Let $A$ be a square matrix such that $a_{i,j}=|j-i|$, show that the determinant is $D_n=(-1)^{n-1}(n-1)2^{n-2}$   I have been struggling with this problem for a while now and I was hoping to get the slightest of nudges towards the solution. I would appreciate it if you phrase your hints in forms of questions so I can work through the idea. Please don't post a complete solution unless I asked, and thank you for your time. I wanted to obtain a recurrence relation between $D_n$ and $D_{n-1}$ but I failed. So far, I attempted to use Gaussian elimination on the first and last row and I obtained $(-1,1,\ldots,1)$ and $(1,1,\ldots,-1)$ for the first and last row respectively but I didn't know how to proceed from then on.
A friend of mine asked me to calculate the value of a perticular determinant, but after spending quite some time on it I'm unable to find a solution to it. The given determinant is:  $$         \begin{vmatrix}         0 &amp; 1 &amp; \cdots &amp;n-2&amp;n-1 \\         n-1 &amp; 0 &amp; \cdots &amp;n-3&amp;n-2 \\         \vdots &amp; \vdots &amp; \ddots &amp; \vdots &amp;\vdots\\         1 &amp; 2 &amp; \cdots &amp; n-1 &amp;0 \\         \end{vmatrix} $$  Also my friend is supposed to solve this determinant by more or less elementary methods, like elementary row operations, transpose... and no eigenvalues, nor adjoints.    First of all it seems like the value of the determinant is:  $$         \frac{1}{2n}\begin{vmatrix}         0 &amp; n &amp; \cdots &amp;n&amp;n \\         n &amp; 0 &amp; \cdots &amp;n&amp;n \\         \vdots &amp; \vdots &amp; \ddots &amp; \vdots &amp;\vdots\\         n &amp; n &amp; \cdots &amp; n &amp;0 \\         \end{vmatrix} $$  Anyway I tried adding all the rows to the last one and by taking $\frac{n(n-1)}{2}$ in front to have a row of 1s, but it didn't helped me much.  Then I tried to use the fact that:  $$         \begin{vmatrix}         A &amp; C \\         0 &amp; B \\         \end{vmatrix} = \det(A) \cdot \det(B) $$, where $A,B$ are square matrices. By subtracting the $(\frac{n}{2} + i)$-th column from the $i$-th ($i=0,1,2...$) and then adding the $i$-th row to the $(\frac{n}{2} + i)$-th row, one can obtain the wanted form, but again nothing more. Also this method pretty much fails when $n$ is odd.  My best attepmt was to notice that $A_{ij} + A_{ji} = n$ and using $\det(A^{T}) = \det(A)$, but unfortunatelly  $$\begin{vmatrix}         0 &amp; n &amp; \cdots &amp;n&amp;n \\         n &amp; 0 &amp; \cdots &amp;n&amp;n \\         \vdots &amp; \vdots &amp; \ddots &amp; \vdots &amp;\vdots\\         n &amp; n &amp; \cdots &amp; n &amp;0 \\         \end{vmatrix} \not = \begin{vmatrix}         0 &amp; 1 &amp; \cdots &amp;n-2&amp;n-1 \\         n-1 &amp; 0 &amp; \cdots &amp;n-3&amp;n-2 \\         \vdots &amp; \vdots &amp; \ddots &amp; \vdots &amp;\vdots\\         1 &amp; 2 &amp; \cdots &amp; n-1 &amp;0 \\         \end{vmatrix} + \left( \begin{vmatrix}         0 &amp; 1 &amp; \cdots &amp;n-2&amp;n-1 \\         n-1 &amp; 0 &amp; \cdots &amp;n-3&amp;n-2 \\         \vdots &amp; \vdots &amp; \ddots &amp; \vdots &amp;\vdots\\         1 &amp; 2 &amp; \cdots &amp; n-1 &amp;0 \\         \end{vmatrix}\right)^T$$ I tried to decompose the LHS by using the $n-$linearity of the determinants, by then one gets $2^n$ determinants on the RHS similar to the first one, but things are messy.
Given the positive integer numbers $n$,prove that $$\sum_{k=1}^{2^{n-1}}\sigma{(2^n-2k+1)}\sigma{(2k-1)}=8^{n-1}$$  where $\sigma(n)$ is defined as $$\sigma{(N)}=\sum_{d|N}d$$  This problem I think is very interesting, and have nice methods?  Thank you everyone.  and I find some of this     or can see: Show $d_{1}(x_{n},x)\rightarrow{0}$ if and only if $d_{2}(x_{n},x)\rightarrow{0}$. Let $X=(0,\infty)$. Define two metrics on $X$ by   $d_{1}(x,y)=|x-y|$ and   $d_{2}(x,y)=|x-y|+|\frac{1}{x}-\frac{1}{y}|$   for all $x, y \in{X}$.   Let $(x_{n})$ be a sequence in $X$ and $x\in{X}$.   Show  $d_{1}(x_{n},x)\rightarrow{0}$ if and only if $d_{2}(x_{n},x)\rightarrow{0}$.  