Export My Stack Overflow Data Is there a first party (or third party?) way to export just my data out of Stack Overflow?  I realize there's a massive Creative Commons data dump.  I'm interested in knowing if there's a quick and easy way for me to get my posts dumped out in some plain text file format for easy  personal archiving.
Dump of my own Stack Exchange content Is there a way to quickly save all my Stack Exchange content across all Stack Exchange sites, as hundreds of HTML pages? (for me probably 15 or 20 sites: SO, dsp.SE, unix.SE, etc.)?  The ideal way would be one .HTML file for each question that I asked / answered / commented? What would be the ideal tool for that?    Remark: I've already read interesting questions like  but it doesn't exactly answer my concern here.
Which Defiant was shown in ST: First Contact? In Memory Alpha, the page on the  states that two Defiant-class ships had that name: the prototype Defiant, and a successor, the U.S.S. Sao Paulo, which was assigned to DS9 and renamed Defiant after the loss of the original. All of this happened during the DS9 series.  In the Star Trek: First Contact movie, the Defiant participates in the opening battle, where a Borg Cube penetrates the Sol system's defenses and besieges Earth. Worf is in command, and is transported out of the crippled Defiant along with whatever's left of his crew by the Enterprise; out-of-universe, this is a rather transparent method to get Worf back into the TNG side of things after the events of ST: Generations, when he moved to DS9 as tactical officer and unofficial envoy to the Klingon Empire.  The question is, was Worf's Defiant in the movie the original, or the Sao Paulo? This would primarily depend on where the movie falls in the overall canon timeline. According to in-movie dialog, whichever one it was was salvageable and therefore would have survived the movie, so the movie would have had little impact on the canon of any running TV series at the time. What is a less offensive synonym for "hypocrite"? Is there a word that describe a person who doesn't "practice what they preach"? Basically, is there a synonym for "hypocrite" that carries less pejorative connotations?   For example, let's say a friend of mine says he is a Christian. Yet he engages in behaviors considered sins by the Bible, making him a hypocrite. And I would like to tell him I think he is a hypocrite.   What would be a good word to use here, so my criticism seems constructive rather than offensive?     I tried using a , but the synonyms that came up were even worse (ie: they are more offensive than "hypocrite"). Related (but different) question: , which asks what to call someone who knows they are in the group of people they criticize. How to indent subsection in AMS TOC I used AMS LaTeX template with gsm-l package to write my teacher's book.   The table of content that produced has the following form:    in which 1, 2, 3.... are sections and 3.1, 3.2, 4.1, 4.2,.... are the subsection.   I want to indent the subsection with respect to the section, like this (it's ok without dots, though): Gravatar profile image not showing sometimes It seems that with some profiles the Gravatar image is broken. I am not sure if this is an issue with SE or Gravatar, or both, so posting here.  As an example,  shows an error on , which shows a HTTP 500 on Gravatar.  Profile image on SO:        Image on Gravatar:        Switching to the MSE profile of the user, the image is working (but has a different hash).  Is this a bug on the SE side? Can anything be done to make Gravatar fix this?    Update 2016-10-04:  It happened to me again two times today. The last user profile was . Calling the service two or three times fixes the issue. Moving status bar in iOS 7 Bearing in mind I am a layman - with no background in physics - please could someone explain what the "big deal" is with quantum entanglement?  I used to think I understood it - that 2 particles, say a light-year apart spatially, could affect each other physically, instantly.  Here I would understand the "big deal".  On further reading I've come to understand (maybe incorrectly) that the spatially separated particles may not affect each other, but in knowing one's properties you can infer the other's.  