View Single Post
SychoScribler
#2
Dec31-11, 09:36 PM
P: 3
That is where equation [itex](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[/itex]
is inverted to equation [itex](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[/itex].

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

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

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

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