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Do SAS expanders work transparently with SAS controllers? I'm new to SAS expanders and I'm trying to get things right. I understand that SAS expanders, when used with SAS controllers which supports expanders, allows you to utilise the full capacity of supported drives on a controller.  However, say I've got a SAS controller with 4 x SSF-8087 ports, which by itself would support 16 disks. I could then, as far as I understand, instead connect each of these ports to an expander to increase disk capacity.  However, will the SAS controller "see" all disks as if they were connected directly to the controller and not through an expander. Could I look at controllers as switches in a network? Would I be able to create a RAID from a disk on expander 1 and expander 3?  UPDATE:  I don't really have a specific project where I need to use SAS expanders. I'm just curious about how they work, and was thinking about how it would work if I used one or more SAS expanders with a LSI MegaRAID controller (9260-16i) which is running in one of our servers. However, I haven't really put any thought into the distinction between "local" disks and disks in another enclosure. The LSI MegaRAID controller has 4 SFF-8087 ports, which makes it capable of accomodating 16 drives. Why does space expansion not expand matter? I have looked at other questions on this site (e.g. "why does space expansion affect matter") but can't find the answer I am looking for.  So here is my question: One often hears talk of space expanding when we talk about the speed of galaxies relative to ours. Why, if space is expanding, does matter not also expand? If a circle is drawn on balloon (2d plane), and the balloon expands, then the circle also expands. If matter is an object with 3 spatial dimensions, then when those 3 dimensions expand, so should the object.  If that was the case, we wouldn't see the universe as expanding at all, because we would be expanding (spatially) with it.  I have a few potential answers for this, which raise their own problems:   Fundamental particles are 'point sized' objects. They cannot expand because they do not have spatial dimension to begin with. The problem with this is that while the particles would not expand, the space between them would, leading to a point where the 3 non-gravity forces would no longer hold matter together due to distance Fundamental particles are curled up in additional dimensions a la string theory. These dimensions are not expanding. Same problems as 1, with the added problem of being a bit unsatisfying. The answer seems to be (from Marek in the previous question) that the gravitational force is so much weaker than the other forces that large (macro) objects move apart, but small (micro) objects stay together. However, this simple explanation seems to imply that expansion of space is a 'force' that can be overcome by a greater one. That doesn't sound right to me. Why are the longer edges being beveled more? iOS: Your passcode is required to enable Touch ID Occasionally, when unlocking my iPhone or iPad, I get the following message:     Your passcode is required to enable Touch ID   Now this is not after a number of failed attempts or turning the device off. The device was used within the last 12 hours. Two of the devices are less than 2 months old, and one is a year old. Everything is up to date. It appears to be happening more-or-less at random.  Does anybody know why this message might appear? Say you take a dice and you roll it twice so that you have a pair (X,Y) where X represents the first roll, Y represents the second.   When you have a distribution like max(X,Y), what type of distribution is that? It's not uniform, obviously, because the outcomes of max(X,Y) are clearly more favorable towards the higher numbers. I'm being asked to "describe" the distribution. I know that we can look at each distinct outcome, like:  max(1,1) = 1  max(1,2) = 2  max(1,3) = 3  max(1,4) = 4  max(1,5) = 5  max(1,6) = 6  max(2,1) = 2  max(2,2) = 2  max(2,3) = 3  max(2,4) = 4  max(2,5) = 5  max(2,6) = 6  max(3,1) = 3  max(3,2) = 3  max(3,3) = 3  max(3,4) = 4  max(3,5) = 5  max(3,6) = 6  max(4,1) = 4  max(4,2) = 4  max(4,3) = 4  max(4,4) = 4  max(4,5) = 5  max(4,6) = 6  max(5,1) = 5  max(5,2) = 5  max(5,3) = 5  max(5,4) = 5  max(5,5) = 5  max(5,6) = 6  max(6,1) = 6  max(6,2) = 6  max(6,3) = 6  max(6,4) = 6  max(6,5) = 6  max(6,6) = 6  But is that the most effective way to look at a distribution like this? Is there a faster and more concise way to describe this distribution other than saying the probability of each outcome?  