Why is there no progress anymore on H.O.T.D?
Is Highschool of the Dead on hiatus again?
Exercise 3.5.12. Let $ f: \mathbb N \times \mathbb N \to \mathbb N$ be a function, and let c be a natural number. Show that there exists a function $a:\mathbb N\to \mathbb N$ such that $$a(0)=c$$ and $$a(n++)=f(n,a(n))\;for\;all\;n\in \mathbb N,$$ and furthermore that this function is unique. Tao gives the hint: first show inductively, by a modification of the proof of Lemma 3.5.12, that for every natural number $N\in \mathbb N$, there exists a unique function $a_N:\{ n\in \mathbb N:n\lt N\}\to \mathbb N $ such that $a_N(0)=c$ and $ a_N(n++)=f(n,a(n))$ for all $n\in \mathbb N$ such that $n\lt N$.  In my opinion, we could solve it by proving that $\forall n\in \mathbb N,a(n)\in \mathbb N$ according to mathematcial induction.  The property P(n) is:$$\forall n\in \mathbb N, a(n)\;is\;unique$$ P(0) is true. And if P(n) holds, P(n++) is true because different natural numbers must have different successors. $$a(n++)=f(n,a(n))$$ $$a(m++)=f(m,a(m))$$ $$n++=m++\Rightarrow n=m\Rightarrow a(n++)=a(m++)$$ It's similar to the proof of  Proposition 2.1.16 (Recursive definitions): Suppose for each natural number $n$, we have some function $f_{n} : \mathbb{N} \rightarrow \mathbb{N}$ from the natural numbers to the natural numbers. Let $c$ be a natural number. Then we can assign a unique natural number $a_{n}$ to each natural number $n$, such that $a_{0} = c$ and $a_{n{++}} = f_{n} (a_{n})$ for each natural number $n$.  This proof indicates both the existence and uniqueness of the function $a$. I don't know why Tao says it's informal.  But regarding the hint, how do we get the function f even if we've got the conclusion from the hint. Also it seems like the function a(n) can be diffentent considering different N. I don't know if the hint is correct. space on disk running low error logging /var/log/cups/error.log I have Lubuntu 16.10 installed on my disk mounted as /.  The problem is that the CUPS log file( /var/log/cups/error.log)  grows all the time until I no have free space left on the disk...   So when I delete that file, the space on disk freed up again  I already tried:    fsck disk usage analyzer every apt-cleaning/autocleaning, and dependencies fixing.    How should I proceed?  du -sxh:  9,0G    .  Model: ATA ST3160815AS (scsi) Disk /dev/sda: 160GB Sector size (logical/physical): 512B/512B Partition Table: msdos Disk Flags:   Number  Start   End     Size    Type      File system     Flags  1      32,3kB  50,3GB  50,3GB  primary   ext4            boot  2      50,3GB  160GB   110GB   extended  5      50,3GB  158GB   108GB   logical   ext4  6      158GB   160GB   2136MB  logical   linux-swap(v1)   parted --list &amp;&amp; sudo df -h:  Model: ATA ST3160815AS (scsi) Disk /dev/sda: 160GB Sector size (logical/physical): 512B/512B Partition Table: msdos Disk Flags:   Number  Start   End     Size    Type     File system  Flags  1      32,3kB  50,3GB  50,3GB  primary  ext4         boot  2      50,3GB  160GB   110GB   primary  ext4   Filesystem      Size  Used Avail Use% Mounted on udev            985M     0  985M   0% /dev tmpfs           201M  6,3M  195M   4% /run /dev/sda1        46G   43G  709M  99% / tmpfs          1003M  188K 1003M   1% /dev/shm tmpfs           5,0M  4,0K  5,0M   1% /run/lock tmpfs          1003M     0 1003M   0% /sys/fs/cgroup tmpfs           201M   36K  201M   1% /run/user/1000 /dev/sda2       101G   60M   96G   1% /media/aram/ What is vertex and edge shading? Does a non-abelian semigroup without identity exist? Possible Duplicate:          I'm working on a website that displays distance to various locations to the nearest 1/10 mile. One of the developers (a non-native speaker) asked me an interesting question: If the distance is 1.0 mile(s), is that singular or plural? Clearly, anything other that 1.0 is plural, but is it:     1.0 miles away   or     1.0 mile away   The first option sounds more correct to my ear, but the voice of my elementary school language arts tells me that the latter option is correct.  Edit: Added the word 'away' to the example. It doesn't change the question, but might provide clarity for the answer. Also, regarding numbers between 0 and 1, after more thought, I'm not sure that the answer is as clear as I originally thought. See . Electrons have spin 1/2, and as they are charged, they also have an associated magnetic moment, which can be measured by an electron beam splitting up in an inhomogeneous magnetic field or through the interaction of the electrons's magnetic moment with an external magnetic field in spectroscopic measurements.   On the other hand, a photon is neutral - how can one measure its spin if there's no magnetic moment? How do we know it has spin 1? Physical meaning of this integral in $k$-space I am trying to solve an integral in $k$-space where I want to calculate the ratio of the average energy of an arbitrary electronic band ($E(k)=k^n)$ to the square of average group velocity of electrons($v_g   = k^{n-1})$ approximately in the form of this expression given by $$L=\frac{\frac{\sum_{\mathbf{k}}k^{n}}{\sum_{\mathbf{k}}1}}{\left(\frac{\sum_{\mathbf{k}}k^{n-1}}{\sum_{\mathbf{k}}1}\right)^2}\;.