Whenever I boot to Ubuntu and boot into Windows again my time changes to GMT. Why is this and how can I fix this?
Dual boot system Windows XP Pro and Ubuntu 12.04.  I have the bios set for the correct time and Ubuntu set for US Eastern time.  Ubuntu will boot up and the time will be off by -4 hours.  If I correct the time in Ubuntu then when I boot up in Windows XP the time will be off by + 4 hours.  This is a fresh install of 12.04. I did not have this problem before reinstalling Ubuntu.
Difference between Unity and GNOME I often come across the terms Unity and GNOME while reading about Ubuntu. I understand that Unity is the default desktop environment for Ubuntu. What is GNOME and how is it related to/different from Unity? Find a combinatorial argument for the following binomial identity:   $$n(n+1)2^{n-2} = \sum_{k=1}^{n}k^2\binom{n}{k} .$$   Algebraic proofs can be found at  , and a related identity at  . Rectangles around cells in table What is the best collation to use for MySQL with PHP? I'm wondering if there is a "best" choice for collation in MySQL for a general website where you aren't 100% sure of what will be entered? I understand that all the encodings should be the same, such as MySQL, Apache, the HTML and anything inside PHP.  In the past I have set PHP to output in "UTF-8", but which collation does this match in MySQL? I'm thinking it's one of the UTF-8 ones, but I have used utf8_unicode_ci, utf8_general_ci, and utf8_bin before. Experimental observation of matter/antimatter in the universe $F$ is a finite field of order $q$. What is the size of $\operatorname{GL}_n(F)$ ?  I am reading Dummit and Foote "Abstract Algebra".  The following formula is given:  $(q^n - 1)(q^n - q)\cdots(q^n - q^{n-1})$.  The case for $n = 1$ is trivial.  I understand that for $n = 2$  the first row of the matrix can be any ordered pair of field elements except for $0,0$. and the second row can be any ordered pair of field elements that is not a multiple of the first row.  So for $n = 2$  there are $(q^n - 1)(q^n - q)$ invertible matrices.  For $n\geq 3$, I cannot seem to understand why  the formula works.  I have looked at Sloane's OEIS A002884.  I have also constructed and stared at a list of all $168$ $3\times 3$ invertible matrices over $GF(2)$.  I would most appreciate a concrete and detailed explanation of how say $(2^3 - 1)(2^3 - 2)(2^3 - 2^2)$ counts these $168$ matrices. Solving Keplerian two-body problem numerically? I'm trying to solve the two-body problem numerically, setting up G, m1 and m2 =1.0.  The masses are placed at the positions -10 and 10 respectively along the x-axis and gave them both 0 on the y-axis. I am having some real issue with the initial conditions fitting for a keplerian orbit with a given eccentricity (e=0.9)! Also how would the initial condition change for different mass ratios? I need to find the intersection point of 2 line segments (lines are finite, i.e., they have end points).   e.g. segment 1 from $(x_1, y_1)$ to $(x_2, y_2)$ -- segment 2 from $(x_3, y_3)$ to $(x_4, y_4)$  you can assume $m_1$ and $m_2$ are the gradients of segment 1 and segment 2 respectively  similarly, $c_1$ and $c_2$ being the y-intercepts of segment 1 and segment 2 respectively  Using $y=mx+c$ I can easily find the equations of both lines and then derive the intersection point from those equations and check if that point actually lies on both the line segments (intersection point may not lie between the end points).  My problem is when I calculate the gradients $m_1$ and $m_2$ I do $\frac{y_2-y_1}{x_2-x_1}$ there for if one of the lines is vertical then I am going to have problems. How can I deal with this?  Another problem is that if the two segments intersect at a point which is also the same as one of the end points then I want to assume that they are not intersecting.  e.g. if segment 1 is from $(2, 3)$ to $(3, 7)$ and segment 2 is from $(3, 7)$ to $(7, 3)$ then I want to assume they don't intersect. Is there an MD5 Fixed Point where md5(x) == x? Is there a fixed point in the MD5 transformation, i.e. does there exist x such that md5(x) == x? How can one find the zeroes of $f(x)=ae^{bx}+cx+d$? A certain physics problem I have been working on has turned into a math problem. Particularly, I want to find the solutions of some equation of the form   $$f(x)=ae^{bx}+cx+d = 0$$  where $a, b, c,$ and $d$ are constant, real numbers that come from the physics problem and $x$ will be a real number. I do not know how to find the solutions of an equation of this form. In the special case where $c=0$, the solution is simply  $$x = \frac{\ln(\frac{-d}{a})}{b}$$  but for other values of $c$, I am stumped. If there is no exact form for the roots, is there a relatively simple way to estimate the roots? Factorizing polynomials over fields other than $\mathbb{C}$ has two corollaries:   The gravitational attraction of a spherically symmetric body acts as if all its mass were concentrated at the center, and The gravitational acceleration inside the cavity of a spherically symmetric body vanishes.   Consider a spaceship floating freely in space. In a homogeneous universe, the combined attraction from all matter should cancel out, and the spaceship should stay motionless. Nevertheless, I'm free to divide the attraction into several parts originating from different parts of the universe: In the figure below, I've divided the universe into a red sphere centered on some arbitrary point (×) with my spaceship located at the edge of the sphere, plus infinitely many shells centered on the same point.  