In every formula I've seen that involves energy, I've also seen a distance element. Is there nothing that doesn't involve movement? For instance if I hold my arm out parallel to the ground with a weight in it, my arm will get tired very quickly but according to the formulas I've done no work (force * distance) since distance is 0 and I'm not even applying any force (mass * acceleration) because the mass is not accelerating.  Where's the flaw in my understanding?
I read the definition of  as  $$W ~=~ \vec{F} \cdot \vec{d}$$  $$\text{ Work  =  (Force) $\cdot$ (Distance)}.$$ If a book is there on the table, no work is done as no distance is covered. If I hold up a book in my hand and my arm is stretched, if no work is being done, where is my energy going?
Printing pointers in C I was trying to understand something with pointers, so I wrote this code:  #include &lt;stdio.h&gt;  int main(void) {     char s[] = "asd";     char **p = &amp;s;      printf("The value of s is: %p\n", s);     printf("The direction of s is: %p\n", &amp;s);      printf("The value of p is: %p\n", p);     printf("The direction of p is: %p\n", &amp;p);      printf("The direction of s[0] is: %p\n", &amp;s[0]);     printf("The direction of s[1] is: %p\n", &amp;s[1]);     printf("The direction of s[2] is: %p\n", &amp;s[2]);      return 0; }   When compiling it with gcc I get these warnings:  $ gcc main.c -o main-bin -ansi -pedantic -Wall -lm main.c: In function ‘main’: main.c:6: warning: initialization from incompatible pointer type main.c:9: warning: format ‘%p’ expects type ‘void *’, but argument 2 has type ‘char (*)[4]’ main.c:11: warning: format ‘%p’ expects type ‘void *’, but argument 2 has type ‘char **’ main.c:12: warning: format ‘%p’ expects type ‘void *’, but argument 2 has type ‘char ***’   (The flags for gcc are because I must be C89)  Why incompatible types of pointer? Isn't the name of an array a pointer to it's first element? So if s is a pointer to 'a', &amp;s must be a char **, no? And why do I get the other warnings? Do I have to cast the pointers with (void *) in order to print them?   And when running I get something like this:  $ ./main-bin The value of s is: 0xbfb7c860 The direction of s is: 0xbfb7c860 The value of p is: 0xbfb7c860 The direction of p is: 0xbfb7c85c The direction of s[0] is: 0xbfb7c860 The direction of s[1] is: 0xbfb7c861 The direction of s[2] is: 0xbfb7c862   How can the value of s and it's direction (and of course the value of p) be the same? New Beamer Version: Transparent Title Page Boxes Consider the following file:  \documentclass{beamer}  \setbeamercolor{title}{bg=red} \setbeamercolor{author}{bg=yellow} \setbeamercolor{date}{bg=} \setbeamercolor{institute}{fg=red,bg=green}  \title{TitelTitelTitel} \author{meiermueller} \institute{Institute} \date{lalala}  \begin{document}  \usebackgroundtemplate{\color{blue}\rule{20cm}{15cm}}    \begin{frame}    \titlepage  \end{frame} \end{document}   On one of my computers (using MiKTeX 2.9 and beamer 2015/01/05 3.36), the result is the following:    Note how, in particular, the date box has a transparent background  On my other computer, I installed a new MiKTeX version (2.9.6800), including beamer 2018/09/02 v3.52, and get this:    Here, the date field has a white background, and I cannot seem to get it to be transparent.   Apparently, the difference has some relation to the beamer version, or maybe some other change in the new MiKTeX version.  But what ist the reason, and how do I get transparent backgrounds? I just browsed through this wonderful book: . The irony is, I am still not able to understand the motivation for using multivariate (regression) models instead of separate univariate (regression) models. I went through stats.statexchange posts  and  that explain (a) difference between multiple and multivariate regression and (b) interpretation of multivariate regression results, but I am not able to tweak out the use of multivariate statistical models from all the information I get online about them.   My questions are:   Why do we need multivariate regression? What is the advantage of considering outcomes simultaneously rather than individually, in order to draw inferences. When to use multivariate models and when to use multiple univariate models (for multiple outcomes).  Take an example given in the  with three outcomes: locus of control, self-concept, and motivation. With respect to 1. and 2., can we compare the analysis when we do three univariate multiple regression versus one multivariate multiple regression? How to justify one over another? I haven't come across many scholarly papers that utilize multivariate statistical models. Is this because of the multivariate normality assumption, the complexity of model fitting/interpretation or any other specific reason? Collection of short stories; in one a spy attempts to escape detection on Mars's moon by moving faster than the spaceship can scan Prove : $p$(n│even number of ODD parts)=$p$(n│distinct parts ,number of ODD parts is even ) After swapping the positions of the hour and the minute hand, when will a clock still give a valid time? At 12 o'clock, the hour hand and minute hand of the clock can be swapped, and the clock still gives the same time, but at 6 o'clock, it can not be swapped. So in what cases when we swap the hour and the minute hand position does a clock still give a valid time? What is the origin of BrEng ‘bird’ meaning “young woman”? Bird:         (Brit.) a girl or young woman, esp one's girlfriend ()      According to  bird:         "maiden, young girl," c.1300, confused with   (q.v.), but felt by later writers as a figurative use of  (n.1). Modern slang meaning "young woman" is from 1915, and probably arose independently of the older word.      also :         , bird was used for ‘girl’, but this was probably owing to confusion with another similar middle English word, burde, which also meant ‘young woman’. The usage crops up from time to time in later centuries, clearly as an independent metaphorical application, but there does not really seem to be an unbroken chain of occurrences leading up to the sudden explosion in the use of bird for ‘young woman’ in the 20th century.   (www.dictionarycentral.com)      Modern usage of bird meaning girl appears to be unrelated to the old Middle Ages meaning. Etymonline dates it back precisely to 1915.    What is the modern origin of bird meaning young woman?  In what context did they start to use it at the beginning of the 20th century? Mac Sierra won't boot - SmartBattery: finished polling type 4 Ignoring invalid input features with Intersection tool in QGIS? In QGIS 2.18, when I used the Vector > Geoprocessing Tools > Intersection tool, I was able to check an option which said "Ignore NULL geometries [optional]".  Now I've downloaded QGIS 3.2 and that option no longer exists. But when I try to create the intersection between two layers I'm working on, I get a message saying that there is an invalid geometry somewhere and that I should either fix it it or "change the Processing setting to the 'Ignore invalid input features' option."   But that option doesn't exist! Or if it does exist, I just can't find it anywhere.  Does anyone know how/where to change this setting in QGIS 3.2? shell: keep trailing newlines ('\n') in command substitution I want to be able to capture the exact output of a command substitution, including the trailing new line characters.  I realise that they are stripped by default, so some manipulation may be required to keep them, and I want to keep the original exit code.  For example, given a command with a variable number of trailing newlines and exit code:  f(){ for i in $(seq "$((RANDOM % 3))"); do echo; done; return $((RANDOM % 256));} export -f f   I want to run something like:  exact_output f   And have the output be:  Output: $'\n\n' Exit: 5   I'm interested in both bash and POSIX sh. I'm currently using the below script to timestamp when an edit is made to column 7 on a sheet named "Sheet 1".  function onEdit(e) {    if ([7].indexOf(e.range.columnStart) != -1) {    e.range.offset(0, 7).setValue(new Date()).setNumberFormat("dd.MM.yyyy at HH:mm");    }   }     I have a second sheet "Sheet 2" which I need to run the same script on but against column 5.  function onEdit(e) {    if ([5].indexOf(e.range.columnStart) != -1) {    e.range.offset(0, 7).setValue(new Date()).setNumberFormat("dd.MM.yyyy at HH:mm");    }   }     I've tried .getSheetByName("Sheet1") but just cant seem to get it working correctly. Version tags reloaded I'm getting sick of version numbers in tag names (aka version tags). Look,      I just want them all to be just . They are all about JSF. I browse and answer questions about JSF. I don't want to browse in all those different tags. I don't want gold badges for both  and . I can't hammer a duplicate  question because I don't have a gold badge for that. Across years I observed that a version tag is usually only picked by the asker when a &quot;relatively new&quot; or &quot;relatively old&quot; version is being treated in the question. When the &quot;currently mainstream&quot; version is being treated,  or when the asker is &quot;more proficient&quot; in the language, then askers often just pick the versionless . While moving in time, this may cause confusion by future readers when the version isn't explicitly specified anywhere in the question. I try to maintain them by adding/removing the version tag depending on its relevance, and by adding the versionless tag when it's absent. It only feels a neverending story. I'm sure other tag caretakers also have mixed feelings on this. I propose to normalize the version from the tag name:  Ban version numbers in tag names in the model (not in the UI). Add an optional single-choice version field to the Tag entity in the model. Users with edit privileges (no suggested edits please) must be able to change versions. Users with gold tagbadges must be able to add and edit versions. I'd imagine the version edit thing to be reachable via that tag info popover (when you hover the tag). Tags themselves however should look exactly the same, the version just separated with e.g. a hyphen from the tag name.  When in place, migration of existing version tags should be piece of cake. There's apparently already a script in place which recognizes version tags, it could be reused to re-match current tags and extract the version from it into the new version field. The alternative would be to automatically add the versionless tag when a version tag is being used. But this is IMO only tag pollution.  