Note: $d_{1}$ and $d_{2}$ are then said to be equivalent metrics on $X$.  Since $d_{1}$ and $d_{2}$ are equivalent metrics, will it be sufficient to show $c*d_{2}(x,y)\leq{d_{1}(x,y)}\leq{C*d_{2}(x,y)}$?  So far I have the following: $d_{1}(x,y)=|x-y|\leq{|x-y|+|\frac{1}{x}-\frac{1}{y}|}=d_{2}(x,y)$. Therefore, $d_{1}(x,y)\leq{C*d_{2}(x,y)}$ where $C=1$. Is there a $c$ that would satisfy the left-hand side of the inequality?  Any help would be appreciated. (this is not homework) Export data as CSV in back end with proper HTTP headers I wrote a plugin that displays all products in a woocommerce store in admin settings option, now I want to add a link to download the products as CSV file.  The problem is, when I click the link I get a permission error saying I don't have permission to view this page.  Here's my code:  function extra_tablenav($which) {     if ($which == "top") {         echo '&lt;h3 style="display:inline"&gt;'         . __('These products are currently in the database:')         . '&lt;/h3&gt;' .         '&amp;nbsp;&amp;nbsp;&amp;nbsp;' .         '&lt;a href="' . admin_url('admin.php?page=download_csv.php') . '"&gt;' . __('Export to CSV') . '&lt;/a&gt;';     } }   How can I fix those permissions? Let $\omega := dx \wedge dy$ denote the standard area form on $\mathbb{R}^2$. As the question title suggests, does there exist a $1$-form $\alpha$ with $d\alpha = \omega$? Cannot log in with Apple ID after performing password reset My wife forgot the password to her Apple ID account. Her Macbook Air (OS X Yosemite) restarted to install some updates. She has her user account set up through her Apple ID account.  Since she forgot the password she reset it from my laptop.  She can log in to iCloud and Apple ID with the new password, but cannot log in with the new password on the Air.  When she tries to log in she is greeted with this message:     "This account uses an iCloud password to log in. If you need to change   the password, use another device to change it at appleid.apple.com." I'm wanting to parse a string into a nullable int in C#.  ie. I want to get back either the int value of the string or null if it can't be parsed.  I was kind of hoping that this would work  int? val = stringVal as int?;   But that won't work, so the way I'm doing it now is I've written this extension method  public static int? ParseNullableInt(this string value) {     if (value == null || value.Trim() == string.Empty)     {         return null;     }     else     {         try         {             return int.Parse(value);         }         catch         {             return null;         }     } }      Is there a better way of doing this?  EDIT:  Thanks for the TryParse suggestions, I did know about that, but it worked out about the same.  I'm more interested in knowing if there is a built-in framework method that will parse directly into a nullable int? Your last flag was declined - no it wasn't I tried to write out the aura if statement but I am getting stuck if someone can explain it to me in detail. thank you  &lt;aura:if isTrue=&quot;{!lessthan(v.Tl.time, v.Tl.End_Date_Time1__c)}&quot;&gt;                            &lt;/aura:if&gt; Flows/Processes, Version Control and CI I've started using Processes and Flows and am really excited about these features -- but am still wrapping my mind about how to manage them in Version Control (which I use) and Continuous Integration (which I don't yet... but am trying to work in that direction). The underlying challenge is that each time a process or flow is edited, the metadata file is renamed to reflect its update version, e.g. MyFlow-1.flow, MyFlow-2.flow, etc. The Challenge for Version Control If each version of the flow has its own file name, this makes it more complicated to track changes to flows.  It seems like there could be two strategies here:  Rename the files to remove the version number.  I'd need a script to run through my src folder pre-commit, and remove the number so that my file name is always MyFlow.flow.  