Derivation of drift velocity
Drift velocity in Drude model
PostgreSQL crash when connection failed Error log is:  016-02-21 21:22:34 GET DETAIL:  Connection matched pg_hba.conf line 83: "host    all             all             ::1/128                 md5" .... .... 2016-02-21 21:22:37 GET DETAIL:  Connection matched pg_hba.conf line 83: "host    all             all             ::1/128                 md5" 2016-02-21 21:22:37 GET FATAL:  password authentication failed for user "postgres" 2016-02-21 21:22:37 GET DETAIL:  Connection matched pg_hba.conf line 83: "host    all             all             ::1/128                 md5" 2016-02-21 21:22:37 GET LOG:  CreateProcess call failed: A blocking operation was interrupted by a call to WSACancelBlockingCall.       (error code 299) 2016-02-21 21:22:37 GET LOG:  could not fork new process for connection: No error 2016-02-21 21:22:37 GET LOG:  incomplete startup packet 2016-02-21 21:22:38 GET FATAL:  password authentication failed for user "postgres" 2016-02-21 21:22:38 GET DETAIL:  Connection matched pg_hba.conf line 83: "host    all             all             ::1/128                 md5" 2016-02-21 21:22:42 GET FATAL:  password authentication failed for user "postgres" 2016-02-21 21:22:42 GET DETAIL:  Connection matched pg_hba.conf line 83: "host    all             all             ::1/128                 md5" 2016-02-21 21:24:26 GET LOG:  could not receive data from client: An operation on a socket could not be performed because the system lacked sufficient buffer space or because a queue was full. 2016-02-21 21:24:26 GET LOG:  incomplete startup packet 2016-02-21 21:24:29 GET FATAL:  password authentication failed for user "postgres" 2016-02-21 21:24:45 GET PANIC:  could not write to log file 000000010000000100000041 at offset 4259840, length 8192: Invalid argument 2016-02-21 21:24:45 GET STATEMENT:  COMMIT 2016-02-21 21:24:53 GET LOG:  server process (PID 470944) was terminated by exception 0xC0000409 2016-02-21 21:24:53 GET DETAIL:  Failed process was running: COMMIT 2016-02-21 21:24:53 GET HINT:  See C include file "ntstatus.h" for a description of the hexadecimal value. 2016-02-21 21:24:53 GET LOG:  terminating any other active server processes   Can anyone help to solve the problem? I spent many hours but didn't find a solution.   Postgres version 9.4 Operating System: Windows  postgresql.conf is:  listen_addresses = '*' port = 5432  max_connections = 100 shared_buffers = 128MB   dynamic_shared_memory_type = windows   pg_hba.conf is:  # IPv4 local connections: host    all             all             127.0.0.1/32            md5 host    all             all             0.0.0.0/0               md5 How to create a generic array in Java? What can you say about $f$ and $g$ in the case that $fg$ is 1) Injective, 2) Surjective - Cohn - Classic Algebra Page 15 Question: Are my proofs below valid?   In both cases we are using: $f:A\to B, g: B\to C$  Notation of your type converted: $(g\circ f)(x)=g(f(x))=xfg$    If $fg$ is injective what can be said about $f$ and $g$?  If $fg$ is injective then $f$ is injective:  Proof:  $(x_1fg = x_2fg \implies x_1=x_2)$  $x_1fg=x_2fg$ is clearly true when $x_1f=x_2f$  Which requires $x_1=x_2$, hence $f$ is injective.  If $f$ were not injective, then $x_1fg\ne x_2fg$  $\blacksquare$    If $fg$ is surjective what can be said about $f$ and $g$?  If $fg$ is surjective then $g$ is surjective:  Proof:  $(\forall c\in C)(\exists a\in A) |( a(fg))=c$  This means for $g$ every element must have had been mapped to. How can this be proved Mathematically?  well $(af)g=c$ is surjective, hence $(af)=(b\in B)$ and $bg=c$  $(\forall c\in g)(\exists b\in B)|(bg=c)$  Hence $g$ is surjective.  $\blacksquare$ Weighting features prior to SVM I'm building an object detector using HOG features and linear SVM. Some of the regions of the object are more "distinctive" so I would like to give more weight to the features extracted from those regions. (e.g. imagine we want to detect between mugs and glasses, then the most "distinctive" part is the mug's handle)  How could I do this? For sure there should be a more intelligent way than just replicating the features of those "distinctive" regions. What is "umask" and how does it work? I believe that umask is something that controls , but do not fully understand it.  