If a contrib module's info.yml contains core: 8.x and not core_version_requirement: ^8.8 || ^9, how can a site be upgraded to Drupal 9 using composer? composer update downloads modules in &quot;require&quot;: { which needs to be before &quot;patches&quot;: { because there needs to be a module before it can be patched. Assume the patch patches the info.yml file with core_version_requirement: ^8.8 || ^9. So composer update can't resolve to an installable set of packages and throws errors like this: - Root composer.json requires drupal/entity_jump_menu 1.1 -&gt; satisfiable by drupal/entity_jump_menu[1.1.0]. - drupal/entity_jump_menu 1.1.0 requires drupal/core ~8.0 -&gt; found drupal/core[8.0.0-beta6, ..., 8.9.x-dev] but it conflicts with your root composer.json require (^9.1).  It looked like the answer was changing drupal/core and drupal/core-dev to a value of &quot;^8.8||^9.1&quot;. composer update will now run, except the site is still building in version 8.9.13.  Following the suggestions on the  blog post, I changed composer.json to the following   &quot;repositories&quot;: [       {           &quot;type&quot;: &quot;composer&quot;,           &quot;url&quot;: &quot;https://packages.drupal.org/8&quot;,           &quot;exclude&quot;: [&quot;drupal/entity_jump_menu&quot;]       },       {           &quot;type&quot;: &quot;git&quot;,           &quot;url&quot;: &quot;git@git.drupal.org:issue/entity_jump_menu-3130218.git&quot;       }   ],   &quot;require&quot;: {       ...       &quot;drupal/entity_jump_menu&quot;: &quot;dev-3130218-drupal-9-deprecated&quot;,       ...      },     Which downloads the fork of the module, into vendor/drupal/entity_jump_menu. Unfortunately Drupal and/or Drush don't see the module in that directory. I'm going to try the  solution next.  The solution from  adds the fork of the module to the modules/contrib directory. Here's the code. &quot;repositories&quot;: [     {         &quot;type&quot;: &quot;package&quot;,         &quot;package&quot;: {             &quot;name&quot;: &quot;drupal/entity_jump_menu&quot;,             &quot;type&quot;: &quot;drupal-module&quot;,             &quot;version&quot;: &quot;dev-3130218-drupal-9-deprecated&quot;,             &quot;source&quot;: {                 &quot;type&quot;: &quot;git&quot;,                 &quot;url&quot;: &quot;git@git.drupal.org:issue/entity_jump_menu-3130218.git&quot;,                 &quot;reference&quot;: &quot;268adc649cfb4ad716751a8643e60f5df4dbb80d&quot;             }         }     },     {         &quot;type&quot;: &quot;composer&quot;,         &quot;url&quot;: &quot;https://packages.drupal.org/8&quot;     } ], &quot;require&quot;: {     ...     &quot;drupal/entity_jump_menu&quot;: &quot;dev-3130218-drupal-9-deprecated&quot;,     ... },  Thanks for pointing me in the right direction.
Given I have Drupal 9 composer installation with composer-patches plugin and given a contrib module with a stable v8 release, but no v9 release (not even dev branch) and given that contrib module has a working v9 patch in the issue queue  is there any method to install that module + patch in composer? Even if I manually add both, the package and patch, to my composer.json, I still can't require or update this module with composer due conflicting versions. I really want to avoid duplicating /contrib code into my project's /custom codebase. My current workaround is:  forking that module to my own, private git repo applying patch there creating a new composer.json in my private git, and changing the package vendor to my custom_private_vendor adding my private git as VCS repo in the D9 project's composer.json and then composer require custom_private_vendor/contrib_module  This fulfills my goal of not duplicating the contrib module in my project's custom codebase, but every time I do this I feel the urge to wash my dirty hands. Is there something more elegant like composer require drupal/contrib_module --apply-patch-first or can I somehow target drupal.org's git with a specific patch included?
