What is the connection with quadratic map While reading Prof. Tao's Wordpress blog. I noticed he mentioned a different function 

$\displaystyle\Lambda_2(n):= \sum_{d|n}\mu(d)\log^2(n/d)\ldots(\ast)$

and said that this function vanishes for numbers with more than two distinct prime factors and this follows from $x\mapsto x^2$ vanishes after being differentiated more that twice.
But how can we reduce the expression $(\ast)$ to differentiating $x^2$? Or am I misunderstanding his statement?



Here's the link to the blog.

If we take equation (2),
$$
\sum_{n\le x}\Lambda(n)\approx x
$$
So aren't we just doing that with $\Lambda_2$?
$$
\sum_{n\le x}\Lambda_2(n)\approx x^2
$$
Specific examples:
$$
\begin{align}
\Lambda_2(6)&=\sum_{k=1,2,3,6}\mu(k)\log^2{\frac{6}{k}}\\
&=(\log3+\log2)^2-(\log^23+\log^22)\\
&=2\log2\log3
\end{align}
$$
$$
\begin{align}
\Lambda_2(30)&=\sum_{k=1,2,3,5,6,10,15,30}\mu(k)\log^2{\frac{30}{k}}\\
&=(\log5+\log3+\log2)^2-(\log3+\log2)^2-(\log5+\log2)^2-(\log3+\log5)^2\\&+\log^22+\log^25+\log^23\\&=0
\end{align}
$$
$\Lambda(abc\cdots d)=0$ where $a$, $b$, $c$ etc are distinct prime factors can be proved using the inclusion-exclusion principle.
