Ubuntu 32 to 64 bit
Is it possible to "upgrade" from a 32bit to a 64bit installation?
Clarify an idiom in a political text Please carefully read the text below:     On 24 November, 1993, a meeting of Leftist intellectuals occurred in   London under the auspices of the Institute for Public Policy   Research (IPPR), which is a Labour-leaning think-tank.  A short document   was circulated in advance of the said meeting to clarify its   purpose.        Among other things, the document declared that the task of the   IPPR was: "To do what the Right did in the seventies, namely to break   through the prevailing parameters of debate and offer a new   perspective on contemporary British politics."    The explanatory document also said, "Our concern is not to engage in a   philosophical debate about foundations of socialism."      If this meant that those foundations were not the appropriate thing to   talk about at the 24 November meeting, then that might have been   right; not every thing has to be discussed at every meeting.  But if   what was meant was that discussion of philosophical foundations is not   what the Left now needs, then I disagree, and if that indeed is what   was meant, then it is curious that the breakthrough by the Right should   have been invoked as an achievement for the left to emulate, for if   there is a lesson for the Left in the Right's breakthrough, it is that   the Left must repossess itself of its traditional foundations on   pain of continuing along its present, politically feeble, reactive   course.  If the Left turns its back on its foundations, it will be   unable to make statements that are truly its own.   Since I've heard only one definition of the part identified in bold, I am at a dead end.  As you know, and according to Oxford's dictionary, the verb "repossess" means: To take back property or goods from SB who has arranged to buy them and can not pay.  Here, the object of verb is a thing not a person, but in the text, first,itself comes before of and its traditional foundations. As a result, I thought that this phrase might be an idiom and so has a different meaning with the original verb.  Please explain the meaning of this phrase. Theorem.  If $\sum_{n=1}^{\infty}c_{n}$ is Cesaro summable, then $c_{n}/n$ tends to $0$.  How to prove it? is it possible to skip immediately preceding time points? So that, if, for example, you are using model order 2, that is, two time points, but not the two immediately previous time points, but rather, say for example, skipping 2 points, and then using those two points.  So, e.g., it would be y=x(t-3) + x(t-4). Find the Values of $p$ and $q$ such that $\int_0^{\infty} x^p\ln(1+x)^q dx$ Converges or Diverges. The question is to find the value values of $p$ and $q$ such that the improper integral converges or diverges.  My friend indeed found some values of $p$ and $q$ such that it converges, but after many trials, I am still getting divergent for all values.  Via L'Hospital, we have the following:  $$\lim_{x \to 0} x^c\ln(1+x)^q = \infty \space \text{when} \space  c &lt; 0$$ $$\lim_{x \to 0} x^c\ln(1+x)^q = 0 \space \text{when} \space  c &gt; 0$$ $$\lim_{x \to \infty} x^c\ln(1+x)^q = \infty \space \text{when} \space  c &gt; 0$$ $$\lim_{x \to \infty} x^c\ln(1+x)^q = 0 \space \text{when} \space  c &lt; 0$$  Thus, $$\int_1^{\infty} x^p\ln(1+x)^q dx = \int_1^{\infty} x^{p-c}x^c\ln(1+x)^q dx &gt; \int_1^{\infty} x^{p-c} dx$$ which diverges when $p &gt; -1$.  Here, we set $c = \frac{1}{2}(p+1) &gt; 0$.  On the other hand, $$\int_0^1 x^p\ln(1+x)^q dx = \int_0^1 x^{p-c}x^c\ln(1+x)^q dx &gt; \int_0^1 x^{p-c} dx$$ which diverges when $p &lt; -1$.  Here, we set $c = \frac{1}{2}(p+1) &lt; 0$.  Thus, the integral diverges whenever $p \ne –1$.  Then, let $p = -1$.  $$\int_0^{\infty} \frac{1}{x}\ln(1+x)^q dx = \int_0^1 \frac{1}{x}\ln(1+x)^q dx + \int_1^{\infty} \frac{1}{x}\ln(1+x)^q dx $$ $$ \ge \int_0^1 \frac{1}{x+1}\ln(1+x)^q dx + \int_1^{\infty} \frac{1}{x+1}\ln(1+x)^q dx $$  The first summand diverges when $q &lt; -1$ and the second diverges when $q &gt; -1$.  Finally, let $ p = -1$ and $q = -1$.  The integral $\int_0^{\infty} \frac{1}{x\ln(1+x)} dx ≥ \int_0^{\infty} \frac{1}{(x+1)\ln(1+x)} dx$ diverges on both ends.  Did I make a mistake? Actually snapping to pixel grid Does anyone know if there's a way to have Illustrator CC (2014) constrain control points to the pixel grid? I understand that "Align to pixel grid" makes it so that a shape will try to match the pixel grid, but the actual control points don't line up to said grid... The image I've attached illustrates my problem. The documentation for the  function states that you pass it a number, and the positions past the decimal to round. Thus it should do this:  n = 5.59 round(n, 1) # 5.6   But, in actuality, good old floating point weirdness creeps in and you get:  5.5999999999999996   For the purposes of UI, I need to display 5.6. I poked around the Internet and found some  that this is dependent on my implementation of Python. Unfortunately, this occurs on both my Windows dev machine and each Linux server I've tried. .  Short of creating my own round library, is there any way around this? Let $S$ be the collection of all groups. Deﬁne a relation on $S$ by $G \sim H$ iﬀ $G ≈ H$. Prove that this is an equivalence relation. So $S$ is partitioned into isomorphism classes.   