0
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 May 4 '10 at 18:55
  • 1
    $\begingroup$ 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 May 4 '10 at 19:12
  • $\begingroup$ (it seems I assumed the Riemann hypothesis there. :)) $\endgroup$ – David Hansen May 4 '10 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 May 5 '10 at 11:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.