According to , the current version of Python 2 is 2.7.10.  On my computer, I have 2.7.6:  $ python Python 2.7.6 (default, Jun 22 2015, 18:00:18)  [GCC 4.8.2] on linux2 Type "help", "copyright", "credits" or "license" for more information. &gt;&gt;&gt; exit()   I figured I'd update to 2.7.10 using apt-get.  However, this doesn't work:  $ sudo apt-get install python2.7 Reading package lists... Done Building dependency tree        Reading state information... Done python2.7 is already the newest version.   apt-get is telling me that I already have the newest version.  What gives?
The version of python2 on my Ubuntu 14.04 machine is Python 2.7.6. How can I upgrade it to the the latest version of Python 2.X? The latest is currently 2.7.11.  I have tried apt-get update/upgrade, but the repository doesn't seem to have the latest version.
How to disable "time travel" entirely? I was looking for alternate proofs of the theorem that "a vector space $V$ of dimension greater than $1$ over an infinite field $\mathbf{F}$ is not a union of fewer than $|\mathbf{F}|$ proper subspaces" and possible generalizations.  A simple measure-theoretic proof over $\mathbb{R}$ is as follows: By countable additivity the sum of the measures of any collection of subspaces is zero since the measure of each subspace is zero, which is a contradiction.  I would like to look at proofs over arbitrary infinite fields and would like to know if similar statements hold for say modules (finitely generated or otherwise) over infinite rings. Where is "Connect to Server" for ssh connections in Unity? Cannot find "connect to server" in 11.04, looks like unity doesn't have it. Is there other very similar way to connect to another Ubuntu computer via SSH? (not in terminal) Maximum power transfer and Thevenin Nice looking progress bar in java Test whether string is a valid integer I need to get rid of the GNOME Desktop Environment which has been installed by default when I updated my pc from 15.10 to 16.04.  After a while, I found myself not only with GNOME DE, but also with Cinnamon alongside (in the login screen, I could choose at least five or six DE!)  Now I am using Unity, and I'm happy with that, and I successfully uninstalled Cinnamon, but now I'm afraid that after uninstalling GNOME I will mess up the Unity DE, since I see many packages running in unity with the packagename starting with gnome-... Would someone mind to help me? Thx in advance In the sentence: "I didn't know she had a son,"   Can I say "I didn't know she has a son" instead, because he is a teenager now?  Or are both correct? What white point temperature should I set my LCD monitor to? I've got the Xrite Eye One Display 2, and tried the advanced calibration today for the first time.  It asked what white point I wanted, and had a default of 6500K.  I didn't realize that was up for discussion!  I would have assumed the daylight temperature, around 5500K, would be desired.  What temperature do I want my whites to be? Why can't white just be white?! Trying to find some simple SQL Server PIVOT examples.  Most of the examples that I have found involve counting or summing up numbers.  I just want to pivot some string data.  For example, I have a query returning the following.  Action1 VIEW   Action1 EDIT   Action2 VIEW   Action3 VIEW   Action3 EDIT     I would like to use PIVOT (if even possible) to make the results like so:  Action1 VIEW EDIT   Action2 VIEW NULL   Action3 VIEW EDIT     Is this even possible with the PIVOT functionality? Analyzing limits problem Calculus (tell me where I'm wrong). I came accross: $$\lim_{x \to\ 0}\frac{x\cos x - \log (1 + x)}{x^{2}}$$  I tried the following please tell me where I 'm wrong:  $$\lim_{x \to\ 0}\frac{x\cos x - \log (1 + x)}{x^{2}} $$  $$\text{(Dividing by }x)$$   $$=\displaystyle\lim_{x \to\ 0}\dfrac{ \cos x - \dfrac{\log (1 + x)}{x}}{x} $$  $$=\lim_{x \to\ 0}\frac{\cos x - 1}{x} $$  $$=\lim_{x \to\ 0}\frac{-2\sin^{2} \dfrac{x}{2}}{x} $$  $$=\lim_{x \to\ 0}\frac{-2x\sin^{2}\dfrac{x}{2}}{(\dfrac{x}{2})^{2}\times 4} $$ $$\lim_{x \to\ 0}\dfrac{-x}{2}=0$$ But Answer given is $\dfrac{1}{2}$  Please Help. The strange thing about the maximum in Planck's law I read that it makes a difference whether you calculate $\frac{dE(\lambda) }{d \lambda}=0$ or $\frac{dE(\omega)}{d \omega}=0$ in the sense that the maximum energy density with respect to the wavelength does not coincide with the frequency maximum that one would assume to be at $\omega_{max} = \frac{c}{\lambda_{max}}$. Actually, I do not understand why this is so ( Now, I am only interested in a pure physical explanation, mathematically this is clear and due to the chain rule). Does anybody know how to explain this odd thing?  Somehow I feel that the core of the question has not been completely answered. Although I can easily look up wikipedia, where the need for the chain rule is explained, I am rather interested in understanding where the following argument breaks down, which seems to be why this question has caused some confusion in the past(as you can see by googling this question):  So we have $E(\omega)$ the energy radiated at a given frequency. Now this function has a maximum somewhere, so there is a frequency where a maximal amount of energy is emitted. In other words: If you add the sum of the photon's energies at each frequency that are emitted, you will notice that the maximum is reached at this frquency. Now $E(\lambda)$ tells you basically the same for the wavelength, but again: We know where at which frequency the maximal amount of energy is radiated, so we know the corresponding wavelength.   