Dumbledore said that Harry must die, and Voldemort himself must do it. If that were to happen, and Harry was to actually permanently die, how would Voldemort be killed?
‘So the boy ... the boy must die?’ asked Snape, quite calmly.   ‘And Voldemort himself must do it, Severus. That is essential.’   Deathly Hallows - page 551 - UK Hardcover - chapter 33, The Prince's Tale   Dumbledore tells Snape it is 'essential' that Voldemort be the one to kill Harry, I'm assuming in order for the piece of Voldemort's soul in Harry to be properly killed as well.  But why?   Hermione destroyed the Hufflepuff cup; Ron destroyed the Slytherin locket; Neville killed Nagini; Harry destroyed the diadem and Tom Riddle's diary. So the Horcruxes were not immune to destruction at the hand of someone other than Voldemort. And as it ended up, Harry himself wasn't even a Horcrux, but rather just an unknowing host to a parasitic bit of Voldemort's soul.   Why was it 'essential' that Voldemort be the one to kill Harry in order for the piece of Voldemort's soul to die?
I’m having trouble finding an official clarification of this, so I’m wondering if anyone has found a ruling:  Solarians gain powers known as stellar revelations. These are either “photon” or “graviton” revelations. It is specifically mentioned they can be used out of combat, though most include improved effects or separate powers that can only be used “while attuned” - a state a Solarian enters during combat. They may be photon or graviton attuned, at the player’s choice, and are able to increase their level of attunement each round until they become fully attuned to one or the other. However, apart from the very powerful “zenith revelations”, nowhere in the rules mentions whether photon revelations may only be used while photon attuned, or graviton revelations only while graviton attuned. The revelations themselves usually say “while attuned or fully attuned” for the more powerful effects, without mentioning the mode, and there’s no general rule in the class feature that answers this question.  Is there an official ruling on whether the “only while attuned” power of a photon revelation can be used when the character is graviton attuned, or vice versa? I suspect the answer is yes but it’s hard to tell based on the core rules. Can I use two different cards with the same name in Commander? I have a mesh and in the region around each triangle, I want to compute an estimate of the principal curvature directions. I have never done this sort of thing before and  does not help a lot. Can you describe or point me to a simple algorithm that can help me compute this estimate?  Assume that I know positions and normals of all vertices. I'm given a inequality as such: $n! &gt; n^3$ Where n > 5,   I've done this so far:  BC: n = 6, 6! > 720 (Works)  IH: let n = k, we have that: $k! &gt; k^3$  IS: try n = k+1, (I'm told to only work from one side)  So I have (k+1)!, but I'm not sure where to go from here.  I've been told that writing out: $(k+1)! &gt; (k+1)^3$ is a fallacy, because I can't sub in k+1 into both sides, but rather prove from one side only.    Any Help on how to continue from here, would be much appreciated. If $f$ is continuous from $ [0,1]$ to itself and $f(0)=0$, $f(1)=1$ and $f(f(x))=x$ then show that $f(x)=x$ for every $ x $ in the domain of function.  I'd like to prove this statement, my idea goes like below:  Suppose that the converse of the statement holds. then without loss of generality, assume that there exists some $c$ in $[0,1]$ such that $d=f(c)&gt;c$. By symmetry, $f(d)=f(f(c))&lt;f(c)=d$. And $f$ is bijective because $f$ has an inverse function in this domain. So $f$ should be strictly increasing function. But on $[c,d]$, $f$ is decreasing. This is a contradiction.  So it should be that $f(x)=x$ on $[0,1]$.  Is this argument right? And I'd like to know another argument which prove this statement. Thank you for your answer in advance :) Area of the Limiting Polygon Start with an equilateral triangle with unit area. Trisect each of the sides and then cut-off the corners. In this case, we get a regular hexagon - see the picture below. Next, trisect each of the sides of the hexagon and cut-off the corners. This will give a dodecagon, but not a regular one.   