I have another counting functions question! My solution is as follows, but I am not 100% sure if I am correct.   Q: I have a function $f$ which maps from $\{1,2,3,4,5,6\}$ to $\{1,2,3\}$. I want to count how many functions $f$ such that for all $i,j \in \{1,2,3,4,5,6\}$, if $ i \leq j $, then $f(i) \leq f(j)$.  My solution:  There are 3 cases:  Case 1: The case that $f(6)$ is 3. Then there are 3 options (namely 1,2 or 3) each for $f(5), f(4), ... f(1)$ so in total $3^5 = 243 $  Case 2: The case that $f(6)$ is 2. Then there are 2 options (namely 2 or 1) for $f(5), f(4), ... f(1)$ so then in total $ 2^5 = 32 $   Case 3 : The case that $f(6) $ is 1. Then $f(5), f(4), ... f(1)$ have 1 choice ( namely 1 ) so 1 such function.  So in total we have 276 such functions.  Am I correct - I have an exam tomorrow so would be very much appreciated if someone could answer :)
I stumbled upon a question given like:  Let $m$ and $n$ be two integers such that $m \geq n \geq 1$.   Count the number of functions $$f: \{1, 2, · · · , n\} \to \{1, 2, · · · , m\}$$ of the following two types:  (a) strictly increasing; i.e., whenever $x &lt; y, f(x) &lt; f(y)$, and  (b) non-decreasing; i.e., whenever $x &lt; y, f(x) \leq f(y)$.  I tried in the following way.  For (a). We can say $f(n)&gt;f(n-1)$ AND $f(n)&gt;f(n-2) \ldots f(n)&gt;f(1)$ (total $n-1$ elements)  Similarly for $f(n-1)&gt;f(n-2), f(n-1)&gt;f(n-3) \ldots f(n-1)&gt;f(1)$ (total $n-2$ elements)...  and $f(2)&gt;1$  Like this $$(n-1) + (n-2) + \ldots + (1) = \frac{n(n-1)}{2}$$  Is this correct?  For (b). Since there's equality, it will be $$n + (n-1) + ... (1) = \frac{n(n+1)}{2}$$  Is my approach correct? Any help is appreciated. Thanks in advance.
Plotting in the Complex Plane number length limit in SOQL Where condition When I execute SOQL in Developer Console:  Select Id, Amount From Opportunity Where Amount &gt; 10000000000   An parsing error occurs:  Unknown error parsing query   Is there length limit of number on SOQL Where condition? I installed kubuntu-desktop just to experiment with its interface.  Then I decided to remove and did the traditional apt-get purge and autoremove but it seems elements of kubuntu are still there. For example interface elements, scrollbars, bootscreen etc are still there.  Is there any method I can completely remove without installing a fresh copy of ubuntu? Detecting an undefined object property Is it possible to enter Russia by train 1-1.5 hours before the visa validity period starts? A friend of mine is taking the Train from Finland to Russia. She has now paid and sent the Tourist visa application, with the requested validity being from 3 to 16 August.  Now she's considering entering the evening before, so I would like to ask the following:  The Train she wants to take arrives at Vyborg (first Russian station) on 2 August at 22:22, then St Petersburg at 23:27. The Russian border guards board the train at Vyborg and do the checks while the Train is travelling between the stations, so sometime between 22:30-23:00  If the visa validity starts on 3 August, and she is checked at 22:30-23:00 the evening before, what would happen? Are they very strict about the starting date if it's within 1-1.5 hours, and how bad would the consequences be?  Although they do the check while travelling between Vyborg and St Petersburg, could they simply wait until arrival in St Petersburg, then hold her in custody for 30 minutes and then stamp her in? Proving which QM interpretation is correct Let's assume that the existence of gravitons is theoretically proven or they are detected by LHC could one interpretaion be proven correct? Formatting section titles Forgive the stupid simplicity of this question, and please feel free to refer me to an older answer. I'm not quite familiar enough with LaTeX terminology to have found it.  All I want to do is set a different default format for my section titles. Instead of just being bolded and slightly larger than the main text, I'd like them to be of normal weight, the same size as the main text, and all caps.   I'm using the article class. Prove that $V = V_1 \oplus V_2$ in the following Let $V$ be a vector space over $\mathbb{C}$ such that $T^2=1_V$. Define  $V_1 = \left\{v\in V |\ T(v) = v \right\}$,  $\ V_2 = \left\{v\in V |\ T(v) = -v\right\}$  prove that $V = V_1 \oplus V_2$  For start, I've shown that $V_1\ ,V2$ are subspace of $V$. However, I'm stuck trying to show that $\forall \ v\in V,\ v = v_1 + v_2$ where $v_1 \in V1, v_2 \in V_2$.  I know at some point I have to make use of  $T^2=1_V$, but cant figure out how. Any help or insight is deeply appreciated. Is there a modal command for MakeLowercase? Analog to  question I've been wondering WHAT is to \lowercase as \bfseries is to \textbf? (... and for uppercase, or \MakeLowercase and MakeUppercase, etc., which also seem to be text-block commands as explained .)  