i.e.,  P(max = 1) = 1/36  P(max = 2) = 3/36 = 1/12  P(max = 3) = 5/36  P(max = 4) = 7/36  P(max = 5) = 9/36 = 1/4  P(max = 6) = 11/36  So, obviously, I understand how to LOOK at a discrete distribution and see what the probabilities are, but not until I look at each possible outcome. But what does it mean to "describe" a distribution? This isn't uniform or any sort of discrete distribution we're taught. Can someone provide clarity? Prove by induction: $n! \ge 2^{(n-1)}$ for any $n \ge 1$ Automatic coloring of numbers according to size Scientific talks have their own culture, reporting the results, typically for me, in a table of the following form  1.7333    1.6472    1.5418    1.4682    1.5740 1.1525    1.2878    1.2652    1.3511    1.5102 1.2181    1.1500    1.2220    1.2425    1.5699 0.9786    1.1051    1.1472    1.3096    1.6757 0.9569    0.9190    1.0433    1.2924    1.4867   There is clearly a(n almost perfectly) monotonic pattern for rows and columns. Now for the naked eye, this is not the most quick way of capturing informing.  I would like numbers that bigger in absolute value, to appear differently (either green vs red or light vs dark) than smaller numbers, such that the pattern becomes clear instantly.  Is there a way of programming this in LaTeX? Square of a second derivative is the fourth derivative I have a simple question for you guys, if I have this:  $$\left(\frac{d^2}{{dx}^2}\right)^2$$  Is it equal to this:  $$\frac{d^4}{{dx}^4}$$  Such that if I have an arbitrary function $f(x)$ I can get:  $$\left(\frac{d^2 f(x)}{{dx}^2}\right)^2 = \frac{d^4 f(x)}{{dx}^4}$$  Sorry if it's a pretty simple question, but I was trying to simplify something like this:  $$\left(a \cdot \frac{d^2}{{dx}^2} - f(x)\right)^2 g(x)$$ that's where it came up. How does he have those memories? Plurality of verb depending on plurality of list elements This question is also open on EL&amp;U:   Considering the amount of controversy it aroused, I believe it's past "Learners'" level.    An edit was suggested to my sentence.     There was were an orange, some grapes, two apples and a small pile of cherries on the plate.   In my native language plurality of the verb always follows plurality of the first element on the list. There were an orange,... sounds awkward to me, no matter what follows up. My simple solution was reordering:     There were some grapes, an orange, two apples and a small pile of cherries on the plate.   But that's not the first time I faced this situation and I'd like to know what the rules of grammar say about that — was my editor overzealous or am I trying to copy rules of my language that don't apply in English? InputStream returning NullPointerException (Java) Modelling the "Moving Sofa" I believe that many of you know about the moving sofa problem; if not you can find the description of the problem .     In this question I am going to rotate the L shaped hall instead of moving a sofa around the corner. By rotating the hall $180^{\circ}$ what remains between the walls will give the shape of the sofa. Like this:    The points on the hall have the following properties:  \begin{eqnarray} A &amp; = &amp; \left( r\cos { \alpha  } ,t\sin { \alpha  }  \right)  \\ { A }' &amp; = &amp; \left( r\cos { \alpha  } +\sqrt { 2 } \cos { \left( \frac { \pi  }{ 4 } +\frac { \alpha  }{ 2 }  \right)  } ,t\sin { \alpha  } +\sqrt { 2 } \sin { \left( \frac { \pi  }{ 4 } +\frac { \alpha  }{ 2 }  \right)  }  \right)  \\ { B } &amp; = &amp; \left( r\cos { \alpha  } -\frac { t\sin { \alpha  }  }{ \tan { \left( \frac { \alpha  }{ 2 }  \right)  }  } ,0 \right)  \\ { B }' &amp; = &amp; \left( r\cos { \alpha  } -\frac { t\sin { \alpha  }  }{ \tan { \left( \frac { \alpha  }{ 2 }  \right)  }  } -\frac { 1 }{ \sin { \left( \frac { \alpha  }{ 2 }  \right)  }  } ,0 \right)  \\ C &amp; = &amp; \left( r\cos { \alpha  } +t\sin { \alpha  } \tan { \left( \frac { \alpha  }{ 2 }  \right)  } ,0 \right)  \\ { C }' &amp; = &amp; \left( r\cos { \alpha  } +t\sin { \alpha  } \tan { \left( \frac { \alpha  }{ 2 }  \right)  } +\frac { 1 }{ \cos { \left( \frac { \alpha  }{ 2 }  \right)  }  } ,0 \right)  \end{eqnarray}  Attention: $\alpha$ is not the angle of $AOC$, it is some angle $ADC$ where $D$ changes location on $x$ axis for $r\neq t$. I am saying this because images can create confusion. Anyways I will change them as soon as possible.  