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 P: 3 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?