Is there a "switch"/"declaration"/"modal command" (what is the correct term here?, is there a difference between them?) that would allow one to turn something into lowercase/uppercase in the same way that one would use \scshape for example, i.e., not as text-block command with the curly braces?  Or how to create such a command?  I couldn't find any clue on the forum. I found this question in a book.   How many terms of the Maclaurin expansion of $(1+x)^{82}$ are needed to guarantee finding a value of $1.1^{82}$ to an accuracy of $10^{-6}$?  This is how I tried to do it:  We want the Lagrange error to be less than $10^{-6}$:  $$R_{n}(0.1)&lt;10^{-6}$$  Now to find $R_{n}(0.1)$:  $$R_{n}(0.1)=\frac{f^{(n+1)}(0.1)\cdot0.1^{n+1}}{(n+1)!}$$  After writing the values of ${f^{(n+1)}}(x)$, I find a pattern and say that  $${f^{(n+1)}}(x)=\frac{82!}{(82-n)!}(1+x)^{82-n}$$  $${f^{(n+1)}}(0.1)=\frac{82!}{(82-n)!}(1.1)^{82-n}=\frac{82!(1.1)^{82-n}}{(82-n)!}$$  Now just substitute all of this back to $R_{n}(0.1)&lt;10^{-6}$:  $$\frac{\frac{82!(1.1)^{82-n}}{(82-n)!}\cdot0.1^{n+1}}{(n+1)!}&lt;10^{-6}$$  Great! Now I have an inequality that I can solve. I can't graph this so I just use trial and error (which seems like quite an intensive way to do it, so if anyone has a better way I'd love to know).  $$\frac{\frac{82!(1.1)^{82-n}}{(82-n)!}\cdot0.1^{n+1}}{(n+1)!}-10^{-6}&lt;0$$  This seems to work first for $n=28$. That means the highest power of $x$ that I need is 28, so the number of terms in the expansion needs to be $n+1=29$. But the book's answer is       30   What am I doing wrong here? I'm still quite new to the power series expansions so if there's anything unclear in my working, please let me know (also, as a side note, the book itself may be wrong, though that isn't very likely I'd assume). quotation mark (quotation sign) xelatex + polyglossia + csquotes the problem I have is the "?" that shows instead of appropriate quotation marks (signs) when citing/quoting text in foregin language. I have searched manuals for the packages such as polyglossia or csquotes.   Unfortunately I haven't found a command that would allow me to declare quotation marks (signs) for the languages I am intending to use in my doc.   In the plain text I could of course use the ,, or '' to mimic the behavior of enquote but the bibliographic entries ale also affected by this unpleasant phenomena :(  % !TEX TS-program = arara % !TEX encoding = UTF-8 Unicode  % arara: xelatex: { shell: true } % arara: biber % arara: xelatex: { shell: true }  \documentclass[12pt]{article}  \usepackage[no-math]{fontspec} \defaultfontfeatures{Ligatures=TeX} \newfontfamily\greekfont[Script=Greek,Scale=MatchUppercase]{Linux Libertine O} \newfontfamily\cyrillicfont[Script=Cyrillic,Scale=MatchUppercase]{Linux Libertine O}  \usepackage{polyglossia} \setdefaultlanguage{polish} \setotherlanguages{english,latin,greek,russian,german}  \usepackage[strict=false,autostyle=true,english=american,german=guillemets]{csquotes}  \PassOptionsToPackage{%         natbib=true,         style=authoryear-comp,         hyperref=true,         backend=biber,         maxbibnames=99,         firstinits=true,         uniquename=init,         maxcitenames=1,         citetracker=true,         parentracker=true,         backref=true,         backrefstyle=two,             }   {biblatex} \usepackage{biblatex}  \usepackage{filecontents} \begin{filecontents}{bibliography.bib} @article{Author2014,     author      = {Joe, Smith},     title       = {Publication title in polish language},     journal     = {Journal of Journals},     pages       = {1},     year        = {2014},     month       = {March}     } \end{filecontents}  \addbibresource{bibliography.bib}  \begin{document}      \noindent{}{\enquote{\languagename{} \today}}\\     \textenglish{\enquote{\languagename{} \today}}\\     \textgerman{\enquote{\languagename{} \today}}\\     \textlatin{\enquote{\languagename{} \today}}\\     \textgreek[variant=ancient]{\enquote{\languagename{} \today}}\\     \textrussian{\enquote{\languagename{} \today}}\\     \textpolish{\enquote{\languagename{} \today}}\\     \cite{Author2014}      \printbibliography  \end{document} Is there a case insensitive version of the  jQuery selector or should I do the work manually by looping over all elements and comparing their .text() to my string? When you reach 10k rep, you get to see other users deleted answers. I just think it's dishonest as a system to call a feature "delete," but show that to totally random strangers in the backroom. Due to the nature of the rep, given enough time, all of the regular users would eventually reach 10k rep, no matter how insensitive or jerk they are.  