This is an annoying extra step, but makes version control simpler.  On the other hand, it probably makes deployment much harder.  Track each file in version control with its original name.  We'd need to be diligent about deleting MyFlow-1 in the commit where we add MyFlow-2 so that git would recognize this as a rename/move and show helpful diffs.  Otherwise, the whole file appears to change with every change, which makes version control rather less useful.   The Challenge for Deployment/Continuous Integration From my experiments so far, once a flow has been activated in a given environment, the metadata file can't be deployed again (either it is currently active, or has been active in the past and superseded). That means that I need to prevent my previous flow versions from being included in a deployment package (either by not tracking them at all, or by using a script to delete earlier versions).  I also need some way to determine whether the current version (in my repo) should be included in the deployment payload for a given environment or not. I could imagine a script that (1) downloads flows from the target environment, and removes *.flow files from the payload if they already exist in the target environment, but that feels complicated.  And it's even harder if I've removed the version numbers from my flow files. Are you using flows in VS or CI?  How are you managing these issues? Note:  I'm not sure whether VC and CI should be one question or two - depends on how integrated the strategy for dealing with these two challenges.  If you think these should be separate questions, comment and I'll re-post. How to capture Control+D signal? Let $G$ be a group and $H$ be a normal subgroup of index $p$ ( a prime ) ; suppose $K$ is a subgroup of $G$ not contained in $H$ , then is it true that $G=HK$ ? ( I know that the fact is true if $p=2$ , but I don't know for the general case . Please help . Thanks in advance ) Erasing a Linux laptop I have a laptop with Debian on it, and I am going to sell this laptop.   Would it suffice to erase the Debian installation before selling it to completely clean up my laptop from my personal data, and if yes how can I uninstall Debian (so that there isn't any operating system on the laptop)? Biblatex enumerating sorted bibliography (using authoryear-ibid style) How can I set biblatex to insert "1. " before first bibliography item and so on? The bibliography is sorted by author and the numbers are not meant to be used anywhere else. How to correctly do a division using a slash? What's the correct way to do division using a backslash? If I write $a/bc + d$, will that be equal to $(a/(bc))+d$?   Basically, if I place a slash, will I divide by what's directly behind the slash $(b)$, the term that's behind the slash $(bc)$, or everything after the slash $(bc+d)$? How to achieve a super leavened cake as the Pierre Herme ones? I made several cakes in my life. I know the difference between different leavening agents. I know that low temperature tends to make flatter cakes, but I never could get this kind of shape. If I use a super hot oven I burn the surface and I don't get this shape, there must be something else. Does anyone have experience doing this? Does anyone have a recipe from any modern pastry chef showing this kind of cake? Maybe gluten content in the flour could help?  Thank you Show this equality (The factorial as an alternate sum with binomial coefficients). OrderBy and OrderByDescending are stable? Permission Denied To Delete File First of all, I'm not looking for an answer on how to remove the files. I am looking for a reason why this doesn't work - and a suggestion on how to remove the files without sudo.  I have a directory in my Desktop that has a few inner directories (don't judge the naming - I just extract what I need from files I'm given) and after getting the files I try and remove the files with rm -rf Bengal\ Tiger\ Creative/, I am surprised to see the following warning for each image.     