Open tunnel only during local command run
Kill backgrounded SSH when shell exits
How can I have  for Ubuntu? Strategy for "Which is better" questions In questions that are asking:    "Which is better option A or option B?"  It seems that several close reasons immediately come to mind:   Primarily Opinion Based.  In many cases the choice between A and B is simply a personal preference, a "6 vs. half dozen", either approach will work just as well and answers to this type of question will be almost entirely opinion based. Too Broad.  In some cases A is better and in others B is better, to go through every possible use case would lead to too many possible answers or one answer that reads like a short text book. Off-topic, Minimal Understanding, include attempted solutions.  This is perhaps the the most important one. If the OP needs to know "Which is better", they should probably have tried both and decided which is better, based on a practical trial.   Yet another option would be to downvote and/or comment with a "Try it and find out".  So...  Are there cases when these questions are permissible?   Could we add some more specific guidance in the help center?  Or should we continue to handle these case by case and just downvote/close appropriately when they pop up?  Strongly related, but hopefully not a duplicate of:   "Which is faster" isn't the only case where these pop up. Let $(a_n)$ be a sequence of positive terms and suppose that $\sum_\limits{n=0}^{\infty} a_n$ converges. Show that $\sum_\limits{n=0}^{\infty} a_n^2$ converges.   This is in the section on the Comparison Test so that must be what I'm supposed to use.  But I don't see how.  $(a_n)^2$ might be smaller or larger than $a_n$ depending on $a_n$.  And I can't use the Comparison Test with some other series because there's no info here about how fast $\sum a_n$ converges.  Hmm.  Any hints? How do you choose Spells Known when creating a new character above 1st level? It's not clear to me how to choose Spells Known when creating a new character at a higher level.  Let's take the Bard as a working example first:  Bard level 2:   Spells Known: 5 Slots per Spell level: 1st &times; 3   Bard level 3:   Spells Known: 6 Slots per Spell level: 1st &times; 4, 2nd &times; 2   I'm interpreting this to mean that, if an existing 2nd-level Bard gains a new level (level 3), he's gaining 1 (one) new spell known. He can then choose the new spell known to be either 1st or 2nd, correct?  What I don't understand then, is the case of creating a new character starting at a higher level; say, creating a new Bard at level 3.  In that case you're forced to choose 6 spells, which must be from the list of 1st- and 2nd-level spells, correct?  When making this 3rd-level Bard from scratch, though, can I select 2nd level spells for all 6 spells known?  As another example, say you're creating a very high level Bard from scratch, starting at level 15. The book says your Spells Known is 19, and your spell slots are 1st × 4, 2nd × 3, 3rd × 3, 4th × 3, 5th × 2, 6th × 1, 7th × 1, and 8th × 1.  So, for those 19 Spells Known, you can choose any spells from 1st through 8th spell level. Am I free then to select as many spells as I can at 8th spell level (5 in number), working backwards, and in the end ignoring the 1st, 2nd, 3rd level spells entirely?  As another example, the Ranger doesn't know any spells at 1st level, nor have any spell slots, but the book says “You know two 1st-level spells of your choice from the ranger spell list.” Does it mean at 2nd level and above?  Long story short: When choosing Spells Known for a spellcasting character created above 1st level, am I allowed to choose my Spells Known at whatever spell level I like, just so long as I have at least one spell slot of that level?  I'm working on a D&amp;D 5e character generator and need to clarify this. Why are 50mm Lenses Cheaper? How do I get the linksys WUSB6300 wireless adapter to work on linux Is it possible to have a rate of change of acceleration? I know this may seem a weird question, but it always bothers me. My physics book (Resnick,Halliday,Walker), and also various sites never say anything beyond acceleration.   But when a moving body is being acted by a variable force , its acceleration will definitely change: it will either increase or decrease. Then there will be rate of change of acceleration with respect to time. So, why don't books mention this? What is the cause for not measuring $\frac{d\vec{a}}{dt}$ ? If it exists, what is the use of it? How do I get the member to which my custom attribute was applied? Get image pixels of prescribed color This is probably a dumb question. I guess what I'm trying to ask is if radio waves travel the same speed as gamma rays, how do gamma photons carry more energy than radio photons? Do they spin faster? What other energy sources could they carry if they are moving the same speed through space? How to compare two different files line by line in unix? File1:  123 234 345 456   File2:  123 234 343 758   Expected output: File3:  TRUE TRUE FALSE FALSE   so the code should compare two files and print 'TRUE' if it matches otherwise it should print 'FALSE' in the new file. Could anyone please provide the solution for this? (Reference : Fraleigh, A first course in abstract algebra)     Prove that a nonempty set $G$, together with an associative binary operation * on $G$ such that $a*x=b$ and $y*a=b$ have solutions in G, $\forall a,b\in G$, is a group.   