I have an Android 4.4.2 phone and I enabled ART instead of Dalvik.  When I go to the Google Play Store and try to install Facebook, it can't install the app.  I get two errors, sometimes I get     Invalid installation package   and sometimes     Error -24   Is there a way to know what's going on?
I just updated my Moto X to latest and greatest Android 4.4 KitKat (I even did full factory reset just to be sure) and now I'm trying to install  but unable to do so, I'm getting following error:     Unknown error code during application install "-24"   Please advise.
I want to turn a photo in a engraving style picture, like  using GIMP and, if possible, then make an SVG out of it with Inkscape.  Can you recommend any filters/tools/plugins/tutorials for this? TOC entries and labels for included PDF pages I am embedding full pages from another PDF document (using the pdfpages package), but I would like to be able to declare my own section headings before each embedded document (for e.g. hyperlinking / TOC purposes). pdfpages has an experimental feature which lets me do this, but the problem is that I want to put all of these embedded documents inside a section, so I need to insert a section heading and a subsection heading on the first document I embed. Unfortunately pdfpages doesn't seem to be able to do this. Is there a way for me to do this?  Edit: Thank you for the suggestions. I am also in need of something similar for general labels — at the moment I am abusing the addtolist option of \includePDF but the objects I'm labelling aren't really figures or tables. Which one is correct, or are both of them fine?    "It would take the average person 10 days to read this novel"    Or    "It would take an average person 10 days to read this novel" I want to show that for $S,T\in B(H)$ bounded operators on Hilbertspace with $S\geq 0,T\geq 0$ and $ST=TS$,  we have $S+T\geq 0$, and $ST\geq 0$.  $T\geq 0$ means $(Tx,x)\geq 0$.    To me it seems that $((S+T)x,x) = (Sx,x)+(Tx,x)\geq 0$. But for $ST$ i like some help.   There is a theorem in Rudin (12.32) that says that for every $T\in B(H)$ we have an equivalence:   $T$ is positive $\Leftrightarrow$ $T$ is self adjoint, i.e. $T=T^*$ and $\sigma(T)\subset [0,\infty)$.   Since $S, T$ are self adjoint we know that $ST=TS = S^*T^* = T^*S^*$ ($ST$ is selfadjoint since $S,T$ commute). Also I read somewhere that if $S,T $ commute then $\sigma(ST)\subset \sigma(S)\sigma(T)$, but I am not sure if this is true. Clearly I cant prove it, but if it were true by this ''Theorem'' it would follow that $ST \geq 0$.   Is there a more straightforward way? Short story that takes place off planet, involving sun and rain I read this short story about 20 years ago, and I'd love to find the title and author. This is what I can recall about the story:   It takes place in a classroom full of kids (around 10 years old?) The planet always gets rain The kids lock the main character (a girl, I think) in the closet during the very rare occurrence of sunlight and she misses the entire event.    I know it's not much to go on, but I hope it rings a bell for someone! the converse of Pythagoras Theorem Let $\left| {{a_{ii}}} \right| > \sum\limits_{i \ne j} {\left| {{a_{ij}}} \right|} $.Why does $A$ is nonsingular? . How can I kill a thread? without using stop(); Linux / Bash, using ps -o to get process by specific name? Are there uncountably many surjections from $\mathbb N$ to $\mathbb N$? Statement: There are uncountably many surjections from $\mathbb N$ to $\mathbb N$.  Is the following proof valid:  The real numbers are uncountable.   There exists a bijection between the set of real numbers and the set of all functions from $\mathbb N$ to $\{0,1\}$.   Given a function, $f$, from $\mathbb N$ to $\{0,1\}$, construct a surjection from $\mathbb N$ to $\mathbb N$, denoted $g$, as follows:  We map the $n$th composite number to $n$, and the nth prime, to $f(n)$.   Clearly this is a surjection since the composite numbers are countable. Clearly different functions $f$ lead to different surjections $g$. Since there are uncountably many $f$s, there are uncountably many surjections $g$. Merging two incoming broadband lines for faster Internet We have a broadband line coming into our building which delivers around 8-10Mbps, which is okay but not great. Fiber is available in our area (30-100Mbps) but it's not available where we are just yet.   Is there some way we could connect two lines (which we would take from two different providers so we would have redundancy if one failed) and merge them into one signal which we could distribute?  Is there some sort of setup that can do this?  Of course we have the option of a leased line but for 12Mbps symmetrical upload/download we'd be looking at around £500 per month compared to £15 per month for normal or fiber broadband.  I made a small pseudo flow chat of the setup below: Applescript run from menu bar? Does $\sum \limits_{n=1}^\infty\frac{\sin n}{n}(1+\frac{1}{2}+\cdots+\frac{1}{n})$ converge (absolutely)? I've had no luck with this one. None of the convergence tests pop into mind.  