then some identities that follow by immediate application of the inversion formula; but I don't see an inversion of summations like the J to Prime Pi.

Perhaps someone knows some chapters of books or papers for me to read. What I would really like to do is find a bunch of examples and questions to answer where I have to invert some 'sums over the integers' of various functions to give experience on the techniques necessary to achieve such a thing. I would need solutions too but the forum could help with that I suppose.

I understand that the Mobius Inversion only inverts 'summatory functions' summing over the divisors of a number.

I have a particular 'summation over the integers' of a function to invert if possible but I would like to have a go at it myself first.

Thanks.

 

[SychoScribler]

That is where equation $(1)\hspace{1.5em}J(x)=\pi(x)+\frac{1}{2}\pi(x^{1/2})+\frac{1}{3}\pi(x^{1/3})+\cdots+\frac{1}{n}\pi(x^{1/n})+\cdots$
is inverted to equation $(2)\hspace{1.5em}\pi(x)=J(x)-\frac{1}{2}J(x^{1/2})-\frac{1}{3}J(x^{1/3})-\frac{1}{5}J(x^{1/5})+\frac{1}{6}J(x^{1/6})+\cdots+\frac{\mu(n)}{n}J(x^{1/n})+\cdots$.

The explanation given at the bottom of the page is as follows.

"Very simply this inversion is effected by performing successively for each prime $p=2,3,5,7,\ldots$ the operation of replacing the functions $f(x)$ on each side of the equation with the function $f(x)-(1/p)f(x^{1/p})$. This gives successively

$\begin{eqnarray}J(x)-\frac{1}{2}J(x^{1/2})&=&\pi(x)+\frac{1}{3}\pi(x^{1/3})+\frac{1}{5}\pi(x^{1/5})+\cdots,\\J(x)-\frac{1}{2}J(x^{1/2})-\frac{1}{3}J(x^{1/3})+\frac{1}{6}J(x^{1/6})&=&\pi(x)+\frac{1}{5}\pi(x^{1/5})+\frac{1}{7}\pi(x^{1/7})+\cdots,\end{eqnarray}$​

etc, where at each step the sum on the left consists of those terms of the right side of $(2)$ for which the factors of $n$ contain only the primes already covered and the sum on the right consists of those terms of the right side of $(1)$ for which the factors of $n$ contain none of the primes already covered. Once $p$ is sufficiently large, the latter are all zero except for $\pi(x)$."

Can anyone help me understand the "missing bits" please?

 

[SychoScribler]

O.K., it looks like I'm on my own for now.

$\begin{eqnarray}J(x)=\sum_{n=1}^\infty{\frac{1}{n}}\pi(x^{1/n})\end{eqnarray}$​

Let $f_1(x)=J(x)=\sum_{n=1}^\infty{\frac{1}{n}}\pi(x^{1/n})$.

$\begin{eqnarray} f_1(x)-\frac{1}{2}f_1(x^{1/2})&=&f_1(x)-\frac{1}{2}f_1(x^{1/2})\\ J(x)-\frac{1}{2}J(x^{1/2})&=&\sum_{n=1}^\infty{\frac{1}{n}}\pi(x^{1/n})-\frac{1}{2}\sum_{n=1}^\infty{\frac{1}{n}}\pi((x^{1/2})^{1/n})\\ J(x)-\frac{1}{2}J(x^{1/2})&=&\sum_{n=1}^\infty{\frac{1}{n}}\pi(x^{1/n})-\sum_{n=1}^\infty{\frac{1}{2n}}\pi(x^{1/2n}) \end{eqnarray}$​

The multiples of 2 are all gone from the sum on the right hand side.

Let $f_2(x)=J(x)-\frac{1}{2}J(x^{1/2})=\sum_{n=1}^\infty{\frac{1}{n}}\pi(x^{1/n})-\sum_{n=1}^\infty{\frac{1}{2n}}\pi(x^{1/2n})$.