I came across a list of functions in my calculus textbook whose indefinite integral cannot be found. It was written that the integral $$\int \frac{\sin(x)}{x} dx$$ cannot be evaluated without any explanation as to why.  I did some research over the internet and found out that the definite integral $$\int_{0}^{\infty} \frac{\sin(x)}{x} dx$$ can be evaluated using Laplace transformation and is equal to $\pi /2$. But I still couldn't find answer to my original question. I read somewhere that the integral cannot be expressed using 'elementary functions'. A little help is appreciated, I'm in Calc 1 going advanced than my course but I am sorry if my post shows lack of research. Thank you! Need to print out iPhone text messages I have a court case and have text messages on my iPhone I need to transfer to my iPad email for evidence.  Every time I try it doesn't work, it will send them to my iMessage but not my email.  I need to print them out How do I do this? Let $a,b,c$ are non-negative numbers, such that $a+b+c = 3$.  Prove that $\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca$    Here's my idea:  $\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca$  $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(ab + bc + ca)$  $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) - 2(ab + bc + ca) \ge 0$  $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) - ((a+b+c)^2 - (a^2 + b^2 + c^2) \ge 0$  $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) - (a+b+c)^2 \ge 0$  $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge (a+b+c)^2$  $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3^2 = 9$  And I'm stuck here.  I need to prove that:   $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge (a+b+c)^2$ or  $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3(a+b+c)$, because $a+b+c = 3$  In the first case using Cauchy-Schwarz Inequality I prove that:  $(a^2 + b^2 + c^2)(1+1+1) \ge (a+b+c)^2$  $3(a^2 + b^2 + c^2) \ge (a+b+c)^2$  Now I need to prove that:   $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3(a^2 + b^2 + c^2)$   $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(a^2 + b^2 + c^2)$   $\sqrt{a} + \sqrt{b} + \sqrt{c} \ge a^2 + b^2 + c^2$   I need I don't know how to continue.  In the second case I tried proving:  $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(a+b+c)$ and  $a^2 + b^2 + c^2 \ge a+b+c$  Using Cauchy-Schwarz Inequality I proved:  $(a^2 + b^2 + c^2)(1+1+1) \ge (a+b+c)^2$  $(a^2 + b^2 + c^2)(a+b+c) \ge (a+b+c)^2$  $a^2 + b^2 + c^2 \ge a+b+c$  But I can't find a way to prove that $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(a+b+c)$  So please help me with this problem.  P.S   My initial idea, which is proving:  $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3^2 = 9$  maybe isn't the right way to prove this inequality. I have a question concerning the time ordering operator. Let's suppose we have a time evolution generated by some Hamiltonian $H(t)$ given by $$ U(t)=T_\leftarrow\exp\left(-\mathrm{i}\int_0^t\mathrm{d}s\,H(s)\right)\tag{1}. $$ In Breuer and Petruccione, it's said that if the commutator of the Hamiltonian at some time $t$ with itself at some other time $t^\prime$ is a $c$-number function/ a complex function, i.e. $$\left[H(t),\,H(t^\prime)\right]=f(t,t'),$$ then the time evolution is given by $$ U(t)=\exp\left(-\frac{1}{2}\int_0^t\mathrm{d}s\int_0^t\mathrm{d}s^\prime\,\left[H(s),\,H(s^\prime)\right]\Theta(s-s^\prime)\right)$$ $$\times \exp\left(-\mathrm{i}\int_0^t\mathrm{ds}\,H(s)\right),\tag{2} $$ where $\Theta(s-s^\prime)$ is the Heaviside function. No proof or reference is given and I could not find any explanation anywhere, this is why I ask here. Any help would be very appreciated. DEFINE vs Variable in PHP Are there any age limits (formal, informal, or guidelines) that schools use when deciding to accept someone into a PhD program? I'm most curious about the upper age limits. For example, will most schools accept someone in their 40's? How about their 50's? Do gravitational lenses work on gravitational waves? Do gravitational lenses work on gravitational waves? Could we get an Einstein cross of gravity? Rounded caps on a Bezier Curve pipe This may be a simple question.  Essentially I am procedurally generating random 3D figures using a bezier curve and bezier circle bevel object, however, one of the only things I can't figure out how to do is how to round the end caps of the object.   I have tried using a taper object like suggested  but have had only limited success.  This is about as close as I've gotten:   I should note that all of this is done through Python and I'm not modeling any of this by hand. So any suggestions for how I can make the rounded caps programmatically would be greatly appreciated. how to add object to a keyframe range? lets assume we have an iron man pointing his hand to a target and shooting few frames later as shown in the picture below:    frame 0 to 10 he will point frame 11 to 15 he will shoot, which is just a long cube with one larger face to give idea or a laser projectile   how can I add the projectile cube only on frame 11? I don't what it to exist before that  thanks How to append to a file with fstream fstream::app flag seems not to work How do I create a line graph which ignores zero values? I want to create a graph showing trends over time, calculated from a formula of various data.  