Proof: Let $S$ be a relation on S defined by $G \sim H$ iﬀ $G ≈ H$. Show reflexive, symmetric, transitive. It is obvious that $G≈G$ so $G\sim G$.  Assume $G\sim H$, thus $G≈H$, which implies that $H≈G$, so $H\sim G$, thus symmetric.  Assume $G\sim H$ and $H\sim K$, for $K$ a group in $S$. Thus $G≈H$ and $H≈K$. (I have already shown that if $G≈H$ and $H≈K$, then $G≈K$). From earlier this implies that $G≈K$, so $\sim$ is an equivalence relation.  Thus $S$ is partitioned into isomorphism classes. QED. I'm looking for the method by which the partial derangement formula $D_{n,k}$ was derived. I can determine the values for small values of N empirically, but how the general case formula arose still alludes me.  Any links/books or an explanation will be appreciated.  The formula is:  $D_{n,k} = {n \choose k}!(n-k)$  Links: I'm stuck in full screen in the node editor. Escape isn't working, and I have no menus and no idea how to leave except to quit Blender and reopen the program. Charging lead-acid batteries? The  recommends that images have Copy-On-Write disabled on the directory when using btfs.  I do see that it would be a good idea if you had a lot of file read/writes.   question explores this idea.  I know that VMWare will grow into different files and will write snapshots out which could be problematic when using Copy-On-Write.  For QEMU the file is left open for the VMs existance so there is a potential problem in the write after the VM is shutdown, but I would see slow I/O after a VM is shutdown not being an issue.  What pitfalls would I be avoiding by performing this step for QEMU.    Additionally: I have assumed that the image is raw for this question.  Is there a possible stability issue as qcow2 already has copy-on write. How Does Kylo Ren not stop Chewbacca's blaster? How does Kylo Ren stop the blaster shot in the beginning when being shot at by Poe, only to be unaware of the coming blaster shot from Chewbacca, which seems much more predictable.   I am sure he knew Chewbacca, Rey and Finn where there as Chewbacca turns with the opening of the doors when Rey and Finn come in.  EDIT in regards to Richards Answer: I just want to expand a little more on question. What I don't understand is that Kylo Ren uses the force, so it is just not a matter him not hearing. When Poe shoots him from behind, he makes no sound, but yet he turns and stops the blaster and Poe. That has to be due to the power he commands in the Force.  And this is from Obi-Wan Kenobi:     "Well, the Force is what gives a Jedi his power. It's an energy field   created by all living things. It surrounds us and penetrates us; it   binds the galaxy together."―Obi-Wan Kenobi —   So it seems to me that it would be very hard for him even in a weakened state to not even notice that he was being shot at by virtue of the force being a part of Sith and Jedi's alike. Even if he was too weak to actually stop the blaster, the Force should have been able to warn him. How do I make an object face the mouse position, but only on one plane? The real fastest gun in the west Are the questions in the about page taken from real posts? If not, who designs them?  Particularly, why does this specific answer have a date from before the question was asked? At college we use ArcGIS Desktop 10, but that does not function on my Mac so I have QGIS there instead.  I want to do my exercises with QGIS, but in ArcGIS Desktop we use .lyr files with ArcMap and .lyrx files with ArcGIS Pro, and they are not supported by QGIS.  Does anyone know, how I can connect the *.lyr files to my *.shp files to use them on my Mac with QGIS? Prolog count the number of times a predicate is true Mannequin Duplication glitch Creating polygons as % of original area using ArcGIS Desktop? I have a polygon shapefile and need to create new polygons, within the originals, as % of original area.  I used buffer by field (Field= new area as %) but the buffers are created in circles around the polygons.  What am I doing wrong?  I am using ArcGIS Desktop 10. How to add Lat/long columns to a point layer which is not in WGS84? I need to extract the Lat/long coordinates of the centroids of districts. My QGIS project SCR and the different layers have all been set on "WGS 84 / UTM zone 18N".   I've created the centroids using the usual tool. I could then extracted the coordinates using the "Vector > Geometry Tools > Export/Add Geometry Columns", or saving the layer in CSV typing "GEOMETRY=AS_XY" in the "Layers" part of the "OGR creation options".  Nevertheless, both methods output Cartesian coordinates, whereas I need Lat/long coordinates. My regex is matching too much. How do I make it stop? I have this gigantic ugly string: J0000000: Transaction A0001401 started on 8/22/2008 9:49:29 AM J0000010: Project name: E:\foo.pf J0000011: Job name: MBiek Direct Mail Test J0000020: Document 1 - Completed successfully  I'm trying to extract pieces from it using regex. In this case, I want to grab everything after Project Name up to the part where it says J0000011: (the 11 is going to be a different number every time). Here's the regex I've been playing with: Project name:\s+(.*)\s+J[0-9]{7}:  The problem is that it doesn't stop until it hits the J0000020: at the end. How do I make the regex stop at the first occurrence of J[0-9]{7}?