- #1
Whovian
- 652
- 3
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}##?
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}##?