Continue this process ad infinitum.  What is the area of the limiting "polygon"?  My first thought was that it would be the circle tangent to the midpoint of each side. However, by making some pictures, it becomes clear quite quickly that the limiting polygon is not a circle and, in fact, might perhaps be an ellipse. The problem with that is that while there is a unique circle passing through any three points in general position, I need five generic points to specify a conic.   Can anyone suggest any hints as to how to find the area? By "hint" I don't mean "what you think might work", I mean "what you have tried and you know will work". Thanks in advance. Dollar-cost-averaging interval based on decay of autocorrelation? $ \lim_{n \to \infty} \left(\frac 1{n^2+1}+\frac 2{n^2+2}+\frac 3{n^2+3}+\cdots +\frac n{n^2+n}\right)$ So basically I want to show:  $\aleph_{\alpha}^{\aleph_{1}} = \aleph_{\alpha}^{\aleph_{0}}\cdot2^{\aleph_1}$ , Where $\omega\leq\alpha&lt;\omega_1$  There is a proof that I found;    But I don't understand his proof for the limit case of $\alpha$.  To be more specific, I dont understand why $ \left( \sup_{\omega \le \beta &lt; \alpha} \aleph_{\beta}\right)^{\aleph_1} \le \left( \prod_{\omega \le \beta &lt; \alpha} \aleph_{\beta}\right)^{\aleph_1}$  Any insights or help is deeply appreciated.  Cheers Is there a way to force a C# class to implement certain static functions? I am developing a set of classes that implement a common interface. A consumer of my library shall expect each of these classes to implement a certain set of static functions. Is there anyway that I can decorate these class so that the compiler will catch the case where one of the functions is not implemented.  I know it will eventually be caught when building the consuming code. And I also know how to get around this problem using a kind of factory class.  Just curious to know if there is any syntax/attributes out there for requiring static functions on a class.  Ed Removed the word 'interface' to avoid confusion. Python does not find packages installed without Anaconda I'm trying to install kivy for python 3 using apt-get:      sudo apt-get install python3-kivy   And it is installed in      /usr/lib/python3/dist-packages   But, when I tried to load it in python, I get a ModuleNotFoundError      Python 3.6.3 |Anaconda custom (64-bit)| (default, Oct 13 2017, 12:02:49)      [GCC 7.2.0] on linux     Type "help", "copyright", "credits" or "license" for more information.     &gt;&gt;&gt; import kivy     Traceback (most recent call last):       File "&lt;stdin&gt;", line 1, in &lt;module&gt;     ModuleNotFoundError: No module named 'kivy'   When I check the system path from python, I only get packages in Anaconda:      &gt;&gt;&gt; import sys     &gt;&gt;&gt; print('\n'.join(sys.path))      /home/santiago/anaconda3/lib/python36.zip     /home/santiago/anaconda3/lib/python3.6     /home/santiago/anaconda3/lib/python3.6/lib-dynload     /home/santiago/anaconda3/lib/python3.6/site-packages     /home/santiago/anaconda3/lib/python3.6/site-packages/torchvision-0.2.0-py3.6.egg   Is there any way to include the '/usr/lib/python3' path to Anaconda?  Thanks. User suggested great corrective edit to my answer, but the edit was rejected. What can I do to approve it? A nice user suggested  to . The edit, though very good and desirable, was rejected. I do not understand the reasons, and I did not find a way to contact those people who rejected it. What can I do to re-approve the edit? Privilege Notifications Finding what's using all the space in *nix Which sequence of commands will tell me which files are the largest starting from a particular directory, including all sub directories? I want to know where all the space went.  Preferably just with regular unix'y commands.  If possible, try to keep it compatible with Sun OS 5.10 (perhaps in addition to regular Linux answers, not as a replacement, to keep it as general as possible.) How can I switch back to Nvidia card from Intel with Nvidia-prime So I updated to Ubuntu 14.04 and found that the primusrun command no longer works, so ditched bumblebee in favour of nvidia-prime. This lets you switch video card from nvidia-settings rather than running through the command line (I prefer the old method, but want the primus performance).  