I've just purchased a new Olympus E-M1 Mark II from Jessops as Brand New.  It's my first semi-expensive camera so I'm looking for opinions. At the moment there is a promotion that gives you cashback when you purchase this camera, as with all my products I'll register the warranty as soon as I receive it, did the same with the cashback.   In this case I decided to check the shutter actuation's as recommended by a friend but slipped my mind until I've submitted the warranty etc.  The MS: 347  Have spoken to Jessops and they say there is nothing they can do (In terms of a refund or replacement) as I've registered the product and can only offer me £15 for the trouble.  So my question's are:   Is this normal? If it isn't, how much should I be asking for back? (Body was £1400).   Thank you all.
How many Shutter Clicks on a New Camera considered normal?    I purchased a quotes new Nikon D750 and checked the shutter count and it has 264 Shutter Clik count on it.    Is that suspect of a returned purchase or used camera?
Why does a Lipschitz function $f:\mathbb{R}^d\to\mathbb{R}^d$ map measure zero sets to measure zero sets? Why does a Lipschitz function $f:\mathbb{R}^d\to\mathbb{R}^d$ map measure zero sets to measure zero sets?   It is easy to prove this statement if the domain is bounded. Is there any way to extend the argument to unbounded domains? Can anyone give me some hint, or show me a simple proof sketch? I installed Android studio some time ago using the following suggested method:     To set up Android Studio on Linux:      Unpack the downloaded Tar file,   android-studio-ide--linux.zip, into an appropriate location   for your applications. To launch Android Studio, navigate to the   android-studio/bin/ directory in a terminal and execute studio.sh. You   may want to add android-studio/bin/ to your PATH environmental   variable so that you can start Android Studio from any directory.   Follow the links to install the SDK outside of the Android Studio   directories.   For whatever reason it has become corrupt and insists I must un-install and re-install. I can't for the life of me find how to un-install it however.  I didn't use apt-get so I can't do apt-get remove.  Does anyone have any suggestions? Tracking email bounces, opens, clicks Where did the "Magic Blood" of the Harry Potter Universe Originate From? What is a wizard that makes them non-Muggle? Or, more directly, what ancestry does the witch/wizard gene follow? Is there a patriarchal wizard (or matriarchal witch) that Muggles have no part in? Elves perhaps? Stop reading here.  Purpose of the Question/History/Don't Read There seem to be solid consequences associated with being Muggle, or Muggle-related. A dominant theme in the novels is discrimination against the non-gifted race, but, when read into, the wizard bloodline doesn't appear all that well defined. JKR indicates that wizard-ness is genetic, or inherited:  &quot;Muggle-borns will have a witch or wizard somewhere on their family tree, in some cases many, many generations back. The gene resurfaces in some unexpected places.&quot;  By the above, it stands that a family that produces a squib (non-magical born of magical parents) must be a half-blooded family in some way, since half-blooded only requires one non-magical ancestor. Of course, we are dealing in magic genetics. Any other form of genetics would make this inherited, seemingly recessive trait (or squibs couldn't exist), never appear at all. Unless they were XMen. If it is so important, what does the wizard-bloodline follow? Other Mythos In other literature a wizard can be nearly anything, but magical powers are typically inherited as well as learned. Merlin of the Arthurian Legend (also mentioned in HP) is Cambian, meaning he is born of demonic intervention; this explains his source of enchantment. Gandalf of Middle Earth is a Maiar, as are all Wizards of Middle Earth. Wizards in Harry Potter seem to follow a bloodline, and it has been mentioned (in the question Gabe linked) that genetics have a heavy play in it, meaning that squibs and parents of Muggle-born wizards should have identical genetics. This bloodline must come from somewhere. Magical Creatures In Harry Potter it is common for magical creatures to &quot;mingle&quot; with humans, these beings are typically somewhat magical. Elves have been described to have very strong magic, but are for some unreasonable reason subservient to humans. I may be mistaken, but believe it was hinted that elves and the wizarding lineage had intersected at one time. Maybe I'm imagining this. Genocidal Allusions Voldemort hates muggles, and half-bloods, and whatever else is non-wizard. The presence of this theme, suggests that