where is the oil in a cumin seed?
Do ground seeds contain oil?
Is there an intelligent way to left-align the parts right of &amp; of the two cases environments shown below (if x >= 0, otherwise, if y &lt; 0, otherwise). I wonder if there is a better way than using manual spaces:  \begin{align*}  u(x) &amp;=   \begin{cases}    0        &amp; \text{if } x \geq 0 \\    1        &amp; \text{otherwise}   \end{cases}   \\  \tau &amp;=   \begin{cases}    a+b+c+d  &amp; \text{if } y &lt; 0 \\    1        &amp; \text{otherwise}   \end{cases} \end{align*} Can a wizard using an arcane focus quarterstaff and a shield still cast somatic component spells? If there is a wizard that, by some means, has acquired proficiency in the use of shields, can said wizard use his shield in one hand and his arcane focus quarterstaff in the other and still cast somatic component spells?  Since the quarterstaff has been crafted to also be an arcane focus, and the Sage Advice Compendium states that you can provide the somatic components of spells with the same hand wielding the arcane focus, in this case an arcane focus quarterstaff, could that wizard still be able to perform somatic component spells while using a shield in the other?  I understand War Caster is the feat that allows you to cast somatic component spells even while using a shield and weapon or two weapons in each hand, but is this a simple way around those rules seeing as this "weapon" is your arcane focus? I would like to believe it is not, but I'm not fully fluent in the ways of spellcasting. Can we use curved spacetime to measure the one-way speed of light? This is related to  and  The idea is to shine a beam of light from a clock towards a heavy mass, like a black hole, in some manner such that the curvature of spacetime around the mass is sufficient to &quot;bend&quot; the light directly back to the clock. That is, to send the light along a self-intersecting geodesic using a sufficiently curved spacetime. Note that in this case the source and the detector are the same object, so no clock synchronization is required. Hypothetically, this should measure the one-way speed of light. None of the given refutations from the similar posts on this site are convincing. Here are some of the main refutations, as I understand them, and my counter to them: (1) In being curved by spacetime, the light is actually changing direction, hence this is not the one-way speed. To my understanding of General Relativity, the light travels along a geodesic; that is, a curve such that, ignoring technicalities of Differential Geometry, its tangent direction is constant. There is no reflection or acceleration. So in this sense, how can the light with a constant tangent direction be said to change direction? One might say that the path deviates from a straight line in Euclidean space, but given that spacetime is not Euclidean, has no , and has no global notion of direction, this is the only definition of straight or &quot;one direction&quot; that makes sense. Thus, either the light does not change direction and is &quot;one-way,&quot; or we need to be clearer about what exactly we mean by &quot;one-way&quot; in the context of General Relativity. (2) Such an experiment would be impossible or impractical. I think some posters were saying this because the original questions implied that the clock would have to be at the event horizon of a black hole, which is of course impossible or impractical; however, I don't think such an extreme closeness to the black hole is required. Given that the  of a black hole, where light orbits in a circle, is outside of the event horizon, shining the light at some distance outside of the photon sphere should be sufficient to curve the light back to the clock, regardless of the distance of the clock from the black hole (see my rough illustration). At the extreme, we could place the clock directly on the photon sphere, which should be theoretically possible. Regardless, the question is not about practicality, but whether the one-way speed is theoretically measurable.  