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Apparent indiscrepancy in the average value of phi function

  1. Sep 21, 2014 #1
    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. jcsd
  3. Sep 25, 2014 #2
    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##.
     
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