$$  Now, I want to do this for $d$-dimensions, so I convert this sum to integral.  $$L=\frac{\dfrac{\int_{0}^{k_f}k^{n} d^d k}{\int_{0}^{k_f}1d^d k}}{\left(\dfrac{\int_{0}^{k_f}k^{n-1}d^d k}{\int_{0}^{k_f}1d^d k}\right)^2}$$ where $d^dk  = \dfrac{d \pi^{d/2}}{\Gamma(\frac{d}{2}+1)}k^{d-1}dk$ is the volume element in $d$-dimensional space.  I want to get a simple expression for $L$ and how it depends with $k_f,n$ and $d$. any physical insight into this integral could also be helpful. Solve the integral $\int_0^\infty x/(x^3+1) dx$ I'm new here!  The problem: integrate from zero to infinity x over the quantity x cubed plus one dx. I checked on wolfram alpha and the answer is that the indefinite integral is this:  $$\int \frac{x}{1+x^3} dx = \frac{1}{6}\left(\log(x^2-x+1)-2 \log(x+1)+2 \sqrt{3} \arctan((2 x-1)/\sqrt{3})\right)+\text{constant}$$  and the definite integral is this:  $$\int_0^\infty \frac{x}{1+x^3} dx = \frac{2 \pi}{3 \sqrt{3}}\approx 1.2092$$  I am trying to figure out all the steps in between. I see that there are logs, which are equivalent to the ln variant that i am more familiar with, which means it was integrating $1/x$ at some point; I also see an inverse tangent in there.  I started with long division to simplify and I got (which could be wrong because I am very tired right now) $x/(x^3+1) = x^2+ 1/(x^3+1)$ which seems to be a step in the right direction. Wolfram Alpha thinks I definitely did that step wrong. The two equations do not evaluate as equal.  Then I cheated and took the wolfram alpha factorization of $(x^3+1) = (x+1)(x^2-x+1)$...I probably should have known that but didn't offhand. Now it is looking like $x^2$ plus the partial fraction decomposition $1/(x^3+1) = A/(x+1)+(Bx+C)/(x^2-x+1)$. Am I heading in the right direction with this? At this point do I just plug in and crunch? Let moderators navigate the review queues at will This post is an expansion of  For many graduated sites (eg Physics), moderators do not want to apply the unilateral close vote in all cases--it is better to let some community votes pile on first.  We only use the unilateral in obvious cases then. This renders the CV queue useless to us. If we &quot;skip&quot; something, we can't get back to it. If we say &quot;do not close&quot;, it's removed from the queue and our community can't see it. If we use &quot;close&quot;, we are using a unilateral mod vote--we don't want to do that. What I propose is a set of &quot;next/previous&quot; buttons, that let us navigate the queues, looking for closeable questions, and coming back to the others later. Thoughts? Update and bounty reason Due to the recent Great Flag Overhaul, mods no longer see close flags in the flag queue. Which means that the only way to get to them is via the review queue 1 Shog's solution is ideal in this case; let mods view the close queue as a list. (In fact, a list similar to the old flag queue -- which shows a list of mods who have seen the post and what mod actions have already been taken -- would be great) On Physics, the majority of the flags that we get are VTC flags. And the community isn't always able to close things on their own, so we have to step in. However, powering through the close queue isn't an option; because &quot;skip&quot;ing a post2 leads to it being dumped into the void (I can't get back to it unless I have a link). Extremely inconvenient. Could the workflow here be improved? 1. The /tools listing for posts with outstanding close votes doesn't work because it doesn't list posts which have been flagged but not VTCd 2. which I may want to take a closer look at later; gotta be careful with the binding votes. I have some text in a table and I want to add a forced line break. I want to insert a forced line break without having to specify the column width, i.e. something like the following: \begin{tabular}{|c|c|c|} \hline Foo bar &amp; Foo &lt;forced line break here&gt; bar &amp; Foo bar \\ \hline \end{tabular}  I know that \\ inserts a line break in most cases, but here it starts a new table row instead.  A similar question was asked before: There is a way to use my Galaxy tab s like a wacom table, on linux? How do I create a very long delay with Redstone? The only things I've found so far are a cart method (which ceased working in 1.6), a dispenser method (utilizing the decay on dropped items, but this requires occasional feeding), and a binary counter (which won't work for a single input but rather a fluctuating one).  I want to have a circuit that when I apply an input signal, the output signal doesn't change for 2 to 5 minutes.  When the input turns off, I again want a 2 to 5 minute delay.  