this wizardly bloodline must be at least defined if the main antagonist is so obsessed with keeping that mysterious bloodline pure. Since Slytherin himself was said to be a supremacist (circa a really long time ago) it suggests that this bloodline should have existed long before he was around, especially if he founded a school with people he was not related to, but were also magical. The square roots of different primes are linearly independent over the field of rationals Homemade coconut milk separates into different layers after a day in the fridge Why does the natural ring homomorphism induce a surjective group homomorphism of units? I'm trying an old problem here:   Suppose $n\mid m$, and I have a natural ring homomorphism $\varphi\colon \mathbb{Z}/m\mathbb{Z}\to\mathbb{Z}/n\mathbb{Z}$ defined by $\varphi(j)=j\mod{n}$. I can verify that this is a ring homomorphism, but why does it induce a surjective group homomorphism on the unit groups $(\mathbb{Z}/m\mathbb{Z})^\times\to(\mathbb{Z}/n\mathbb{Z})^\times$?  My question in particular is why is it surjective? I take $k\in(\mathbb{Z}/n\mathbb{Z})^\times$. Then there exist integers $s,t$ such that $sn+tk=1$. Since $n\mid m$, I can also write $m=na$. So multiplying through I get $sm+tka=a$. I'm lost here. How can I find a unit in $(\mathbb{Z}/m\mathbb{Z})^\times$ which maps to $k$ to show surjectivity? I read that it is a best practice to double quote all of your variable expansions in Bash. I also read that one cannot use a shell glob (wildcard(*)) right after a double quoted variable expansion. These situations conflict when one wants use the value of a regex pattern, expanded out of a variable.   The reason I would sometimes want to combine regex with a wildcard is to keep my regex minimal, neater, in my personal taste.  My particular problem  I downloaded phpmyadmin into my document root directory and unzipped it, but I fail to rename it with mv by a regex pattern I put in a variable, and available when expanding its variable. Here's the exact trace:  userName@compName:/var/www/html# ll total 11336 drwxr-xr-x  3 root root     4096 Feb 14 07:04 ./ drwxr-xr-x  3 root root     4096 Feb 14 06:56 ../ -rw-r--r--  1 root root      612 Feb 14 06:57 index.nginx-debian.html drwxr-xr-x 12 root root     4096 Dec 23 08:50 phpMyAdmin-4.7.7-all-languages/ -rw-r--r--  1 root root 11589684 Dec 23 14:08 phpMyAdmin-latest-all-languages.zip userName@compName:/var/www/html# echo $pma [pP][hH][pP][mM][yY][aA][dD][mM][iI][nN] userName@compName:/var/www/html# mv "$pma"*/ phpmyadmin/ mv: cannot stat '[pP][hH][pP][mM][yY][aA][dD][mM][iI][nN]*/': No such file or directory   If I'll unquote the variable expansion to ${pma} I would indeed be able to combine the variable expansion and the regex, as in ${pma}*, but it is important to me to follow the best practice without exceptions, if I can.  My question  How could I keep the variable expansion double quoted, but still using the extracted value with a wildcard? Sometimes I want to learn about a new programming related issue—topics such as DB ORM, search indexer, identity providers, logging, etc.  When I don't know anything about a subject, I ask for benchmarks or comparisons. I got some serious answers like links to tests, pros/cons, and Gartner's reviews. I found these very constructive and helpful. But then the question was blocked or deleted. Sometimes this happens even before I get an answer.  So where would it be more appropriate to ask questions about third party products and/or open source frameworks with which I am completely unfamiliar? Define $\sigma(m) = \sum$ d : d|n.  Prove that $p^j$$q^i$ cannot be a perfect number for $p, q$ odd, distinct primes.  Attempt at Solution: I have shown that $p^k$ can never be a perfect number, and im trying to use the multiplicative property of $\sigma$ to generalize to $p^j$$q^i$ What's the name of each pseudo-rectangle in a spherical surface? If $\omega \in \Omega^q(M)$, and $(\omega|_{U})_p = 0$, is $\omega_p = 0$? If I have a manifold $M$ and $\omega \in \Omega^q(M)$, considered an open subset $U$ of $M$, and choose a point $p \in M$, is it true that $(\omega|_{U})_p=0 \rightarrow \omega_p=0$? With $\omega|_{U}$ I mean the pull-back of $\omega$ by the embedding $F:U\to M$. The converse is obviously true, but this implication? Another word for a troll not on the internet The definition of an internet troll:     In Internet slang, a troll (/ˈtroʊl/, /ˈtrɒl/) is a person who sows   discord on the Internet by starting arguments or upsetting people,   by posting inflammatory,[2] extraneous, or off-topic messages in an   online community (such as a forum, chat room, or blog), either   accidentally[3][4] or with the deliberate intent of provoking readers   into an emotional response[5] or of otherwise disrupting normal   on-topic discussion.