Update: Essentially what this argument boils down to, I think, is that the average speed of light along a single self-intersecting geodesic may differ from the average speed of light along a non-geodesic self-intersecting path, which is always c. Put that way, however, it does seem a little ridiculous. It is hard to see how the anisotropy of the speed could be defined in such a way as to be consistent with this pathological case. Presumably, we could approximate a single self-intersecting geodesic with a non-geodesic path arbitrarily closely, so, logically, the average speeds over each path should be approximately the same, i.e., c. Furthermore, how would the &quot;preferred direction&quot; of the speed of light behave around a black hole, for instance? Regardless, I still think this experiment does actually solve the problem for the given notion of &quot;direction&quot;. Though the problem is usually posed in the flat spacetime of Special Relativity, with a global notion of direction, so the apparent unanswerability of this question should be seen as a limitation of Special Relativity only, which is then solved by General Relativity. Computing the Lie bracket on the Lie group $GL(n, \mathbb{R})$ Consider the Lie group $GL(n, \mathbb{R})$. Since $GL(n, \mathbb{R})$ is an open subset of the space $M_{n,n}(\mathbb{R})$ of $n \times n$ matrices, we can identify the tangent space (Lie algebra) $T_{e}GL(n, \mathbb{R})$ with $M_{n,n}(\mathbb{R})$, where $e$ is the identity matrix. I'm trying to understand the following:  For all $x, y \in M_{n,n}(\mathbb{R}) = T_{Id}GL(n, \mathbb{R})$ we have $[x, y] = xy -yx$ where $[\cdot, \cdot]$ is the Lie algebra bracket.  According to the question , this seems to be the case for matrix Lie groups in general.  So far, I've worked out the following: Let $x \in M_{n,n}(\mathbb{R}) = T_{Id}GL(n, \mathbb{R})$ and let $X \in \Gamma^{inv}(TG)$ be the left invariant vector field corresponding to $x$, i.e. with $X(e) = x$. By openness, we can choose $\epsilon &gt; 0$ such that the curve $(-\epsilon, \epsilon) \ni t \mapsto e + tx$ through the identity is contained in $GL(n, \mathbb{R})$. For $g \in GL(n, \mathbb{R})$ let $L_g$ be the left multiplication by $g$ in $GL(n, \mathbb{R})$. Using left invariance of $X$ calculate: $$X(g) = ((L_g)_{*}X)(g) = D_eL_g(X(e)) = D_eL_g(x) = \frac{d}{dt}|_{t=0}g(e+tx) = gx$$ Now, the above implies for $f \in C^{\infty}(GL(n, \mathbb{R}))$ and $g \in GL(n, \mathbb{R})$: $$X(g)(f) = (D_gf)(gx)$$ where we identify the tangent vector $X(g)$ with the corresponding derivation at $g$.  Now how do I calculate $[x,y] \stackrel{def.}{=} [X, Y](e) \stackrel{\text{WHY?}}{=}xy-yx$ from this? How do I set up a Cron job? Is there any app for Stack Exchange? Is there any app available in the Android Market for Stack Exchange sites? The  (ESC), states that for every natural number $n \geq 2$, there exists a set of natural numbers $a, b, c$ , such that the following equation is satisfied:  $$\frac{4}{n}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\tag{1}$$  The basic approach to solving this problem outlined by Mordell [Ref1] is described below  By defining $t$ and $m$ as positive integers greater than zero and $q$ a positive integer greater than one we can observe that  a) There is always a solution for even $n$, since if $n=2^qt$ we have the trivial solution $$\frac{4}{4t}=\frac{1}{t}$$  In the remaining case $n=2(2t+1)$, a solution in the form of two Egyptian fractions can always be found  e.g. $$\frac{4}{2(2t+1)}=\frac{2}{2t+1}=\frac{1}{t+1}+\frac{1}{(t+1)(2t+1)}$$  b) If $(1)$ is a solution for some particular prime $n$ then all composite numbers $mn$ divisible by $n$ are also solutions, thus  $$\frac{4}{mn}=\frac{1}{ma}+\frac{1}{mb}+\frac{1}{mc}$$  will also be a solution. This means that we can simplify the analysis to the cases where $n$ is a prime greater than 2.  Using Mordell's approach we have just shown that we only need to consider the cases where $n$ is prime and where $n \equiv 1 \pmod{2} \;\;[meaning \;\;n=2t+1]$  The argument continues...  