Show that if $f$ is a non-constant entire function,it cannot satisfy the condition: $$f(z)=f(z+1)=f(z+i)$$  My line of argument so far is based on Liouville's theorem that states that every bounded entire function must be a constant.  So I try-to no avail-to show that if $f$ satisfies the given condition, it must be bounded. I haven't made much progress with this, so any hints or solutions are welcome.
Let $f: \Bbb C \to \Bbb C$ be an entire (analytic on the whole plane) function such that exists $\omega_1,\omega_2 \in \mathbb{S}^1$, linearly independent over $\Bbb R$ such that: $$f(z+\omega_1)=f(z)=f(z+\omega_2), \quad \forall\,z\in \Bbb C.$$Prove that $f$ is constant.   The intuition seems clear to me, we have the three vertices of a triangle given by $0$, $\omega_1$ and $\omega_2$. All points in the plane are one of the vertices of that triangle under a suitable parallel translation. The constant value will be $f(0)$, fine. Throwing values for $z$ there, I have found that $$f(n\omega_1) = f(\omega_1) = f(0)=f(\omega_2) = f(n\omega_2), \quad \forall\, n \in \Bbb Z.$$  I don't know how to improve the above for, say, rationals (at least). Some another ideas would be:   Checking that $f' \equiv 0$. I don't have a clue of how to do that. Write $w = a\omega_1+b\omega_2$, with $a,b \in \Bbb R$, do stuff and conclude that $f(w) = f(0)$. This approach doesn't seem good, because I only have a weak result with integers above. Finding that $f$ coincides with $f(0)$ on a set with an accumulation point. This seems also bad: the set on with $f$ coincides with $f(0)$ by which I found above is discrete.    Nothing works and this is getting annoying... And I don't see how analyticity comes in there. I'll be very thankful if someone can give me an idea.  (On a side note.. I know that this title is not informative at all. Feel free to edit if you come up with something better.)
How do I reboot a Ubuntu Server from the command line? sudo reboot now never seems to work for me. Because its a server, I don't have a window manager installed, so I can't do it from a GUI (which always seems to work).    I've had the problem for years on different versions of the operating system. It happens in virtual machines and on real metal. The same problem exists with sudo shutdown now, too.  I've lived with it in the past. The difference now is I don't want to just hit the power button because I might lose data. Is there any way to actually impose disadvantage (not just cancel advantage) on saving throws against spells for a creature with Magic Resistance? Many creatures in D&amp;D 5e have a feature called Magic Resistance. One such is the :  Magic Resistance. The archmage has advantage on saving throws against spells and other magical effects.  So the Archmage has advantage on saving throws against magical effects. Suppose I wanted to weaken the Archmage's resistance to magic - I use some effect that gives disadvantage on the saving throw for my spell. Normally, this would balance out to a straight roll, as the  say:  If circumstances cause a roll to have both advantage and disadvantage, you are considered to have neither of them, and you roll one d20. This is true even if multiple circumstances impose disadvantage and only one grants advantage or vice versa. In such a situation, you have neither advantage nor disadvantage.  This seems to set a baseline for a straight roll on saves against magical effects, that the Archmage can never roll at disadvantage against magical effects. But is there a way to get around this? Keep in mind, such an ability must respect the  rule:  That said, many racial traits, class features, spells, magic items, monster abilities, and other game elements break the general rules in some way, creating an exception to how the rest of the game works. Remember this: If a specific rule contradicts a general rule, the specific rule wins.  The general rule here is that advantage and disadvantage balance out to a single die roll, so getting around Magic Resistance must either specifically override the rule for advantage and disadvantage, or eliminate Magic Resistance entirely. Is there any way (e.g. magic item, class feature, spell, etc.) to force a creature with Magic Resistance to make a save against a magical effect with disadvantage?  While writing this up I did find  which asks generally how to combat creatures with Magical Resistance, I intend this to be a (hopefully) more focused version of that question. Is it possible in Latex to make a triangular hat? The one that is used in math over three letters representing points to indicate a triangle.  Something like: Representative point of a cluster with L1 distance The representative point of a cluster or cluster center for the k-means algorithm is the component-wise mean of the points in its cluster. The mean is chosen because it helps to minimize the within cluster variances (which is to say that it is minimizing within cluster squared Euclidean distance, since its the same). Now, when I want to use a different distance metric, like L1 (Manhattan distance), I would be minimizing the within cluster absolute deviations (if I choose to). I was looking at what the centers calculation would be like, because the mean won't reduce the new objective function, and I found this on the Matlab documentation of their kmeans function for the distance parameter     cityblock - Sum of absolute differences, i.e., the L1 distance. Each centroid is the component-wise median of the points in that cluster.   