I want to prove that if $\alpha$ is a root of $x^{p^{d}} - x$ and $n=d\cdot s$, then $\alpha$ is also a root of $x^{p^{n}} - x$. So far I know that $x^{p^{d}-1} =1$ and I've tried this: $$x^{p^{n} -1} = x^{p^{ds}-1}$$ I have also tried: $$x^{p^{n}} = x^{p^{ds}} = (x^{p^{d}})^s=\alpha^s$$ but I can't conclude that $\alpha^s=\alpha$. Can someone help me?
I've only been able to resolve this far  $$ d \mid n \Rightarrow \exists l \in \mathbb{Z}\; \textrm{such that} \; n=dl \Rightarrow x^{q^n} - x = x^{q^{dl}} - x.$$  I can't move forward.
Exclude lstinputlisting I want to save paper when I print my LaTeX file. That's why I want to hide all my code, inserted wtih lstinputlisting{}  I already know how to exclude tables and images or math but the lstinputlisting{} is a problem.   Hope someone can help me  \usepackage{comment} \excludecomment{figure} \let\endfigure\relax \excludecomment{table} \let\endtable\relax \excludecomment{align*} \let\endalign\relax \excludecomment{lstlisting} \let\endlstlisting\relax   edit: added lstlisting I believe in Windows Vista, we could use linkd command. However, I cannot find this command in Windows 7. I know I can use Windows Explorer and delete the junction, but I wonder if this can be done in Command Prompt.  I can use mklink command to create a junction from a Command Prompt, so, what is the opposite of it (something like linkd)? Can abilities with a [Click] cost be used during a run? Some abilities have a [Click] cost to them, such as Crypsis' ability to add a virus counter to it.  Does it count as a "paid ability" for the purposes of the ? I.e. could I, mid-run, add a virus counter to Crypsis for the cost of one [Click]? How many ways can you divide $9$ students into three unlabeled teams where one team contains $4$ people, one contains $3$ people and the last contains $2$ people? Unlabeled, meaning that groups with abc = bca = cba, etc  I understand how to do this if the teams are labeled:  $$\frac{9!}{4!3!2!}$$  But there is a term missing in the denominator when the teams are unlabeled and I'm having difficulty understanding how to calculate how many ways the teams can be organized.  There are $3!$ ways to organize the same first group of $3$, $4!$ ways to organize the same second group of $4$ and $2!$ ways to organize the last group of $2$. Why wouldn't you multiply $3!4!2!$ in the denominator?   For instance:  (ABC, DEFG) = (ABC, DEGF)  = (ABC, DFEG) = (ACB, DFGE), etc Spy from Earth studies Galactic Empire nobility and prevents plot to depose Emperor In Triangle, $\sin\frac A2\!+\!\sin\frac B2\!+\!\sin\frac C2\!-\!1\!=\!4\sin\frac{\pi -A}4\sin\frac{\pi -B}4\sin\frac{\pi-C}4$ To prove  $$\sin\frac A 2+\sin\frac B 2+\sin\frac C 2-1=4\sin\frac{\pi -A}4\sin\frac{\pi -B}4\sin\frac{\pi-C}4$$  My approach :   $$ \begin{align} \text{L.H.S.} &amp; = \sin\frac{A}{2}+\sin \frac{B}{2}+\sin\frac{C}{2} -1 \\[8pt] &amp; = 2\sin\frac{A+B}{4}\cos\frac{A-B}{4}+\cos\left(\frac{\pi}{2}-\frac{C}{2}\right)-1 \\[8pt] &amp; = 2\sin\frac{\pi -C }{4}\cos\frac{A-B}{4} - 2\sin^2\left(\frac{\pi -C}{4} \right) \\[8pt] &amp; =2\sin\frac{\pi -C }{4}\left\{ \cos\frac{A-B}{4} - \sin\left(\frac{\pi -C}{4} \right)\right\} \end{align} $$  Unable to move further please help. thanks. Can a Ranger select two humanoid races as their favored enemies at 6th and 14th levels? In D&amp;D 5e, the Ranger can choose at first level to have a Favored Enemy of a single broad type (i.e. fiends, giants, aberrations, etc.) or of two different specific humanoids (such as orcs and gnolls). They gain additional Favored Enemies at 6th and 14th level.  At each of those levels, could they opt to pick two humanoid races? Or would the choice be between one type and one specific humanoid?  Relevant text:     Choose a type of favored enemy: aberrations, beasts, celestials,   constructs, dragons, elementals, fey, fiends, giants, monstrosities,   oozes, plants, or undead. Alternatively, you can select two races of   humanoid (such as gnolls and orcs) as favored enemies.   