If that it the case, I don't see what the big deal is... 2 things have some properties set in correlation to each other at the point of entanglement, they are separated, measured, and found to have these properties...?  What am I missing? Is it that the particles properties are in an "un-set" state, and only when measured do they get set?  (i.e. the wave-function collapses).  If this is true - why do we think this instead of the more intuitive thought that the properties were set at an earlier time? I have seen many articles that talk about a star's "habitable zone" as the ring around it which is just the right temperature for planets to have liquid water, which is necessary for life. My question is, why does a planet need to have liquid water in order to sustain life? Who says that extraterrestrial life can't live off of some other kind of sustenance? In  I prove, using generating functions, a statement equivalent to  $$\sum_{k=0}^n \binom{2k}{k} \binom{2n-2k}{n-k} (-1)^k = 2^n \binom{n}{n/2}$$ when $n$ is even.  (Clearly the sum is $0$ when $n$ is odd.)  The nice expression on the right-hand side indicates that there should be a pretty combinatorial proof of this statement.  The proof should start by associating objects with even parity and objects with odd parity counted by the left-hand side.  The number of leftover (unassociated) objects should have even parity and should "obviously" be $2^n \binom{n}{n/2}$.  I'm having trouble finding such a proof, though.  So, my question is     Can someone produce a combinatorial proof that, for even $n$, $$\sum_{k=0}^n \binom{2k}{k} \binom{2n-2k}{n-k} (-1)^k = 2^n \binom{n}{n/2}?$$   Some thoughts so far:   Combinatorial proofs for $\sum_{k=0}^n \binom{2k}{k} \binom{2n-2k}{n-k}  = 4^n$ are given by  and by .  The proofs are basically equivalent.  In Phira's argument, both sides count the number of paths of length $2n$ starting from $(0,0)$ using steps of $(1,1)$ and $(1,-1)$.  By conditioning on the largest value of $2k$ for which a particular path returns to the horizontal axis at $(2k,0)$ and using the facts that there are $\binom{2k}{k}$ paths from $(0,0)$ to $(2k,0)$ and $\binom{2n-2k}{n-k}$ paths of length $2n-2k$ that start at the horizontal axis but never return to the axis we obtain the left-hand side.  With these interpretations of the central binomial coefficients $2^n \binom{n}{n/2}$ could count (1) paths that do not return to the horizontal axis by the path's halfway point of $(n,0)$, or (2) paths that touch the point $(n,0)$.  But I haven't been able to construct the association that makes these the leftover paths (nor do all of these paths have even parity anyway).  So perhaps there's some other interpretation of $2^n \binom{n}{n/2}$ as the number of leftover paths.    Update. Some more thoughts:  There's another way to view the identity $\sum_{k=0}^n \binom{2k}{k} \binom{2n-2k}{n-k}  = 4^n$.  Both sides count the number of lattice paths of length $n$ when north, south, east, and west steps are allowed.  The right side is obvious. The left side has a similar interpretation as before: $\binom{2k}{k}$ counts the number of NSEW lattice paths of length $k$ that end on the line $y=0$, and $\binom{2n-2k}{n-k}$ counts the number of NSEW lattice paths of length $n-k$ that never return to the line $y =0$.  So far, this isn't much different as before.  However, $2^n \binom{n}{n/2}$ has an intriguing interpretation: It counts the number of NSEW lattice paths that end on the diagonal $y = x$ (or, equivalently, $y = -x$).  So maybe there's an involution that leaves these as the leftover paths.  (Proofs of all of these claims can be found on , for those who are interested.) System.LimitException: Too many query rows: 50001 What I am trying to do is to set a date in my date field and also update the object. I am using the class below to do so which works just fine in my sanbox but in production I am getting this error:      System.LimitException: Too many query rows: 50001   public with sharing class updateObject{     public  List&lt;NRProducts__c&gt;  values;     public PageReference updateObject() {         values=[SELECT Refreshed_Date__c FROM NRProducts__c ];         for (NRProducts__c s :values){            s.Refreshed_Date__c =Date.today();          }          update values;         return null;                       }     }   VF Page:  &lt;apex:page showheader="false" controller="updateObject" sidebar="false"&gt; &lt;apex:form&gt;  &lt;apex:commandButton action="{!updateObject}" value="Update"/&gt; &lt;/apex:form&gt; &lt;/apex:page&gt; You can use the range feature of unicode-math to set the math font for a specific unicode slots or the macro accessing the glyph. I tried this with the sum sign &amp; Latin Modern, see my example:  \documentclass{scrartcl}  \usepackage{amsmath} \usepackage{unicode-math} % use unicode in math &amp; setup math fonts \setmathfont{Asana Math} %\setmathfont[range=\sum]{Latin Modern Math} % substitute Sum sign  \begin{document}  Notice the different sum sign $\sum$, as intended.   Notice also, the difference in spacing, not intended: \begin{equation}     \begin{bmatrix}         0           &amp; \frac{1}{2}   &amp; \frac{2}{3} \\         \frac{3}{4} &amp; 0             &amp; \frac{4}{5} \\         \frac{5}{6} &amp; \frac{6}{7}   &amp; 0     \end{bmatrix}     \begin{pmatrix}         A \\         B \\         C     \end{pmatrix} \end{equation} \end{document}   However, not only the sum sign, but also the spacing in the matrices changes. With Latin Modern commented out (as in the example):  With Latin Modern active (remove % before relevant line):  Why is that? BTW, I'm using Lualatex from TeXLive 2012. How do I transfer a file from a guest account on a MacBook Air? show that $\sum_{i=0}^l \dfrac{{l\choose i}}{{n\choose k + i}} = \dfrac{n+1}{(n-l+1) {n-l\choose k}}$ Show that $\displaystyle\sum_{i=0}^l \dfrac{{l\choose i}}{{n\choose k + i}} = \dfrac{n+1}{(n-l+1) {n-l\choose k}}$ for all $k,l, n\geq 0$ with $k+l \leq n.$  The $k+l\leq n$ part doesn't seem that important. The claim clearly holds for $n=0,$ and in that case $k$ and $l$ must both be zero. I think the claim can be shown by induction, but I'm not sure how to do so; I'm not sure how to use the inductive hypothesis to prove the inductive step. The sum on the LHS is not always an integer, so finding a combinatorial argument doesn't seem very feasible. For $n = 1,$ we can have $(k,l) = (0,0), (1,0), (0,1),$ and in each case the sum holds. Perhaps multiplying both sides by a number (idk ${n-l\choose k}$) might be useful? Using the expression "the same" for a previously mentioned item Given a DateTime object, how do I get an ISO 8601 date in string format? Given:  DateTime.UtcNow   How do I get a string which represents the same value in an -compliant format?  Note that ISO 8601 defines a number of similar formats. The specific format I am looking for is:  yyyy-MM-ddTHH:mm:ssZ The the title is self explanatory, I guess.  Why can two (or more) electric  never cross? Everytime I try to sudo something at the terminal, I get a warning message:  alexandre@XPS-15Z $ sudo mount file.iso /mnt/ -o loop [sudo] password for alexandre:  no talloc stackframe at ../source3/param/loadparm.c:4864, leaking memory   It's just annoying to get that all the time. How do I solve that? I'm on Ubuntu 14.04 (upgraded from 13.10) Evaluation of $\displaystyle \int_{0}^{1}\frac{x\ln (x)}{\sqrt{1-x^2}}dx$   $\bf{My\; Try::}$ Let $\displaystyle I = \int_{0}^{1}\frac{x\ln x}{\sqrt{1-x^2}}dx\;,$ Put $x=\cos \phi\;,$ Then $dx = -\sin \phi d\phi$  and Changing Limit, We get  $$\displaystyle I = -\int_{\frac{\pi}{2}}^{0}\cos \phi \cdot \ln(\cos \phi )d\phi = \int_{0}^{\frac{\pi}{2}} \ln(\cos \phi)\cdot \cos \phi d\phi$$  Now Using Integration by parts, We get  $$\displaystyle I = \left[\ln(\cos \phi)\cdot \sin \phi\right]_{0}^{\frac{\pi}{2}}+\int_{0}^{\frac{\pi}{2}}\frac{\sin^2 \phi}{\cos \phi}d\phi$$  So $$\displaystyle I = \left[\ln(\cos \phi)\cdot \sin \phi\right]_{0}^{\frac{\pi}{2}}+\int_{0}^{\frac{\pi}{2}}\frac{(1-\cos^2 \phi)}{\cos \phi}d\phi$$  So $$\displaystyle I = \left[\ln(\cos \phi)\cdot \sin \phi\right]_{0}^{\frac{\pi}{2}}+\int_{0}^{\frac{\pi}{2}}\sec \phi d\phi-\int_{0}^{\frac{\pi}{2}}\cos \phi d\phi$$  So $$\displaystyle I = \left[\ln(\cos \phi)\cdot \sin \phi\right]_{0}^{\frac{\pi}{2}}+\left[\ln\left|\sec \phi+\tan \phi\right|\right]_{0}^{\frac{\pi}{2}}-\left[\sin \phi\right]_{0}^{\frac{\pi}{2}}$$  So $$\displaystyle