I need to know the answer to this question to find out why bees use hexagonal cells in hives. I know that a circle takes up the most area using the least perimeter, so bees would try to make shapes as close to a circle as possible to use the most space without wasting too much material on walls. However, bees don't use circles because using circles creates a lot of waste space between cells, so bees use a shape that can tile a plane without overlap. The shape that meets the conditions of having a lot of sides and being regular to look like a circle, and being able to tile a plane  is the hexagon, so bees use this in their hives. However, I want to know whether or not this is the polygon with the most sides that fits this requirement. I try to prove there is no bigger polygon like this. First, I note that an integer number of interior angles must meet at a site and add up to $360$ degrees to tile a plane. For example, squares can tiles a plane because they all have $90$ degree angles, and $4$ of these make $360$. All polygons have exterior angles adding up to $360$, so a polygon with $n$ sides has $\frac{360}n$ degrees per side. However, interior and exterior angles are supplementary, thus the interior angle of an $n$ side polygon measures $180- \frac{360}n$ degrees. Now, let's test whether this quantity divides $360$ degrees. We get $360 \over 180- \frac{360}n$, which simplifies down to $\frac {2n}{n-2}$. Now, I need to prove that $n=6$ is the biggest number such that this quanitity is an integer, but I am not sure how I would go about  doing that.
I read in several places, including Wikipedia, that a tessellation of the plane by a single, convex, $n-$sided polygon is not possible for $n\geq7$. I was not able to locate a proof, or a paper that discusses this. Maybe it is too trivial, but I am not able to figure it out myself.   So I would like to ask you for help. Where can I find a proof for this?  Also, I am interested in the mathematical background behind plane tiling (especially periodic), If you can suggest papers on the topic it would really help.
Function whose inverse is also its derivative? What are some good examples of a function  $f : \mathbb{R} \to \mathbb{R}$ where its derivative is equal to its inverse? I attempted to find a monomial that satisfied it by starting with $f(x) = ax^b$ and showing that $f^{-1}(x) = f'(x) \implies b-1=\frac{1}{b} \implies b=\phi$ and got $$f(x) = \frac{x^\phi}{\sqrt[\phi]{\phi}}$$ Which seems to work according to WolframAlpha, but I'm having trouble double-checking it. Any other ideas? How should I refer to a friend who is a girl but not a girlfriend? When I'm talking about my friend, who is a girl, but not a girlfriend, what word or phrase should I use? If the gender was unimportant, it would not be a problem. But if I want to note that the friend is female, not male, how should I say that, to avoid ambiguity? I just came across this glowy ore, and my 'Sharp Silver Pickaxe' won't mine it. I'm playing on the PC version of Terraria, if that helps.  How do I mine the glowy ore block? Convergence of $\int\limits_1^{\infty}x^\alpha\sin(x^\beta)dx$ return eats exception I found the following behavior at least weird:  def errors():     try:         ErrorErrorError     finally:         return 10  print errors() # prints: 10 # It should raise: NameError: name 'ErrorErrorError' is not defined   The exception disappears when you use return inside a finally clause. Is that a bug? Is that documented anywhere?  But the real question (and the answer I will mark as correct) is: What is the python developers' reason to allow that odd behavior? Cylindrical vs azimuthal symmetry What is the difference between cylindrical and azimuthal symmetry, are they the same? I have not been able to find a clear explanation of these terms wrt. a cylinder. I need to know which length I should integrate over to solve the problem below.  A cylinder of radius $R$ and length $L$ has a cylindrically symmetric volume charge density of $\rho(x)$. Calculate the total charge on the cylinder. Changing Default Apache version on MacOS Sierra A DHS sweep of the network showed that one of the researchers Macs is running a non-safe version of Apache (2.4.25) and they would like him to update to at least 2.4.27.  He is having difficulty getting MacOS to use the new version of Apache rather than the one installed by default with Sierra. His issues are:  Mac OS X 10.12.6 Sierra   I have installed httpd 2.4.27 . . .  /usr/local/apache2/bin/ but the default httpd running is located at /usr/sbin/ I cannot install a new httpd into /usr/sbin/  - no write persmissions for sudo I made a root user active - but "root" does not have write permissions for /usr/sbin/ either   I cannot find any source on the web that describes how to change the httpd path from the default system version (/usr/sbin/httpd) to a new USER installed version, like /usr/local/apache2/bin/httpd  What steps am I missing? Show address bar w. full path in Nautilus instead of breadcrumbs bar How to install Blender from the official website I installed Blender from official Ubuntu repository but its version is always behind the version available from the official Blender website.  