I have an USB mouse and keyboard, both of which have a cable 1.8m long. However for my purposes this is too short.  I read that USB standard limits the length of a cable to 5 meters for high speed devices and 3 meters for low speed devices. I guess that both mouse and keyboard belong into the low speed ones.  I measured that I need 1.2 more meters. However there are no 1.2m USB cables available in the shops I have access to, only 1m or 1.8m.  When I buy the 1.8m USB cable, will it work? Or do I have to look for active USB extenders with own power supply?    And What about the standard USB hubs? What if I connect both mouse and keyboard to an hub and that one will be connected to the extending USB cable, will it extend the range at least a bit?
I'm planning to use USB port for an application(from a device to a PC) but the distance between the device and the PC will be around 15m. The recommended maximum USB cable length is 5m.  How can I buffer the data without loosing any information, safely? Would a USB hub work?
As we all know, going near a black hole (Gargantua in this case) slows down time, according to the theory of relativity.  So when Cooper comes out of the black hole, after spending a fair bit of time there, shouldn't he be much older than Brand who hasn't gone inside Gargantua? I have a normal door, and left to its own devices it will simply rest about 20-30 degrees. It won't swing shut on its own, nor stay open on its own. It hangs a bit in the middle. Convenient enough to catch you elbow when you're not looking.  I'd like the door to stay put wherever we leave it.  I've heard idea such as slipping cardboard in to the hinge to rebalance the door, etc. but no guidelines as to the thickness of that cardboard, or which hinge to change to move it in which direction.  I would hope you can appreciate how I'd rather not make this a task of trial and error and continually removing the hinge and putting it back on, etc.  I've also heard something about lightly bending one of the hinge pins, but not sure how I should go about that, or "how much" to bend it.  So, I'm looking for some tips on how to fix this (and other) door. US Election results 2016: What went wrong with prediction models? First it was , now the US election. Many model predictions were off by a wide margin, and are there lessons to be learned here? As late as 4 pm PST yesterday, the betting markets were still favoring Hillary 4 to 1.  I take it that the betting markets, with real money on the line, should act as an ensemble of all the available prediction models out there. So it's not far-fetched to say these models didn't do a very good job.  I saw one explanation was voters were unwilling to identify themselves as Trump supporters. How could a model incorporate effects like that?  One macro explanation I read is the rise of . The question then is how could a statistical model capture a macro trend like that?  Are these prediction models out there putting too much weight on data from polls and sentiment, not enough from where the country is standing in a 100 year view? I am quoting a friend's comments. Problem: If   $$C_0+\frac{C_1}{2}+\cdots + \frac{C_{n-1}}{n}+\frac{C_n}{n+1} =0,$$  where $C_0,...,C_n$ are real constants, prove that the equation  $$C_0+C_1x+\cdots +C_{n-1}x^{n-1}+C_nx^n=0$$  has at least one real root between $0$ and $1$.  Source: W. Rudin, Principles of Mathematical Analysis, Chapter 5, exercise 4. Proof that every metric space is normal Ability to give away iOS application, usually for sale I am wondering if it's possible to do "mass giveaways" on the AppStore, without lowering, changing or otherwise changing the listed application.  Is it possible to distribute hundreds to, potentially, thousands of promo codes, legally, while still leaving the "normal" distribution channel of purchasing the app open?  An example of the above would be selling physical items in a store which have a bundled digital component available. Essentially you buy a product in a store, which comes "bundles with a free app" (BB and Android have already been dealt with, iOS is unclear). It would be nice to give the option to have a promo code in the box, that the user can then use to get the application for free. Is this possible? How to sort a list of objects based on an attribute of the objects? I have an app that's been out for a couple of weeks. Its not exactly selling like hot cakes yet, but I'm pretty sure I've gotten a rating or a review by now. But nothing shows when viewing either in iTunes or directly on the AppStore on the iPhone.  