And:     You choose one additional favored enemy, as well as an associated   language, at 6th and 14th level.   (Both quotes from PHB, pg 91.)  Specifically, I'm wondering whether "one additional favored enemy" means one absolutely, or whether the "Alternatively..." sentence means that two humanoid races can be substituted for one type whenever you select a favored enemy. Question is :  Suppose $X$ is a topological space of infinite cardinality which is homeomorphic to $X\times X$. Then which of the following is true:   $X$ is not connected. $X$ is not compact $X$ is not homeomorphic to a subset of $\mathbb{R}$ None of the above.   I guess first two options are false.  We do have possibility that product of two connected spaces is connected.  So, $X\times X$ is connected if $X$ is connected. So I guess there is no problem.  We do have possibility that product of two compact spaces is compact.  So, $X\times X$ is compact if $X$ is compact. So I guess there is no problem.  I understand that this is not the proof to exclude first two options but I guess the chance is more for them to be false.  So, only thing I have problem with is third option.  I could do nothing for that third option..  I would be thankful if some one can help me out to clear this.  Thank you :) I have a ton of MP3s, but their ID tags are not filled in.  Going in manually to populate them is simply too much work.    I was wondering if there is a way to populate them automatically.  Perhaps by feeding the MP3s to Shazam or Midori or something like.  I am not above writing some code to make it happen, if someone can suggest a way. Update: Failed to lock /var/cache/apt/archives/lock I am running the upgrade(12.10 to 13.04) through terminal and it keeps getting stuck at this:  Failed to lock /var/cache/apt/archives/lock   What do I do? I have root access. I have no idea what else to try other than a clean install.  Checking package manager Reading package lists... Done     Building dependency tree           Reading state information... Done Building data structures... Done   Calculating the changes Calculating the changes  Do you want to start the upgrade?   3 packages are going to be removed. 155 new packages are going to be  installed. 1204 packages are going to be upgraded.   You have to download a total of 595 M. This download will take about  1 hour 15 minutes with a 1Mbit DSL connection and about 23 hours with  a 56k modem.   Fetching and installing the upgrade can take several hours. Once the  download has finished, the process cannot be canceled.    Continue [yN]  Details [d]y  Fetching  Could not download the upgrades   The upgrade has aborted. Please check your Internet connection or  installation media and try again. All files downloaded so far have  been kept.   Failed to lock /var/cache/apt/archives/lock    Restoring original system state  Aborting Reading package lists... Done     Building dependency tree           Reading state information... Done Building data structures... Done  root@chaitanya-XPS-Ubuntu:~# Why does observation collapse the wave function? In one of the first lectures on QM we are always taught about Young's experiment and how particles behave either as waves or as particles depending on whether or not they are being observed. I want to know what about observation causes this change? A few weeks back I was using std::ifstream to read in some files and it was failing immediately on open because the file was larger than 4GB. At the time I couldnt find a decent answer as to why it was limited to 32 bit files sizes, so I wrote my own using native OS API.  So, my question then: Is there a way to handle files greater than 4GB in size using std::ifstream/std::ostream (IE: standard c++)  EDIT: Using the STL implementation from the VC 9 compiler (Visual Studio 2008).  