Is it safe to delete lost.dir files to save space in an android cellular phone, or is it better to move it to the sd card?
I've always had a LOST.DIR folder on my external SD card, which until now has been empty.  Today my phone complained that my SD card was damaged and wouldn't mount it, so I pulled it out and looked at it on my computer.  There's now a 70237 file inside LOST.DIR which is 128 MB.  It starts with something like "&euml;&lt;MSWIN4.1", seemingly to indicate that it's a dump of a  (which makes no sense).  What did Android dump here and and why?  (I think I may have disconnected my phone from my PC yesterday without ejecting the drives first, so that might be the "why", though it was from Windows XP and not Windows 98.)  Is there a simple way to recover the information in this file?
Position vs Part Currently my spouse and I rent a small apartment.  Eventually we'd like to expand the family, and a one bedroom rental won't cut it forever - even if we continue renting we'll be expanding to a detached single family home eventually.  In our area the rent ratios work out to about 16, which is under the cut for buying being a better financial decision over the long haul - i. e. I don't think it'd be economically an awful decision to buy (NYT calculator shows 10 years at the current prices and rents for buying to make sense).  I've also read enough of the rental propagandists to know that home ownership isn't the best way to grow my money.  My reasons for wanting to own are the intangibles: knowing we aren't going to have to move due to a landlord, being able to modify the place to our own liking, and that intangible sense of "this is mine" (even when it's the bank's for a while).  The things we do have:   Around (at or over) a 30% down payment depending on purchase price, and we won't be under 30% Closing costs separate from that fund Around 1% of the home value in reserve for emergency repairs separate from those funds A six month emergency fund separate A fund to replace an ailing car that will probably die before we can save up after this expense   At some point in the year or so after the purchase we plan to have enough saved up to pay down the principal and change our monthly payment somewhat, but not significantly.  So it seems like we've got the financial ducks in a row.  We can continue to fully fund our retirement accounts and make the monthly payments (PITI).  The only hesitation we have is that we can't make our payments comfortably on one income alone - it'd be over 30% of one income to do so, by just a bit.  So neither of us can completely stop working, although the second income does not need to be large.  Are we missing any important financial data about this step?  Should we be concerned about job loss (even though no signs of it are imminent for either of us)? Changing Posts Per Page and offset with pre_get_posts I have a filter using pre_get_posts to change the number of posts per page from 4 on the news landing page to 8 on all other pages (page 2, 3, etc). However I don't seem to have got the offset correct, or something, because my get_next_posts_page_link doesn't show a page 3, even though it does exist if I go to the url manually. The current offset is something I somehow managed to conjure up accidentally, I thought it worked as page 2 was correct but as I added more test posts, I realised it didn't work.  function page_2_posts_per_page_filter( $query ) {      $paged = (get_query_var('paged')) ? get_query_var('paged') : 1;      if ( !is_admin() &amp;&amp; $query-&gt;is_paged &amp;&amp; $query-&gt;is_main_query() ) {          $offset = 4 + (($paged - 2) * 8);         $query-&gt;set( 'offset', $offset );         $query-&gt;set( 'posts_per_page', 8 );      } else if (!is_admin() &amp;&amp; $query-&gt;is_main_query()) {          $query-&gt;set( 'posts_per_page', 4 );      }     return; } add_action( 'pre_get_posts', 'page_2_posts_per_page_filter', 1 );   EDIT: SOLUTION  function page_2_posts_per_page_filter( $query ) {      $paged = (get_query_var('paged')) ? get_query_var('paged') : 1;     $offset = 4;      if ( !is_admin() &amp;&amp; $query-&gt;is_paged &amp;&amp; $query-&gt;is_main_query() ) {          $offset = (($paged - 1) * 8) - $offset;         $query-&gt;set( 'posts_per_page', 8 );         $query-&gt;set( 'offset', $offset );      } else if (!is_admin() &amp;&amp; $query-&gt;is_main_query()) {          $query-&gt;set( 'posts_per_page', 4 );      }     return; } add_action( 'pre_get_posts', 'page_2_posts_per_page_filter');  function found_offset( $found_posts, $query ) {      $offset = 4;      if( !is_admin() &amp;&amp; $query-&gt;is_main_query() ) {         $found_posts = $found_posts + $offset;     }     return $found_posts; } add_action( 'found_posts', 'found_offset', 10, 2 ); Jordan normal form for a characteristic polynomial $(x-a)^5$ Write down all the possible Jordan normal forms for matrices with characteristic polynomial $(x-a)^5$. In each case, calculate the minimal polynomial and the geometric multiplicity of the eigenvalue $a$. Verify that this information determines the Jordan normal form.  