\varphi(d))^{\varphi(d)}$? [closed]

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17 events

when toggle format	what		by	license	comment
Feb 18, 2015 at 20:23	history	closed	Daniel Loughran Peter Crooks Joonas Ilmavirta Lucia Alex Degtyarev		Needs details or clarity
Feb 18, 2015 at 13:33	review	Close votes
Feb 18, 2015 at 20:23
S Feb 18, 2015 at 2:43	history	suggested	Ethan Splaver	CC BY-SA 3.0	Formatting Error
Feb 18, 2015 at 2:08	review	Suggested edits
S Feb 18, 2015 at 2:43
Aug 18, 2014 at 21:13	comment	added	Terry Tao		I think I made a number of sign errors in my previous comments and are no longer able to edit them to correct this, but the broad point of my comments are still valid even if the specific formulae should not be taken literally.
Aug 18, 2014 at 20:02	comment	added	Emil Jeřábek		In view of the recent edits and the fact that it’s a bit odd for $\varphi(d)$ to appear in an exponent, I wonder whether that’s not a typo for $\mu(d)$ or better yet, $\mu(n/d)$.
Aug 18, 2014 at 19:56	comment	added	Terry Tao		Asymptotics are more interesting; writing $(1 - \frac{\mu(d)}{\phi(d)})^{\phi(d)}$ as $\exp( \mu(d) - \frac{\mu(d)^2}{2\phi(d)} ) (1 + O(\frac{1}{\phi(d)^2} ))$, the product becomes $\exp( -\frac{1}{2} - \sum_{d\|n} \frac{\mu(d)^2}{2\phi(d)}) ( 1 + O( \sum_{d\|n; d>1} \frac{1}{\phi(d)^2} ))$ which can simplify to $\exp(-\frac{1}{2}-\frac{1}{2}\prod_{p\|n} (1+\frac{1}{2(p-1)})) (1 + O( \sum_{p\|n} \frac{1}{p^2} )))$. One can be a bit more accurate about the contribution of the small primes $p$ to the error term if one wants more precise asymptotics.
Aug 18, 2014 at 19:50	comment	added	Terry Tao		If for instance $n=pq$, then the product is $(\frac{p-2}{p-1})^{p-1} (\frac{q-2}{q-1})^{q-1} (\frac{pq-p-q+2}{(p-1)(q-1)})^{(p-1)(q-1)}$ which has no evident cancellation or further factorisation, and only a small amount of like terms to collect. Doesn't seem like there is much else to be done here.
Aug 18, 2014 at 19:30	history	edited	Robert	CC BY-SA 3.0	edited title
Aug 18, 2014 at 18:57	history	edited	Robert	CC BY-SA 3.0	deleted 42 characters in body
Aug 18, 2014 at 17:47	comment	added	Robert Israel		One thing you might note is that your product is $0$ if $n$ is even.
Aug 18, 2014 at 17:45	comment	added	Robert		Yes, I am searching for a "simpler" form of the function for $n$ square-free and I doubt that this can be done. On the other hand, any other information concerning asymptotic behavior of the function is welcomed.
Aug 18, 2014 at 16:40	comment	added	Daniel Loughran		It is not entirely clear to me exactly what you want; you ask for an explicit formula, yet the function is already given to us by a fairly explicit formula. Do you want to know if there is a "simpler" formula? Your function is not multiplicative, so standard tricks to simplify it do not work here. Alternatively, would information about the growth rate/average order/Dirichlet series of your function suffice?
S Aug 18, 2014 at 15:20	history	suggested	Daniel Soltész	CC BY-SA 3.0	The formula loks nicer this way.
Aug 18, 2014 at 15:18	review	Suggested edits
S Aug 18, 2014 at 15:20
Aug 18, 2014 at 15:06	review	First posts
Aug 18, 2014 at 15:18
Aug 18, 2014 at 15:06	history	asked	Robert	CC BY-SA 3.0

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