I tried to show that the set $X = \{(x, y)\in \mathbb{R}^2 : x^2 + y^2&lt;1\}$ is convex.
Let $X$ be a normed linear space, $x\in X$ and $r&gt;0$. Define the open and closed ball centered at $x$ as  $$ B(x, r) = \{y \in X : \Vert x − y\Vert &lt; r\} $$  $$ \overline{B}(x, r) = \{y \in X : \Vert x − y\Vert \leq r\}. $$ Then $B(x, r)$ and $\overline{B}(x, r)$ are convex.  I tried to prove this, but either my calculation is incorrect, or I am on the wrong path:  I aim to show for the closed ball $\overline{B}(x,r)$ (for open ball I assume the proof is similar). Suppose $y,z \in \overline{B}(x, r)$. Then $\Vert x − y\Vert \leq r$ and $\Vert x − z\Vert \leq r$. We must show that $t \in [0,1]$ implies $ty + (1-t)z \in \overline{B}(x,r)$. But $t \in [0,1]$ implies $$ \Vert ty + (1-t)z - x\Vert = \Vert t(y-z) + z - x\Vert \leq |t| \Vert y-z\Vert + \Vert z-x\Vert \leq |t|(\Vert y-x\Vert + \Vert x-z\Vert) + \Vert z-x\Vert &lt; |t|(2r) + r = r(2|t| + 1), $$ which is not necessarily $\leq r$. We probably wanted to end up with $&lt; |t|r$ or $\leq |t|r$ as our final inequality.  Thanks in advance.
I guess the proof of the identity $$ \sum_{n = 0}^{\infty} \frac{1}{n!} \equiv \lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x $$ explains the connection between such different calculations. How is it done? How do I move child windows without moving or minimizing parent in GNOME 3? I have Ubuntu 17.10 with GNOME 3.  What happens: I have (for example) LibreOffice Calc open, with some data columns in it.  I highlight the columns I want to chart, then I select INSERT > CHART.  The chart child window dialog pops up on top of my data, so I want to move it to the side.  When I grab the top bar and move it, the parent (spreadsheet) window minimizes (I think it actually disappears).  Continuing the above example, if I click FINISH on the chart window, I can then see the minimized spreadsheet application, which is the window title bar plus the menu bar just below it.  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So far I have this:  If $A$ is similar to $B$, then      $B=P^{-1}AP$ and $A=PBP^{-1}$   This implies that:     $\operatorname{trace}(B) = \operatorname{trace}(P^{-1}AP)$ and $\operatorname{trace}(A) = \operatorname{trace}(PBP^{-1})$   Not sure where to go from here Graph Proof by induction. Can you prove via induction that there exists a node in a directed graph of n nodes that can be reached in at most two edges from every other node in the graph. Every node in the graph is required to have an edge with every other node that either points out or in (For every pair of nodes X and Y, either X points to Y or Y points to X). I am using PHP and JavaScript. My JavaScript code contains a function, get_data():  function get_Data(){     var name;     var job;     .....      return buffer; }   Now I have PHP code with the following.  &lt;?php     $i=0;     $buffer_data;      /* Here I need to get the value from JavaScript get_data() of buffer;        and assign to variable $buffer_data. */ ?&gt;   How do I assign the JavaScript function data into the PHP variable? Move incoming Outlook emails to folder containing previous conversation I file email conversations in various folders. But every new email in that conversation goes to inbox, so I have to manually move it to the folder that contains the rest of the conversation. Is there a way for Outlook to do this automatically with all new emails?  What I want: Outlook will do the following automatically as long as it's running:   For all unread emails in inbox... ...if email is not the first in the conversation, search for folders where that conversation is stored. If found, move the new unread email to the first* folder it finds (except for Inbox, Trash, or Sent). Optionally: If there is more than one folder containing messages in the same conversation, then do nothing.   **If there is more than one folder containing messages in the same conversation, then do nothing*. (This extra rule is not necessary if it complicates the solution too much)  Thanks! What causes the dimensions of a star increase when its hydrogen fuel is exhausted? Is the Echo of a Echo Knight actually a creature? I've always been certain that the echo created by an Echo Knight fighter using their 3rd-level feature Manifest Echo is not defined in game terms as a creature, since the description of the echo never explicitly tells it is a creature. But the presence of people believing the echo to be a creature in lieu of the fact that it is immune to all conditions and might make saving throws (game characteristics typically associated with creatures only) brought this doubt to my mind.  from Jeremy Crawford implies it is just an image and not a creature, further confirming that the feature should tell you, but  implies that the echo is an object whilst the feature doesn't explicitly say so. (Included for reference and insight, since tweets are not eligible as official ruling) Is the echo a creature? There were so many discussions about an Ubuntu with RR. They were so many pros and contras. At the end ubuntu wont be changed, the 6 month cycle system wont be changed. Thats fine!   But because there are so many people who would like to use an ubuntu with RR Im asking myself why isnt there an Ubuntu Derivate with RR. I couldnt found one, there are a loooot of Ubuntu Derivates but no RR. Is it just impossible? I mean Ubuntu = 6 months cycle, so the derivate have to be a 6 month cycle too? Is there no other option? Or are there already some smart guys trying to solve this problem? Thank you! I quite often add a question as a favourite when there is something of interest to me in it or its answers. Looking back through my favourites, I cannot remember why I added some of them as a favourite.  What I'd like to be able to do is to add a note/comment when I add a favourite as a reminder as to why I've added it. This will show on my favourites page, but nowhere else - and possibly to no one else.  In addition, it would be good to be able to favourite an answer to a question, thereby adding a favourite to both the question and a particular answer with something of note in it. This would only show as a single favourite, but when clicking its link would go to the favourited answer.  Possible? Or is there another way of doing this? I've already got too many favourites to remember why I added a lot of them.