I found this question in a textbook that I'm using for a test I have tomorrow. I think that I need to use the method for finding the Jordan normal form of a matrix but I can't see how to apply it and I don't have much intuition about the answer...  I'm guessing that there are 5 possibilities since the minimal polynomial can be any factor of the characteristic, but I don't know how to prove this.  Some help would be great for my test tomorrow! Thanks Putting a table next to a figure I am working with a basic  which is a popular modal used on many sites. I noticed that the default template comes with both a close button on the bottom, as well as an x on the top, these two buttons essentially having the same purpose. Are the two redundant and the close button completely unnecessary and just taking up extra space? Or, is having multiple usage options more user friendly?   To make this even more fun, suppose I throw in a Save button:   Is it now clearer that if you don't press save and just press the x, your changes will be canceled, or is there now more of a need for a cancel button for users who are scared that a mistake will become permanent? Update: To clarify for those who feel this question is similar to  - That question is asking if a clickable background is sufficient to close a popup. I'm using buttons and it is just a matter of how many buttons to use. (I do also have a clickable overlay and ESC key enabled, but that is irrelevant to my question) 0 ram free on ICS Why can I use a function before it's defined in JavaScript? Impact of hyperref, when varioref and cleveref are used Book identification about boy who gets computer zapped into head When $d$ is in $\mathbb{Z}$, $\mathbb{Z}[\sqrt{d}] = \{a_0 + a_1 \sqrt{d} \mid a_0, a_1 \in \mathbb{Z}\}$  However, when $d$ is in $\mathbb{Q} \setminus \mathbb{Z}$, I cannot prove it.  Of course, when I see an example, I can understand that this is not a finitely generated abelian group, but I want to know the proof in general. What are the pros and cons of engineered hardwood? Are there any downsides to engineered hardwood? I'm looking to install hardwood in dining room and kitchen areas. Prove that $\lim \limits_{n \to \infty} \frac{x^n}{n!} = 0$, $x \in \Bbb R$. Answer appears automatically converted as a comment I posted an answer (pointing to another question already asked) but when I hit submit, I saw my answer appear as a comment to the question instead.   I tried it a couple of times and still the same result. The answer gets eaten up and is transformed into a comment.   Is this a bug? Can I enter the Schengen area with a German long-term visa through a different country? I am an Indian national. Right now I am staying in Israel. I will be moving to Germany on the last week of this month with german national visa directly from tel aviv. Ticket price of direct flight is high. So I am thinking to buy either Air France or LOT for travelling having layover at Paris or Poland. Am I allowed to do that? I mean is it mandatory to enter in Germany first? May I travel with these airlines with proper transit visa? Precise connection between complexification of $\mathfrak{su}(2)$, $\mathfrak{so}(1,3)$ and $\mathfrak{sl}(2, \mathbb{C})$ I'm desperatly confused by notations and formulations so if someone could clarify the following things a little Í would be deeply grateful. The Lie algebra $\mathfrak{so}(1,3)_+^{\uparrow}$ of the proper orthochronous Lorentz group $SO(1,3)_+^{\uparrow}$ is given by \begin{equation} [J_i,J_j]=i \epsilon_{ijk} J_k \end{equation} \begin{equation} [J_i,K_j]=i \epsilon_{ijk} K_k \end{equation} \begin{equation} [K_i,K_j]=- i \epsilon_{ijk} J_k \end{equation} We can now define new generators with the old ones $N^{\pm}_i= \frac{1}{2}(J_i \pm i K_i)$ which satisfy \begin{equation} [N^{+}_i,N^{+}_j] = i \epsilon_{ijk} N^{+}_k ,\end{equation} \begin{equation} [N^{-}_i,N^{-}_j] = i \epsilon_{ijk} N^{-}_k ,\end{equation} \begin{equation} [N^{+}_i,N^{-}_j] = 0. \end{equation} where we can see that $N^{+}_i$ and $N^{-}_i$ make up a copy of the Lie algebra $\mathfrak{su}(2)$ each. My problem is to get what is going one here mathematically precise. Are the following statements