Now, how did they come up with the component-wise median? How does a median minimize absolute deviations? How to check if sed has changed a file Why do first order languages have at most countably many symbols? Every proof that I read seems to assume that $|L|\leq\aleph_0$. But then how do you model things like field over $\mathbb{R}$ without running out of variable symbols?  More importantly, how can I prove that $|L|\leq\aleph_0$? Prove that if $n$ is a natural number and $(3n+1)$ &amp; $(4n+1)$ are both perfect squares, then $56$ will divide $n$.   Clearly we have to show that $7$ and $8$ both will divide $n$.  I considered first $3n+1=a^2$ and $4n+1=b^2$. $4n+1$ is a odd perfect square.  - so we have $4n+1\equiv 1\pmod{8}$; from this $2|n$ so $3n+1$ is a odd perfect square.  - so $3n+1\equiv 1\pmod{8}$ so $8|n$ but I can't show $7|n$. How do I show this?  Thanks for the help. If $p$ is congruent to $2 \pmod 3$, how can I prove that all $a$, $1 \le a \le p-1$ are cubic residues $\mod p$?  Here's what I've done:  $1^3$ congruent to $1 \pmod p$ thus, 1 is a cubic residue, Also $(p-1)^3=(p-1) \pmod p$ implies  $$(p-1)^3 - (p-1) = (p-1)[(p-1)^2 -1] = (p-1)(p^2 -2p) = p(p-1)(p-2) $$ implies $$p|(p-1)^3 -(p-1)$$ Thus  $p-1$ is a cubic residue $\mod p$. Now I don't know how to show for $1\lt a \lt p-1$ are all cubic residues when $p$ is congruent to $2\pmod 3$. I have two areas of my house which are cantilevered. One is a bay area which extends about 2'. The other is the bottom of a fireplace box (gas fireplace). I have placed two layers of R-13 (I think) insulation in this area, but inside the house it's like a heat sink where all of the heat is sucked out ... very cold to the touch in the winter time. Obviously there must be a better way to insulate this area.  My question is, what is the best way to insulate a cantilevered area so as to allow the inside area of the house to not be so cold in the winter time? I want to have more than one user logged on at the same time, on the same computer. Each user should have a full X session. It is easy enough to log in to multiple concurrent non-X Linux terminals, but I want the X factor. Maybe I've missed something, but I only see a Main Menu option to Log out of the current user, and nothing for Switch User. Is there a simple way to do this? What was the joke that Geordi told Data in TNG: "Encounter at Farpoint"? states:     Apparently at some point during this mission, Geordi La Forge told Data a joke, which the humorless android did not "get" until years later, when he activated his emotion chip during the events of Star Trek: Generations.    Is it known what the joke is? Allow cross-site queries in data explorer I think it would be useful if we could join across databases in the data-explorer.  For example, I'd like to query up a list of questions that I had a hand in migrating away from SO, and see what percentage of them were subsequently closed or re-migrated.  I realise I could download the various data dumps and do this on my own machine, but it would be nice if it were allowed on the online version.  Is this possible/practical? I just updated all my sites using the patch method of resolving the Drupal SA-CORE-2014-005 exploit. I just read reports that just yesterday there is someone from a Russian IP infiltrating drupal sites.     My main concerns are now:   How do I tell if my sites have been comprised? What should I search for in my apache access logs to detect if my site was a victim or not? So far what are these hackers doing to comprised sites? Converting a drupal/drupal site to Work With Composer Moderator actions on items in the review queues don't count towards the review totals I received some text that is encoded, but I don't know what charset was used. Is there a way to determine the encoding of a text file using Python?  deals with C#. What stops other benders from mastering the four elements? How do I configure Nautilus to open at my Documents folder instead of my home folder?  I've scoured this site and found answers instructing users to edit the /usr/share/applications/nautilus.desktop file, replacing %U with the path to the desired default directory. I've edited all the variants of nautilus.desktop in the said directory, but none of the edits has an effect. I've checked all the files after editing them to make sure that I saved the edit effectively, and all look good to me.   I've edited every file to read as follows:  Exec=nautilus --new-window /home/judy/Documents   I've also edited the org.gnome.Nautilus.desktop file, just to be sure, with no joy.   I'm running Ubuntu 15.10.  Note: if I enter nautilus /home/judy/Documents from the command line, Nautilus properly opens in the desired directory.  Is there a simple way that I can configure an icon in the launcher bar to run this command and have the Nautilus icon that looks like a metal file drawer?  Thanks. I'm pulling my hair out over this one. Apparently, you can speed up your computer by adding more RAM. Is the amount of ram to be added is directly or indirectly related to the amount of hard-disk  space, processor speed, Motherboard etc... Can the memory of a RAM be compared with Virtual Memory and Cache Memory?  Note: About my PC Configuration: Intel core i5 processor, Motherboard DP55WB 4GB RAM, 700 GB Hard-disk space. How can you prove that the normal equations: $(X^TX)\beta = X^TY$ have one or more solutions without the assumption that X is invertible?  My only guess is that it has something to do with generalized inverse, but I am totally lost.