[6]      This sense of the word troll and its associated verb trolling are   associated with Internet discourse, but have been used more widely.   Media attention in recent years has equated trolling with online   harassment. For example, mass media has used troll to describe "a   person who defaces Internet tribute sites with the aim of causing   grief to families."[7][8]     What is the equivalent for a person who does this type of thing in real life (non-cyber) troll? I'm downloading some images from a service that doesn't always include a content-type and doesn't provide an extension for the file I'm downloading (ugh, don't ask).  What's the best way to determine the image format in .NET?  The application that is reading these downloaded images needs to have a proper file extension or all hell breaks loose. I know from the  that it is not possible to introduce direct abbreviations like \newcommand{\bal}{\begin{align}} % doesn't work \newcommand{\eal}{\end{align}}   % doesn't work  or \newcommand{\bal}{\align}    % doesn't work \newcommand{\eal}{\endalign} % doesn't work  What is puzzling me is that \newenvironment{foo}{\align}{\endalign}  does work, at least if one is well-behaved enough to diligently type \begin{foo}...\end{foo}, since \foo...\endfoo won't work either. I'd really appreciate some help to shed some light on what is going on behind the scenes :-) How do I delete all Folders with a specific name (e.g. 128Kbps_Songs) without deleting its files  For example if I have directory "MP3_SONGS" with subdirectories "A", "B", "C" and each subdirectory has MANY directory Contains Film Name "FILM_NAME1", "FILM_NAME2", "FILM_NAME3" in it, In That Each Film NAme I have specific Folder NAme (128Kbps_Songs) in this "128Kbps_Songs" Folder I have That Film mp3 Songs Files  how can I remove the Folder "128Kbps_Songs", in Each Film_Name directory and Have to get these mp3 songs in that FILM_NAME..   ( Have to Delete 128Kbps_Songs Folder in all FILM_NAME without deleting the mp3 Songs in it, have to get the mp3 in FILM_NAME Directory..  MP3_SONGS (Parent Directory) ------ A , B, C  A (Sub Directory1) ------ FILM_NAME1 , FILM_NAME2 B (Sub Directory2) ------ FILM_NAME3 , FILM_NAME4 c (Sub Directory3) ------ FILM_NAME5 , FILM_NAME6  FILM_NAME1 (Directory) ---- 128Kbps_Songs (Directory) ------ FILM_NAME1 (MP3 Files) FILM_NAME2 (Directory) ---- 128Kbps_Songs (Directory) ------ FILM_NAME2 (MP3 Files)  FILM_NAME3 (Directory) ---- 128Kbps_Songs (Directory) ------ FILM_NAME3 (MP3 Files) FILM_NAME4 (Directory) ---- 128Kbps_Songs (Directory) ------ FILM_NAME4 (MP3 Files)  FILM_NAME5 (Directory) ---- 128Kbps_Songs (Directory) ------ FILM_NAME5 (MP3 Files) FILM_NAME6 (Directory) ---- 128Kbps_Songs (Directory) ------ FILM_NAME6 (MP3 Files)  Need Files in: FILM_NAME1 (Directory)  ------ FILM_NAME1 (MP3 Files) FILM_NAME2 (Directory)  ------ FILM_NAME2 (MP3 Files)  FILM_NAME3 (Directory)  ------ FILM_NAME3 (MP3 Files) FILM_NAME4 (Directory)  ------ FILM_NAME4 (MP3 Files)  FILM_NAME5 (Directory)  ------ FILM_NAME5 (MP3 Files) FILM_NAME6 (Directory)  ------ FILM_NAME6 (MP3 Files) Why are hand-built wheels considered to be better than machine-built wheels? Why are hand-built wheels considered to be better than machine-built wheels? I would think that a machine-built wheel would be better and cheaper, since machines generally do a more precise job than humans. A groupoid is defined to be a category where every morphism is an isomorphism. So sometimes a group is said to just be a groupoid with one object.   When I try to make sense of this, I denote the single object as $G$. I view the morphisms as the analogue of "elements." The identity $1_G$ is the analogue of the usual identity $e$, we can compose any two morphisms since they are all arrows on $G$, and for any arrow $f$, we have some $f^{-1}$ such that $f\circ f^{-1}=f^{-1}\circ f=1_G$, so the idea of inverses is still there.  So I informally associate the elements of the group in the usual definition to be the arrows in the groupoid. But what does the sole object $G$ in the groupoid "correspond to" if I were to try to informally make sense of a group in the usual sense? Does it even correspond to anything? If one is asking for a person, what should s/he say&mdash;1 or 2?   Could I speak to Kristina? Could I speak with Kristina? I don't have administrator account on my mac On my mac I have 2 accounts and the both are normal account. So in this moment I don't have a Administrator account, only one that was in the past a administrator, but now is not.  How I can have again a Administrator account?