I don't seem to find any tools in the mobile app while writing answers and questions; I only see them in the browser view.   Are there no tools in the Android app, or do I just overlook them?
When creating a question/answer on the Android app, it gives you a little bar above the keyboard to bold/italicize, add a link easily, add a picture etc.  When editing however, this bar isn't shown. To ease editing on mobile, it would be exceedingly helpful to have this bar show up. I find myself resorting to the full site to add a embedded link if I can't at the moment remember the exact syntax.
In Ubuntu ISO MBR, why does a GPT partition overlap the boot partition, and make MBR uneditable? "grey" vs "gray" are both common equally in use? Yesterday I wrote "gray" in the meaning of the color, but my friend corrected me and told me to write "grey". Today I checked in the  and I found that they are both correct for the same meaning. So, now I just have doubt which one of them is more common in use. Prove $4(x + y + z)^3 > 27(x^2y + y^2z + z^2x)$ Prove that, for all positive real numbers $x$, $y$ and $z$, $$4(x + y + z)^3 &gt; 27(x^2y + y^2z + z^2x)$$  I've tried expressing it as a sum of squares, but haven't got anywhere.   Hints are also welcome. Is there way to move one vertex and make another move with it proportionally? I'll try to explain in this gif. I am doing this manually in 2 steps - rotate, scale. And I select all vertexes. Is there instrument rotate them in one step or can I move one and make other move too? Does a particle annihilate only with its antiparticle? If yes, why? Or to put the question another way - what is the result of a proton-positron collision, or an up quark-charm antiquark collision, etc.? As far as I know, annihilation happens only between particles of opposite charge and same mass, but perhaps I am wrong?  And if the types of annihilation mentioned above cannot happen, what are the reasons? Let $\mathbb F$ be a field and $\mathbb  K $ be an extension field of $\mathbb F$  such that $\mathbb K$ is algebraically closed.   Let $\mathbb L$ be the field of all elements of $\mathbb K$ which are algebraic over $\mathbb F$. Then $\mathbb L_{|\mathbb F}$ is an algebraic extension.   My question is : Is $\mathbb L$ algebraically closed ?  I am trying to prove the existence of algebraic closure, so please don't assume that every field has an algebraic closure. Why sum of two little "o" notation is equal little "o" notation from sum? Why sum of two little "o" notation is equal little "o" notation from sum?  $o( f(n) ) + o( g(n) ) = o( f(n) + g(n) ) ?$  For example:   $f(n) = n^3$  $g(n) = 1/n$   so    $o(f(n)) = n^2$  $o(g(n)) = 1/n^2$   and    $o( f(n) ) + o( g(n) ) = n^2 + 1/n^2$ $o( f(n) + g(n) ) = n^2$   Of course, I could write it like   $o( f(n) ) + o( g(n) ) = n^2 + o( g(n) )$ $o( f(n) + g(n) ) = n^2 + o( g(n) )$   My question is why? I don't understand it, because in first we always get two parameters. Under which assumptions a regression can be interpreted causally? First, don't panic. Yes, there are many similar question on this site. But I believe none gives a conclusive answer to the question below. Please bear with me.  Consider a data generation process $\text{D}_X(x_1, ... , x_n|\theta)$, where $\text{D}_X(\cdot)$ is a joint density function, with $n$ variables and parameter set $\theta$. It is well known that a regression of the form $x_n = f(x_1, ... , x_{n-1}|\theta)$ is estimating a conditional mean of the joint distribution, namely, $\text{E}(x_n|x_1,...,x_{n-1})$. In the specific case of a linear regression, we have something like $$ x_n = \theta_0 + \theta_1 x_1 + ... + \theta_{n-1}x_{n-1} + \epsilon $$ The question is: under which assumptions of the DGP $\text{D}_X(\cdot)$ can we infer the regression (linear or not) represents a causal relationship? It is well known that experimental data does allow for such interpretation. For what I can read elsewhere, it seems the condition required on the DGP is exogeneity: $$ \text{E}(x_1, ... x_{n-1}|\epsilon) = 0$$ The nature of randomisation involved in experimental data ensures the above is the case. The story goes then to argue why observational data normally fails in achieving such condition, reasons including omitted variable bias, reverse causality, self-selection, measurement errors, and so on. I am however uncertain about this condition. It seems too weak to encompass all potential arguments against regression implying causality. Hence my question above. UPDATE: I am not assuming any causal structure within my DGP. I am assuming that the DGP is complete, in the sense that there must be some causality (an ontological position it could well be debated), and all relevant variables are included. The key is to identify the set of assumptions which ensure me causality goes from certain variables to the other, without assuming from the outset such direction of causality.  Many similar posts on the site spend time mentioning why correlation does not imply causation, without providing hard arguments about when it does. That is the case, for instance, of  very popular post. Additionally, in the  on the site about the topic, the accepted answer gives the very vague answer:  Expose all hidden variables and you have causation.  