I tried looking at it in this form $\sum \sin n\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}$ and apply Dirichlets test. I know that $\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n} \to 0$ but not sure if it's decreasing.  Regarding absolute convergence, I tried:  $$|\sin n\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}|\geq \sin^2 n\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}=$$  $$=\frac{1}{2}\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}-\frac{1}{2}\cos 2n\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}$$  But again I'm stuck with $\cos 2n\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}$. Assuming it converges then I've shown that $\sum \sin n\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}$ doesn't converge absolutely. Every integer is congruent to the sum of its digits mod 9 This is a question from the free Harvard online abstract algebra .  I'm posting my solutions here to get some feedback on them.  For a fuller explanation, see  post.  This problem is from assignment 6.  The notes from this lecture can be found .     Prove that every integer $a$ is congruent to the sum of its decimal digits modulo 9.   Let $s$ be an $(n+1)$-digit integer given by $s=a_n\cdot10^n+\dots+a_2\cdot10^2+a_1\cdot10^1+a_0\cdot10^0$ with $a_i\in\mathbb{Z}$ such that $0\leq a_i\leq9$.  Then $ \begin{align} s\,\mathrm{mod}\,9&amp;=\bar{s}\\ &amp;=\overline{a_n\cdot10^n+\cdots+a_2\cdot10^2+a_1\cdot10^1+a_0\cdot10^0}\\ &amp;=\overline{a_n\cdot10^n}+\overline{\cdots}+\overline{a_2\cdot10^2}+\overline{a_1\cdot10^1}+\overline{a_0\cdot10^0}\\ &amp;=\overline{\overline{a_n}\cdot\overline{10^n}}+\overline{\cdots}+\overline{\overline{a_2}\cdot\overline{10^2}}+\overline{\overline{a_1}\cdot\overline{10^1}}+\overline{\overline{a_0}\cdot\overline{10^0}}\\ &amp;=\overline{\overline{a_n}\cdot1}+\overline{\cdots}+\overline{\overline{a_2}\cdot1}+\overline{\overline{a_1}\cdot1}+\overline{\overline{a_0}\cdot1}\\ &amp;=\overline{a_n}+\overline{\cdots}+\overline{a_2}+\overline{a_1}+\overline{a_0}\\ &amp;=a_n\,\mathrm{mod}\,9+\cdots+a_2\,\mathrm{mod}\,9+a_1\,\mathrm{mod}\,9+a_0\,\mathrm{mod}\,9 \end{align} $  Again, I welcome any critique of my reasoning and/or my style as well as alternative solutions to the problem.  Thanks. Why do Google search results include pages disallowed in robots.txt? I have some pages on my site that I want to keep search engines away from, so I disallowed them in my robots.txt file like this: User-Agent: * Disallow: /email  Yet I recently noticed that Google still sometimes returns links to those pages in their search results.  Why does this happen, and how can I stop it? Background: Several years ago, I made a simple web site for a club a relative of mine was involved in.  They wanted to have e-mail links on their pages, so, to try and keep those e-mail addresses from ending up on too many spam lists, instead of using direct mailto: links I made those links point to a simple redirector /  script running on my own site.  This script would return either a 301 redirect to the actual mailto: URL, or, if it detected a suspicious access pattern, a page containing lots of random fake e-mail addresses and links to more such pages.  To keep legitimate search bots away from the trap, I set up the robots.txt rule shown above, disallowing the entire space of both legit redirector links and trap pages. Just recently, however, one of the people in the club searched Google for their own name and was quite surprised when one of the results on the first page was a link to the redirector script, with a title consisting of their e-mail address followed by my name.  Of course, they immediately e-mailed me and wanted to know how to get their address out of Google's index.  I was quite surprised too, since I had no idea that Google would index such URLs at all, seemingly in violation of my robots.txt rule. I did manage to submit a removal request to Google, and it seems to have worked, but I'd like to know why and how Google is circumventing my robots.txt like that and how to make sure that none of the disallowed pages will show up in their search results. Ps. I actually found out a possible explanation and solution, which I'll post below, while preparing this question, but  in case someone else might have the same problem.  Please do feel free to post your own answers.  I'd also be interested in knowing if other search engines do this too, and whether the same solutions work for them also. "What kind of a person" vs. "what kind of person" I often hear people saying what kind of [singular noun] rather than what kind of a [singular noun].  Are we not supposed to use an article (a) before noun? ${!FOO} and zsh ${!FOO} performs a double substitution in bash, meaning it takes the (string) value of FOO and uses it as a variable name. zsh doesn’t support this feature.  Is there a way to make this work the same in bash and zsh?  Background:  I’ve got a list of environment variables, like  PATH MAIL EDITOR   and want to first print the variable names and afterwards their values.  