I'm posting this on meta.se instead of meta.so because version numbers aren't only on SO a problem, but on basically all HW/SW related sites. The word "quad" generally means 4. Quadratics don't have 4 of anything. Can anyone explain where the name comes from? I cannot remember anyone writing the letter epsilon in any other way than \varepsilon in any math class; but in LaTeX \epsilon and \varepsilon are different symbols. Do any of you know why there are two different symbols? (I.e. if \epsilon is the correct way to write the letter epsilon, why aren't mathematicians using it, and when is, according to the standards today, the correct situation to use each of the symbols?) How can top principal components retain the predictive power on a dependent variable (or even lead to better predictions)? How do I use sudo to redirect output to a location I don't have permission to write to? I've been given sudo access on one of our development RedHat linux boxes, and I seem to find myself quite often needing to redirect output to a location I don't normally have write access to.  The trouble is, this contrived example doesn't work:  sudo ls -hal /root/ &gt; /root/test.out   I just receive the response:  -bash: /root/test.out: Permission denied   How can I get this to work? Extension of degree 2 with characteristic not 2 I am told that $K \leq L$ is a field extension of degree 2 and that $K$ has characteristic not 2. I am then asked to show that $\exists \: \alpha \in L$ such that $L = K(\alpha), \; \alpha^2\in K $  So far I have that:  $\exists \; \alpha_1, \alpha_2 \in L$ such that $L = K( \alpha_1, \alpha_2)$  $\Rightarrow |L:K| = |L:K(\alpha_1)| |K(\alpha):K| $ so $|K(\alpha):K| = 1$ or $2$  $|K(\alpha_1):K| = 2 \Rightarrow L = K(\alpha_1)$ and $deg(f_{\alpha_1}(t)) = 2  $ where $f_{\alpha_1}(t)$ is the minimal polynomial of $\alpha_1$ over $K$   So $f_{\alpha_1}(t) = t^2 + at + b$ for some $a,b \in K$  Also, $|K(\alpha_1):K|=1 \Rightarrow K(\alpha_1) = K$ and $L = K(\alpha_2)$ which means $degf_{\alpha_2}(t) = 2$ where $f_{\alpha_2}(t)$ is the minimal polynomial of $\alpha_2$ over $K$.   So $f_{\alpha_2}(t) = t^2 + at + b$ for some $a,b \in K$  Given that we consider the same polynomial each time, relabel to consider:   $f_{\alpha}(t) = t^2 + at+ b $ for some $a,b \in K$  We have already demonstrated that $L=K(\alpha)$ for some $\alpha \in L$ so now we much show that $\alpha^2 \in K$  $\alpha^2 \in K \Leftrightarrow a = 0_K$  I now have no idea how to show that this coefficient of $f_{\alpha}(t)$ is necessarily $0$.   I know that $f_{\alpha}(t)$ has two roots in $L$ and that because $K$ and thus $L$ has characteristic not 2, they cannot be the same root, but I don't know how to then conclude that they must sum to $0$  Any help would be greatly appreciated. Is there an upside down caret character? I have to maintain a large number of classic ASP pages, many of which have tabular data with no sort capabilities at all. Whatever order the original developer used in the database query is what you're stuck with.  I want to to tack on some basic sorting to a bunch of these pages, and I'm doing it all client side with javascript. I already have the basic script done to sort a given table on a given column in a given direction, and it works well as long as the table is limited by certain conventions we follow here.    What I want to do for the UI is just indicate sort direction with the caret character ( ^ ) and ... what?  Is there a special character that is the direct opposite of a caret?  The letter v won't quite cut it.  Alternatively, is there another character pairing I can use? Brace expansion not working in a script for i in {1..40} do     echo $i done   I got  {1..40}   and I would like to have something like  1 2 3 and so on   so I can use the variable i inside a command's parameter. Why is the inclusion of the tensor product of the duals into the dual of the tensor product not an isomorphism? Let $V$ and $W$ be vector spaces (say over the reals). There is a linear injection $V^* \otimes W^* \to (V \otimes W)^*$ which sends $\sum_i f_i \otimes g_i \in V^* \otimes W^*$ to the unique functional in $(V \otimes W)^*$ sending $v \otimes w \mapsto \sum_i f_i(v) \cdot g_i(w)$ for all $(v,w) \in V \times W$. In the finite dimensional case it is easy to see this is an isomorphism by comparing dimensions on both sides. I'm aware that this is not an isomorphism in the infinite-dimensional case (see Relation with the dual space in the  on tensor products) but I must admit that I'm not sure why. Cardinal arithmetic does not seem help here and, even if it did, I would be much happier to see an explicit example of a functional in $(V \otimes W)^*$ outside the range of this map. I'm having trouble cooking one up myself.  Added: Let me elaborate on my comment about cardinal arithmetic. Suppose that $X$ and $Y$ are infinite dimensional vector spaces over a field $k$. I'm confident that $\dim (X \otimes Y) = \dim X \cdot \dim Y$ holds. It is clear that $|X^*| = |k|^{\dim X}$ since we may identify a functional on $X$ with a function from a basis for $X$ to $k$. Also I think I've convinced myself that if $\dim X \geq |k|$ then in fact $|X| = \dim X$ ie. the cardinality and dimension of a vector space agree when the dimension is larger than the cardinality of the ground field. Consequently, assuming say that $|k| \leq \dim X \leq \dim Y$ it seems we have $$ \dim(X \otimes Y)^* = |k|^{ \dim X \cdot \dim Y} = |k|^{ \dim Y} = \dim Y^* = \dim X^* \cdot \dim Y^* = \dim (X^* \otimes Y^*)$$  which implies that in fact $(X \otimes Y)^*$ and $X^* \otimes Y^*$ are isomorphic but, rather frustratingly, the obvious map does not do the job. Did I make a mistake here? If not, is it generally true (ie without making assumptions about $|k|$) that there exists some isomorphism $(X \otimes Y)^* \to X^* \otimes Y^*$?