So far I have  $${{\phi(n)^2} \over n} = n \prod_{i = 1}^k \left(1 - {1 \over p_i}\right)^2$$  We know that $$\left(1 - \frac{1}{p_i}\right) \geq \frac{1}{2}$$  So $$n \prod_{i = 1}^k \left(1 - {1 \over p_i}\right)^2 \geq \frac{n}{4^k} = \frac{n}{2^{2k}}$$  Want to show that $$\frac{n}{2^{2k}} \geq \frac{1}{2}$$  $n \geq  2^{2k - 1}$  Got stuck here. Any help to prove this?  EDIT: Ok so I looked at the answer in the thread provided and I don't understand how $\frac{\phi(n)^2}{n}= \prod_{p|n \ \text{prime}} \frac{(p^{a_p-1}(p-1))^2}{p^{a_p}}$  There's a factor of n missing since the $\phi$ function is being squared
So I'm trying to prove the following two inequalities: $$\frac{\sqrt{n}}{2} \leq \phi(n) \leq n.$$ The upper bound we get from simply noting that $\phi(n) = n \prod_{p | n}\left( 1 - \frac{1}{p}\right)$ and the fact that $(1 - \frac{1}{p}) \leq 1$. But how can we get the lower bound? I tried using the same expression for $\phi(n)$ but it seems to not really give me the inequality. Can you help?
$\gcd(a,b,c)=\gcd(\gcd(a,b),c)\,$ [GCD Associative Law] Deleting self-answered question gets "Delete this answered question" warning Accidentally ejected External drive while downloading file, now having problems reading I'm hoping someone can help! I recently downloaded a torrent directly onto an external hard drive, (using a 2009 MacBook Pro and 1TB Seagate Expansion portable Drive) somehow during the download the drive was accidentally ejected, since this incident when I plug the drive back in it’s having problems reading.  The drive works fine for about 10 minutes, I can read/copy/paste and access files but then starts to wreak havoc; apps hangs, it hangs the Finder, etc.  I’ve tried repairing it in Disk Utility, which reads and repairs this disk but problems inevitably continue as described above.  I’ve deleted the Torrent file also. All my files on this external drive so re-formatting is not an option until I can retrieve them. Removing all fraction symbols like “¼” and “½” from a string SQL Server nvarchar(max) vs nvarchar(n) affects performance A specified Include path is not valid. The EntityType does not declare a navigation property with the name * Word for "willing to try new and unfamiliar things" Example:  A: Hey, do you like to do bungee jumping? B: Um, OK. Haven't done it before, but let's try it.  I'm not looking for adventurous or something similar. That word should also fit in less action-packed activities:  A: I have a new board game. Do you like to play? B: I don't know this game and, actually, I don't like these varieties of games but I'll play. Maybe it's fun.  First I thought about open-minded but I think this means more willing to listen to people, think about their ideas, arguments, ... and accept them (not necessarily agree). What word describes person who is willing to try activities he doesn't know or hasn't experienced before? In grade school we learn to rationalize denominators of fractions when possible. We are taught that $\frac{\sqrt{2}}{2}$ is simpler than $\frac{1}{\sqrt{2}}$.  says that "there is a bias against roots in the denominator of a fraction". But such fractions are well-defined and I'm failing to see anything wrong with $\frac{1}{\sqrt{2}}$ - in fact, IMO it is simpler than $\frac{\sqrt{2}}{2}$ because 1 is simpler than 2 (or similarly, because the former can trivially be rewritten without a fraction).  So why does this bias against roots in the denominator exist and what is its justification? The only reason I can think of is that the bias is a relic of a time before the reals were understood well enough for mathematicians to be comfortable dividing by irrationals, but I have been unable to find a source to corroborate or contradict this guess. Which is correct: "he don't" or "he doesn't"? Preventing applications from stealing focus Should questions include "tags" in their titles? How can I put a bit mask on /dev/zero so that I can get bytes other than zero? How can I put a bit mask on /dev/zero so that I can have a source not only for 0x00 but also for any byte between 0x01 and 0xFF? I'd love to see if there are trends over time, patterns of improving behaviour after being flagged, consistent reoffenders, most misused types of flag, etc.  I understand the who-said-what-about-who aspect would most likely cause this to be denied, but even anonymised data would be interesting to dig through. Are there experiments that are banned from being done at the LHC? I am quite proficient with the keyboard and try to only use the mouse when necessary. Given that context I was very familiar with the shortcut keys in MS Outlook 2003 and never needed to use the mouse to use that product. Now my company has recently upgraded to MS Outlook 2010 and I am finding it impossible to operate it without having to use the mouse. In particular replying to meeting requests with the old shortcut keys no longer seems possible.   I have held down the ALT key and sometimes it shows some shortcuts on the ribbon but this seems to be limited to ALT+H which tends to alternate between designating "Accept" and "Tentative" without any keys for the other options. Also as this appears to switch between "Accept" and "Tentative" one can't even use this consistently without first having to check what it means for each particular instance.  Any help with this would be greatly appreciated! A conjectured closed form of $\int\limits_0^\infty\frac{x-1}{\sqrt{2^x-1}\ \ln\left(2^x-1\right)}dx$ Calculating $\int_{|z|=2}\frac{e^{1/z^2}}{1-z}dz$ Find the value of $\displaystyle\int_{|z|=2}\frac{e^{1/z^2}}{1-z}dz$   The first idea I had was to apply Cauchy's integral formula. But can this be done? $e^{1/z^2}$ is not holomorphic on the interior of $|z|=2$. Still, if we would apply this we would obtain:  $\displaystyle\int_{|z|=2}\frac{e^{1/z^2}}{1-z}dz=-2\pi ie$.  Another way I think, is to apply the residue theorem. For $1-z$ has a simple zero at $z=1$ and $e^{1/1^2}\neq 0$, then the residue of $e^{1/z^2}/(1-z)$ at $z=0$ is $-e$. Again we obtain $2\pi iRes(g;0)=-2\pi ie$, where $g=\frac{e^{1/z^2}}{1-z}$.  But again, it seems to me this is not right because $g$ has two singularities, at $z=1$ and $z=0$. What we might do about $z=0$? The answer is not $-2\pi i e$? Can I see things that I've said in chat that have been starred? I was reviewing the star on chat.SE list this morning and I noticed that quite a few of the items on the list were mine.  I was curious as to how often people enjoy the things I post there enough to star them, but I can't seem to find statistics about this anywhere.  Mana told me that no such feature exists.  (the best we could come up with is ) So here's my request - can I filter/search the chat history for things I've said that other people have starred? I have heard a few times that one way of describing quantum computers is that they essentially use the computing power of their counterparts in alternate realities that they access through superposition. My first question is, of course,    is this actually an accurate description of how quantum computers work, or just a misrepresentation?   Also, if it were to be assumed true and taken literally, then presumably all the possible outcomes of any given computation done would be experienced by each of these alternate realities. I have a few questions regarding the implications of this:   Would it be correct to say that our universe essentially "spawns" these alternate realities to use for each computation, or that they all exist simultaneously in a higher dimension and only come into contact via the initiation of superposition by each computer at the same time? Would it be possible to create contact between universes via this connection? I'd expect that, if it was, it would be entirely random, and so not only would not cause any information transfer, it also would be completely undetectable, because the deviations would be within the expected probabilities of error rates - but is there any way to circumvent this? Calculating average raster value for area in ArcGIS Desktop? I have Raster data for a country which has Evaporation Data in its pixels. I also have Vector layer which has the 30 cities of the country.   What I need is the average Evaporation value of each city from the raster data using ArcGIS?  I have access to the spatial analyst extension.