Hello can anyone please help me with the following, I try to find a delphi routine which allows me to find which other program is holding a comport. (A comport I want to open is occupied by another application which application does that).  I am aware of the tool to find it but I would like to do that programmatically in my application, please can someone give me a hint? Thanks
The company I work for makes hardware that communicates to the computer though a serial port. Third party companies write software that communicates with our hardware.   There are times when I need to diagnose our hardware. However, a third party software app connects to the serial port when Windows starts up, blocking any other connection. I don't know the name of this application/service and it's not always the same one.  Is there any way to either:   Find the name/pid of the app/service that is currently using a given serial port or Steal the serial port connection from another app.   vb.net preferably, but I'll take a language agnostic answer as well.
Example of a finitely generated module with submodules that are not finitely generated What's the difference between using black and white mode in-camera, and converting in post? For black and white photography, will there be any quality difference between using the built-in black and white mode, or shooting in color initially and then removing the color later using Photoshop (or similar application)?  I have a high-end compact camera, the . How can I make my ad hoc iPhone application's icon show up in iTunes? I've got an iPhone app with icon file Icon.png.  This icon shows up properly when the app is on the phone itself, but it doesn't show up in the applications pane in iTunes.  What do I need to do to get it to show up properly? Is there a single word which means "a person, and all of his ancestors", or "a person, and all of his descendants"? There are a lot of words describing ancestral relationships between people, such as:  parent grandparent great-grandparent any of the above can be described as an "ancestor".  child grandchild great-grandchild any of the above can be described as a "descendant".   I am wondering if there are single words (nouns) which describe any of the following concepts:  A person and all of his/her ancestors A person or any of his/her ancestors A person and all of his/her descendants A person or any of his/her descendants   Any ideas? Strong Induction Proof: Fibonacci number even if and only if 3 divides index The Fibonacci sequence is defined recursively by $F_1 = 1, F_2 = 1, \; \&amp; \; F_n = F_{n−1} + F_{n−2} \; \text{ for } n ≥ 3.$ Prove that $2 \mid F_n \iff 3 \mid n.$   Proof by Strong Induction : $\bbox[5px,border:1px solid green]{\color{green}{n = 1 }}$ $2 \mid F_1$ is false. Also, $3 \mid 1$ is false. The implication [False $\iff$ False] is vacuously true.  $\bbox[5px,border:1px solid green]{\color{green}{\text{Induction Hypothesis}}}$ Assume that $2 \mid F_i \iff 3 \mid n$ for every integer $i$ with $1 ≤ i ≤ k$.  $\bbox[5px,border:1px solid green]{\color{green}{k + 1 \text{th Case}}} \;$  To prove:  $\quad 2 \mid F_{k + 1} \iff 3 \mid k + 1.$   $\bbox[5px,border:1px solid green]{\color{green}{n = k + 1 = 2}} \;$ $2 \mid F_2$ is false. Also, $3 \mid 2$ is false. So [False $\iff$ False] is vacuously true.  Hence assume that $k + 1 ≥ 3.$ We now consider three cases:  $\bbox[5px,border:1px solid green]{\color{green}{\text{Case 1: } k + 1 = 3q}}$ Thus $3 \require{cancel}\cancel{\mid} k$ and $3 \require{cancel}\cancel{\mid} (k − 1)$. By the ind hyp, $3 \require{cancel}\cancel{\mid} k    \iff F_k$ odd &amp; $3 \require{cancel}\cancel{\mid} (k − 1) \iff F_{k - 1}$ odd. Since $F_{k+1} = F_k + F_{k−1}$, thus $F_{k+1}$ = odd + odd = even.  $\bbox[5px,border:1px solid green]{\color{green}{\text{Case 2: } k + 1 = 3q + 1}}$  Thus $3 | k$ and $3 \require{cancel}\cancel{\mid} (k − 1).$ By the ind hyp, $3 | k \iff F_k$ even &amp; $3 \require{cancel}\cancel{\mid} (k − 1) \iff F_{k - 1}$ odd. Thus $F_{k+1}$ odd.  $\bbox[5px,border:1px solid green]{\color{green}{{\text{Case 3: }} k + 1 = 3q + 2}}$ Thus $3 \require{cancel}\cancel{\mid} k$ and $3 | (k −1).$ By the ind hyp, $3 \require{cancel}\cancel{\mid} k \iff F_k$ odd and $3 \mid (k − 1) \iff F_{k - 1}$ even. Thus $F_{k+1}$ odd. $\blacksquare$  $\Large{1.}$ Does the proof clinch the $(\Leftarrow)$ of the $(k + 1)$th case?   $\Large{2.}$ Since the recursion contains $n, n - 1, n - 2$, thus the recursion "time lag" is $3$ here. So shouldn't $3$ base cases be checked?   $\Large{3.}$ Further to #2, shouldn't "assume $k + 1 \geq \cancel{3} 4$" instead?   $\Large{4.}$ Shouldn't the $n = k + 1 = 2$ case precede the induction hypothesis?  I referenced . Source: Exercise 6.35, P152 of Mathematical Proofs, 2nd ed. by Chartrand et al    Supplement to peterwhy's Answer:  $\Large{1.1.}$ I wrongly believed that all 3 Cases proved the $\Leftarrow$. I now see that Case 1 is $\Leftarrow$ via a Direct Proof. Cases 2 and 3 are $\Rightarrow$ via a Proof by Contraposition. Nonetheless, how would one foreknow/prevision to start from $3 \mid n$  for both directions of the proof? Proof that any simple connected graph has at least 2 non-cut vertices. I'm trying to prove that any simple connected graph with at least $3$ vertices ($|V| \ge 3$) has at least $2$ vertices whose removal will not lead to the increment of number of components. In other words, there are at least $2$ vertices that are non-cut vertices.  Attempt at solution:   I tried to prove it using induction by the number of vertices: $n - t \ge 2$, where $n$ is the number of vertices our graph has, and $t$ is a number of cut vertices.   Base  Let $n =3$, then the graph will look like this:  $О-О-О$  Clearly, it has one cut vertex (the middle one). So, $3-1 \ge 2$ holds.  Or it can be a full graph, then $3 - 0 \ge 2$ also holds. Assumption  Let $n = k$, and the formula holds, $k - t &gt;= 2$ Step  Prove for $n = k + 1$ If we add one vertex to the graph, it could either be a cut vertex or non-cut vertex. In case it is a cut vertex, we have $$\begin{align*}&amp;(k+1) - (t+1) \ge 2\\ &amp;k + 1 - t - t\ge 2\\ &amp;k\ge 2\end{align*}$$   In case it is not a cut vertex...and I'm stuck here, because anything can happen. The number of cut vertices might increase, might decrease, or stay the same.  I'm sure there is better and shorter proof to this, and I really hope someone could help me. Looking at moving my laundry upstairs, and the right answer is to combine two adjacent closets to get the width and move one of their walls a foot or two into the kitchen to get the depth.  The kitchen floor is 10-inch-square tiles. It would be sort of nice if that continued on into the laundry closet. I do have a full bundle of spare tiles in the basement.   Question: Could/should I plan on the new wall coming in on top of the existing tiles (presumably having to drill to secure the footer without shattering them), or should I plan on removing the ones under the footer and fitting cut pieces back in afterward? Evaluate $\mathop {\lim }\limits_{x \to 0} \left( {{1 \over {{{\sin }^2}x}} - {1 \over {{x^2}}}} \right)$ I tried l'Hospital but that will require a lot (and I mean A LOT!!!) of differentiating  Is there a shortcut? $$\mathop {\lim }\limits_{x \to 0} \left( {{1 \over {{{\sin }^2}x}} - {1 \over {{x^2}}}} \right)$$  Thanks in advance How to copy all files of a certain type into a folder from command prompt? I want to copy all files of a certain type (like .docx, .pdf, etc) on my PC into a folder. How can I do that? Evans PDE Problem 5.15: Poincaré inequality for functions with large zero set I have trouble proving the following problem (Evans PDE textbook 5.10. #15). Could anyone kindly help me solving the problem? I know that I should somehow use Poincaré inequality but I still cannot solve it.   