Because * is an associatie operation, we have to prove that: there exists $e\in G$ such that $e*a=a=a*e$ for any $a\in G$, and that for any $a\in G$ there exists $\alpha \in G$ such that $a*\alpha=e=\alpha *a$.  This is very confusing, first I tried to prove the identity exists. If we consider a=b, then $a*x=a=y*a$, and because of the hypothesis $x,y\in G$ always exists, now we have to show that $x=y$, but I just can't do this with out inverses, so I tried to prove that the inverses exists, however I have no idea how the identity looks like, I'm stuck.  And I tried to get unstuck, so I thought particular cases, first if $G$ had only one element, then what would happend if it had just two elements... but I remembered that we don't know if G is finite or not, and I think this makes things a little bit more complicated.  What can I do? My text introduces multi-quibt quantum states with the example of a state that can be &quot;factored&quot; into two (non-entangled) substates. It then goes on to suggest that it should be obvious1 that the joint state of two (non-entangled) substates should be the tensor product of the substates: that is, for example, that given a first qubit $$\left|a\right\rangle = \alpha_1\left|0\right\rangle+\alpha_2\left|1\right\rangle$$ and a second qubit $$\left|b\right\rangle = \beta_1\left|0\right\rangle+\beta_2\left|1\right\rangle$$ any non-entangled joint two-qubit state of $\left|a\right\rangle$ and $\left|b\right\rangle$ will be $$\left|a\right\rangle\otimes\left|b\right\rangle = \alpha_1 \beta_1\left|00\right\rangle+\alpha_1\beta_2\left|01\right\rangle+\alpha_2\beta_1\left|10\right\rangle+\alpha_2\beta_2\left|11\right\rangle$$ but it isn't clear to me why this should be the case. It seems to me there is some implicit understanding or interpretation of the coefficients $\alpha_i$ and $\beta_i$ that is used to arrive at this conclusion. It's clear enough why this should be true in a classical case, where the coefficients represent (where normalized, relative) abundance, so that the result follows from simple combinatorics. But what accounts for the assertion that this is true for a quantum system, in which (at least in my text, up to this point) coefficients only have this correspondence by analogy (and a perplexing analogy at that, since they can be complex and negative)? Should it be obvious that independent quantum states are composed by taking the tensor product, or is some additional observation or definition (e.g. of the nature of the coefficients of quantum states) required?  1:  (bottom of p. 18) &quot;so the state of the two qubits must be the product&quot; (emphasis added). Let $n&gt;0$ and assume that $\chi^{(n)} \in \mathrm{Irr}(G)$ for every $\chi \in \mathrm{Irr}(G)$. Show that $G = H \times A$, where $A$ is abelian and $(|H|, n) = 1$.  $\chi^{(n)}$ is defined by $\chi^{(n)}(g) = \chi(g^n)$. $\mathrm{Irr}(G)$ is the set of all irreducible characters of $G$.  Here are the hints in the book:   Let $d = (|G|,n)$. Show that it is no loss to assume that $(|G|/d,n) = 1$. Let $A = \bigcap_{\chi \in \mathrm{Irr}(G)} \mathrm{ker} \chi^{(n)}$. Show that $A = \{g \in G | g^n=1\}$ and $|A| = d$. Let $H = \bigcap \{\mathrm{ker} \chi | \chi \in \mathrm{Irr}(G), \chi^{(n)} = 1_G\}$. Show $|G : H| = d$.   Hint 1 and 2 are easy, and it is easy to deduce the result from the hints. As for hint 3, I have proven that $\{ g \in G | (o(g), n) = 1 \} \subseteq H$, and that $|G : H| \mid d$. And I am stuck here.  Can anyone help me? Removing duplicate rows from table in Oracle I'm testing something in Oracle and populated a table with some sample data, but in the process I accidentally loaded duplicate records, so now I can't create a primary key using some of the columns.  How can I delete all duplicate rows and leave only one of them? How to stream a video with VLC in a LAN I am trying to get from a particular point more informations about the surroundings (which are also points). So I wrote this:  SELECT *    FROM osm_point    WHERE ST_DWithin(way,(ST_SetSRID(ST_MakePoint(12.2385578688105, 51.8435615931295), 4326, 1000);   But there is an error     ERROR: FEHLER:  LINE 3: ...T_MakePoint(12.2385578688105,   51.8435615931295), 4326, 1000) ^ SQL state: 42601 Character: 133   What is wrong? How can I get debug log in Mavensmate Determine the Galois group of $\mathbb{Q}(\sqrt{a+b\sqrt{d}})$ Suppose $L=\mathbb{Q}(\sqrt{a+b\sqrt{d}})$,($d$ and $a+b\sqrt{d}$ are square free algebraic integers) when is $L/\mathbb{Q}$ a normal extension? When does $Aut(L/\mathbb{Q})=\mathbb{Z}/4\mathbb{Z}$ or $Aut(L/\mathbb{Q})=\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$? What's the keyboard shortcut for keyboard layout switch in Mojave? Shift + Control + Space doesn't seem to work anymore.