I = \left[\ln(\cos \phi)\cdot \sin \phi\right]_{0}^{\frac{\pi}{2}}+\left[\ln\left|\sec \phi+\tan \phi\right|\right]_{0}^{\frac{\pi}{2}}-1$$  Now How can I solve after that, Help Required, Thanks Just as background, I should say I am a mathematics grad student who is trying to learn some physics. I've been reading "The Theoretical Minimum" by Susskind and Hrabovsky and on page 134, they introduce  transformations. Here's the first example they use:  Consider a particle moving in the x,y plane under the influence of a potential, $V$, which depends only on the radius, with Lagrangian:   $L=\frac{m}{2}(\dot{x}^2+\dot{y}^2)-V(x^2+y^2)$  This is clearly invariant under rotations:  $x \rightarrow x\cos \theta + y\sin \theta$  $y \rightarrow -x\sin \theta + y\cos \theta$  All well and good. Now they say "consider what happens ... when the angle $\theta$ is replaced by an infinitesimal angle $\delta$." Already I could say "What the heck is $\delta$ really?", but I'm willing to play along with my intuition. Since $\delta$ is infinitesimal, we work to first order and say   $\cos \delta=1$ and $\sin \delta= \delta$.  Plugging this into our rotation formulas above, we obtain:  $x \rightarrow x+y\delta$  $y \rightarrow y-x\delta$  By differentiating, we see that:  $\dot{x} \rightarrow \dot{x}+\dot{y}\delta$  $\dot{y} \rightarrow \dot{y}-\dot{x}\delta$  Plugging these into the Lagrangian and ignoring terms higher than first order, we see that the Lagrangian is invariant under this transformation.  My main problem with all of this is that I don't understand what the physical nature of an infinitesimal transformation actually is. All I got from the above was that if you do this formal calculation by following rules like "only work to first order in $\delta$," then the Lagrangian is invariant. This is in contrast to the case where we have an actual transformation, like a rotation, where there is no question about what is going on physically.  I would also like to know how all of this relates to rigorous mathematics. In mathematics, I can't recall ever using infinitesimals in an argument or calculation, so it would be useful if there was some way of formulating the above in terms of limits/ derivatives/ differential forms (for example). I sense a connection to Lie Algebras, as the infinitesimal version of the rotation is $(I+A)$ where $I$ is the identity matrix and $A$ is an element of the Lie Algebra of $SO(2)$.   Here are some questions whose answers I believe may be useful to me (feel free to answer some or all):  -What is an infinitesimal quantity like $\delta$ to the physicist?   -Why do physicists argue using infinitesimals rather than "standard" calculus?  -What is the physical meaning of infinitesimal transformation? How does it relate to Lie Algebras?   -Is there a rigorous theoretical apparatus for justifying the computations shown above?  -What is meant by the Lagrangian being invariant under infinitesimal transformations?  If any of the questions seem too vague, please say so. Thanks in advance for your insights! Is the product of symmetric positive semidefinite matrices positive definite? I see on Wikipedia that the product of two commuting symmetric positive definite matrices is also positive definite. Does the same result hold for the product of two positive semidefinite matrices?  My proof of the positive definite case falls apart for the semidefinite case because of the possibility of division by zero... Why are image thumbnails cropped proportionally (not per dimensions)? For some reason WordPress is ignoring my thumbnail sizes and keeping thumbnails cropped proportionately instead of specific dimensions. Here is what I have in functions.php:  add_image_size( 'main-thumbnail', 728, 410, true );   and in content.php:  &lt;div class="thumbnail"&gt;     &lt;a href="&lt;?php the_permalink(); ?&gt;"&gt;        &lt;?php the_post_thumbnail( 'main-thumbnail' ); ?&gt;     &lt;/a&gt; &lt;/div&gt;   I added these two things and regenerated my images with the  plugin, but it doesn't seem to matter. The images don't change.  For some reason WordPress isn't recognizing the hard crop and is keeping the images proportional, so some of my images are way taller than 410 pixels. Why is this?