Fix $\alpha&gt;0$ and let $U=B^0(0,1)\subset \mathbb{R}^n$. Show there exists a constant $C$ depending only on $n$ and $\alpha$ such that $$ \int_U u^2 dx \le C\int_U|Du|^2 dx,  $$ provided that $u\in W^{1,2}(U)$ satisfies $|\{x\in U\ |\ u(x)=0\}|&gt;\alpha$. Render has strange artifacts I've got a problem with my rendering! I'm trying to render some objects:    But, this is how it renders:    Can somebody help me out? So I know that two particles can be entangled in a quantum way, but is it possible that more than two particles be entangled in a quantum way? Most descriptions provide with two-particles cases, so I wonder. (It's hard to think of three particles entangled in spin, or so.) Between vs. Among - "Difference [between | among] one thing and other things." Consider the following two versions of a question:1   Version 1: What is the difference in meaning between "[to be] not invited" and similar negation forms?  Version 2: What is the difference in meaning among "[to be] not invited" and similar negated forms?    Personally, I think Version 1 might subtly suggest a Star Topology (see below) with the target sentence "[to be] not invited"at the center.  And Version 2 might subtly suggest a more free-form comparison like the Mesh or Fully Connected, with the target sentence simply "one among many":   Between approx equal "Star" and among approx equal "Fully Connected"?  Don't get me wrong: I do think this is quite subtle, not always applicable, and I don't think anyone would intuit such a "network topology" meaning without further explanation.   So the main question is, are both grammatically correct?  If yes, is there a semantic difference to choose one over another, perhaps as I suggested?  Or does the plurality aspect of between and among trump other semantic and stylistic issues?    1. In my question , another user provided a very helpful edit. However, the change of the word between to among prompted this question. in questions, answers, and comments. Can we get something like this going on Stack Overflow?  I think it’d be appropriate as I at least pretty often want to write something like n^2 and would benefit greatly from prettier markup. How can I reference the script element that loaded the javascript that is currently running?  Here's the situation. I have a "master" script being loaded high in the page, first thing under the HEAD tag.  &lt;!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"  "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"&gt; &lt;html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en"&gt; &lt;head&gt; &lt;script type="text/javascript" src="scripts.js"&gt;&lt;/script&gt;   There is a script in "scripts.js" which needs to be able to do on-demand loading of other scripts. The normal method doesn't quite work for me because I need to add new scripts without referencing the HEAD tag, because the HEAD element hasn't finished rendering:  document.getElementsByTagName('head')[0].appendChild(v);   What I want to do is reference the script element that loaded the current script so that I can then append my new dynamically loaded script tags into the DOM after it.  &lt;script type="text/javascript" src="scripts.js"&gt;&lt;/script&gt; loaded by scripts.js--&gt;&lt;script type="text/javascript" src="new_script1.js"&gt;&lt;/script&gt; loaded by scripts.js --&gt;&lt;script type="text/javascript" src="new_script2.js"&gt;&lt;/script&gt; Obtain screen status using ADB What is the formula for percent of frame filled at a specific distance and focal length? Proper capitalization of commonly used acronyms and initialisms When going to any page of  with a new reputation gain, the achievements dropdown keeps showing a new badge green highlight:    And then trying to dismiss the notification by clicking on it gets rid of the green badge notification, but the +5 remains even after I've clicked away from the dropdown. Tensor products of simple modules over algebras Let $A$ and $B$ be $\mathbb{C}$-algebras. Suppose that $M$ and $N$ are respectively simple $A$ and $B$ modules.   We can regard $M$ and $N$ as $A\oplus B$-modules in natural way, namely, $AN=0$ and $BM=0$. Let $M\otimes_{\mathbb{C}} N$ be the tensor product of $A\oplus B$-modules $A$ and $B$.  ${\bf My\ Question:}$ Is $M\otimes_{\mathbb{C}} N$ still a simple $A\oplus B$-module? What conditions on $A$ and $B$ will make this result true? Thanks very much in advance!