However, some data is missing, resulting in zeroes in the calculated data which the chart uses.  When plotting this data against as a line graph, it's drawing a line down to the axis, which I don't want.  I'd like either a gap in the line, or skip the data point and draw from directly previous to next.  i.e. given this data:  00.0    61.6    39.5    74.6    86.8    72.0    66.5    99.5    85.6    80.6    95.0    122.4 81.0    83.1    88.2    00.0    94.3    60.5    00.0    99.5    95.8    86.2    00.0    111.4 84.1    110.8   78.9    62.0    75.9    102.2   89.9    87.1    81.6    73.5    00.0    116.9   All the 00.0 items mean "no data", so I'd like them ignored (not treated as 0 value).   This is the chart I currently have:   This is how I'd like the chart to come out:   Again, I don't mind if the dotted lines are blank or solid, so long as they don't go down to zero.  The data is dynamic/calculated, so I can't just say =AVERAGE(prev,next) since I don't know where missing data will occur in advance - I need a solution which takes this into account.  Any ideas? How to get over "device or resource busy"? I tried to rm -rf a folder, and got "device or resource busy".  In Windows, I would have used LockHunter to resolve this. What's the linux equivalent? (Please give as answer a simple "unlock this" method, and not complete articles like . Although they're useful, I'm currently interested in just ASimpleMethodThatWorks™) Is a transit visa required to pass through Bangkok Airport? Given an instance of System.Reflection.Assembly. Theorem  Let $\{a_n\}$ and $\{b_n\}$ be convergent real sequences. Assume that there exists a $N\in\mathbb{N}$ so  $a_n\le b_n$ (eq. 1)  for all $n\ge N$. Then  $\lim_{n\to\infty}a_n\le \lim_{n\to\infty}b_n$.  My attempt at proving  Let $a=\lim_{n\to\infty}a_n$ and $b=\lim_{n\to\infty}b_n$. $a$ and $b$ are real numbers because the limit of a sequence of real numbers is a real number (which is a corollary in my textbook). For proof by contradiction, assume  $b&lt;a$.  Let $\epsilon=\frac{a-b}{2}$. Then we can find $N_a,N_b\in\mathbb{N}$ so  $|a-a_n|\le\frac{\epsilon}{2}$ when $n\ge N_a$, (eq. 2)  $|b-b_n|\le\frac{\epsilon}{2}$ when $n\ge N_b$. (eq. 3)  In a textbook I found that I should do the following:  Choose a $n$ so $n\ge\max(N,N_a,N_b$), then eqs. (1-3) are satisfied. Then  $a-\epsilon\le a_n+\frac{\epsilon}{2}-\epsilon$  How do I get that last thing? Is it necessary to do a proof by contradiction or can one do a direct proof? Do I have to care about bad boxes? When I compile my PDF from LaTeX source, do I have to care about all the messages (currently 28) about 'bad boxes'?  The PDF seems fine to me. Generic Parse Method without Boxing I am trying to write a generic Parse method that converts and returns a strongly typed value from a NamedValueCollection.  I tried two methods but both of these methods are going through boxing and unboxing to get the value.  Does anyone know a way to avoid the boxing?  If you saw this in production would you not like it, how bad is it for performance?  Usuage:  var id = Request.QueryString.Parse&lt;int&gt;("id");   Attempt #1:  public static T Parse&lt;T&gt;(this NameValueCollection col, string key) {     string value = col[key];      if (string.IsNullOrEmpty(value))         return default(T);      if (typeof(T) == typeof(int))     {         //return int.Parse(value); // cannot convert int to T         //return (T)int.Parse(value); // cannot convert int to T         return (T)(object)int.Parse(value); // works but boxes     }     if (typeof(T) == typeof(long))     {         return (T)(object)long.Parse(value); // works but boxes     }     ...      return default(T); }   Attempt #2 (using reflection):  public static T Parse&lt;T&gt;(this NameValueCollection col, string key) {     string value = col[key];      if (string.IsNullOrEmpty(value))         return default(T);      try     {         var parseMethod = typeof(T).GetMethod("Parse", new Type[] { typeof(string) });          if (parseMethod == null)             return default(T);          // still boxing because invoke returns an object         var parsedVal = parseMethod.Invoke(null, new object[] { value });         return (T)parsedVal;     }     // No Proper Parse Method found     catch(AmbiguousMatchException)      {     }      return default(T); } Does $\sum_{n=3}^\infty \frac {1}{(\log n)^{\log(\log(n)}}$ converge? Does the following series converge:  $$\sum_{n=3}^\infty \frac {1}{(\log n)^{\log(\log(n)}}$$  I've tried all test I know... Any ideas ? preventDefault when clicking bounty "cancel" link When you click "start a bounty" then "cancel," href="#" exiles you to the top of the page and you lose your spot near the end of the question's comments. Clicking anywhere else on the page simply closes the bounty popup; clicking "cancel" ought to do the same.  In case that wasn't clear:    Dramatization aside, it's a nuisance. Can the click event on that anchor be preventDefaulted? "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?