I'm aware that I can launch the program within the downloaded folder, but i want to replace the install version with a new version downloaded from the website?   It must be present in the unity launcher.  I'm also aware of the available PPA's that are available, however I just want to install Blender manually.  Thanks in advance Shell Shortcuts in Windows OS I have an elliptic curve $E$ over $\mathbb{F}_{11}$ defined by $y^2=x^3+4x$ with the point at infinity $\mathcal{O}$  I have a divisor of $E$, defined by $$D=\left[(0,0)\right]+\left[(2,4)\right]+\left[(4,5)\right]+\left[(6,3)\right]-4\left[\mathcal{O}\right]$$  I know that $\text{deg}(D)=1+1+1+1-4=0$ and it is clear that $\text{sum}(D)=\infty$  Therefore $D$ is the divisor of a function - we want to find this function  I am given \begin{align}\text{div}(y-2x)&amp;=\left[(0,0)\right]+2\left[(2,4)\right]-3\left[\mathcal{O}\right]\\ \text{div}(x-2)&amp;=\left[(2,4)\right]+\left[(2,-4)\right]-2\left[\mathcal{O}\right] \end{align}  I am then told that we can express $D$ as follows:  $$D = \left[(2,4)\right]+\text{div}\left(\frac{y-2x}{x-2}\right)+\left[(4,5)\right]+\left[(6,3)\right]-3\left[\mathcal{O}\right]$$  However, when I expand this back out, I get the following:  \begin{align}D &amp;= \left[(2,4)\right]+\text{div}\left(\frac{y-2x}{x-2}\right)+\left[(4,5)\right]+\left[(6,3)\right]-3\left[\mathcal{O}\right]\\ &amp;= \left[(2,4)\right]+\text{div}(y-2x)-\text{div}(x-2)+\left[\left(4,5\right)\right]+\left[(6,3)\right]-3\left[\mathcal{O}\right]\\ &amp;= \left[(2,4)\right]+\left(\left[(0,0)\right]+2\left[(2,4)\right]-3\left[\mathcal{O}\right]\right)-\left(\left[(2,4)\right]+\left[(2,-4)\right]-2\left[\mathcal{O}\right]\right)+\left[\left(4,5\right)\right]+\left[(6,3)\right]-3\left[\mathcal{O}\right]\\ &amp;= \left[(0,0)\right]+2\left[(2,4)\right]+\left[(4,5)\right]+\left[(6,3)\right]+2\left[\mathcal{O}\right]-\left[(2,-4)\right] \end{align}  which clearly isn't equivalent to the original $D$ unless $$\left[(2,4)\right]+2\left[\mathcal{O}\right]-\left[(2,-4)\right]=-4\left[\mathcal{O}\right]$$  Can anyone either explain what I've done wrong or show me how I can obtain the above equation please    This question comes in three parts:   Step 1 - This one Step 2 -  Step 3 - I'm wanting to use Xen Hypervisor to be able to run both Ubuntu 12.04 and Windows 7 at the same time using virtualization. I can find plenty of XCP tutorials but I'm not looking to install the cloud stack, I just want the server confined to my machine.  All I need is a link to a decent tutorial for this purpose. I've read a ton of articles that lead me to believe I'm on the right track, but not enough to feel comfortable getting it right if I just dive in. I'm getting caught up on whether I need to install Ubuntu and then install Xen into it, or do I need to install Xen on a small partition and then leave the rest of the hard drive open for the virtualization... I dunno.  Basically, I need the virtualization of Windows 7 for Adobe Premiere and After Effects (until Lightworks for Linux comes out and I can afford NUKEX). If I didn't need those, I'd be rocking out a straight up Ubuntu machine or just using VirtualBox for my virtualization. However, I need Windows to run as an HVM so it runs like I would expect rather than having the issues I may run into with VirtualBox (video drivers, etc).  So, any suggestions? In electrostatics, why the electric field inside a conductor is zero? In electromagnetism books, such as Griffiths or the like, when they talk about the properties of conductors in case of electrostatics they say that the electric field inside a conductor is zero.  I have 2 questions:    We know that conductors (metallic) have free electrons which randomly moves in all directions, so how come we can talk about electrostatics which by definition means stationary charges? When the textbooks try to show why the electric field inside a conductor is zero they say let us put our conductor in an electric field. What happens then is that there will be an induced surface charge density which consequently induces an electric field within the conductor such that the total electric field within the conductor will be zero. That is perfectly understood, but my problem is the following: the original claim was that the electric field within a conductor is 0, not the electric field after putting the conductor in an external electric field it became zero. I do not understand the logic! I am creating an animated sprite list in Adobe Illustrator. I finished creating the sprites and used the Horizontal/Vertical Distribute Center commands to distribute the objects equally in some area.  But, if the objects are spread too far apart, the Distribute Center tool leaves, albeit equal, much distance between the objects. The opposite is true: if the objects are too close then distributing them will just overlap them.  My question is, is there any tool that just puts objects side by side?  