Mordell goes on to show in turn that the search can be reduced further to the cases when $$n \equiv 1 \pmod{4} \;\;[meaning \;numbers \;\;n=4t+1]$$ $$n \equiv 1 \pmod{8} \;\;[meaning \;numbers \;\;n=8t+1]$$ $$n \equiv 1 \pmod{3} \;\;[meaning \;numbers \;\;n=3t+1]$$ $$n \equiv 1,2,4 \pmod{7} \;\;[meaning \;numbers \;\;n=7t+1,n=7t+2 \;or\;n=7t+4 ]$$ $$n \equiv 1,4 \pmod{5}  \;\;[meaning \;numbers \;\; n=5t+1 \;or\;n=5t+4]$$  Assembling these results together, Mordell showed that the conjecture can be proved in this context except for the cases when $$n \equiv 1,11^2,13^2,17^2,19^2,23^2 \pmod{840}$$  Mordell stated that since the first prime meeting this condition is 1009, this is proof that the conjecture holds for $n&lt;1009$.  This basic approach can be pursued further. Other workers have shown that the conjecture holds for much higher values of $n$ using similar methods as can be seen on the above Wikipedia page.   Note that other intermediate results can be constructed from the above congruence's, e.g. $n \equiv 1 \pmod{24}$.  The question is:  Are there any other elementary approaches to solving this problem than the one outlined by Mordell (and described above)?  [Ref1] Louis J. Mordell (1969) Diophantine Equations, Academic Press, London, pp. 287-290. How to draw open-close interval in LaTeX I am using the following commands to have half-open interval in a real line:   \begin{tikzpicture}[scale=7] \draw[-&gt;, thick] (-0.1,0) -- (1.7,0); \foreach \x/\xtext in {0/8,0.2/$S_1$,0.4/$T_4$,0.6/$S_3$,0.8/$T_2$,1/$S_3$,1.2/14,1.4/$S_3$,1.6/16}     \draw[thick] (\x,0.5pt) -- (\x,-0.5pt) node[below] {\xtext}; \draw (0.2,0.5pt) node[above] {$c$}; \draw[[-), ultra thick, blue] (0.2,.01) -- (.4,0.01); \draw[(-, ultra thick, green] (0.6,.02) -- (1,0.02); \end{tikzpicture}   Now, I need to have an open-close interval, but I am not able to have that close interval at the end, using the following command:  \begin{tikzpicture}[scale=7] \draw[-&gt;, thick] (-0.1,0) -- (1.7,0); \foreach \x/\xtext in {0/8,0.2/$S_1$,0.4/$T_4$,0.6/$S_3$,0.8/$T_2$,1/$S_3$,1.2/14,1.4/$S_3$,1.6/16}     \draw[thick] (\x,0.5pt) -- (\x,-0.5pt) node[below] {\xtext}; \draw (0.2,0.5pt) node[above] {$c$}; \draw[(-], ultra thick, green] (0.6,.02) -- (1,0.02); \end{tikzpicture}   May I ask for anybody help? Proper capitalization of commonly used acronyms and initialisms I was always under impression, that acronyms ought to be written all caps. However reading  site very often I see some of the common acronyms written as proper names (first cap). For example "Nato" instead of "NATO".   I know that BBC is quite strict in proper English usage, so what would be the reason behind this? When do "well" and "good" mean the same? How to change this matrix with respect to standard basis? Given the basis $\beta = \{ (1, 1, 0),\ (1, 0, -1),\ (2, 1, 0)\}$ and the matrix:  $$ A =         \begin{bmatrix}         1 &amp; 1 &amp; 0 \\         1 &amp; 1 &amp; 0 \\         0 &amp; 0 &amp; 1 \\         \end{bmatrix} $$  with respect to  $\beta$.  How to change this matrix with respect to standard basis? I'm trying to follow the  (on my Ubuntu 10.04 VM).  The first step is:  sudo add-apt-repository ppa:linaro-maintainers/toolchain   But when I run that I get:  Error reading https://launchpad.net/api/1.0/~linaro-maintainers/+archive/toolchain: &lt;urlopen error [Errno 110] Connection timed out&gt;   I'm guessing the problem is with my configuration for my employer's proxy server, particularly for HTTPS.  When I opened  in Firefox, I got "The connection has timed out".  I then went into Firefox Preferences -> Advanced -> Network -> Settings... and set HTTP Proxy to "gateway" and Port to 8080, and checked "Use this proxy server for all protocols".  Then the page loaded.  This supports my theory.  