After running umask 0644 in a , I cannot read the files I create with the command-line text editor nano. I noticed that the permissions of that file are set to 0022 instead of the default 0755.  How does umask work? I thought I could just remove the each digit in the umask from 0777, 7 - 6 = 1 and 7 - 4 = 3, so I expect the permissions to be 0133, but apparently, this is not the case.   What is umask exactly? Explain it to me like I was a "Linux noob" How do I calculate with umask? What are use cases for umask? Creating non-reverse-engineerable Java programs Usage of -n and -z in test built in - Bash In the Prisoner of Azkaban, Harry and Lupin could practice the patronus on the boggart, but when Harry was in the final task in the Goblet of Fire, he tried to use the patronus on the boggart, it didn't work and he then moved on to use ridikkilus.  Why could he use a patronus against the boggart, but then not in another? Compute the mean of $(1 + X)^{-1}$ where $X$ is Poisson$(\lambda)$ Question  Let $X$ be Poisson with parameter $\lambda$.  Compute the mean of $(1 + X)^{-1}$. (Introduction to Probability Theory, Hoel, pp. 104)  Answer Key  The answer key shows that the mean is $\lambda^{-1}(1 - e^{-\lambda})$  My Solution  $$E(1 + X)^{-1} = \sum\limits_{j = 1}^\infty (1 + j)^{-1} \dfrac{\lambda^{(1 + j)^{-1}}e^{-\lambda}}{(1 + j)^{-1}!}$$ $$= e^{-\lambda} \sum\limits_{j = 1}^\infty \dfrac{\lambda^{(1 + j)^{-1}}}{((1 + j)^{-1} - 1)!}$$ $$= \lambda^{-1} e^{-\lambda} \sum\limits_{j = 1}^\infty \dfrac{\lambda^{(1 + j)^{-1} - 1}}{((1 + j)^{-1} - 1)!}$$ $$= \lambda^{-1} e^{-\lambda}(e^{\lambda} - 1)$$ $$= \lambda^{-1}(1 - e^{-\lambda})$$  EDIT: I found why I got the answer incorrect.  Thanks for those who answer the question! :) I would like to treat a user as a non-gender noun and refer to it with the gender-neutral pronoun, it. E.g.,     The user defines two variables, x and y. It then multiplies each variable by a prime number.    However, on  I found this:      The word "it", however, has an extremely impersonal connotation, even offensive, in common usage and is rarely used in English to refer to an unspecified human being or person of unknown gender. This is because the word "it" connotes that the person being specified is inferior to a person or is an object.   Is to appropriate to refer to a person of unknown sex as it?  Should I rephrase my sentence as follows:     The user defines two variables, x and y. The user then multiplies each variable by a prime number. I have a USB drive that I can see/access/modify/etc in Linux/Ubuntu but it doesn't show up on my Mac OSX.  I did sudo fdisk /dev/sdb (where my USB is) and then g to create a new GPT partition table, n to make a new partition, used all default settings, then w to save and exit.  Then I did sudo mkfs.fat -F32 /dev/sdb1 to format the partition with a FAT32 filesystem.  I ejected the USB and put it in my Mac and it doesn't show up / get detected.  What did I miss? Review of a proof for De Moivre's theorem using mathematical induction The following is a rigorous proof of De Moivre's theorem by means of mathematical induction. The theorem put simply is that:  Any complex number, $z = a+bi$, on a cartesian plane can be expressed in polar form, where $a=r\cos\theta$ and $b=r\sin\theta$ and $r$ is the absolute distance from the origin to the point $z$. In light of this, the expression for the complex number $z$ in polar form is $z=r(\cos\theta+i\sin\theta)$ In order to find the $n^{th}$ power of $z$, the following rule applies: $$\displaystyle z^n=(r(\cos\theta+i\sin\theta))^n=r^n(\cos n\theta+i\sin n\theta)$$ where $n\in\mathbb N$ (I have confined $n$ to the natural numbers given that I intend on using mathematical induction to prove the theorem.  