How would you go about showing that $p(x)=\frac{x^5-1}{x-1}=x^4+x^3+x^2+x+1$ is irreducible over $\mathbb{Q}$. I'm having trouble seeing how one can show whether this kind of polynomials are irreducible or not  Thanks for your help.
As opposed to the generic polynomial form for utilizing the Eisenstein Criterion ($a_nx^n+a_{n-1}x^{n-1}+\dots+a_0\in\mathbb{Z}[x]$ is irreducible in $\mathbb{Q}$) how do we prove that if $p$ is a prime, $x^{p-1}+x^{p-2}+\dots+x+1$ is irreducible over $\mathbb{Q}$?
Looking for a 1000 most frequent english words Is there a free list of 1000 most frequent words used in everyday English? If there is a list with usage examples (as each of those words has multiple meanings for sure, not to cound phrasal verbs), it would be even greater.  Is there something alike, but with phrases? (E.g. 1000 most frequent phrases) What bends fabric of space-time? I know that mass can bend fabric of space-time, which causes gravity by making an object curve around a planet or star but is there anything else that can bend it?  Other energy sources, forces anything? Summing the power series $\sum\limits_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1}\prod\limits_{k=1}^n\frac{2k-1}{2k} $ I'd like to determine the function corresponding to the following power series: $$x + \sum_{n=1}^\infty (-1)^n\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots2n} \frac{x^{2n+1}}{2n+1}, $$ where $|x|&lt;1$. How to make a field NULL using MySQL GUI Tools? I'm using the MySQL Query Browser (part of the ) and need to change a field to NULL, but I can't figure out how to do it - if I delete the value it tries to update it to ''. Typing "NULL" makes it try to update to 'NULL' (a string).  I know I could just write a query to do it, but that defeats the entire purpose of the tool, no? Center a column in align I have the following type of code in my document:  \begin{align*}        (-m)+m &amp; \stackrel{\mathrm{Ax 1.1 (i)}}{=}  m+(-m)\\   &amp;\stackrel{\mathrm{Ax 1.4}}{=}  0 \   I want to align on the = but that is complicated by the fact that there is something hovering over it.  It would seem to be easier to align their centers (no matter what, that is where the = will be).  How can I center a column in align (or align* to be honest, I am not that clear on the difference) Shadows along edges of mesh in 3D View Variable importance from GLMNET Battle of Hoth: how did the AT-ATs manage to disable the shield generator? (This question is related to , and was originally posted as an answer there, but it's not the same)    In , the Empire attacks Rebel Echo Base located in Hoth. The Rebels are alerted before the attack, and raise some sort of shield to prevent orbital bombardment of their base. The Empire is then forced to deploy land forces to first disable the shield generator, before destroying the base. Presumably, the Imperial forces are deployed using dropships.  My question is about how the Rebel shield generator works, and how the Imperials manage to bypass it.  I think the shield generator is something of a plot hole. At first I incorrectly thought it was a planetary shield (i.e. a shield around all of Hoth) preventing orbital bombardment, and also preventing any dropships (with their AT-AT payload) from landing, but I think : the  states it is an energy field around just an area of Hoth, presumably Echo Base:     VEERS: My lord, the fleet has moved out of light-speed. Com-Scan has   detected an energy field protecting an area around the sixth planet of   the Hoth system. The field is strong enough to deflect any   bombardment.   So my initial objection ("how did the Imperial dropships bypass the shield?") has been answered: they didn't. They landed outside the energy field to deliver the AT-ATs. But now I have another question: does an AT-AT have to walk inside the field to fire at the generator, or can it fire into the field from outside?  Maybe the field simply cannot stop matter, only energy weapons (therefore preventing orbital bombardment), in which case the AT-ATs would be able to just walk inside. However, the script also states:     LEIA: All troop carriers will assemble at the north entrance. The   heavy transport ships will leave as soon as they're loaded. Only two   fighter escorts per ship. The energy shield can only be opened for a   short time, so you'll have to stay very close to your transports.   So it's clear that the energy field is also capable of stopping matter, in this case troop transports and fighters. It follows that it should also prevent AT-ATs from walking through it. So the only way I can see the walkers disabling the generator is by firing into the energy field!  This would mean the field is strong enough to stop matter such as ships and fighters, and energy weapons used for orbital bombardment (and also the Ion Cannon the Rebels use as an anti-orbit battery: Leia also states they must lower the shield to use it), but not able to stop whatever energy weapons the AT-ATs use!  So how on Earth did Imperial Walkers manage to disable the shield generator? Is spacetime discrete or continuous? After upgrading to 12.10 quantal, the packaged version of fglrx no longer works. I discovered that this is because there is a separate 'legacy' fglrx driver for the HD 2k-4k series cards, but it is incompatible with the xorg server on 12.10.  