Dual boot issues win8/ubuntu16.04
How can I repair grub? (How to get Ubuntu back after installing Windows?)
At the end of my animation, the animation pauses a bit then loops again, how can I stop the slow pause, and let the loop work fluently? Mark a comment as the answer to a question I have a question and have seen others to which the best and most relevant answer (at least the one that solved my particular problem) is placed in the comments to the question as opposed to the answers:   I would like to recognize the commenter as the person who correctly answered the question, but there doesn't seem to be any way to do that. The only thing I could think of was to ask the commenter to repost his comment as an answer, but I cannot figure out a way to contact him. In the meantime the question is sitting unanswered and I am getting other "answers" (leading me to believe that people are not reading the comments and answers already posted before trying to answer it themselves).  How are we supposed to do this?  Note: I don't want to accept the comment as an answer (I agree completely with the answers to this question that comments are comments and answers are answers and they should stay that way). I do wonder if there should be some way to contact people in a reasonable manner (comments on their profile that they must opt in to receiving in email?). Is there such a thing as a deposit-only bank account? My employer wants me to give my routing-numbers, social-security numbers, and basically everything else required to take money out of my account, to several third-party direct-deposit websites (one for weekly paychecks, one for mileage reimbursements, etc.).  I am not comfortable giving my identity information away to all these places.  I'd like to give them access to an account which they can only deposit into, never withdraw (either a separate account, or, even better, a deposit-only alias to my real account).  Does such a thing exist?    [Edit] I'm not worried about the companies so much as I am worried about bad guys who inevitably hack those companies' databases, or get hired to work there.  This type of account would also be handy for giving read-only access to sites like mint.com or quicken.com, easing my mind about their hackers/unscrupulous employees. I'm a photographer, who dabbles in graphic design from time to time as well. What are the differences between the various color spaces? Is there a way to "rescue" mesh that has very bad topology? update-initramfs missing /lib/modules/4.4.0-13-generic rsync over ssh "error in protocol data stream" (code 12). ssh works I tried:  rsync -v -v -e 'ssh -p YY' ./testfile me@XXXXX:/home   Error Message  opening connection using: ssh -p YY -l me 146.6.84.206 rsync -&gt;-server -vvve.s . /home  [sender] make_file(testfile,*,0) send_file_list done send_files starting server_recv(2) starting pid=17537 rsync: connection unexpectedly closed (9 bytes received so far) [sender] rsync error: error in rsync protocol data stream (code 12) at io.c(600)[sender=3.0.6] _exit_cleanup(code=12, file=io.c, line=600): about to call exit(12)   This works fine:  ssh -p YY me@XXXXX   Which suggests to me the problem is neither that sshd is not running nor that port YY is firewalled. I have checked anyway.  What other problems could there be?  EDIT: The problem seems to have been "self resolving." I could not replicate the following day. I started my local computer up. Perhaps notably I had a different IP address than last time. And now rsync magically works. I'd appreciate guesses as to what it could have been in light of it going away. mknod /tmp/oracle.pipe p  sqlplus / as sysdba &lt;&lt; _EOF set escape on  host nohup gzip -c &lt; /tmp/oracle.pipe &gt; /tmp/out1.gz \&amp; spool /tmp/oracle.pipe select * from employee; spool off  _EOF  rm /tmp/oracle.pip   I need to insert a trailer at the end of the zipped file out1.gz , I can count the lines using  count=zcat out1.gz |wc -l   How do i insert the trailer  T5 (assuming count=5)   At the end of out1.gz without unzipping it. Let $f:G\to K$ be a surjective group homomorphism and $H$ be a subgroup of $G$ satisfying $N_G(H)=H$. Then $N_K(f(H))=f(H)$. I'm trying to prove or disprove it. perhaps, this proposition is true. But, how can we prove it? Edit: I thought it is true because WIKI said it is true without proof Please provide me with the guidance to install oracle 11gr2 on Ubuntu 14.04. A financial planner treated us in a dismissive fashion and I want to say that he "gave us the back of his hand" but that phrase doesn't seem to have the meaning of dismissive. Does it or is there a similar idiom? So I've been thinking through some test cases. If $a_n = n$ then $\sum_n \frac{a_{n+1} - a_n}{a_n}$ is the harmonic series which diverges. And if $a_n = \sum_{k=1}^n 1/k$ then $\sum_n \frac{a_{n+1} - a_n}{a_n}$ diverges like $\sum_n 1/(n \log n)$. So that got me thinking, if $a_n$ is a strictly increasing unbounded sequence, does $\sum_n \frac{a_{n+1} - a_n}{a_n}$ necessarily diverge? Since a while, 2 weeks, Dropbox does not startup anymore on my laptop running Ubuntu 12.04 LTS with GNOME shell.  