I imagine that this tool would work if the objects aren't even equal in size.   I have viewed  and answer but it is not what I need. Can there be only one extension to the factorial? Usually, when someone says something like $\left(\frac12\right)!$, they are probably referring to the Gamma function, which extends the factorial to any value of $x$.  The usual definition of the factorial is $x!=1\times2\times3\times\dots x$, but for $x\notin\mathbb{N}$, the Gamma function results in $x!=\int_0^\infty t^xe^{-t}dt$.  However, back a while ago, someone mentioned that there may be more than one way to define the factorial for non-integer arguments, and so, I wished to disprove that statement with some assumptions about the factorial function.    the factorial   is a  for $x\in\mathbb{C}$ except at $\mathbb{Z}_{&lt;0}$ because of singularities, which we will see later. is a monotone increasing function that is concave up for $x&gt;1$. satisfies the relation $x!=x(x-1)!$ and lastly $1!=1$     From $3$ and $4$, one can define $x!$ for $x\in\mathbb{N}$, and we can see that for negative integer arguments, the factorial is undefined.  We can also see that $0!=1$.  Since we assumed $2$, we should be able to sketch the factorial for $x&gt;1$, using our points found from $3,4$ as guidelines.  At the same time, when sketching the graph, we remember $1$, so there can be no jumps or gaps from one value of $x$ to the next.  Then we reapply $3$, correcting values for $x\in\mathbb{R}$, since all values of $x$ must satisfy this relationship.  Again, because of $1$, we must re-correct our graph, since having $3$ makes the derivative of $x!$ for $x\in\mathbb N$ undefined.  So, because of $1$ and $3$, I realized that there can only be one way to define the factorial for $x\in\mathbb R$.  Is my reasoning correct?  And can there be only one extension to the factorial?    Oh, and here is a  to how I almost differentiated the factorial only with a few assumptions, like that it is even possible to differentiate.  Putting that in mind, it could be possible to define the factorial with Taylor's theorem? How to fix Finance::Quote to pull quotes in GnuCash I made the jump from Jumsoft's Money app to GnuCash a while back. It appealed to all my OCD tendencies. I don't know how to use Perl, or stuff of the sort, but with detailed instructions I manage to get things done... when I can find them.  I set up Finance::Quote to pull quotes after I set up GnuCash. I even managed to get it to pull quotes for Korean funds. And every thing had been working well, until tonight. I only get back a pop-up that states "There was an unknown error while retrieving the price quotes." That's where I need help. What can I do?  I dug around and found that it seems that it has something to do with Yahoo Finance ending it's API support. I also learned that Finance::Quote can pull data from many sources, so I don't really understand why one missing API would break it. I also saw that someone out there is trying to use AEX to pull the quotes instead for GnuCash specifically. I found Finance::Quote on cpan (I also first heard of cpan), but I don't understand what I'm looking at. But that's all beside the point. I don't know how to fix this.  Any help would be appreciated. I'm on Windows 10, GnuCash version 2.6.17, and have Strawberry Perl 5.18.2.2-32bit installed.  edit Thanks everyone who responded. It seems to be working more or less. The currency and Korean funds still haven't updated, but you've shown this horse to water. I'll do my best to drink it now. Analogue of the Cantor-Bernstein-Schroeder theorem for general algebraic structures The Cantor-Bernstein-Schroeder theorem states that if there are two sets $A$ and $B$ such that there exist injective (alternatively, surjective, assuming choice I think) maps $A \to B$ and $B \to A$, then $A$ and $B$ are in bijection.  It then seems natural to try to strengthen the result to other structures.  If $A$ and $B$ are, say, groups (or rings, modules, etc) such that there are injective homomorphisms from each to the other, then $A$ and $B$ are isomorphic.  I'm wondering if there are any results on this, or if there are known counterexamples.  The theorem for sets alone is nontrivial, so I feel like any results about how it holds for other structures would be quite interesting. Absolute path & Relative Path What's the difference between absolute path &amp; relative path when using any web server or Tomcat? How to find a 4D vector perpendicular to 3 other 4D vectors? In 3 dimensions it is possible to find a vector c (one of infinitely many) perpendicular to two vectors a and b using the cross product. Is there any way of extending this to 4 dimensions, i.e. given three vectors a, b, and c finding a vector d perpendicular to all of them?  I realize that this can be done by solving the equation system:  dot(a, d) = 0 dot(b, d) = 0 dot(c, d) = 0   and imposing an additional constraint, for instance setting the length of d to or one of its components to 1, but is there a way of doing this that does not involve solving an equation system? What does the Asterisk * mean in Objective-C?