By corollary #1, the gravitational attraction of the red sphere equals that of all its mass centered at the point ×. By corollary #2, the combined acceleration of the spaceship from all mass in the green shell vanishes. The same can be said for the blue shell, the orange shell, and so on ad infinitum.  Hence my spaceship should start accelerating toward ×. By choosing the sphere large enough, I should be able to make it accelerate arbitrarily fast, and by choosing the location of × I can make it accelerate in any direction.    Of course this doesn't work, but why?  My best guess is that, even in an infinite universe, you can't keep adding spheres because you'll exit the observable universe, in which case there's no way to feel the gravity in part of the shell so that it's no longer symmetric. Perhaps also the expansion of the universe matters. But see the last two points below.  A few more things to consider:   The mass of the red sphere increases with the chosen radius $r$ as $r^3$, while the acceleration it generates is proportional to $r^{-2}$; hence the acceleration increases linearly with the chosen $r$. Our universe — the "Universe" — has an average density of some $10^{-29}\,\mathrm{g}\,\mathrm{cm}^{-3}$. Hence if I set $r$ equal to the radius of the observable Universe (46.3 billion light-years), the acceleration is a minuscule $10^{-7}\,\mathrm{cm}\,\mathrm{s}^{-2}$. If that bothers you, choose another universe where $\rho$ is ten orders of magnitude higher. Our Universe is not really homogeneous, but on large enough scales ($\gtrsim$ half a billion light-years) it seems it is. Still, the acceleration of the spaceship will be dominated by nearby sources. If that bothers you, choose a sufficiently homogeneous universe. On the scales we're considering, the Universe is not governed by Newtonian dynamics, but by general relativity. If that bothers you, use  instead — I think the issue is the same. If the issue is really that the size of the observable universe matters, then my intuition tells me that I can just choose an arbitrarily old universe where the asymmetric contribution from the most distant shells is arbitrarily small. If the issue is that the universe expands (so that gravity from the far side of a shell is somehow weakened, or "redshifted"), then my intuition tells me that I can just choose a sufficiently static universe. Removing las2dem streaks from raster DEM? I've created a series of DEM files using the  tool from LAStools; however since I don't have a licensed version they have black streaks running across them. Does anyone know a good method for removing these streaks? Inversion/non-inversion in wh-questions with long phrases after the wh-words Can a sentence like this:      "I don't know who the first man that made such and such thing in such and such place was,"    be grammatically correct if we don't put "was" at the end of the long phrase, that is, if we write:      "I don't know who was the first man that made such and such thing in such and such place"?    I can see in Google Books examples that in such cases the verb is often put after the wh-word, but I don't know if there is a grammar rule to support this. Some examples:  "We do not know who was the first man who ascended above a poor and humble people to become Egypt's first king ..."  "... we do not know what was the ultimate judgment of the various members of the community ..."  "I do not know who was the first to suggest a connection between the problem of free will and the breakdown ..."  "I do not know what was the date of this change in me, nor of the train of ideas ..."  "We do not know what was the primitive text from which Codex Bezae derived its Latin or its Greek ..."  "We do not know what was the practice in the days of the monarchy, but the story of Athaliah shows ..." biblatex: remove commas between last and first names in bibliography I want to install the PyCharm IDE by JetBrains for Python (free community edition).  I managed to , but I do not know how to install it. Can you please help me? I'm facing this frustrating "feature" that I can't seem to remedy; when I duplicate an object (and now, even when I create two completely new objects), their transformations/scaling is synchronized. This is extremely annoying when trying to use any sort of transformation keyboard shortcut.  It should be noted by going to the properties menu (N) and manually setting the properties works just fine.  For instance, here is just the camera selected (trust me, I made sure):   And when I try to transform by clicking+dragging on the X axis OR by hitting G->X, this happens:   They are not grouped (verified by going to the Groups list up in the Scene panel) and they do not have parents (verified by going to Object->Relations).  I have tried the unlink command as well, to no avail.  I have seriously run out of options; what did I change to have my G->X shortcuts transform multiple objects? It seems completely at random, too; certain objects seemed to be "linked" to other random objects, and this "linking" is guaranteed when duplicating even when I do a non-linking duplicate (Shift+D).  What gives? Can non-magical persons see the Hogwarts Express during its journey to and from Hogwarts? Difference between (function(){})(); and function(){}(); Possible Duplicate:          This is something I haven't quite figured out yet, but I have been using function(){}() just because my VIM syntax highlight screws up if I add the parenthesis, although I've seen (function(){})() around many times, maybe its an IE thing?  edit:  var singleton = function() {     // code }();  var singleton = (function() {     // code })(); apt: how to be sure that a specific package gets installed/updated from specific repository and not from the default one?