Minor UI glitch: The button to award a bounty is active on a deleted answer. Clicking on it results in the "THIS CANNOT BE UNDONE!" alert.         After confirming, I get the following error though:        So not a fatal bug, but still better not show the +50 next to deleted posts.
Clearly this is an edge case and I probably shouldn't have even tried awarding the bounty to a deleted answer, but why is the bounty award button even available?  It's not a bug since someone added in the notification seen in the final image.  But that sure seems like a roundabout way to let the user know that they can't award a bounty to a deleted answer.  The second step is particularly strange since the system really ought to know that the warning is pointless: no bounty can be awarded.  My suggestion is to rip-off the  and let bounties be awarded without the warning.  Then remove the single click button from deleted answers.
I've seen the phrase "sh compatible" used usually in reference to shells. I'm not sure if it also applies to the programs that might be run from within shells.  What does it mean for a shell or other program to be "sh compatible"? What would it mean to be "sh incompatible"?  Edit: This question asking the difference between bash and sh is very relevant:   I'd still like a direct answer to what it means to be "sh compatible". A reasonable expectation might be that "sh compatible" means "implements the Shell Command Language" but then why are there so many "sh compatible" shells and why are they different? De Rham Cohomology of the tangent bundle of a manifold I would like to compute the de Rham cohomology of the tangent bundle $TM$ of a manifold $M$. It seems to me that we can just homotopy each fibre to a point, and that this would give a homotopy equivalence between $TM$ and $M$. Since de Rham cohomology is a homotopy invariant, this would imply that $H^*_{\text{dR}}(TM) \cong H^*_{\text{dR}}(M)$.     Is it always true that the tangent bundle of a manifold is homotopic to the manifold itself by the homotopy of each fibre with a point? Is ∅ equivalent to {∅}? How to determine the most likely reading frame of a DNA sequence? This question is from a past exam paper for an introductory bioinformatics module. I'm a computer scientist doing biology for the first time.  "A short bacterial gene has been sequenced, giving the following DNA sequence. Write out the 6 possible reading frames for this sequence and indicate which is the most likely protein translation of this sequence. Explain your reasoning for picking the given translation, and write all the translations in single letter amino acid code form.  5' - ttattcatccgccagcgccatgcgcgccat - 3'  "  I think I understand the 6 possible reading frames: three from the 5' end starting tta, tat and att; and three from the 3' end starting tac, acc and ccg.  I could also write out the translations for a given reading frame with a codon usage table. It's the most probable reading frame part I don't get. I thought perhaps I was looking for the longest open reading frame. I can only see one start codon (atg). What is the correct way to recognize the most probable reading frame?  I can't find this problem exactly on Biology Stack Exchange. I did find:     It made me wonder if I'm getting confused between "reading frame" and "open reading frame", this question only asks or the former.  Thanks! How to get back a declined Outlook invitation Question on showing a bijection between $\pi_1(X,x_0)$ and $[S^1, X]$ when X is path connected. I am trying to do this question taken from Hatchers algebraic topology and I am struggling to understand the notation and the concepts.  As far as I know $\pi_1(X,x_0)$ is the set of end point preserving homotopy classes of loops in X based at $x_0$ and a loop is just a path    $f:I \rightarrow X$ with $f(0)=f(1)$. The question says we can regard $\pi_1(X, x_0)$ as the set of basepoint preserving homotopy classes of maps $(S^1, s_0) \rightarrow (X, x_0)$.   This confuses me, does it mean the set of basepoint preserving homotopy class on maps $g:S^1 \rightarrow X$ which map loops in $S^1$ based at $s_0$ to loops in X based at $x_0$?  Then how can $\pi_1(X,x_0)$ be regarded as this set?  If someone could explain this it would be really useful Recovering accidentally deleted files independent variables in Random forest model Intuitive explanation of entropy I have bumped many times into entropy, but it has never been clear for me why we use this formula:  If $X$ is random variable then its entropy is:  $$H(X) = -\displaystyle\sum_{x} p(x)\log p(x).$$  Why are we using this formula? Where did this formula come from? I'm looking for the intuition. Is it because this function just happens to have some good analytical and practical properties? Is it just because it works? Where did Shannon get this from? Did he sit under a tree and entropy fell to his head like the apple did for Newton? How do you interpret this quantity in the real physical world? Edit: The previous title didn't really ask the same thing as the question (sorry about that), so I've changed it. To clarify, I understand that the action isn't always a minimum. My questions are in the points 1. and 2. below.  I understand that &quot;principle of least action&quot; is somewhat of a misnomer, since we find that to determine the path taken by a system, we need only impose the condition that the action be stationary, i.e., that $\delta S$ should vanish to first order for small variations of the path, and this leads to the Euler-Lagrange equations. In The Classical Theory of Fields, Landau &amp; Lifshitz discuss the relativistic action for a free particle:  So for a free particle the action must have the form $$S = -\alpha \int_a^b ds$$ (...) It is easy to see that $\alpha$ must be a positive quantity for all particles. In fact, as we saw [earlier], $\int_a^b ds$ has its maximum value along a straight world line; by integrating along a curved world line we can make the integral arbitrarily small. Thus the integral $\int_a^b ds$ with the positive sign cannot have a minimum; with the opposite sign it clearly has a minimum, along the straight world line.  