The problem is that after switching to Intel from nvidia-settings, I can no longer open it up again to switch back to the nvidia card. When I open nvidia-settings I get a segmentation fault (according to the terminal).  Is there any way to open nvidia-settings switch back? I would like the ability to be able to switch between the two since having the nvidia card on all the time kills my battery and heats up the laptop a lot. In music, when two or more pitches are played together at the same time, they form a chord. If each pitch has a corresponding wave frequency (a pure, or fundamental, tone), the pitches played together make a , which is obtained by simple addition. This wave is no longer a pure sinusoidal wave.  For example, when you play a low note and a high note on a piano, the resulting sound has a wave that is the mathematical sum of the waves of each note. The same is true for light: when you shine a 500nm wavelength (green light) and a 700nm wavelength (red light) at the same spot on a white surface, the reflection will be a superposition waveform that is the sum of green and red.  My question is about our perception of these combinations. When we hear a chord on a piano, we’re able to discern the pitches that comprise that chord. We’re able to “pick out” that there are two (or three, etc) notes in the chord, and some of us who are musically inclined are even able to sing back each note, and even name it. It could be said that we’re able to decompose a Fourier Series of sound.  But it seems we cannot do this with light. When you shine green and red light together, the reflection appears to be yellow, a “pure hue” of 600nm, rather than an overlay of red and green. We can’t “pick out” the individual colors that were combined. Why is this?  Why can’t we see two hues of light in the same way we’re able to hear two pitches of sound? Is this a characteristic of human psychology? Animal physiology? Or is this due to a fundamental characteristic of electromagnetism? I basically need to highlight a particular word in a block of text. For example, pretend I wanted to highlight the word &quot;dolor&quot; in this text: &lt;p&gt;     Lorem ipsum dolor sit amet, consectetuer adipiscing elit. &lt;/p&gt; &lt;p&gt;     Quisque bibendum sem ut lacus. Integer dolor ullamcorper libero.     Aliquam rhoncus eros at augue. Suspendisse vitae mauris. &lt;/p&gt;  How do I convert the above to something like this: &lt;p&gt;     Lorem ipsum &lt;span class=&quot;myClass&quot;&gt;dolor&lt;/span&gt; sit amet, consectetuer adipiscing elit. &lt;/p&gt; &lt;p&gt;     Quisque bibendum sem ut lacus. Integer &lt;span class=&quot;myClass&quot;&gt;dolor&lt;/span&gt; ullamcorper     libero. Aliquam rhoncus eros at augue. Suspendisse vitae mauris. &lt;/p&gt;  Is this possible with jQuery? Edit: As  , this is quite possible without jQuery - but I was hoping there might be a special method of jQuery which would let you do selectors on the text itself. I'm already using jQuery heavily on this site, so keeping everything wrapped up in jQuery would make things perhaps a bit more tidy. Is the inverse of a linear transformation linear as well? A question from the field of Linear Algebra. If I have a linear transformation $T$ that is one-to-one and onto, would that mean that the $T^{-1}$ will also be linear? If so, is there any general proof for it?  Thanks Direct product of Sylow subgroups Proposition II.7.5 of Hungerford's Algebra goes     Proposition 7.5. A finite group is nilpotent if and only if it is the direct product of its Sylow subgroups.   Let $G = (\mathbb{Z}_6, +)$. $G$ is abelian, so it is nilpotent and thus by the proposition is the direct product of its Sylow subgroups.  The Sylow subgroups of $G$ are $H = (\mathbb{Z}_3, +)$ and $K = (\mathbb{Z}_2, +)$, so the proposition says $G = H \times K$. In particular, $0 \in H$ and $0 \in K$, so $(0,0) \in H \times K = G$... but that's not true, since $G = \{0,1,2,3,4,5\}$.  Where is/are my misunderstanding(s)? Why can't I destroy or place blocks?