There's usually a good reason why people hit "delete" to their own work. They regret writing something totally bullshit or maybe they spoke too much about their own work that they shouldn't have.  First of all, is there really a practical reason why 10k rep user should view self-deleted answers? Does the benefit really outweigh the creepiness? (Imagine if total strangers could see your deleted messages in Gmail?)  What suggestions could we make to change it? Borrowing from the email metaphor, I think it makes sense for users to "empty trash" so the items gets seriously deleted. Imho, however the deletion is implemented is not important. It could be a soft delete, SQL Delete, or physical destruction of metal plates as long as the user or the 10k rep user no longer sees the deleted item.  What are the workarounds, if we can't change the behavior?  The least we could do is stop calling it "delete" if it's actually not a "delete." Call it "hide from public" or "actually not delete" or something.  Related user voice:        Edit: Recovery of a deleted-but-good answer is a practical benefit for the community, but I see that as creepy dangerous feature. IMHO, each user should have the rights to take back his or her perfectly good answer, and that the protecting privacy outweighs the benefit or creepiness.   I understand that it's all possible to find things out from Google cache or wherever, but why promote this as default behavior? As 10k rep user, I see others deleted answers by default. @codinghorror wrote:     But like any version control system, deletions are illusory.   Anything and everything that's implemented by human is more or less illusory including phone system and the medical records. Should the government start tapping into phone lines and scanning for words? Remember, this is a question of should had we given the opportunity to design a system. Can we have a high-speed low-drag close vote review queue? I am doing some homework exercises and stumbled upon this question. I don't know where to start.      Prove that the union of countably many countable sets is countable.   Just reading it confuses me.   Any hints or help is greatly appreciated! Cheers! There is an UI issue for the ranking div when a user crossed 200 reputation When a user just crossed 200, or 200+ reputation, the ranked div (i.e., top 10% overall) is trying to display. But rank calculation can be done on end of the UTC day. So instead of the ranking div, there is a blue colored hyphen is visible to users, it may be the initial loader for the ranking div.  I saw this UI issue previously too for the same scenario. Can this be fixed? OSVR has a right-handed system: x is right, y is up, z is near.  I need to convert orientation data from a sensor in OSVR to a different right-handed system in which x is forward, y is left, and z is up.  I need to calculate the transformation that will convert a quaternion from one system to the other.  I naively tried:  void osvrPoseCallback(const geometry_msgs::PoseStampedConstPtr &amp;msg) { // osvr to ros     geometry_msgs::PoseStamped outputMsg;      outputMsg.header = msg-&gt;header;     outputMsg.pose.orientation.x = msg-&gt;pose.orientation.y;     outputMsg.pose.orientation.y = msg-&gt;pose.orientation.z;     outputMsg.pose.orientation.z = msg-&gt;pose.orientation.x;     outputMsg.pose.orientation.w = msg-&gt;pose.orientation.w;      osvrPosePub.publish(outputMsg);     ros::spinOnce(); }   But that made things really weird. Facing north pitch is pitch, facing west, pitch is yaw, facing south, pitch is negative pitch...  How can I convert my OSVR quaternion to the a corresponding quaternion in the new coordinate system? What is the limit of:  $$\lim_{n\to \infty} \frac{n^1+\dots+n^n}{1^n+\dots+n^n} = ?$$  I did some computation with big numbers, I guess it is in the interval $\left(\frac{1}{2},1\right)$. For example, "dude," "man," "buddy," "pal," etc, when used to stand in for someone's name.  "Hey, pal, how's it going?"  Is there a word for terms like these?  Or is "colloquialism" as close as we can get? I am moving from one job to another (postdoc to tenure track) this summer and I have papers in several stages of being written, these go from handwritten notes with all the necessary arguments to under revision at a journal. All the papers will have my new contact information put on them but where is the line for changing affiliation? It seems obvious to me that the one that is still in the handwritten notes stage will get the new affiliation and that the ones under revision won't. So somewhere in between is a line.   Where is it?  There are some similar questions so far (see  and  but they do not seem to be quite the same situation since I am not changing fields and the appointments are not concurrent.