Base step: We must prove that what is stated is true for $n=1$, hence $$z^1=r^1(\cos 1\cdot\theta+i\sin1\cdot \theta)$$ $$=r(\cos\theta+i\sin\theta)$$ which is true. Next we assume it is true for $n=k$, and therefore proceed to the inductive step. Inductive step: When calculating $z^{n+1}$, it's the same as calculating $z^nz$, hence $$z^{n+1}=(r^n(\cos n\theta + i\sin n\theta))(r(\cos \theta+i\sin \theta))$$ $$=r^nr(\cos n\theta+i\sin n\theta)(\cos\theta +i\sin \theta)$$ $$=r^{n+1}(\cos n\theta\cdot\cos\theta+\cos n\theta \cdot i\sin\theta+i\sin n\theta\cdot\cos\theta+i^2\sin n\theta\cdot\sin\theta)$$ from here, $i^2=-1$, and therefore $+i^2\sin n\theta\cdot\sin\theta$ becomes $-\sin n\theta\cdot\sin\theta$. In addition to this, the following trigonometric identities will be used: $$\sin(n\theta + \theta) = \sin n\theta\cdot\cos\theta+\cos n\theta\cdot\sin\theta$$ $$or$$ $$k\sin(n \theta+\theta)=k\sin n \theta\cdot\cos \theta + \cos n\theta\cdot k\sin\theta$$ $$and$$ $$\cos(n \theta + \theta)=\cos n\theta\cdot\cos\theta-\sin n\theta\cdot\sin\theta$$ by using these, the original equation has now become $$=r^{n+1}(\cos(n \theta + \theta) + i\sin(n \theta + \theta))$$ then, after factoring out $\theta$, the theorem becomes $$=r^{n+1}(\cos((n+1)\theta)+i\sin((n+1)\theta))$$ which completes the proof. Question: is this proof correct all the way through? Or have I missed any conditions? If all is well in the case of this proof, could somebody aid me in proving it for $\{n|n\in\mathbb R,n\geq1\}$ How do I detect running processes using a library package? This command is doing the rounds for detecting currently running processes using glibc:  lsof | grep libc | awk '{print $2}' | sort | uniq   I find it intensely annoying, since /libc/ matches not only libc, but on my system:  /lib/x86_64-linux-gnu/libcap.so.2.24 /lib/x86_64-linux-gnu/libcgmanager.so.0.0.0 /lib/x86_64-linux-gnu/libcom_err.so.2.1 /lib/x86_64-linux-gnu/libcrypt-2.19.so /lib/x86_64-linux-gnu/libcrypto.so.1.0.0 /usr/lib/libcamel-1.2.so.45.0.0 /usr/lib/unity-settings-daemon-1.0/libcolor.so /usr/lib/unity-settings-daemon-1.0/libcursor.so /usr/lib/x86_64-linux-gnu/colord-plugins/libcd_plugin_camera.so /usr/lib/x86_64-linux-gnu/colord-plugins/libcd_plugin_scanner.so /usr/lib/x86_64-linux-gnu/gtk-3.0/modules/libcanberra-gtk3-module.so /usr/lib/x86_64-linux-gnu/libcairo-gobject.so.2.11301.0 /usr/lib/x86_64-linux-gnu/libcairo.so.2.11301.0 /usr/lib/x86_64-linux-gnu/libcanberra-0.30/libcanberra-pulse.so /usr/lib/x86_64-linux-gnu/libcanberra-gtk3.so.0.1.9 /usr/lib/x86_64-linux-gnu/libcanberra.so.0.2.5 /usr/lib/x86_64-linux-gnu/libcap-ng.so.0.0.0 /usr/lib/x86_64-linux-gnu/libck-connector.so.0.0.0 /usr/lib/x86_64-linux-gnu/libcolordprivate.so.1.0.23 /usr/lib/x86_64-linux-gnu/libcolord.so.1.0.23 /usr/lib/x86_64-linux-gnu/libcroco-0.6.so.3.0.1 /usr/lib/x86_64-linux-gnu/libcupsmime.so.1 /usr/lib/x86_64-linux-gnu/libcups.so.2 /usr/lib/x86_64-linux-gnu/samba/libcliauth.so.0 /usr/lib/x86_64-linux-gnu/samba/libcli_cldap.so.0 /usr/lib/x86_64-linux-gnu/samba/libcli-ldap-common.so.0 /usr/lib/x86_64-linux-gnu/samba/libcli-nbt.so.0 /usr/lib/x86_64-linux-gnu/samba/libcli_smb_common.so.0 /usr/lib/x86_64-linux-gnu/samba/libcli_spoolss.so.0 /usr/lib/x86_64-linux-gnu/samba/liblibcli_lsa3.so.0 /usr/lib/x86_64-linux-gnu/samba/liblibcli_netlogon3.so.0   I understand that libc is used by pretty much everything, but I'd like to refine this command. Consider its applications for other libraries. How can I do this?  The flaw in this approach, in my opinion, is that the package providing libc may include a whole host of other shared library files (all of which may or may not be covered under the umbrella term that is the library name, like glibc, or libboost-python). Using a word boundary or - to mark the end of libc doesn't really fix this. This may not be a flaw, perhaps, as Braiam points out, since many or all of these .so files may link to a core file in which the flaw resides, as is the case for libc6 on Debian.   If a specific OS/distro is needed, assume Debian Linux.  