Is there a standard delay as to when ratings/reviews show up, that Apple imposes, especially for new apps? Or do you think there is some problem with Apple's system? Story about building up speed leading to increased quality of life In , one learns two definitions of free groups associating with sets.        Let $A$ be a set, the free group of $A$, $F(A)$ is the initial object in the category $\mathcal{C}$, where \begin{equation} \operatorname{Obj}(\mathcal{C})=\{A\xrightarrow{g}G\}, \end{equation}where the codomain $G$ are groups, and \begin{equation} \operatorname{Hom}(\mathcal{C})=\{\text{Commutative Diagrams } A\xrightarrow{g_1}G_1\xrightarrow{\phi}G_2\xleftarrow{g_2}A\}  \end{equation} where $\phi$ are group homomorphisms.   Also $F(A)$ has the concrete construction with elements being non-redundant words with alphabet $A$, and the group multiplication being juxtaposition and reduction.      But what good are free groups? Why are they useful?  says they are useful in topology, but does not explain why explicity.   Can someone give some examples that an undergraduate can understand? I am asking about examples that can show the usefulness of this abstract construction. So I guess the identification of $\mathbb{Z}=F({a})$ does not really count. Help with Spivak Calculus Ch3 Problem 6a Yet again I find myself stuck on a Spivak question. This time it is simply the question that isn't clear to me. It states:  If $x_1, ..., x_n$ are distinct numbers, find a polynomial function $f_i$, of degree $n-1$ which is 1 at $x_i$ and 0 at $x_j$ for $j\neq i$. Hint: the product of all $(x-x_j)$ for $j\neq i$, is 0 at $x_j$ if $j\neq i$. This product is usually denoted by: $$\prod_{\begin{smallmatrix}{j=1}\\ {j\neq i} \end{smallmatrix}}^n(x-x_j)$$  To be honest I don't know where to begin. I understand how to create a polynomial equation for a given set of results but this is quite strange. I'm not sure exactly what the author is expecting.  The answer book shows:  $$f_i(x)={\prod_{\begin{smallmatrix}{j=1}\\ {j\neq i} \end{smallmatrix}}^n(x-x_j)}/{\prod_{\begin{smallmatrix}{j=1}\\ {j\neq i} \end{smallmatrix}}^n(x_i-x_j)}$$  Which I have expanded out just fine but don't know what to do with. If $f:[a,b]\to \mathbb{R}$ is continuous and nonnegative and $\int_a^b{f}=0$, then $f(x)=0$ for all $x\in [a,b]$ Suppose that a function $f:[a,b]\to\mathbb{R}$ is continuous and nonnegative. Prove that if $\int_a^b{f}=0$, then $f(x)=0$ for all $x\in [a,b]$.   I've been trying to prove it using the extreme value theorem, continuity and the upper and lower sums but can't come up with something tight enough. More specifically, I've been trying to relate $|f(x)-f(t)|&lt;\epsilon$ to the sum of $M_k-m_k(x_k-x_{k-1})$ Best practice: card expiry date on checkout Is abstract algebra (mostly?) restricted to $2$-ary operators? Session ID from scheduled jobs Here's the scenario.  I have a remote application which is invoked from an HTTP callout.  In this callout, I am passing the session ID.  The application can then do its stuff without needing a username and password.  However, I want to be able to invoke the application from a scheduled job, and getsessionid() returns null in this case.  Is there any way around this?  I am trying very hard to avoid storing sfdc usernames and passwords on my remote server, and I want to avoid the hassle of using oAuth with all the complications of client secrets, etc. What does increased caster level affect? How to get rid of update notifications for a given app in Google Play Store? General question I have an app X installed on my device in version Y. I don't want to update it to any newer version (Y+z) for some reason. Still, the Google Play Store app always lists updates for X, which keeps me from a simple &quot;update all&quot;. My question, put to a single sentence:  How to hide updates for X permanently from the &quot;available updates&quot; list?   