This works in bash but not zsh:    for VAR in LIST do         echo $VAR         echo ${!VAR} done   It should be somehow possible “the old way” with eval, but I can’t get it to work:  for VAR in LIST do         echo $VAR         echo `eval \$$VAR` done   I’m never going to understand why I can’t simply do arbitrary deep substitutions like ${${VAR}} or even ${${${VAR}}} if need be, so an explanation for that would be nice, too. Calculating the limit $\lim((n!)^{1/n})$ Find $\lim_{n\to\infty} ((n!)^{1/n})$. The question seemed rather simple at first, and then I realized I was not sure how to properly deal with this at all. My attempt: take the logarithm,  $$\lim_{n\to\infty} \ln((n!)^{1/n}) =  \lim_{n\to\infty} (1/n)\ln(n!) =  \lim_{n\to\infty} (\ln(n!)/n)$$ Applying L'hopital's rule:  $$\lim_{n\to\infty} [n! (-\gamma + \sum(1/k))]/n! = \lim_{n\to\infty} (-\gamma + \sum(1/k))=  \lim_{n\to\infty} (-(\lim(\sum(1/k) - \ln(n)) + \sum(1/k))  = \lim_{n\to\infty} (\ln(n) + \sum(1/k)-\sum(1/k)  = \lim_{n\to\infty} (\ln(n))$$ I proceeded to expand the $\ln(n)$ out into Maclaurin form $$\lim_{n\to\infty} (n + (n^2/2)+...) = \infty$$ Since I $\ln$'ed in the beginning, I proceeded to e the infinity  $$= e^\infty  = \infty$$  So am I write in how I approached this or am I just not on the right track? I know it diverges, I was just wanted to try my best to explicitly show it. Callout Mock Responses & multiple HTTP Status Code values I'm trying to unit test the behavior of some code which performs callouts and its logic is dependent on the HTTP status code of the response - 200 vs 401.  The pattern being exercised is familiar:   Make callout Response is (unauthorized) HTTP 401 Logic checks Status value on the response if (status == 200) { /* yay! */ } if (status == 401) { /* make request for auth /* }   retry original callout with newly acquired authorization yay!    I am attempting to test the code's "try, auth, retry" behavior using the MultiStaticResourceCalloutMock but it appears that only a single status code can be set for all requests that the mock provider can respond to using the .setStatusCode(int) method. Is this accurate?  Is there a mechanism for using a single MultiStaticResourceCalloutMock and having it respond with different status codes per URL in the test context?  Is the only solution here to use a Dispatcher such as what Andrew Fawcett documents here in this answer?   Is there another method/mechanism that allows for different status codes to be returned by the mock provider on the HTTPResponse?    Some  showing the one-time usage pattern of the .setStatusCode(200) method and the multimock class.  @isTest private class CalloutMultiStaticClassTest {     @isTest static void testCalloutWithMultipleStaticResources() {         // Use MultiStaticResourceCalloutMock to         // specify fake response for a certain endpoint and          // include response body in a static resource.             MultiStaticResourceCalloutMock multimock = new MultiStaticResourceCalloutMock();         multimock.setStaticResource(             'http://api.salesforce.com/foo/bar', 'mockResponse');         multimock.setStaticResource(             'http://api.salesforce.com/foo/sfdc', 'mockResponse2');         multimock.setStatusCode(200);         multimock.setHeader('Content-Type', 'application/json');          // Set the mock callout mode         Test.setMock(HttpCalloutMock.class, multimock);          // Call the method for the first endpoint         HTTPResponse res = CalloutMultiStaticClass.getInfoFromExternalService(             'http://api.salesforce.com/foo/bar');         // Verify response received          System.assertEquals('{"hah":"fooled you"}', res.getBody());          // Call the method for the second endpoint         HTTPResponse res2 = CalloutMultiStaticClass.getInfoFromExternalService(             'http://api.salesforce.com/foo/sfdc');         // Verify response received          System.assertEquals('{"hah":"fooled you twice"}', res2.getBody());     } } Where should collision detection logic be placed? I am developing a small 2D game engine. The characters have a paint method which currently does the following:   Calculate the new position of the character as per its speed, etc. Update the collision grid cell ** Draw the character at the new position   ** I have created a collision grid to reduce the number of intersection checks  Now the basic algorithm which I have thought for detecting collision is:   For Each Character     Check intersection with characters in surrounding 8 cells   I can simply place this code in the paint method. But here is the problem I anticipate.  Suppose two characters A and B lie in adjacent cells in the collision grid. Now as per the above algorithm in the iteration of character A it will detect that it has collided with B. In the iteration for character B it will detect that it has collided with A.  But I have an idea that when A detects that it has collided with B, it should inform B that it has collided with A. This would save lot of comparisons when there are more than 2 actors colliding. But I am not sure how to handle this. I think instead of every character checking for its collision, I should check for the collision inside the game loop.   Would this approach be correct? How have you handled this kind of problem? I thought of the collision grid thing myself. Are there any alternatives to collision grid logic?