I've tried setting the http_proxy and https_proxy environment variables (both lower and upper case):  $ set | grep -i proxy HTTPS_PROXY=https://gateway:8080 HTTP_PROXY=http://gateway:8080 http_proxy=http://gateway:8080 https_proxy=https://gateway:8080   I also tried changing them to:  $ set | grep -i proxy HTTPS_PROXY=gateway:8080 HTTP_PROXY=gateway:8080 http_proxy=gateway:8080 https_proxy=gateway:8080   And I tried adding this second line to /etc/apt/apt.conf:  Acquire::http::proxy "http://gateway:8080/"; Acquire::https::proxy "https://gateway:8080/";   But I continue to get the same error.  How do I troubleshoot this?  Update: I followed the instructions in the  to  and got further:  $ sudo add-apt-repository ppa:linaro-maintainers/toolchain Executing: gpg --ignore-time-conflict --no-options --no-default-keyring --secret-keyring /etc/apt/secring.gpg --trustdb-name /etc/apt/trustdb.gpg --keyring /etc/apt/trusted.gpg --primary-keyring /etc/apt/trusted.gpg --keyserver keyserver.ubuntu.com --recv 1DD749B890A6F66D050D985CF1FCBACA7BE1F97B gpg: requesting key 7BE1F97B from hkp server keyserver.ubuntu.com gpgkeys: HTTP fetch error 7: couldn't connect to host gpg: no valid OpenPGP data found. gpg: Total number processed: 0   I wonder if the "HTTP fetch error" means I also need to add "Defaults env_keep = http_proxy" to /etc/sudoers...  Update #2: I added:  Defaults env_keep="http_proxy" Defaults env_keep="https_proxy"   ...as in , but I still get the same "gpgkeys: HTTP fetch error 7: couldn't connect to host" error.  And this is curious:  $ sudo env | grep proxy https_proxy=https://gateway:8080   ...because http_proxy isn't there.  Any advice is appreciated.  Update #3: Since the host of my VM is a laptop, I took it home and tried there (with no proxies):  $ sudo add-apt-repository ppa:linaro-maintainers/toolchain Executing: gpg --ignore-time-conflict --no-options --no-default-keyring --secret-keyring /etc/apt/secring.gpg --trustdb-name /etc/apt/trustdb.gpg --keyring /etc/apt/trusted.gpg --primary-keyring /etc/apt/trusted.gpg --keyserver keyserver.ubuntu.com --recv 1DD749B890A6F66D050D985CF1FCBACA7BE1F97B gpg: requesting key 7BE1F97B from hkp server keyserver.ubuntu.com gpg: key 7BE1F97B: public key "Launchpad Linaro Overlay PPA" imported gpg: Total number processed: 1 gpg:               imported: 1  (RSA: 1)   It worked.  So problem is related to proxy configuration.  But I suppose I have a workaround. Is there a reason to use multiple question marks instead of an asterisk in filename expansion? If I have a number of directories, named, for example 10001 through 10025 is there any reason to use ls 1*/foo vs. ls 100??/foo?  I have a lot more than 25 of them, so I mostly curious if there's any differences in speed.  I know the difference in use between the two, that the asterisk will match longer file names, like 10001.backup. But let's say I don't have any files that don't follow my conventions. Is there any behind-the-scenes differences? How and when does the tracking of shared links work? Given a prime $p$ and number $a$, how do I find the smallest $n$ such that $n^a\equiv 1 \pmod p$? Is there a non trial-and-error method?  Edit: obviously removing $n=1$! How to fix a stripped pedal crank The threads on my bike's pedal crank just became fully stripped and the pedal fell off, though the pedal itself is fine. Is there any way I can fix this without fully replacing the pedal crank? How to Fix Broken Packages? How to change the desktop wallpaper to a photo of my own selection in Ubuntu 18.04? I'm not interested in downloading some wallpaper changer application, nor am I interested in coding something up. There has to be an easy way to select a photo as a custom wallpaper in Ubuntu 18.04. Suppose you have a history containing the three commits A, B and C:  A-B-C   I would like to combine the two commits A and B to one commit AB:  AB-C   I tried  git rebase -i A   which opens up my editor with the following contents:  pick e97a17b B pick asd314f C   I change this to  squash e97a17b B pick asd314f C   Then Git 1.6.0.4 says:  Cannot 'squash' without a previous commit   Is there a way or is this just impossible? Optimizing a screenshot for adding in a document and printing