I run a site where multiple photographers upload and sell their photos to my users.  Some of them just want to buy a JPEG, some of them want to buy a JPEG to print.  You and I both know that a 6000x4000 72dpi file is essentially "identical" to a 6000x4000 300dpi file, but  a) users don't always b) more importantly, some online printers don't seem to.  I've come across more than one printer who requires photos to be 300dpi, even though simply changing a file's dpi metadata makes no difference to the image data.  If I were to set the dpi EXIF of all photos bought on the site would I end these headaches for people?  Would it cause any problems anywhere else?  Do some computers/phones/softwares treat 300dpi images on a screen differently to the "standard" 72dpi image on a screen?  I don't believe there SHOULD be any problem with doing this... but I wanted to tap into the collective knowledge here to double check.  Thanks
I often see statements in discussions about web displayed image quality that go something like this: "I only upload pictures at 72 dpi and 1200 pixels on the long edge, so if someone copies and prints the image it won't look as good as if it were 300 dpi."  What? Have I missed something here?  Does an image edited and saved/exported with 1200x800 pixels at 300 ppi look any different online than the same image saved/exported with 1200x800 pixels at 72 ppi? Other than the metadata regarding ppi, is there any difference in the two images at all? If so, what is the difference? If I tell my printer to print the images at 4R (4X6 inches) will there be any difference at all in how the printer creates a print from the 1200x800 @ 300 dpi image versus the 1200x800 @ 72 dpi image?
Getting equation from differential equations I have:  $\dfrac {dx} {dt}$=$-x+y$  $\dfrac {dy}{dt}$=$-x-y$  and I am trying to find $x(t)$ and $y(t)$ given that $x(0)=0$ and $y(0)=1$  I know to do this I need to decouple the equations so that I only have to deal with one variable but the decoupling is what I am having trouble with  Do I set them equal to each other and then just move like terms to separate sides getting two different equations and then integrate? In another  regarding a difference between short on and short of, the  says the following:     ... But this is not always the case. A person can be short on/of money,   but lacking a specific amount, someone can only be €10 short or short   €10. This construction never uses a preposition.   but is that really true? I can see sentences like "I am short of 10 dollars" quite often. Does DTIME(O(n)) = REGULAR? So I know that two particles can be entangled in a quantum way, but is it possible that more than two particles be entangled in a quantum way? Most descriptions provide with two-particles cases, so I wonder. (It's hard to think of three particles entangled in spin, or so.) Regular expression: match start or whitespace Can a regular expression match whitespace or the start of a string?  I'm trying to replace currency the abbreviation GBP with a £ symbol. I could just match anything starting GBP, but I'd like to be a bit more conservative, and look for certain delimiters around it.  &gt;&gt;&gt; import re &gt;&gt;&gt; text = u'GBP 5 Off when you spend GBP75.00'  &gt;&gt;&gt; re.sub(ur'GBP([\W\d])', ur'£\g&lt;1&gt;', text) # matches GBP with any prefix u'\xa3 5 Off when you spend \xa375.00'  &gt;&gt;&gt; re.sub(ur'^GBP([\W\d])', ur'£\g&lt;1&gt;', text) # matches at start only u'\xa3 5 Off when you spend GBP75.00'  &gt;&gt;&gt; re.sub(ur'(\W)GBP([\W\d])', ur'\g&lt;1&gt;£\g&lt;2&gt;', text) # matches whitespace prefix only u'GBP 5 Off when you spend \xa375.00'   Can I do both of the latter examples at the same time? Mint 18 will boot only to a black screen The optical density of a medium is quantified by its refractive index $n$. Does it have any correlation with the mass/number density of the medium? In other words is an optically denser medium must have a greater mass density compared to an optically rarer medium? How Many Genius Mixes Are There in iTunes? Is it always better to use 'DbContext' instead of 'ObjectContext'? How to remove an element from a list by index Yesterday, my laptop caught a virus which caused massive damage. Since them, I have been trying to recover important files before reformatting my computer, a task the virus has not made easy.   Restoration points predating the attack have been deleted.  Most of my folders show empty.  My Start menu is essentially empty, with the exception of Trillian and Mirror's Edge. The same goes for my Desktop, which only has programs which were installed after the attack.  Searching for files through my computer is pretty much useless, as it only rarely brings up anything. I suspect most of my files have not been deleted. While my folders show empty, uTorrent still does display them and I can open them from here. Unfortunately, when I select Open Containing Folder, the folder still shows as completely empty even if I'm currently watching a video from that very folder. Further adding evidence to the not-deleted-just-missing theory, the data recovery software I'm using (Restoration) cannot find only find an handful of the missing files.   If they were deleted, I could do a forensic recovery to get them back but since they're probably still somewhere on my computer, just out out of my reach, I can't find them.   Under those circumstances, is there a way I can recover those files? Execute a large SQL script (with GO commands) Finding all solutions for a system of equations with constraint on the sum of absolute values I am looking for a fast algorithm to find all integer solutions for the following system of linear equations ($c_1,\ldots,c_n \in \mathbb{Z}$ and $r\in \mathbb{N}$ ):  $x_1+x_2=c_1$  $x_3+x_4=c_2$  ...  $x_{2n-1}+x_{2n}=c_n$  $|x_1|+|x_2|+\ldots+|x_{2n}|=r$  $x_1,\ldots,x_{2n} \in \mathbb{Z}$   Also can we find the number of solutions before solving the system? How to schedule jobs in Sql server 2008 for less than 10 sec? I want to run a job every 3 seconds, however in SQL Server 2008 we cannot define an interval of less than 10 seconds.  The job is used to insert/update visitor information, and segmentation information into a database which is tracked by google search.   There are up to about 100 rows inserted in a 2 or 3 seconds. That job inserts and update the table in a database. Is there any way to schedule it using sp job scheduling configuration? What happens if I am afflicted with a condition that worsens on a failed save and choose not to try to save? Do I:   Automatically suffer the effects of the failed save? Keep my current condition? Have no choice and must roll the save regardless?   This came up in a game earlier this afternoon when we decided to just stop saving against the effect that was causing us to fall unconscious. Uses of the 3D Cursor I need an appearing effect for text like the one (in javascript) here .  However I don't know how to insert js code into LaTeX (apart from using the AcroTeX Bundle) and that would also force me to use Acrobat Reader, so I'd like to implement a similar effect in Beamer. Is it possible?  (of course I would love to see alternative approaches to this) My internet connection is very slow. I suspect someone in the network is using a BitTorrent client which takes huge amount of bandwidth.  Is there a way to detect who is using my bandwidth (network sniffing tools etc)?  I should have pointed out that I have no access to the router, what I am looking for is a way of detecting someone's MAC/IP if she uses Bittorrent. Entries I can safely exclude doing backups If $(a,b)=1$ then prove $(a+b, ab)=1$. Let $a$ and $b$ be two integers such that $\left(a,b\right) = 1$. Prove that $\left(a+b, ab\right) = 1$.   $(a,b)=1$ means $a$ and $b$ have no prime factors in common  $ab$ is simply the product of factors of $a$ and factors of $b$.  Let's say $k\mid a+b$ where $k$ is some factor of $a$.  Then $ka=a+b$ and $ka-a=b$ and $a(k-l)=b$.  So $a(k-l)=b, \  a\mid a(k-1)$ [$a$ divides the left hand side] therefore $a\mid b$ [the right hand side].  But $(a,b)=1$ so $a$ cannot divide $b$.  We have a similar argument for $b$.  So $a+b$ is not divisible by any factors of $ab$.  Therefore, $(a+b, ab)=1$.  Would this be correct? Am I missing anything?