How to fix black screen after install nvidia driver?
My computer boots to a black screen, what options do I have to fix it?
Problem to launch java at Debian: "error while loading shared libraries: libjli.so" This code always works, even in different browsers:  function fooCheck() {   alert(internalFoo()); // We are using internalFoo() here...    return internalFoo(); // And here, even though it has not been defined...    function internalFoo() { return true; } //...until here! }  fooCheck();   I could not find a single reference to why it should work, though. I first saw this in John Resig's presentation note, but it was only mentioned. There's no explanation there or anywhere for that matter.  Could someone please enlighten me? C's strtok() and read only string literals char *strtok(char *s1, const char *s2)      repeated calls to this function break string s1 into "tokens"--that is   the string is broken into substrings,   each terminating with a '\0', where   the '\0' replaces any characters   contained in string s2. The first call   uses the string to be tokenized as s1;   subsequent calls use NULL as the first   argument. A pointer to the beginning   of the current token is returned; NULL   is returned if there are no more   tokens.   Hi,   I have been trying to use strtok just now and found out that if I pass in a char* into s1, I get a segmentation fault. If I pass in a char[], strtok works fine.   Why is this?  I googled around and the reason seems to be something about how char* is read only and char[] is writeable. A more thorough explanation would be much appreciated. Induced sequence of global sections I'm reading Differential Analysis on Complex Manifolds by Raymond O. Wells. It states the following in the beginning of section 3 of chapter 2 on page 51:  Consider a short exact sequence of sheaves:  $$0 \to \mathcal{A} \to \mathcal{B} \to \mathcal{C} \to 0$$  Then it is easy to verify that the induced sequence  $$0 \to \mathcal{A}(X) \to \mathcal{B}(X) \to \mathcal{C}(X) \to 0$$  is exact at $\mathcal{A}(X)$ and $\mathcal{B}(X)$ but not necessarily at $\mathcal{C}(X)$. After that they give an example of failing exactness at $\mathcal{C}(X)$.  My problem is that Ii dont see why the induced sequence has to be exact at $\mathcal{A}(X)$ and $\mathcal{B}(X)$. I tried to find an answer in other books. In the book Sheaf Theory by Bredon they make the same statement in proposition 2.2 but they also state it is easy to verify.  I tried to come up with an argument of my own and got something like this (the argument is wrong i think): If we have the zero element  $0 \in \mathcal{B}(X)$ this should induce a section $\tilde{0} $ of the stalks of $\mathcal{B}$ and should take the value of the zero element in every stalk. Then since the sequence of sheaves is exact (and thus exact in stalk level) we get that around every $x \in X$ there should be an open $V$ such that $\tilde{0}|_{V}$ is the image of the zero section in $\mathcal{A}(V)$ and since $\mathcal{A}$ is a sheaf we can patch these together to get a global zero section.  Also while i get the example that shows inexactness at $\mathcal{C}(X)$ i dont get why this should generally be the case.  Any help would be much appreciated Why is methane considered an important greenhouse gas? Methane’s absorption bands are centered at 3.2 and 7.2 microns -- far off the peak of the Planck spectrum for a 290 K blackbody near 16 microns.  Moreover, its absorption bands overlap with the water absorption region.  So why is methane considered a stronger greenhouse gas than carbon dioxide, which has an absorption band at 15 microns? On Cesàro convergence: If $ x_n \to x $ then $ z_n = \frac{x_1 + \dots +x_n}{n} \to x $ I have this problem I'm working on. Hints are much appreciated (I don't want complete proof):  In a normed vector space, if $ x_n \longrightarrow x $ then  $ z_n = \frac{x_1 + \dots +x_n}{n} \longrightarrow x $   I've been trying adding and subtracting inside the norm... but I don't seem to get anywhere.  Thanks! Okay, maybe I'm missing something obvious but consider $h_n(x) = x^{ 1+ \frac{1}{2n-1}}$ on the set $[-1,1]$. Now consider a fixed $x \in [-1,1]$, we have:  $$\lim_{n\rightarrow \infty} h_n(x) = x \lim_{n \rightarrow \infty} x^\frac{1}{2n-1} = x$$  But my book says it's $|x|$? Why? Enumerate after \label Definition of the Infinite Cartesian Product Apparently I cannot count. I think there are three files in /media  $ tree /media /media ├── foo ├── onex └── zanna 3 directories, 0 files   However, ls -l finds 12.  $ ls -l /media total 12 drwxr-xr-x  2 root root 4096 Jul 31 20:57 foo drwxrwxr-x  2 root root 4096 Jun 26 06:36 onex drwxr-x---+ 2 root root 4096 Aug  7 21:17 zanna   And, if I do ls -la I get only . and .. in addition to the above, but the count is total 20  What's the explanation? I would like to think that some of the software I'm writing today will be used in 30 years. But I am also aware that a lot of it is based upon the UNIX tradition of exposing time as the number of seconds since 1970.  #include &lt;stdio.h&gt; #include &lt;time.h&gt; #include &lt;limits.h&gt;  void print(time_t rt) {     struct tm * t = gmtime(&amp;rt);     puts(asctime(t)); }  int main() {     print(0);     print(time(0));     print(LONG_MAX);     print(LONG_MAX+1); }   Execution results in:   Thu Jan  1 00:00:00 1970 Sat Aug 30 18:37:08 2008 Tue Jan 19 03:14:07 2038 Fri Dec 13 20:45:52 1901           The functions ctime(), gmtime(), and localtime() all take as an argument a time value representing the time in seconds since the Epoch (00:00:00 UTC, January 1, 1970; see time(3) ).      I wonder if there is anything proactive to do in this area as a programmer, or are we to trust that all software systems (aka Operating Systems) will some how be magically upgraded in the future?  Update It would seem that indeed 64-bit systems are safe from this:  import java.util.*;  class TimeTest {     public static void main(String[] args) {         print(0);         print(System.currentTimeMillis());         print(Long.MAX_VALUE);         print(Long.MAX_VALUE + 1);     }      static void print(long l) {         System.out.println(new Date(l));     } }    Wed Dec 31 16:00:00 PST 1969 Sat Aug 30 12:02:40 PDT 2008 Sat Aug 16 23:12:55 PST 292278994 Sun Dec 02 08:47:04 PST 292269055   But what about the year 292278994? Why does $1+2+3+\cdots = -\frac{1}{12}$? How to create wifi hotspot in Ubuntu 17.10 Do elevator upgrades increase your passive credit earnings, too? I have HTML code like this :  &lt;div&gt;        &lt;a&gt;Link A1&lt;/a&gt;        &lt;a&gt;Link A2&lt;/a&gt;        &lt;a&gt;Link A3&lt;/a&gt; &lt;/div&gt;  &lt;div&gt;        &lt;a&gt;Link B1&lt;/a&gt;        &lt;a&gt;Link B2&lt;/a&gt;        &lt;a&gt;Link B3&lt;/a&gt; &lt;/div&gt;   When user clicks a link from above HTML, I want to get the jQuery object of the corresponding &lt;a&gt; element, and then manipulate its sibling. I can't think of any way other than creating an ID for each &lt;a&gt; element, and passing that ID to an onclick event handler. I really don't want to use IDs.   Any suggestions? What is the advantage of accepting the axiom of choice over other axioms (for e.g. axiom of determinacy)?  It seems that there is no clear reason to prefer over other axioms..  Thanks for help. Suppose $k$ is a field that is finitely generated as a ${\mathbb Z}$-algebra.  (That is, $k$ is a quotient of ${\mathbb Z}[X_1,\dots,X_n]$ for some $n$). Does it follow that $k$ is finite? Is there a way to fake a dual (second) monitor Is there a way to fake a dual monitor for testing.    I have only one monitor and I have to make an application that is suitable for two monitors.  The second monitor screen should appear as a window on my primary monitor. Situation: when a user self-marks their post as a duplicate, it shows as closed by "community" rather than the original poster.  Problems: this is not clear at all to most users of SE what has happened. It looks like the question was "mod closed" as a duplicate against the asker's wishes.    Suggestion: instead of the community user "closing" the post, have the final close user be the asker (or make it "Community &amp; Asker"). This will help clear up confusion and streamline the process for understanding what happens. about Linux's zip utility says this about the -u flag:     Replace (update) an existing entry in the zip archive only if it has been modified more recently than the version already in the zip archive. For example:          zip -u stuff *   will add any new files in the current directory, and update any files which        have been modified since the zip archive stuff.zip was last created/modified.   My question is, if a particular file has been removed since the last time it was zipped, whether -u will recognize that the file has been removed and thus remove it from the *.zip file?  Or, lacking timestamp information (since the file is now removed), it will leave the backed up file in *.zip, since there is no "newer" file?