When I run dropbox start -i in the console I get the following output.    Starting Dropbox...Traceback (most recent call last):   File "/usr/bin/dropbox", line 1387, in &lt;module&gt;     ret = main(sys.argv)   File "/usr/bin/dropbox", line 1376, in main     result = commands[argv[i]](argv[i+1:])   File "/usr/bin/dropbox", line 1247, in start     if not start_dropbox():   File "/usr/bin/dropbox", line 731, in start_dropbox     stderr=sys.stderr, stdout=f, close_fds=True)   File "/usr/lib/python2.7/subprocess.py", line 679, in __init__     errread, errwrite)   File "/usr/lib/python2.7/subprocess.py", line 1249, in _execute_child     raise child_exception   OSError: [Errno 8] Exec format error   I already removed Dropbox and reinstalled it. What should I do to fix this? I've looked through a number of the "Vote too old to be changed" questions, but I think this one's different.  I gave a downvote to an answer I thought had a critical defect, and I left a comment describing the defect. The answer was edited to satisfy my criticism, but it was done within the short grace period where edits aren't recorded as such. When I try to cancel my downvote I get the error - "Your vote is now locked in unless this answer is edited" - but it was edited!  In reference to this answer: Probability of asymmetric random walk returning to the origin Consider the random walk $S_n$ given by $ S_{n+1} =  \left\{   \begin{array}{l}     S_n+2 &amp; \text{with probability $p$}\\     S_n - 1 &amp; \text{with probability $1-p$}   \end{array} \right. \ $  Assume that $S_0 = n &gt;0 $. What is the probability of eventually reaching the the point $0$?  Seems that the probability for this must be less than one; since it is not symmetric, it is not guaranteed to reach the origin if $p&gt;0.5$. It also seems like the positive starting value not being at the origin matters too, but I'm not sure how to compute the probability of eventually reaching the origin. Redirect, Change URLs or Redirect HTTP to HTTPS in Apache - Everything You Ever Wanted to Know About mod_rewrite Rules but Were Afraid to Ask Converting expressions to polynomial form Do there Exist Proper Classes that aren't "Too Big" Some proper classes are "too big" to be a set in the sense that they have a subclass that can be put in bijection with $\alpha$ for every cardinal $\alpha$. It is implied in  post that every proper class is "too big" to be a set in this sense, however I have been unable to prove it. It's true if every two proper classes are in bijection, but it's consistent with ZFC for there to be a pair of non-bijective classes.  So, is the following true in ZFC:     For all proper classes, $C$, and $\alpha\in\mathbf{Card}$, $\exists S\subset C$ such that $|S|=\alpha$?   If not, is there something reasonable similar that preserves the intuition about classes that are "too big to be sets"? I was wondering, why there are passport checks at the check-in counter, Immigration and even at the gate?    For example, I was at Istanbul flying to either UK or US. There's a security check at the check-in counter asking a bunch of questions about my immigration status - even more than the Immigration officer at the border. Also, at the gate, there's also Security officer checking my immigration status and passport asking the same questions.    I thought that would be the responsibility of an Immigration officer rather than some third party company?   The thing I'm curious the most is they don't even know which countries this passport can go to with or without visa. They have to go back to the counter and check with the system which takes a long time. I was wondering why they hire these kind of companies to slow down the process? Find all functions $f$ such that $f:\mathbb{N}\rightarrow\  \mathbb{N}$ and $f(f(n))+f(n+1)=n+2$   Let us plug in $n=1$  $f(f(1))+f(2)=3$  Since the function is from $\mathbb{N}$ to $\mathbb{N}$, $f(2)$ can only take the values $1,2$. Now we divide the problem into cases.  Case-1: $f(f(1))=2,f(2)=1$  We can assume that $f(1)=c$ for the time being. Then plugging in $n=3$ and using $f(2)=1$ gives $$f(3)=4-c$$ and again since the range of the function is positive integers,then $4-c$ has to be positive and hence $c$ belongs to  {$1,2,3$}. Now, $$f(1)=c$$$$\implies f(f(1))=f(c)$$$$\implies 2=f(c)$$ by the assumption that $f(f(1))=2$ . Now,since $c$ can only take the values $1,2,3$,we start treating cases. If $c=1$,$$f(c)=2$$$$\implies f(1)=2$$ but we know from the deinition of $c$ that $f(1)=c=1$,a contradiction.If $c=2$,then $2=f(c)=f(2)$ but $f(2)=1$ by assumption. Finally,if $c=3$ $$2=f(c)=f(3)$$ but $$f(3)=4-c=4-3=1$$ which is once again a contradiction. Therefore there are no such functions in this case.  Case-2: $f(f(1))=1,f(2)=2$  Again assuming $f(1)=c$ and using $f(n)\le n$ along with plugging $n=c-1$ will give us that $f(1)=1$ and then it is easy to prove that such a function exists by recursion. I can only give a "sort of recursive" way to define the function. Here it goes $$f:\mathbb{N}\rightarrow \mathbb{N}$$$$f(1)=1$$$$f(n)=n+1-f(f(n-1))$$  But this case is harder to deal with.Some help will be appreciated.