I have an iPhone 7plus. How can I simultaneously connect music to my headphone jack and my phone to a Bluetooth device? I am trying to play my music on my motorcycle through the stereo using the headphone jack while using my Sena Bluetooth headset to talk on the phone.
Me and my friends want to listen to music together. Two of us have Bluetooth Headphones (Beats Solo2 Wireless, Beats Studio Wireless) and one has a wired headset. Is there any way to use all three of these together?
Quotient rings of Gaussian integers I used this isomorphism today but now I'm having trouble justifying it. The norm function isn't additive so I can't come up with a ring isomorphism to prove the following:     For any $\,a+bi\in\Bbb Z[i],\,\gcd(a,b)=1$, we have a ring isomorphism   $$\Bbb Z[i]/\langle a+bi\rangle\,\cong\Bbb Z/(a^2+b^2)\Bbb Z=:\Bbb Z_{a^2+b^2}.$$   Could someone show me an isomorphism between these rings to prove this? iPhone 7 Plus takes days to turn on showing Apple logo or black screen I’ve had this issue for months but in the last couple of days it’s got very very bad.  When my phone runs out of battery it needs like 2–3 days to turn back on. At the beginning of this problem the phone would need about 1h. Every time when my phone goes off again the time to switching on goes up.  What does this all look like? Well, when my phone died I put it on the charger, then the logo appears and goes off. When it goes off, a black screen pops out with a little white line on in it, then it turns on back to Apple logo. It keeps doing this until it turns on which is 1 or 2 even now 3 days.  I tried to restore it and also update it but nothing happens, literally nothing happened. Is it possible to run Microsoft Office 2010? I am able to evaluate the limit $$\lim_{x \to 1} \frac{x + x^2 + \dots + x^n - n}{x - 1} = \frac{n(n + 1)}{2}$$ for a given $n$ using l'Hôspital's (Bernoulli's) rule.  The problem is I don't quite like the solution, as it depends on such a heavy weaponry. A limit this simple, should easily be evaluable using some clever idea. Here is a list of what I tried:   Substitute $y = x - 1$. This leads nowhere, I think. Find the Taylor polynomial. Makes no sense, it is a polynomial. Divide by major term. Dividing by $x$ got me nowhere. Find the value $f(x)$ at $x = 1$ directly. I cannot as the function is not defined at $x = 1$. Simplify the expression. I do not see how I could. Using l'Hôspital's (Bernoulli's) rule. Works, but I do not quite like it.   If somebody sees a simple way, please do let me know.    Added later: The approach proposed by Sami Ben Romdhane is universal as asmeurer pointed out. Examples of another limits that can be easily solved this way:   $\lim_{x \to 0} \frac{\sqrt[m]{1 + ax} - \sqrt[n]{1 + bx}}{x}$ where $m, n \in \mathbb{N}$ and $a, b \in \mathbb{R}$ are given, or $\lim_{x \to 0} \frac{\arctan(1 + x) - \arctan(1 - x)}{x}$.   It sems that all limits in the form $\lim_{x \to a} \frac{f(x)}{x - a}$ where $a \in \mathbb{R}$, $f(a) = 0$ and for which $\exists f'(a)$, can be evaluated this way, which is as fast as finding $f'$ and calculating $f'(a)$.  This adds a very useful tool into my calculus toolbox: Some limits can be evaluated easily using derivatives if one looks for $f(a) = 0$, without the l'Hôspital's rule. I have not seen this in widespread use; I propose we call this Sami's rule :). I decided to make sure that I know this important difference between American and British English, so I wrote what I have found out so far and I would be grateful to anyone who reads this and tells me whether I am wrong, or not.  In British English when people say  to hospital or in hospital when  talking  about somebody being there as a patient they don't use the definite article : "I had to go to hospital", "She spent two weeks in hospital". And the meaning is that somebody was there as a patient.   If then for some other reasons British English speakers  will use the definite article which will change the meaning itself, I noticed that, in American English, native speakers often use the  the and if they need to show that somebody is in church to pray,  in school as a student, in hospital as a patient, in prison as a prisoner, they use 'in' and not 'at'. Do American English speakers use 'at' like British English speakers use 'the' to give the sentences a different meaning?    Are my sentences correct? Do they show American English usage?   