I do not know how to translate that to my question above. Neither is the second most upvoted answer. And so on. That is why I believe this post does not have an answer elsewhere. How do I prevent and/or handle a StackOverflowException? I would like to either prevent or handle a StackOverflowException that I am getting from a call to the XslCompiledTransform.Transform method within an Xsl Editor I am writing. The problem seems to be that the user can write an Xsl script that is infinitely recursive, and it just blows up on the call to the Transform method.  (That is, the problem is not just the typical programmatic error, which is usually the cause of such an exception.)  Is there a way to detect and/or limit how many recursions are allowed?  Or any other ideas to keep this code from just blowing up on me? I have been thinking about the following formulation of the Axiom of Choice.:     For any relation $R$, there is a function $H$, which is a subset of $R$, and such that $\operatorname{dmn}H=\operatorname{dmn}R$.   At first, it seemed surprising to me that this did not follow from Comprehension (i.e., from the schema  $\forall x_1, x_2, ..., x_n \forall C\exists B \forall x ( x \in B \Leftrightarrow x \in C \land \phi )$, where $B$ does not occur free in $\phi$).    I thought that for any relation, it seems that we should be able to define a subset of it with the same domain meeting a "functional condition".  But when I tried to do this, I couldn't.  However, I felt like I was getting close when I let $B$ be free in $\phi$, i.e., when I let myself say that $w$ is in $R$ and also anything in $H$ meets such and such conditions, where $H$ is the set (function) whose existence I'm trying to prove.    But my understanding is that Choice does not even follow from Naive Comprehension.  So, I wondered if someone could explain the following precisely.   Why does the above formulation of Choice does not follow from regular Comprehension? Why does it not even follow from Naive Comprehension (assuming that it does not)? Is it possible to add attributes at runtime or to change the value of an attribute at runtime? What's the quickest Schengen consulate to issue a visa in London? How do I read from /proc/$pid/mem under Linux? The  tells me that /proc/$pid/mem “can be used to access the pages of a process's memory”. But a straightforward attempt to use it only gives me  $ cat /proc/$$/mem /proc/self/mem cat: /proc/3065/mem: No such process cat: /proc/self/mem: Input/output error   Why isn't cat able to print its own memory (/proc/self/mem)? And what is this strange “no such process” error when I try to print the shell's memory (/proc/$$/mem, obviously the process exists)? How can I read from /proc/$pid/mem, then? How can I make environment variables "exported" in a shell script stick around? How can I sync the new "people" (face recognition) metadata across devices in Photos (iOS 10 and MacOS Sierra)? The new Photos updates for iOS 10 and MacOS Sierra includes a much more powerful and integrated approach to organizing photos by who's in them.  It's built around powerful, on-device facial recognition.  But once it does its automated thing (finding the faces it thinks are the same person), there's still a lot of manual merging and labeling to be done if you have a lot of photos and are as obsessive about organization as I am.  For example, if you have, say... three thousand photos of your daughter, Photos will find almost all of them, but may guess they're 5 or 6 different people.  So you have to merge the buckets that all are one person, assign them a name (so you can search by it), etc.  But unlike other changes (edits, deletions, organization into albums, etc.), manual changes to "people" data don't seem to sync at all by default.      Is there really no way to do this?  It seems bizarre that one of the flagship features Apple's touting in this Photos update doesn't play at all with what was Photos primary selling point over iPhoto - everything synced on all your devices. What is the difference between "why are you worried" and "why you are worried"? What is the difference between these two sentences?     Why are you worried?   Why you are worried?   Which one is correct? Can I load a .NET assembly at runtime and instantiate a type knowing only the name? Some webpages forces mobile version even I change user-agent. I changed user-agent in built-in browser and "Dolphin Browser HD",  I am checking it on page www.user-agent-string.info  Main problem is engadget.com, always in mobile version. I tried to clean cookies, restart browsers etc.  This is my results:   in built-in browser, user-agent - "desktop"   () doesn't work, redirects to mobile version   in "Dolphin Browser HD", user-agent - "desktop"    (NSFW: redtube.com) doesn't work, redirects to mobile version  (allegro.pl) works fine  in "Dolphin Browser HD", user-agent - "IPad"   () doesn't work, redirects to mobile version  (allegro.pl) works fine (NSFW: redtube.com) works fine  in "Dolphin Browser HD", user-agent - "custom": Mozilla/5.0 (Windows NT 6.0; rv:7.0.1) Gecko/20100101 Firefox/7.0.1   (engadget.com]) doesn't work, redirects to mobile version  (NSFW: redtube.com) doesn't work, redirects to mobile version  (allegro.pl) works fine    Maybe this pages using "adobe flash" user-agent detection ?  (I'am using Asus Transformer TF101, Android 3.2.1, Adobe Flash Player 11) I have the string   a.b.c.d   I want to count the occurrences of '.' in an idiomatic way, preferably a one-liner.  (Previously I had expressed this constraint as "without a loop", in case you're wondering why everyone's trying to answer without using a loop). Photoshop resize circularly