I could accomplish this with (roughly) 240 to 600 redstone repeaters, but this is far in excess of being useful.  The only method I can think of, but would require a XOR gate (on the button-input and final output) to start and stop a clock attached to a binary counter.  While I can make room for a device of this size, I was hoping for something simpler. I'm currently making a Minesweeper in C#. I've already done the basics (generating minefield, rules and ui) but I don't know how I should implement when the user clicks a cell and the cell has no mines nearby and clicks nearby cells with also no mines nearby, like this:    And after click    This is my current code on this:  if (MineField(x,y) == false) // If the clicked cell isn't a mine {     if (GetNeighbours(x,y) == 0) // GetNeighbours returns the amount of mines in the closest 8 cells     {         // To be honest, no idea...     } } Does the shape of a containter affect the apparent mass of liquid inside it when measured by a scale? For both containers it is true that:  $p_{bottom}=p_{atm}+\rho gh$ and $F_{bottom}=a*p_{bottom} \iff F_{bottom}=a*p_{atm} + a\rho gh$  I do not know how a scale functions, but I guess that the only force it can count is the force by the liquid on the bottom of each container (the only surface that is in touch with the scale is the bottom of the container). Thus, if we measured the mass of these two masses of water (lets assume that the mass of the containers is negligible) it would appear to be the same in each case. Am I right? Selenium 2: Open link in new tab and close tabs Show $\,p\mid 9^q -1 \implies q\mid p-1$ for odd primes $p$ and $q$ If $p$ and $q$ are odd primes show that $$p|(9^q -1) \implies q|(p-1).$$ ....  I am having difficulty to prove this.  Yes that was the question I have misunderstood between 9 and q!!! I have 4 servers with Debian Wheezy OS. I have Apticron installed that informs me about updates. Debian updates are realized so often that when I finish to update the last of 4 servers I get new email about new updates on the first server. I try to update all servers when I get a notification but I never know if there is a need to reboot the servers. I have read that if the directory "/var/run" contains file "reboot-required" I have to reboot the server. But I never have seen this file in "/var/run". How can I know when reboot is required? I don't want reboot my servers every time when I install new updates if it's not needed.  I understand that if I update PHP or MySQL, etc I don't need to reboot the server but updates usually contain many "lib...".  Below are 9 updates (I have received this week).  krb5-locales 1.10.1+dfsg-5+deb7u3 libdbus-1-3 1.6.8-1+deb7u6 libgssapi-krb5-2 1.10.1+dfsg-5+deb7u3 libk5crypto3 1.10.1+dfsg-5+deb7u3 libkrb5-3 1.10.1+dfsg-5+deb7u3 libkrb5support0 1.10.1+dfsg-5+deb7u3 libruby1.8 1.8.7.358-7.1+deb7u2 libxml2 2.8.0+dfsg1-7+wheezy3 ruby1.8 1.8.7.358-7.1+deb7u2   I have no idea what is "libkrb,libgssapi", etc. How can I detect if reboot is needed? Please do not suggest to install UnattendedUpgrades to let the servers update automatically because this can cause websites going offline if something updates not correct. When searching on a site with  enabled (), I saw some  rendering issues: (and currently it's also in the )        Even though the LaTeX rendering is fine when opening the :        The LaTeX code is:  $f(n) = \begin{cases} 0,  &amp; \text{$ \omega&lt;0, \omega=1, \omega&gt;3 $} \\[2ex] -\omega + 1,  &amp; \text{$ 0\le\omega&lt;1 $} \\[2ex] \omega^2 - 1,  &amp; \text{$ 1&lt; \omega \le 2 $} \\[2ex] 3 \omega,  &amp; \text{$2 &lt; \omega \le 3 $} \end{cases}$ In English, the number 0 is treated as plural. It is then:   0 seconds 1 second    1.2 seconds   2 seconds     Shouldn't it be "nobody want to go there," instead of "nobody wants to go there"?  I also saw in TOEFL that "any __" should be used with a singular.  But I see it very common that it is a plural also.  Why is that?  (updated: example, "we don't have any apples any more" vs "If you get any apple, please let me know.") Loop-cut won't go all the way around Alright so I've seen some postings about loop cuts and done some research. I am very new to blender and modeling overall. I am trying to loop cut this concrete barrier. Now I'm following a tutorial and kind of did my own thing in some areas. I cleaned it up following some instructions in another answer here on stack exchange.  I can't for the life of me figure out why I can loop cut around the whole bottom:   ...but not the whole middle:   I see this marked as answered by another question. That question refers to ngons and inside faces. Now once again I'm totally new to this. I looked on the inside. I don't believe I have inside faces or any ngons by the terms I've read online in my research.  Here is a wireframe picture and the inside view:     Maybe I'm missing something the answers on other posts are saying. If so and you know what it is I'd be very appreciative of what it is.