Specific examples First, I know there are alternatives as e.g. the , which can handle this quite well (skip this update, skip all future updates). I prefer a &quot;generic solution&quot;, which does not depend on a 3rd party's market app. Second, I know that  includes a &quot;Market doctor&quot;, which can be used to disconnect an app from Google Play.1 However, in my specific case that only worked for a few hours: I disconnected GTalk, as I don't want it replaced by Hangouts. The update notification disappeared for a few hours, then it was back. No idea what restored the &quot;link&quot;. Third, just to make sure: I'm not talking about the &quot;auto update&quot; feature as discussed e.g. in . I was rather hoping for something close to , but without the need to &quot;freeze&quot; the app in question (which I still want to use, but in the version installed, not in a &quot;future version&quot; available). Fourth: No, uninstalling the app in question is no option here, though it would perfectly hide any future updates from being displayed :) Nor is &quot;freezing&quot;, as the app is used regularly. Fifth: There is a related question for my GTalk example:  However, I want the opposite: No update.  It might very well be that &quot;Google Apps&quot; (and other pre-installed apps) are handled in some specific way, and restore their &quot;market links&quot; themselves somehow. For this case, my question includes: How to prevent that from happening?  Update There's a  discussing the very same issue. Until recently, a work-around for &quot;user installed apps&quot; was to uninstall them and then sideload the .apk – but even that stopped working. Well, re-signing the .apk with a different key might work for user-apps. But both would fail for pre-installed apps (can't sideload them to override; different signature would refuse that install completely). Moreover, there's an  on this by some &quot;joel.bou..&quot;. Not hard to guess, though blanked out: that would be Joël Bourquard. If it's really him, then that's really an issue (for those who don't know: Joël is the dev behind Titanium Backup). Let me quote from the issue:  For the record: in the Android Market days it was possible to &quot;Attach&quot; or &quot;Detach&quot; apps (on rooted devices) with a system tool such as Titanium Backup. But nowadays (with the Google Play Store app) this has become nearly impossible to do, because the complete app list seems to sync from Google's servers and the user has no control over that.  Which explains why detaching with apps like Titanium Backup is no longer permanent. And I'm afraid I've asked the impossible. But I thought the same in the past with other questions, and surprisingly got solutions here. So I do not yet give up hope on this one!  1: what that does is explained by eldarerathis  (thanks to  for digging that up!) Does 'A countable infinite Cartesian product of countable sets is not necessarily countable' violate induction principle? We have two theorems about Cartesian product:   A finite Cartesian product of countable sets is countable; A countable infinite Cartesian product of countable sets is not necessarily countable.   To prove the first theorem, we can show   Firstly, we can prove $A\times B$ is countable,  Secondly, suppose it is true for $n-1$, i.e. $X=A_1\times A_2\times \ldots \times A_{n-1}$ is countable, we prove the case for $n$ by the fact $A_1\times A_2\times \ldots \times A_{n-1}\times A_{n}=X\times A_{n}$ is in fact the product of two countable sets, in above we know it is still countable.   The above method to prove the first theorem is exactly the mathematical induction, which should show the theorem is true for all $n\in\mathbb{Z}_{+}$, so it should works for countable infinite unions.  However, we may use Cantor diagonal method to prove at least $\{0,1\}^w$ is not countable, thus the second theorem is also right. Does this means that the mathematical induction is wrong?    EDIT: After the following discussions, I have the answer:  The key point is that any element $n\in\mathbb{Z}_+$ is always a finite number, thus induction principle only guarantee the theorem is true for any $n\in\mathbb{Z}_+$, i.e. for any finite number.  When discuss a countable infinite union, it is infinite number of unions, this infinite number no more belong to $\mathbb{Z}_+$, thus can not be guaranteed by induction principle. 2D Integration With Quadratic Arg. of Delta Function I think it would be very useful.  I have a question that I have an open bounty on. Putting up said bounty brought my reputation below the point where I am allowed to comment on other peoples' posts. It's so frustrating not to be able to comment and give feedback on the answers given!  Especially since it would be helpful, for them, me, and the readers, to know particularly what is good and isn't good about their answer, and therefore give some better guidelines to future answerers on how they can win the bounty.