rm: cannot remove ‘Bengal Tiger Creative/Bengal Tiger Creative/images/customer-services-landscape.jpg’: Permission denied   Below is some code outputs.  tree  nathan@nathan-pc:~/Desktop$ tree . ├── Bengal Tiger Creative │   └── Bengal Tiger Creative │       └── images │           ├── awards.jpg │           ├── blank.gif │           ├── bullet1-light-green.gif │           ├── bullet2-light-green.gif │           ├── bullet3-light-green.gif │           ├── bullet4-light-green.gif │           ├── customer-services-landscape.jpg │           ├── desktop.ini │           ├── header-img1.jpg │           ├── header-img2.jpg │           ├── header-img4.jpg │           ├── header-img5.jpg │           ├── softhr.gif │           ├── Thumbs.db │           ├── tiger_main-img1.jpg │           ├── tiger_main-img2.jpg │           ├── tiger_main-img3.jpg │           ├── tiger_main-img4.jpg │           ├── Vanquis-benefits-icons_iphone-green.gif │           └── white-arrow-green.gif   ls -laR  nathan@nathan-pc:~/Desktop$ ls -laR Bengal* Bengal Tiger Creative: total 12 drwxrwxr-x 3 nathan nathan 4096 Aug 10 11:30 . drwxr-xr-x 6 nathan nathan 4096 Aug 10 11:34 .. drwxrwxr-x 3 nathan nathan 4096 Aug 10 11:34 Bengal Tiger Creative  Bengal Tiger Creative/Bengal Tiger Creative: total 12 drwxrwxr-x 3 nathan nathan 4096 Aug 10 11:34 . drwxrwxr-x 3 nathan nathan 4096 Aug 10 11:30 .. dr-xr-xr-x 2 nathan nathan 4096 Aug 10 11:36 images  Bengal Tiger Creative/Bengal Tiger Creative/images: total 356 dr-xr-xr-x 2 nathan nathan   4096 Aug 10 11:36 . drwxrwxr-x 3 nathan nathan   4096 Aug 10 11:34 .. -rwxrwxrwx 1 nathan nathan  14910 Mar 18 12:29 awards.jpg -rw-rw-r-- 1 nathan nathan   1093 Mar 21  2014 blank.gif -rw-rw-r-- 1 nathan nathan   1978 May  8  2014 bullet1-light-green.gif -rw-rw-r-- 1 nathan nathan   1969 May  8  2014 bullet2-light-green.gif -rw-rw-r-- 1 nathan nathan   2416 May  8  2014 bullet3-light-green.gif -rw-rw-r-- 1 nathan nathan   2449 May  8  2014 bullet4-light-green.gif -rw-rw-r-- 1 nathan nathan 113138 Apr 14  2014 customer-services-landscape.jpg -r--r--r-- 1 nathan nathan    142 Mar 25 11:58 desktop.ini -rw-rw-r-- 1 nathan nathan  14312 Mar 23 11:39 header-img1.jpg -rw-rw-r-- 1 nathan nathan   1480 Mar 23 11:41 header-img2.jpg -rw-rw-r-- 1 nathan nathan   1278 Mar 23 11:41 header-img4.jpg -rw-rw-r-- 1 nathan nathan   1168 Mar 23 11:41 header-img5.jpg -rw-rw-r-- 1 nathan nathan   1703 Mar 21  2014 softhr.gif -rw-rw-r-- 1 nathan nathan  88064 Aug  4 10:39 Thumbs.db -rw-rw-r-- 1 nathan nathan  23607 Mar 23 11:56 tiger_main-img1.jpg -rw-rw-r-- 1 nathan nathan  23400 Mar 23 11:56 tiger_main-img2.jpg -rw-rw-r-- 1 nathan nathan   6970 Mar 23 11:56 tiger_main-img3.jpg -rw-rw-r-- 1 nathan nathan   8996 Mar 23 11:57 tiger_main-img4.jpg -rw-rw-r-- 1 nathan nathan    275 Jul 31  2014 Vanquis-benefits-icons_iphone-green.gif -rw-rw-r-- 1 nathan nathan   1821 Jul  9  2014 white-arrow-green.gif   Doing chmod 777 and then delete does not work.  Doing rm -rf * inside the directory does not work.  Doing sudo rm file does work.   I can easily remove these files using sudo, however, I cannot see why they cannot be removed without sudo. As a relatively new Ubuntu user, I would like to learn why.  Thanks in advance, Nathan How to deal with bibliography items containing long URLs? Suppose $R\neq 0$ is a commutative ring with $1$. Let $M$ be a free $R$-module. I would like to prove that $M$ is a flat $R$-module. Everywhere I have looked (mostly online) this is proved by first proving that every free module is projective, and then proving that every projective module is flat. Unfortunately, Atiyah &amp; Macdonald's "Introduction to Commutative Algebra" (Chapter 2) does not discuss projective modules. But the result that every free module is flat comes very handy in the exercises.   So my question is,     Is it possible to prove that every free module is flat just by   definitions and without appealing to projective modules?   Thanks!