correct and if not why:  When we build the new operators from the old generators we complexified $\mathfrak{so}(1,3)_+^{\uparrow}$ \begin{equation}(\mathfrak{so}(1,3)_+^{\uparrow})_\mathbb{C} = \mathfrak{so}(1,3)_+^{\uparrow}\otimes \mathbb{C} \end{equation} We saw that $\mathfrak{so}(1,3)_+^{\uparrow})_\mathbb{C}$ is isomorph to two copies of the complexified Lie algebra of $\mathfrak{su(2)}$: $(\mathfrak{so}(1,3)_+^{\uparrow})_\mathbb{C} \simeq \mathfrak{su(2)}_{\mathbb{C}} \oplus \mathfrak{su(2)}_{\mathbb{C}} $. Where exactly did we need that $\mathfrak{su(2)}$ is complexified here? The Lie algebras defined by $N^{\pm}_i$ are exactly those of $\mathfrak{su(2)}$ and we never use complex linear combination of $N^{\pm}_i$ or am I wrong here? $\mathfrak{su(2)}_{\mathbb{C}}$ is isomorph to $(\mathfrak{sl}(2,\mathbb{C}))_\mathbb{C}$: \begin{equation}\mathfrak{su(2)}_{\mathbb{C}} \simeq (\mathfrak{sl}(2,\mathbb{C}))_\mathbb{C} \end{equation} Here $(\mathfrak{sl}(2, \mathbb{C}))_\mathbb{C}$ denotes the complexified Lie algebra of $SL(2,\mathbb{C})$ Is $(\mathfrak{so}(1,3)_+^{\uparrow})_\mathbb{C} \simeq (\mathfrak{sl}(2, \mathbb{C}))_\mathbb{R}$ correct?  Here $(\mathfrak{sl}(2, \mathbb{C}))_\mathbb{R}$ denotes the real Lie algebra of $SL(2,\mathbb{C})$ Is $(\mathfrak{so}(1,3)_+^{\uparrow})_\mathbb{C} \simeq (\mathfrak{sl}(2, \mathbb{C}))_\mathbb{C} \oplus (\mathfrak{sl}(2, \mathbb{C}))_\mathbb{C}$ correct?  I looked this topic up in different books and each seemed to state something different. One book even used three differrent versions of $\mathfrak{sl}(2,\mathbb{C}) $ namely: $\mathfrak{sl}(2,\mathbb{C}) $, $(\mathfrak{sl}(2,\mathbb{C}))_\mathbb{C}$ and $(\mathfrak{sl}(2,\mathbb{C}))_\mathbb{R}$.  states simply that $\mathfrak{sl}(2,\mathbb{C}) $ is the complexification of $\mathfrak{su(2)}$ without making any reference to $SL(2,\mathbb{C})$ which does not help me either. Any help would be great. Exclude from * in command line Derivation of Mode of grouped data A formula to calculate the mode for grouped data's is given in my text book: Mode = $l + \dfrac{(f_1 - f_0)h}{2f_1 - f_0 - f_2} $ Where, $l = $ lower limit of the modal class, $h = $ size of the class interval, $f_1 = $  frequency of the modal class, $f_0 = $ frequency of the class preceding the modal class, $f_2 =$ frequency of the class succeeding the modal class. Can you please explain the derivation of this formula, as it is not given in my T.B. Thanks . I am not able to play "Half-Life-2" on Steam + Ubuntu 14.04 . Error shown in message box on screen:      Could not find required OpenGL entry pint 'glGetError'! Either your video card is unsupported, or your OpenGL driver needs to be updated.   I have a Dell Inspiron Laptop.  My OS is Ubuntu 14.04 and my graphics card is AMD Radeon HD 8600M.  I have installed AMD properietry Driver.    The partial output of "glxinfo" is as follows:      name of display: :0     libGL: screen 0 does not appear to be DRI3 capable     libGL: pci id for fd 4: 8086:0a16, driver i965     libGL: OpenDriver: trying /usr/lib/i386-linux-gnu/dri/tls/i965_dri.so     libGL: OpenDriver: trying /usr/lib/i386-linux-gnu/dri/i965_dri.so     libGL: Can't open configuration file /home/arushi/.drirc: No such file or directory.     libGL error: failed to load driver: i965     libGL: OpenDriver: trying /usr/lib/i386-linux-gnu/dri/tls/swrast_dri.so     libGL: OpenDriver: trying /usr/lib/i386-linux-gnu/dri/swrast_dri.so     libGL: driver does not expose __driDriverGetExtensions_swrast(): /usr/lib/i386- linux-gnu/dri/swrast_dri.so: undefined symbol: __driDriverGetExtensions_swrast     libGL: Can't open configuration file /home/arushi/.drirc: No such file or directory.     libGL: Can't open configuration file /home/arushi/.drirc: No such file or directory.     libGL error: failed to load driver: swrast     display: :0  screen: 0     direct rendering: No (If you want to find out why, try setting LIBGL_DEBUG=verbose) SUID-bit not working for executables within /tmp directory Printing the value of getuid() and geteuid() from an executable with SUID-bit (chmod +s) turned-on, seems to result with the original caller ID instead of the owner id if the executable is located within the /tmp directory.  When compiling the exact same code to home directory (and executing chmod +s), this seems to work as expected.  I've searched a lot and could not find any reference to such behavior.  Does anybody know why is this happening?  this source is simple as that:  #include &lt;stdio.h&gt; #include &lt;stdlib.h&gt; #include &lt;sys/types.h&gt; #include &lt;unistd.h&gt;  int main(){     printf("%d\n%d\n", getuid(), geteuid());     return 0; }