$\begin{eqnarray} f_2(x)-\frac{1}{3}f_2(x^{1/3})&=&f_2(x)-\frac{1}{3}f_2(x^{1/3})\\ \left[J(x)-\frac{1}{2}J(x^{1/2})\right]-\frac{1}{3}\left[J(x^{1/3})-\frac{1}{2}J((x^{1/3})^{1/2})\right]&=&\left[\sum_{n=1}^\infty{\frac{1}{n}}\pi(x^{1/n})-\sum_{n=1}^\infty{\frac{1}{2n}}\pi(x^{1/2n})\right]-\frac{1}{3}\left[\sum_{n=1}^\infty{\frac{1}{n}}\pi((x^{1/3})^{1/n})-\sum_{n=1}^\infty{\frac{1}{2n}}\pi((x^{1/3})^{1/2n})\right]\\ J(x)-\frac{1}{2}J(x^{1/2})-\frac{1}{3}J(x^{1/3})+\frac{1}{6}J(x^{1/6})&=&\sum_{n=1}^\infty{\frac{1}{n}}\pi(x^{1/n})-\sum_{n=1}^\infty{\frac{1}{2n}}\pi(x^{1/2n})-\sum_{n=1}^\infty{\frac{1}{3n}}\pi(x^{1/3n})+\sum_{n=1}^\infty{\frac{1}{6n}}\pi(x^{1/6n}) \end{eqnarray}$​

The multiples of 3 are all gone from the sum on the right hand side and we have added the multiples of 6.

Let $f_3(x)=J(x)-\frac{1}{2}J(x^{1/2})-\frac{1}{3}J(x^{1/3})+\frac{1}{6}J(x^{1/6})=\sum_{n=1}^\infty{\frac{1}{n}}\pi(x^{1/n})-\sum_{n=1}^\infty{\frac{1}{2n}}\pi(x^{1/2n})-\sum_{n=1}^\infty{\frac{1}{3n}}\pi(x^{1/3n})+\sum_{n=1}^\infty{\frac{1}{6n}}\pi(x^{1/6n})$.

$\begin{eqnarray} f_3(x)-\frac{1}{5}f_3(x^{1/5})&=&f_3(x)-\frac{1}{5}f_3(x^{1/5})\\ \vdots\hspace{4em}&=&\hspace{4em}\vdots\\ J(x)-\frac{1}{2}J(x^{1/2})-\frac{1}{3}J(x^{1/3})+\frac{1}{6}J(x^{1/6})-\frac{1}{5}J(x^{1/5})+\frac{1}{10}J(x^{1/10})+\frac{1}{15}J(x^{1/15})-\frac{1}{30}J(x^{1/30})&=&\sum_{n=1}^\infty{\frac{1}{n}}\pi(x^{1/n})-\sum_{n=1}^\infty{\frac{1}{2n}}\pi(x^{1/2n})-\sum_{n=1}^\infty{\frac{1}{3n}}\pi(x^{1/3n})+\sum_{n=1}^\infty{\frac{1}{6n}}\pi(x^{1/6n})-\sum_{n=1}^\infty{\frac{1}{5n}}\pi(x^{1/5n})+\sum_{n=1}^\infty{\frac{1}{10n}}\pi(x^{1/10n})+\sum_{n=1}^\infty{\frac{1}{15n}}\pi(x^{1/15n})-\sum_{n=1}^\infty{\frac{1}{30n}}\pi(x^{1/30n}) \end{eqnarray}$​

The multiples of 5 are all gone from the sum on the right hand side. We have taken the multiples of 30 off as well, and added multiples of 10 and 15. I expected that we would eventually subtract all the multiples of the primes (including the primes themselves) from the sum.

I can see the $\sum_{n=1}^\infty\frac{\mu(n)}{n}J(x^{1/n})$ appearing on the left side, but it is not not clear that the right side will become $\pi(n)$. Also, the book states that "Riemann inverts this relationship by means of the Mobius inversion formula", but I am not seeing where the formula has been applied.

The key must be recognizing that "at each step the sum on the left consists of those terms of the right side of $(2)$ for which the factors of $n$ contain only the primes already covered and the sum on the right consists of those terms of the right side of $(1)$ for which the factors of $n$ contain none of the primes already covered" as stated in the book, but I am not seeing it yet.

Is there anybody out there?