Also annoying is that this command could be condensed to lsof | awk '/libc/{print $2}' | sort -u. A relative of mine has received a multiple-entry UK visa for 2 years. Is there a particular limit on how many months in total per year she can stay in UK? The way I understand the current rule is that such visitors can enter UK as many times as they want as long as they do not stay for longer than 6 months per single visit. However, does this mean that if she stayed with us for 5 months and went back home for 1-2 months, she could come and stay for another 5 months? I heard that it is advisable to spend only 180 day per calendar year in UK and not more, but I could not find that rule in any official UK Home Office documents. Thanks in advance for any info about this. Is it poor form to switch accepted answers? I had a question of the form "does something like X exist?" Someone wrote a response to say that, No, X does not exist, but I could write it if I liked. I had suspected as much, so I marked his answer as the accepted answer. Today someone actually posted the code to do X for me, and I feel that the new poster deserves the "Accepted Answer" checkmark.  Is it poor form for me to pick a new Accepted answer? Does the original answerer get penalized in terms of reputation?  While the original responder's answer was accepted at the time, I feel this new answer is really more helpful.  Edit: Which of these responses should I accept? What does the notion of a single accepted answer mean with a question that is subjective that could have multiple "right" answers? Array List count to users Seeking GeoServer tutorials? What are sources where I can learn how to use GeoServer?  I know of these two sites:    - which in fact is not working (at least at the moment)  - not only (tutorials) for GeoServer   But I am interested in more complex information, and not to read only documentation. script to count files in a directory To display a sensitivity analysis I want to produce a . In the plot a specific base case value is compared to the values produced after changing parameters in the model. I started from this . With this I was able to get this MWE:   \usepackage{pgf} \usepackage{tikz} \usepackage{pgfplots} \pgfplotsset{compat=newest} \usetikzlibrary{positioning} \usetikzlibrary{calc} \usetikzlibrary{intersections,backgrounds} \usetikzlibrary{decorations.pathmorphing}  \pgfplotsset{     TornadoPlot/.style={% Overall settings         width = 1\textwidth,     line width=1pt,     tick style={line width=0.8pt},     xmin = -50, xmax = 50,     xtick       = {-50, -40, -30, -20, -10, 0, 10, 20, 30, 40, 50},     xticklabels = {-50, -40, -30, -20, -10, 0, 10, 20, 30, 40, 50},     xmajorgrids = true,     minor x tick num = 1,     tick align = outside,     xtick pos = left, ytick pos = left,     xbar, bar shift=0pt,     height = 7cm,     enlarge y limits = 0.1, ytick = {0,1,2,3,4,5,6},     nodes near coords={         \pgfkeys{/pgf/fpu=true}%         \pgfmathparse{\pgfplotspointmeta}%         \pgfmathprintnumber{\pgfmathresult}     },     nodes near coords align = {horizontal},     before end axis/.code={         \draw ({rel axis cs:0,0}-|{axis cs:0,0}) -- ({rel axis cs:0,1}-|{axis    cs:0,0});     }              } }  \begin{document} \begin{tikzpicture}     \begin{axis}[TornadoPlot, legend style={at={(0.975,0.25)}}]             \addplot[fill=green!40] coordinates{(-3.5,0) (-16,1) (-17.5,2) (-19,3) (-34,4) (-27,5) (-22,6)};             \addplot[fill=red!40] coordinates{(4,0) (19,1) (12.5,2) (28.5,3) (34,4) (82,5) (38,6)};             \legend{upside, downside}      \end{axis} \end{tikzpicture} \end{document}   However, one of the bars (82, 5) has a positive x value that is outside the range of the domain of the plot. Can I restrict the length of that bar and force a break, while still displaying the original value of x? I have seen  for vertical bar-charts, but so far I have not been able to reproduce this for my tornado plot. The following is an excerpt from a New York Times article of April 6.     “Tonight, I ordered a targeted military strike on the air base in Syria from where the chemical attack was launched,” Mr. Trump said in remarks at his Mar-a-Lago estate in Florida. “It is in this vital national security interest of the United States to prevent and deter the spread and use of deadly chemical weapons.”    Shouldn't that "from where" be either "from which" or else just "where" without "from?"