I think there is a need to explain this.   If anything is unclear, please let me know. Double limit of $\cos^{2n}(m! \pi x)$ at rationals and irrationals I stumbled upon this "relation" (is the name correct?):  $$ \lim_{m \to \infty} \lim_{n \to \infty} \cos^{2n}(m! \pi x) = \begin{cases} 1,&amp;x\text{ is rational}\\ 0,&amp;x\text{ is irrational}\end{cases} $$  How is it called and why is it so? I'm really not asking for a proof since I fear it would be too complicated for me to understand, but rather for an "intuition". Making adult decisions "is" or "are" really not fun: Which is correct? Making adult decisions are/is really not fun.     What is the proper answer and why? My gut says "is" but I can't explain why. How to mix three different shaders on the same material? So I came across some difficulties with the &quot;Mix shader&quot; node in cycles. I have 3 textures on a label: gold, paper and rough black letters.  How can I mix these effects? I can make 2 textures work together, but not the 3rd one. Here's the node graph.  Here's the reference. Metric is continuous, on the right track? Let $X$ be a metric space with metric $d$. Show that $d:X\times X\rightarrow \mathbb{R}$ is continuous.   The problem is taken from Munkres Topology second edition, Section 20.  I know that if $d$ is a metric on $X$ then $d:X\times X\rightarrow \mathbb{R}$. My thinking is that the topology that is on $X$ is the topology induced by the metric $d$, and that the topology on $X\times X$ is the product topology on that space where we take the basis to be $$\mathcal{B}=\{ U \times V \mid \text{ $U,V$ both open in $X$}\}.$$ Am I on the right track to say that we define some new metric on $X\times X$ and show that this metric induces the same topology as the product topology and then work with the function $d$ as a function between metric spaces to show continuity? The question doesn't mention anything about defining some new metric and I've tried to solve the problem by looking at $X\times X$ as having the product topology, but in picking some point $(x,y) \in X\times X$ and some neighborhood around $d(x,y)$ in $\mathbb{R}$, I haven't yet found the way to make a neighborhood around $(x,y)$ which maps into the neighborhood around $d(x,y)$. Story whose final line is "Let there be light!" I read this series around 5-10 years ago. There were two parallel universes, ours and the universe of demons, and much of the series revolved around attacks in which demons created gateways to cross into our universe to kill people and wreak havoc, and had to be stopped by human magicians. Three youngsters (one guy from modern times, one from a few decades ago, and a girl from ancient Celtic times, all with funny names) end up being the key to the war when it's revealed that they are the three components of a mythical superweapon. These three are the only point-of-view characters in the series, each book being told from the PoV of just one of them. At the very end, they   destroy our entire universe so that, after killing all the demon masters, they can rebuild it and make sure everything happens as before up until the very last moment.  As they begin recreating the universe, the current PoV character (the one from modern times) says something like the following, the very last lines of the series. (sic on the quotation marks)  &quot;In the beginning, there was nothing but darkness. Then [name] said, 'Let there be light!'&quot; And there was light. Coolio! What is the relation between electromagnetic wave and photon? How to distinguish Nexus 7 (2012) and Nexus 7 (2013) I am having trouble with Bluetooth Low Energy (BT 4.0) on a Nexus 7.  I think I may have a 2012 model.  How to tell if I have a 2012 or 2013 version of the Nexus 7? Combinatorics Question Asking How Many Integer Solutions Given Different Restrictions How many integer solutions of $x_1 + x_2 + x_3 + x_4 = 28$ are there with:  (a) $0 ≤ x_i ≤ 12$?  (b) $−10 ≤ x_i ≤ 20$?  (c) $0 ≤ x_i, x_1 ≤ 6, x_2 ≤ 10, x_3 ≤ 15, x_4 ≤ 21$?  \ \  My attempts (I was really only able to solve part (a) but I would appreciate if you could check my work for this one as well):  (a) First I took the total number of integer solutions with no restrictions: There would be ${28+3\choose 3}$ = ${31\choose 3}$ total solutions with no restrictions.   Then I used the inclusion exclusion principle and knew I would have to subtract [P(A) + P(B) + P(C) + P(D)] where A --> $x_1$ is greater than or equal to 13, ..., D--> $x_4$ is greater than or equal to 13. The probabilities of all these cases would equal ${15+3\choose 3}$ = ${18\choose 3}$ because once we assign 13 to one of the $x_i$ there would be $28 - 13 = 15$ remaining to distribute among the $x_i$ values. We would also need to choose which of the $x_i$ to give the 13 to so we would multiply the ${18\choose 3}$ by ${4\choose 1}$ = ${4\choose 1}{18\choose 3}$  We would then have to add back, since we "oversubtracted", the case in which there is an intersection between two of the $x_i$ or in other words two of the $x_i$ receive 13 or more. P$(A\bigcap B)$ = $28-2(13) = 2$ remaining to distribute among the $x_i$ values. So we would also have to choose which of the $x_i$ to give the 13 to so we would have ${4\choose 2}$ ways to choose them. So there would be ${4\choose 2}{2+3\choose 3}$ = ${4\choose 2}{5\choose 3}$ solutions in this case.   Thus the number of solutions would be: ${31\choose 3} - {4\choose 1}{18\choose 3} + {4\choose 2}{5\choose 3}$.  I'm not sure if this is correct and I'm really stuck on b and c so I would appreciate any help or feedback. Thank you!