I could consider $r=f(\alpha)$ and $t=g(\alpha)$ but for this question I am going to take $r$ and $t$ as constants. If they were functions of $\alpha$ there would appear some interesting shapes. I experimented for different functions however the areas are more difficult to calculate, that's why I am not going to share. Maybe in the future.  We rotate the hall for $r=t$ in the example above: In this case:   point A moves on a semicircle  The envelope of lines between A' and C' is a circular arc. One has to prove this but I assume that it is true for $r=t$.   If my second assumption is correct the area of sofa is $A= 2r-\frac { \pi r^{ 2 } }{ 2 } +\frac { \pi  }{ 2 } $. The maximum area is reached when $r = 2/\pi$ and it's value is: $$A = 2/\pi+\pi/2 = 2,207416099$$  which matches with Hammersley's sofa. The shape is also similar or same:    Now I am going to increase $t$ with respect to $r$. For $r=2/\pi$ and $t=0.77$:    Well, this looks like .   I believe thearea can be maximized by finding the equations of envelopes above and below the sofa. Look at  where @Aretino has computed the area below $ABC$.   I don't know enough to find equations for envelopes. I am afraid that I will make mistakes. I considered to calculate area by counting number of pixels in it, but this is not a good idea because for optimizing the area I have to create many images.  I will give a bounty of 200 for whom calculates the maximum area. As I said the most difficult part of the problem is to find equations of envelopes. @Aretino did it.  PLUS: Could following be the longest sofa where $(r,t)=((\sqrt 5+1)/2,1)$ ?   If you want to investigate further or use animation for educational purposes here is the Geogebra file:      Ok, I had some free time and I count number of pixels in the sofa and I am sure that I have something bigger than Hammersley's constant.  First, I made a simulation for Hammersley's sofa where $r=t=2/\pi$ and exported the image to png in 300 dpi (6484x3342 pixels) and using Gimp counted number of pixels which have exactly same value. For Hammersley I got $3039086$ pixels.   For the second case $r=0.59$ and $t=0.66$ and I got $3052780$ pixels. To calculate area for this case:  $$\frac{3052780}{3039086}(2/\pi + \pi/2)=2.217362628$$  which is slightly less than Gerver's constant which is $2.2195$. Here is the sofa: Given fields $K\subseteq L$, why does $f,g$ relatively prime in $K[x]$ imply relatively prime in $L[x]$? vinegar pickles safety Prove that $\phi(n) \geq \sqrt{n}/2$ So I'm trying to prove the following two inequalities: $$\frac{\sqrt{n}}{2} \leq \phi(n) \leq n.$$ The upper bound we get from simply noting that $\phi(n) = n \prod_{p | n}\left( 1 - \frac{1}{p}\right)$ and the fact that $(1 - \frac{1}{p}) \leq 1$. But how can we get the lower bound? I tried using the same expression for $\phi(n)$ but it seems to not really give me the inequality. Can you help? How to Hide sections of InfoPath form based on Group membership I need to Hide "Sections" of InfoPath form using SharePoint or AD group membership (either is fine). How can I achieve this?  InfoPath 2010 in SharePoint 2010 Solvability of $a \equiv x^2 \mod b$ $A$ abelian variety. Is the multiplication by $n_A$ surjective? Running command on multiple files within multiple directories I want to run the command   python quast.py -o quast contigs.fasta   but I want to do this for multiple contigs.fasta files in multiple directories.  I'm running OSX. I have problems with my Tilgin Vood Router. As I try to start my router by turning the power on (captain obvious), it generally doesn't work the first 3-4 times. This is getting very annoying.  Five minutes after turning the power on the router's signal LEDs don't blink in the way they should do in a connected state. I can connect to my routers web configuration interface through my PC connected to it via LAN though I can't access the internet.  It usually takes the router five minutes to get to the point where it should be connected to the internet but as it doesn't work the first times.  So I turn on my router 3-5 times, let him work 5 minutes and then suddenly, after turning the power off and on again it all works.  I also have to restart my PC too in order for everything to work.   How can I solve this problem? Just leave the router turned on all time? I prefer a daily IP switch, though. May the problem have something to do with my PC? There is another one connected to the router too and it doesn't work there either.