This is the most current version of the driver for HD 2000 through HD 4000 series cards. You can't use the non-legacy fglrx driver, but you can use the open-source radeon driver if you prefer your WM compositing to be laggy and your YouTube videos to play like they would on a Pentium MMX series:    Usually this driver can be installed in the following way, necessary because apt-get install fglrx would pull in the non-legacy driver:  wget http://www2.ati.com/drivers/legacy/amd-driver-installer-12.6-legacy-x86.x86_64.zip unzip amd-driver-installer-* sudo sh ./amd-driver-installer-*.run --buildpkg Ubuntu/quantal sudo dpkg -i fglrx*.deb sudo aticonfig --initial -f   If you use a different version of fglrx (for example, a newer 12.9 that doesn't support those cards) then the final command will give you an error no supported hardware detected or something similar. However, everything works at this point and you will get a reasonable xorg.conf:  ... other stuff  Section "Device"         Identifier  "aticonfig-Device[0]-0"         Driver      "fglrx"         BusID       "PCI:1:5:0" EndSection  ... other stuff   At this point you're supposed to reboot and everything will be working with the fglrx driver. However, upon rebooting, you'll be treated to the following errors in Xorg.0.log when fglrx attempts to load:  (EE) Failed to load /usr/lib/xorg/modules/drivers/fglrx_drv.so: /usr/lib/xorg/modules/drivers/fglrx_drv.so: undefined symbol: noXFree86DRIExtension   Some searching around will show that this is a problem with the legacy ATI drivers not supporting xserver 1.13 or newer. () ATI has released a fixed driver for its most recent (HD 5000 series or later) cards, but not for the 'legacy' cards yet. The non-legacy ATI drivers can't be used with the old cards.  What should an Ubuntu user, using one of these HD 2000-4000 series cards, do?    Wait for an updated 'legacy' ATI driver that properly works with xserver 1.13?   Downgrade back to 12.04 Precise, which uses xserver 1.11? Try to downgrade xserver on 12.10 Quantal to 1.12, which could possibly break Unity and GNOME? Forced upgrade to HD 5000 series or later card? (Not possible with integrated graphics...) Some other 1337 action that fixes this problem painlessly? What is the fastest way to convert float to int on x86 What is the fastest way you know to convert a floating-point number to an int on an x86 CPU.  Preferrably in C or assembly (that can be in-lined in C) for any combination of the following:   32/64/80-bit float -> 32/64-bit integer   I'm looking for some technique that is faster than to just let the compiler do it. Wiener Process $dB^2=dt$ Why is $dB^2=dt$? Every online source I've come across lists this as an exercise or just states it, but why isn't this ever explicitly proved? I know that $dB=\sqrt{dt}Z$, but I don't know what squaring a Gaussian random variable means. Let $A\in M_{n\times n}(\Bbb R)$ so that $I\notin span(A,A^2,...,A^n)$. Prove that $\det(A)=0$.   I was thinking of showing $A$ is not invertible, meaning it has an eigenvalue of $0$. Since no matter the power you give to $A$, it is still not the identity, you can deduce that at least one eigenvalue is indeed $0$. However this doesn't work if you have different eigenvalues in the Matrix, so I got stuck.  Help will be most appreciated. Word for the behavior of people who complain at everything you do What is the Big-O time complexity of the following nested loops: for (int i = 0; i &lt; N; i++) {     for (int j = i + 1; j &lt; N; j++) {         System.out.println(&quot;i = &quot; + i + &quot; j = &quot; + j);     } }  Would it be O(N^2) still? Looking for old book, SF, artificial world, different zones for different species that visitors "inhabit" I read this book at least ten years ago.  Any help identifying it would be appreciated.  Characters are teleported to an artificial world where each take over the mind of a different species.  I remember a moving sentient plant, a faun-like character, and a wasp-like character.  The planet is subdivided into different zones designed for each species.  One of the characters had long sharp fingernails with poison in them.  Thanks for any help, steve strptime() equivalent on Windows? Is there a good equivalent implementation of strptime() available for Windows?  Unfortunately, this POSIX function does not appear to be available.   - summary: it converts a text string such as "MM-DD-YYYY HH:MM:SS" into a tm struct, the opposite of strftime(). Continuity of absolute value of a function Let $f(x)$ be a continuous function. Prove that $\left|f(x)\right|$ is also continuous.  Is it correct to say that, by the reverse triangle inequality, $\left|f(x)-f(c)\right| \geq \left|f(x)\right|-\left|f(c)\right|$ in all cases, so we will always have that for all $\left|x-c\right|&lt;\delta$ implies $\left|f(x)-f(c)\right|\leq \epsilon$?  I am not sure if my solution adequately uses the functional form, since I just used the equation and not my actual function. Please help with the proper solution! Proving Pascal's Rule : ${{n} \choose {r}}={{n-1} \choose {r-1}}+{{n-1} \choose r}$ when $1\leq r\leq n$ I'm trying to prove that ${n \choose r}$ is equal to ${{n-1} \choose {r-1}}+{{n-1} \choose r}$ when $1\leq r\leq n$.  I suppose I could use the counting rules in probability, perhaps combination= ${{n} \choose {r}}=\frac{n!}{r!(n-r!)}$.  I want to see an actual proof behind this equation.  Does anyone have any ideas? I am given this problem: let $a\ge0$,$b\ge0$, and the sequences $a_n$ and $b_n$ are defined in this way: $a_0:=a$, $b_0:=b$ and $a_{n+1}:= \sqrt{a_nb_n}$ and $b_{n+1}:=\frac{1}{2}(a_n+b_n)$ for all $n\in\Bbb{N}$    To prove is that both sequences converge and that they have the same limit. I don't know how to show this. I have spent 2 hours on this, no sign of success