# Apparent indiscrepancy in the average value of phi function

1. Sep 21, 2014

### Whovian

So, while solving a problem a friend came up with involving the Totient function, I ended up doing a bit of research into the average asymptotics of the function. On page 268 of Introduction to the Theory of Numbers, it's mentioned that "The average of order of $\phi\left(n\right)$ is $\frac{6\cdot n}{\pi^2}$. More precisely, $\Phi\left(n\right)=\phi\left(1\right)+\ldots+\phi\left(n\right)=\frac{3\cdot n^2}{\pi^2}+O\left(n\cdot\log\left(n\right)\right)$."

My question is, wouldn't this mean $\frac{\phi\left(1\right)+\ldots+\phi\left(n\right)}{n}\approx\frac{3\cdot n}{\pi^2}$, so the average value of the phi function up to n would be approximately $\frac{3\cdot n}{\pi^2}$ and not, as previously stated, $\frac{6\cdot n}{\pi^2}$?

2. Sep 25, 2014

### Petek

Recall that an arithmetic function $f$ is said to be of the average order of a function $g$ if
$$f(1)+f(2) + \cdots + f(n) \sim g(1) + g(2) + \cdots + g(n)$$
Let $f(n)=\phi(n)$ and $g(n)=\frac{6n}{\pi^2}$.
Then $\phi(n)$ has average order $\frac{6n}{\pi^2}$ means that
$$\phi(1)+\phi(2)+\cdots+\phi(n)\sim\frac{6}{\pi^2}(1+2+\cdots+n)$$
and $1+2+\cdots+n\sim\frac{1}{2}n^2$.