He is in the school. (enrolled as a student) He is at the school. (for some different reasons) He is in the hospital. (as a patient) He is at the hospital. (visiting somebody) He is in the church. (to pray) He is at the church. (for some different reasons) He is in the university. (as a student) He is at the university. (not as a student) He is in the college. (as a student) He is at the college. (Not as a student) He is in the prison. (as a prisoner) He is at the prison. (not as a prisoner) This came up in  , but I can't see good documentation or examples that explain how the feature works. List of interesting math videos / documentaries This is an offshoot of the question on . I have listed a few videos/documentaries I have seen. I would appreciate if people could add on to this list.  $1.$ Story of maths      $2.$ Dangerous Knowledge    $3.$   $4.$   $5.$ What do *args and **kwargs mean? What exactly do *args and **kwargs mean?  According to the Python documentation, from what it seems, it passes in a tuple of arguments.  def foo(hello, *args):     print hello      for each in args:         print each  if __name__ == '__main__':     foo("LOVE", ["lol", "lololol"])   This prints out:  LOVE ['lol', 'lololol']   How do you effectively use them? Python and regular expression with Unicode I need to delete some Unicode symbols from the string 'بِسْمِ اللَّهِ الرَّحْمَٰنِ الرَّحِيمِ'  I know they exist here for sure. I tried:  re.sub('([\u064B-\u0652\u06D4\u0670\u0674\u06D5-\u06ED]+)', '', 'بِسْمِ اللَّهِ الرَّحْمَٰنِ الرَّحِيمِ')   but it doesn't work. String stays the same. What am I doing wrong? Is there a way to export large multi-layered Illustrator file to Photoshop?  When I export with preserve layers and text edit-ability, the complex layers are flattened. I've reorganized my layers to have simple groupings but that's time consuming.  Is there a more effective way of doing this? Or script to automate it? My text introduces multi-quibt quantum states with the example of a state that can be &quot;factored&quot; into two (non-entangled) substates. It then goes on to suggest that it should be obvious1 that the joint state of two (non-entangled) substates should be the tensor product of the substates: that is, for example, that given a first qubit $$\left|a\right\rangle = \alpha_1\left|0\right\rangle+\alpha_2\left|1\right\rangle$$ and a second qubit $$\left|b\right\rangle = \beta_1\left|0\right\rangle+\beta_2\left|1\right\rangle$$ any non-entangled joint two-qubit state of $\left|a\right\rangle$ and $\left|b\right\rangle$ will be $$\left|a\right\rangle\otimes\left|b\right\rangle = \alpha_1 \beta_1\left|00\right\rangle+\alpha_1\beta_2\left|01\right\rangle+\alpha_2\beta_1\left|10\right\rangle+\alpha_2\beta_2\left|11\right\rangle$$ but it isn't clear to me why this should be the case. It seems to me there is some implicit understanding or interpretation of the coefficients $\alpha_i$ and $\beta_i$ that is used to arrive at this conclusion. It's clear enough why this should be true in a classical case, where the coefficients represent (where normalized, relative) abundance, so that the result follows from simple combinatorics. But what accounts for the assertion that this is true for a quantum system, in which (at least in my text, up to this point) coefficients only have this correspondence by analogy (and a perplexing analogy at that, since they can be complex and negative)? Should it be obvious that independent quantum states are composed by taking the tensor product, or is some additional observation or definition (e.g. of the nature of the coefficients of quantum states) required?  1:  (bottom of p. 18) &quot;so the state of the two qubits must be the product&quot; (emphasis added). List of lists changes reflected across sublists unexpectedly I needed to create a list of lists in Python, so I typed the following: my_list = [[1] * 4] * 3  The list looked like this: [[1, 1, 1, 1], [1, 1, 1, 1], [1, 1, 1, 1]]    Then I changed one of the innermost values: my_list[0][0] = 5  Now my list looks like this: [[5, 1, 1, 1], [5, 1, 1, 1], [5, 1, 1, 1]]    which is not what I wanted or expected. Can someone please explain what's going on, and how to get around it? I often see questions edited into this form:     Lorem ipsum dolor sit amet blah blah blah paragraph paragraph original question.            UPDATE: The Lorem disappeared and left this message: "new information".   I find this format unaesthetic (since Stack Exchange posts are supposed to be useful in the long term and update boundaries are artifacts of the creation rather than the content) and feel that new information should, whenever possible, be incorporated into the original text “as if it had always been that way” if practical, and pointing out an update should be done mostly in comments.  