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It is well-known that in average $\varphi(n)$ behaves like $\frac1{\zeta(2)}n=\frac{6}{\pi^2}n$. But it looks that in some sense it is ``asymptotically larger''. In particular, the ratio $$ \zeta(2)(1-t)^2\sum_{n=1}^{\infty} \varphi(n)t^{n-1}= \zeta(2)\frac{\sum_{n=1}^{\infty} \varphi(n)t^{n-1}}{\sum_{n=1}^{\infty} nt^{n-1}} $$ seems to be greater then 1 when $t$ increases to 1 (my caclulations say so), and analogous things appear for other averaging procedures involving $\varphi(n)$.

Does it have some sense and/or explanation?

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Write $S(t)= \sum \varphi(n)t^n$. A standard calculation gives

$S(e^{-u})= \frac{1}{2\pi i}\int_{(3)}\frac{\zeta(s-1)}{\zeta(s)}\Gamma(s)u^{-s} ds$,

so pushing the contour to $Re(s)=3/2$ (say) gives $S(e^{-u})=\zeta(2)^{-1} u^{-2}+O(u^{-\frac{3}{2}})$ as $u\to 1$. How many $n$ are you using in your numerical calculations?

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  • $\begingroup$ Yes, thank you, but why this remainder seems to have fixed sign? It suffices to take $1000$ summands fo $t=0.98$ and 2000 for $t=0.99$ in $\sum \varphi(n)t^n$ to get value grater then \zeta(2)^{-1}(1-t)^{-2) , each next summand only increases the sum. Another sum, in which I am more interested, $u^2\sum_{n=1}^{\infty} -\varphi(n) \ln(1-e^{-nu})$ is greater then limit value $\zeta(3)/\zeta(2)$ when $u$ decreases to 0, while analogous sum for $n$ instead $\varphi(n)$ increases for $u\searrow 0$. $\endgroup$
    – Fedor Petrov
    Commented May 4, 2010 at 18:55
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    $\begingroup$ Push the contour to (-1/2) instead. You'll get residues coming from poles of the $\zeta$-function along the 1/2-line, but the first zero is at 1/2+14.1347I, and Gamma(1/2+14.1347I) is of size 10^{-10}. So the contribution of these residues will be invisible to the naked eye. The pole at $s=0$ has residue 1/6. So to the eye it seems that S(e^-u)=zeta(2)^-1 u^{-2} + 1/6 + o(1) as $u\to 0$, but in reality there are wiggly terms of such small magnitude the eye cannot see them. $\endgroup$
    – David Hansen
    Commented May 4, 2010 at 19:12
  • $\begingroup$ (it seems I assumed the Riemann hypothesis there. :)) $\endgroup$
    – David Hansen
    Commented May 4, 2010 at 19:13
  • $\begingroup$ Oh, thank you, now I see, why it may seem so! The little problem is that $$(1-e^{-u})^2(u^{-2}+1/6)$$ is less then 1 when $u\searrow 0$, while evaluation suggest that $\zeta(2)S(t)(1-t)^2$ is greater then 1 when $t\nearrow 1$. But probably this has some stupid explanation, like miscalculation somewhere :) $\endgroup$
    – Fedor Petrov
    Commented May 5, 2010 at 11:52

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