There is also a footnote, addressed a couple of paragraphs earlier, but which is relevant:  Strictly speaking, the principle of least action asserts that the integral $S$ must be a minimum only for infinitesimal lengths of the path of integration. For paths of arbitrary length we can say only that $S$ must be an extremum, not necessarily a minimum.  I have two questions regarding this:  How is the condition &quot;the action must be a minimum for infinitesimal displacements&quot; formulated? I've never heard of that outside of Landau &amp; Lifshitz's books, and in Mechanics they mention it as well but doesn't go into detail. Is this discussed a bit more somewhere?  If for the whole path the action only needs to be stationary, how can we make the argument for the negative sign? If the action had to be a minimum then it would make sense, but surely the fact that $\delta S$ = 0 isn't affected by an overall sign? Reclassifying only part of raster keeping rest unchanged? I'm trying to reclassify only a certain part of a raster...not the entire raster.  Here's the problem.  I have a LULC raster of the entire US.  There are some issues with the raster that I want to fix, namely some cells that should be water are classified as wetlands.  These are mainly off the coast, not inland.  However, some cells that are classified as wetlands should really be wetlands (mainly inland) and I don't want these to change.  I plan on using a ZIP Code boundary file to determine which cells are considered "off the coast" and which ones are not.  I have access to ArcGIS 10.1 and R with the appropriate raster packages I need so a solution in either software would be fine.  Can anyone help me figure this out? If every eigenvector of $T$ is also an eigenvector of $T^{*}$ then $T$ is a normal operator Let $V$ be a finite-dimensional inner product space over $\mathbb{C}$ and $T: V \to V$ a linear transformation.   Show that if every eigenvector of $T$ is also an eigenvector of $T^{*}$ then $T$ is a normal.   We need to show that $\forall v \in V,\ TT^*v=T^*Tv$.  I started by picking $v$ to be an eigenvector of $T$, hence $Tv=\lambda_1 v$, and $T^*v=\lambda_2 v$. Therefore: $$TT^*v=T(\lambda_2 v)=\lambda_2 Tv=\lambda_1 \lambda_2 v=...=T^*Tv$$  But I'm stuck when $v$ is not an eigenvector.  (Also, this question is taken from a chapter about Jordan forms, but I can't see how it relates) Pseudo-terminal message in script Symmetric Inequality in $\mathbb{R}$ Buffering GeoJSON file using Python? Please be kind, I am an electronics nub.  This is in reference to getting an LED to emit photons.  From what I read (Getting Started in Electronics - Forrest Mims III and Make: Electronics) electrons flow from the more negative side to the more positive side.  In an example experiment (involving a primary dry cell, a SPDT switch, a resistor and an LED) it states that the resistor MUST be connected to the anode of the LED.  In my mind, if the electrons flow from negative to positive, wouldn't the electron flow run through the LED before the resistor; thereby making the resistor pointless? I want to encrypt a USB stick, and be able to decrypt it on Windows as well as Ubuntu. How do I do this &amp; what programs do I need to use? Framing a combination of cells in the notebook Why would a slug thrower be more effective against a lightsaber/Jedi than a blaster? From  question about gunpowder in the Star Wars universe it is  that the shot from a  slug thrower is not completely deflectable by a lightsaber. This seems to be listed as an advantage (over blasters) against lightsabers and Jedi.     For example, during the days of the Galactic Republic, mercenaries fearing an intervention by Jedi Knights used rapid-fire slugthrowers that were impossible to completely deflect, unlike blaster bolts.   This doesn't make much sense though. A slug, while maybe not as deflectable should be just as ineffective against a lightsaber as a blaster shot is, due to the nature of a . That is, a blade made of plasma should simply vaporize the slug, should the slug manage to penetrate the containment field.     The weapon consisted of a blade of pure plasma emitted from the hilt and suspended in a force containment field. The field contained the immense heat of the plasma, protecting the wielder, and allowed the blade to keep its shape.   While a slug would seem to be more accurate than a blaster bolt (at least when used by a skilled marksman), I can't imagine that the slug would be faster than a bolt. Or if it is faster I wouldn't imagine it would be so fast that a Jedi would be unable to react with enough speed to block it, either with their saber or with the force. So what is the implied advantage of a slug thrower, versus a blaster, when used against a Jedi? Recursively copy files from one directory to another I have all my music in a folder /media/kalenpw/MyBook/Music/ABunchOfOtherFoldersWithFilesInside. I want to copy all of the mp3s to /media/kalenpw/HDD/Music so I used:  cp -R /media/kalenpw/MyBook/Music/*.mp3 /media/kalenpw/HDD/Music   however this only copied the mp3s in the root music folder and did not open any of the artist subdirectories and copy those files.  I was under the impression -R would recursively copy all the files. How can I achieve said goal?