Is this an appropriate writing style, and if so under what circumstances? If not, should it be removed or discouraged? Story about lost colonized planet sent into orbit around "free" star I was writing test classes for my apex Rest Service that I had built for an Ipad app.  I noted some very interesting things while experimenting. Before doing anything, I just did a google search and found an interesting blog post by Jeff.    @RestResource(urlMapping='/v.9/member/*/results/*')  global with sharing class MemberRestSvc {  @HttpGet  global static ReturnClass doGet() {  String[] uriKeys = RestContext.request.requestURI.split('/'); // get the member name from the uri String memberName = uriKeys.get(uriKeys.size()-3);  // do awesome programming stuff here &amp; catch any exceptions try {    List&lt;Contact&gt; contacts = [Select Id From Contact where member_name__c = :memberName];   return new ReturnClass('true', 'Query executed successfully.', contacts);  } catch (Exception e) {   return new ReturnClass('false', e.getMessage(), null); }   }   global class ReturnClass {  global String success; global String message; global List&lt;Contact&gt; records;    global ReturnClass(String success, String message, List&lt;Contact&gt; records) {   this.success = success;   this.message = message;   this.records = records;  }  }  }   And here is the test class  @isTest private class Test_MemberRestSvc {  static { // setup test data   }  static testMethod void testDoGet() {  RestRequest req = new RestRequest();  RestResponse res = new RestResponse();  req.requestURI = 'https://cs9.salesforce.com/services/apexrest/v.9/member/me/results/today';   req.httpMethod = 'GET'; RestContext.request = req; RestContext.response = res;  MemberRestSvc.ReturnClass results = MemberRestSvc.doGet();  System.assertEquals('true', results.success); System.assertEquals(10, results.records.size()); System.assertEquals('Query executed successfully.', results.message);  }  }   I just commented a few lines and ran and it. I still achieved enough test coverage so my question is why do I need to simply instantiate the Rest request and response objects since I am directly calling the class?    @isTest   private class Test_MemberRestSvc {    static {     // setup test data     }  static testMethod void testDoGet() {  /*RestRequest req = new RestRequest();  RestResponse res = new RestResponse();  req.requestURI =    'https://cs9.salesforce.com/services/apexrest/v.9/member/me/results/today';   req.httpMethod = 'GET'; RestContext.request = req; RestContext.response = res; */Commented the code meant for REST apex intialization and merely calling the method!!  MemberRestSvc.ReturnClass results = MemberRestSvc.doGet();  System.assertEquals('true', results.success); System.assertEquals(10, results.records.size()); System.assertEquals('Query executed successfully.', results.message);  }  }   Can someone explain this behavior? Also what are the best practices for writing Test classes for apex Rest services? Is there any freely available pointcloud data?   I'm especially interested in dense clouds of small areas(5m to 100m radius) but anything else would be fine too.   I've found that these pages have very nice and easy to access point cloud data. How can I report bugs with or request features for the Stack Exchange mobile apps? Let $(a_n)_{n≥0}$ and $(b_n)_{n≥0}$ be two sequences of a nomed algebra such that $\sum{\| a_n\|}$ and $\sum{\| b_n\|}$ converge, and$ $$$\forall n, \  a_n, b_n \neq 0$$  Show that $(\forall  k \in \mathbb N^*,  \ \sum_{n=0}^{\infty}{a_n}^k = \sum_{n=0}^{\infty}{b_n}^k) \implies \exists\ \sigma \in{\mathfrak S}_{\mathbb N} \ ,\ (a_n) = (b_{\sigma(n)})$  The vector-space should be a finite-dimensional $\mathbb R$-vector-space.   I do not see how to deal with the hypothesis. If you have ideas/hints...  It would be relevant to deal with complex sequences even if it is less general. But I don't think one can generalize using real and complex cases. Installing Ubuntu on one of two hard drives I have two hard drives. One is used for Windows and one is empty. I want to install Ubuntu on the empty hard drive. The Windows hard drive is disk0 in Windows and the empty hard drive is disk1. When I install Ubuntu and it asks where I want to install it, I choose "Something Else" but I am stuck there. What do I choose for "boot loader installation"? How to render a background image in 2.8?