EDIT2: Surely there has to be standard way to support file sizes larger than 4GB. Induction proof: $S$ contains powers of 2 and predecessors implies $S={\bf N}$ Let $S$ be a subset of $\mathbb{N}$ such that $$2^k\in S\quad\forall\ k\in\mathbb N$$ and, $$k\in S\Rightarrow k-1\in S\quad\forall\ k\ge2,k\in\mathbb N$$  Show by induction that $S=\mathbb N$. Show that $f_n(z)=z^n$ converges locally uniformly on $|z|&lt;1$ but not uniformly.   I am really confused. I saw a seemingly reliable reference stating that $z^n$ does not converge uniformly to $0$ on $\{|z|&lt;1\}$ but it does converge uniformly on $\{|z|\le r,r&lt;1\}$. What is the nuance between the two former sets? With $r_n&lt;1,r_n\to 1$, shouldn't one conclude that $f_n(z)$ uniformly converges on $\{|z|&lt;1\}$? And besides, this source tells me what I have to prove for Homework is disproved, which really makes it more frustrating.  I would appreciate any help. Why are lamps and mesh lights not illuminating anything? There are several techniques to do a 3 point lighting system.   How do you properly light a scene while using mesh planes as emission light sources?   I'm following , however despite setting up planes to act a lighting sources, the scene is still very dark.      Actual Blender file here: How can I resize (not compress) multiple images in a MS Word document? In a blank document, I insert images (screenshots - all are of same size and same format) from a folder. I want to resize the images to a desired size. All now I am doing is selecting one image by one, setting its size. Word does not seem to have multiselect working for images.  Ideally, I want to select multiple images set its size in one go with out use of a macro. Setting-up SSH-tunnel for VPN? I have a laptop running Ubuntu.  I also rent a VPS running Ubuntu.  I've bought a paid VPN-service, which I use on my laptop.  I was wondering how I can set-up a SSH-tunnel or something similar, from my laptop to the VPS I'm renting, so I then can run VPN (from my laptop) "through" it (to the VPS), to the VPN-server.  The reason I wish to do so, is that a place I often work blocks my VPN-service, but not my SSHing to my VPS...  While I have been using Linux for a while, I'm a novice when it comes to setting-up virtual network-devices, advanced routing, and so on... so please spoonfeed me (using small words and colorful pictures ;-)  I have been googling and looking at various answers on the StackExchange-network, but not seems to be exactly what I'm after.  SSH-tunneling seemed promising, but since I don't have/know an exact domain and port for my VPN, I couldn't figure out how to use it...  I guess I may need to set-up some virtual network-device on my laptop (which ends on the VPS), and connect VPN through it - but how to do this, sadly goes over my head. What is the most "pythonic" way to iterate over a list in chunks? Setting: Let $\sum a_n z^n$ have radius of convergence $R$.  We have that  $$ R = \frac{1}{\underset{n \rightarrow \infty}{\limsup} \left|a_n \right|^{1/n}} $$  via Hadamard's formula for the radius of convergence.  Question: What are the radii of convergence for (i) $\sum a_n z^{2n}$ and (ii) $\sum a_n^2 z^n$?  Attempt for $\sum a_n^2 z^n$:   For ease of notation, let $R_1$ and $R_2$ denote the radii of convergence for power series $\sum a_n z^{2n}$ and $\sum a_n^2 z^n$ respectively. From Hadamard's formula, we then have  $$ R_2 = \frac{1}{\underset{n \rightarrow \infty}{\limsup} \left|a_n^2 \right|^{1/n}} = \left( \frac{1}{\underset{n \rightarrow \infty}{\limsup} \left|a_n \right|^{1/n}} \right)^2 = R^2 $$   But what about $R_1$? I'm looking to install 32-bit windows with IIS to dual-boot with my 64-bit edition of windows on a laptop. I have a ten gig partition on my harddrive - is this large enough for a simple Windows install and a few programs?   I've never dual booted before, so I'm not quite sure how easy it is to mess up the process. Can you recommend any tutorials on how to set up a dual-boot configuration on Windows?