The cardinality of all equivalence relations over $\mathbb{N}$
Cardinality of equivalence relations on N
I have not idea how to approach this differential equation.   $$y'-\int_0^xy(t)dt=2$$.  Basically, I did,  $$F''(t)-F(x)+F(0)=2 \;\;\;\;\;\;\; F'=y$$  I am stuck.  Thank You. On per-site metas, users are given a rank based on meta participation, and this ranking is seen on the default tab of the users page on each per-site meta. How is this participation numerically calculated to give a ranking? Is it by votes? Something else? Are community wiki posts included in this? WIth Vimeo Plus account, how to hide/remove the "play" button in the middle of an embedded video? Divergence of $\sum_{n=1}^{\infty} \frac{1}{2+i^n}$? I notice that,  $i^n = i, \quad if \text{ } n \equiv \text{ } 1 \text{ } mod \text{ } 4$  $i^n = -1, \quad if \text{ }n \equiv \text{ } 2 \text{ } mod \text{ } 4$  $i^n = -i, \quad if \text{ }n \equiv \text{ } 3 \text{ } mod \text{ } 4$  $i^n = 1, \quad if \text{ } n \equiv \text{ } 0 \text{ } mod \text{ } 4$  so the series $\sum_{n=1}^{\infty} \frac{1}{2+i^n}$ seems to be an alternating series that does not converge to any point in the complex plane. Is there a rigorous way to prove this divergence? How would you write a proof to this divergence or convergence (if I am wrong)?  Thanks! Why is the $S_{z} =0$ state forbidden for photons? Is the “sum of all natural numbers” unique? A while ago, there was a great hype about the “identity” $$\sum_{n=1}^{\infty} n = -\frac{1}{12}.$$  Apart from some series manipulations where the validity seems to be at least questionable, the derivation of this always goes through the zeta function:  Where the series converges, the zeta function is defined by $$\zeta(s) = \sum_{n=1}^{\infty}\frac{1}{n^s}$$ and outside that range by analytic continuation. And it turns out that inserting $s=-1$ formally results in $$\zeta(-1) = -\frac{1}{12} = \sum_{n=1}^\infty n$$  However looking at the series in isolation, there is no indication that the zeta function should be chosen.  An obvious way to get an analytic function that at one point gives the sum of all natural numbers is $$f(x) = \sum_{n=1}^{\infty} nx^n$$ at $x=1$, however (not surprisingly) that function diverges at $1$.  Therefore my question:     Is it possible to get another finite value for the series by analytic continuation of another series?   Concretely, do there exist continuous functions $f_1, f_2, f_3, \ldots$ such that   On some non-empty open subset $S$ of $\mathbb C$, $f(x)=\sum_{n=1}^\infty f_n(x)$ converges to an analytic function. At some point $x_0$, $f_n(x_0) = n$ for all positive integers $n$. The analytic continuation of $f$ is well defined and finite at $x_0$. $f(x_0) \ne -1/12$   What if we demand the functions $f_n$ to be analytic rather than just continuous? Convergence of a sequence implies convergence of the norm of the sequence If $\langle X_n \rangle$ is a sequence in $\mathbb R^n$ then $\lim\limits_{n\to \infty}X_n=A$. Show that $\lim\limits_{n\to \infty} \| X_n \| = \| A \|$ .   My thought process, $\| X_i \|= (\sum_{i=1}^n x_i^2)^{(1/2)}$ therefore be the multiplicative property of limits $x_i^2 = a_i^2$. Therefore the limit of norm of the components is equivalent to the limit of the corresponding limit of each component of the sequence. When given $p = 5, q = 11, N = 55$ and $e = 17$, I'm trying to compute the RSA private key $d$.  I can calculate $\varphi(N) = 40$, but my lecturer then says to use the extended Euclidean algorithm to compute $d$. That's where I get stuck.    Here's my work so far:  First I use the Euclid algorithm to calculate:  40 = 2(17) + 6  17 = 2(6) + 5 6 = 1(5) + 1 = gcd   So I know the GCD is 1. Applying the 'extended' section of the algorithm:  6 = 40-2(17) 5 = 17-2(6) 1 = 6-1(5) 1 = 6-1(17-2(6)) 3(6) = 1 (17) 3(40 - 2(17)) - 1(17) 3(40) - 3(17)   I know the answer is $33$, but I have no idea how to get there using the extended Euclidean algorithm. I can't figure out why I'm getting $3(40) - 3(17)$ when I know the answer should contain $33$. Explaining to an 8-year-old why nothing can travel faster than light Diff is a great tool to display the changes between two files. But how to display the similarities of two text files (while ignoring the differences)?  I.e. sample input:  a: Foo Bar X Hello World 42  b: Foo Baz Hello World 23   Pseudo output (something like this):  @@ 2,3 =Hello World   Just sorting both files and using comm is not enough, because in that case the line information is lost. Create a tar archive split into blocks of a maximum size I need to backup a fairly large directory, but I am limited by the size of individual files. I'd like to essentially create a tar.(gz|bz2) archive which is split into 200MB maximum archives. Clonezilla does something similar to this by splitting image backups named like so:  sda1.backup.tar.gz.aa sda1.backup.tar.gz.ab sda1.backup.tar.gz.ac   Is there a way I can do this in one command? I understand how to use the split command, but I'd like to not have to create one giant archive, then split it into smaller archives, as this would double the disk space I'd need in order to initially create the archive. How to uncover a table column wise in Latex beamer I have a table with 7 columns and want to show the first 5 at once, and then  the 6th and 7th on consecutive slides.  This is the code I have for now:  \begin{frame}{Наслов}     \begin{itemize}         \item Ставка.     \end{itemize}         \begin{tabular}{ccccc&lt;{\onslide&lt;2-&gt;}c&lt;{\onslide}c}             $p_1$ &amp; $p_2$ &amp; $\ldots$ &amp; $p_{9}$ &amp; $p_{10}$ &amp; $W$  &amp; $D$\\             $1$   &amp; $1$   &amp; $\ldots$ &amp; $1$      &amp; $1$     &amp; $w_1$&amp; $d_1$\\             $1$   &amp; $1$   &amp; $\ldots$ &amp; $1$      &amp; $0$     &amp; $w_2$&amp; $d_2$\\             $1$   &amp; $1$   &amp; $\ldots$ &amp; $0$      &amp; $1$     &amp; $w_3$&amp; $d_3$\\             $1$   &amp; $1$   &amp; $\ldots$ &amp; $0$      &amp; $0$     &amp; $w_4$&amp; $d_4$\\             $\vdots$ &amp; $\vdots$ &amp; $\vdots$ &amp; $\vdots$ &amp; $\vdots$ &amp; $\vdots$ &amp; $\vdots$\\             $0$   &amp; $0$   &amp; $\ldots$ &amp; $0$      &amp; $1$     &amp; $w_{1023}$&amp; $d_{1023}$\\             $1$   &amp; $1$   &amp; $\ldots$ &amp; $0$      &amp; $0$     &amp; $w_{1024}$&amp; $d_{1024}$         \end{tabular} \end{frame}   This produces 2 slides: the first one with p1, p2, ..., p9, p10 and D columns, and the second with W added, while I want them to appear in the following order:   1st slide: p1, p2, ..., p9, p10  2nd slide: p1, p2, ..., p9, p10, W 3rd slide: p1, p2, ..., p9, p10, W, D   I looked in the manual and found this example:  \begin{frame}     \begin{tabular}{l!{\vrule}c&lt;{\onslide&lt;2-&gt;}c&lt;{\onslide&lt;3-&gt;}c&lt;{\onslide&lt;4-&gt;}c&lt;{\onslide}c}     Class &amp; A &amp; B &amp; C &amp; D \\     217     X &amp; 1 &amp; 2 &amp; 3 &amp; 4 \\     Y &amp; 3 &amp; 4 &amp; 5 &amp; 6 \\     Z &amp; 5 &amp; 6 &amp; 7 &amp; 8     \end{tabular} \end{frame}   which does what is supposed to. However, I don't seem to be able to apply its logic to my problem. :( Is anyone willing to help?   P.S. What are the meanings of &lt; and ! in columns specification? states that the isomorphism type of a finite simple group is determined by its order, except that:   $L_4(2)$ and $L_3(4)$ both have order $20160$ $O_{2n+1}(q)$ and $S_{2n}(q)$ have the same order for $q$ odd, $n &gt; 2$   I think this means that for each integer $g$, there are $0$, $1$ or $2$ simple groups of order $g$.  Do we need the full strength of the  to prove this, or is there a simpler way of proving it? Sigmoid vs Relu function in Convnets Dimension of open subsets of $R^n$ As , the retag privilege serves a lot more confusion and a lot less of its benefit due to the suggested edit system. It must be done away with. It will be done away with. ♪  In light of this decided course of action, we're going to need a new snack for the road between "Create Tags" at 300 reputation and "Established User" at 1000 reputation. We'd like your help in thinking about some ideas on what would be a nice privilege to sit at this level. Since we're... a bit dry on our end.  The primary guideline in deciding what kind of privilege to put here is acknowledging its position in the overall hierarchy of privileges. 500 reputation is an early-intermediate user, one who has started to really get the hang of the site and its function. It can be something that provides convenience in how they currently can act on the site, or it could be some new functionality that fits as an early ability (especially considering there's no new abilities until 2000 reputation's "Edit Everywhat" ability).  I look forward to hearing what creative ideas y'all can provide. Dollar cost averaging - Should I still do it if I have a large pile of cash now? I currently have a large pile of cash savings ($240,000) which I really should have been investing all this time, but the past is past and no point dwelling on it.  I am currently 38 years old (2 kids. Wife who is also a high earner - $100,000), with a high salary ($150,000). I have read "The Intelligent Investor" and am really sold on dollar cost averaging as a strategy, using the index funds offered by Vanguard.  My current dilemma is what to do with the cash I already have. My current plan is to still follow the dollar cost averaging and slowly move it into vanguard over a period of 4 years (~$5000 per month). While leaving the remaining parts of the cash in term deposits. So $50,000 in a 3 year term deposit at 3%, $50,000 in a 2 year term deposit at 2.8% etc.  But in someways it feels overly conservative. Should I just move all the money now? Or perhaps in a shorter time frame (3 years? 2 years? 1 year?).  There are a few other questions on the site similar to mine, but they are from 2013/2014 etc. Given that currently the stock market is at a high and there is a lot of economic uncertainty on the horizon, putting everything in the stock/bond market at once now seems like a risky move. And it seems to violate the principle of dollar cost averaging. Or am I violating dollar cost averaging by trying to "time the market" by looking at it's current value?  Reading this section  seems to indicate that I should just invest everything now. But the first paragraph also seems to suggest giving more weight to that advice if the market was trending upwards (Which I am not certain the current market is/will be for long).  PS: My current investment horizon is long. Basically retirement, about 20 years from now. Problem with IPv6 sudo apt-get update/upgrade I am using Ubuntu 16.04 Beta 2 at the moment and every time I try to update at home via sudo apt-get update or upgrade trough sudo apt-get upgrade I wont get any feedback. Even installing of anything will have the same result. It will just "hang up" on 0% while fetching data.  $ sudo apt-get update [sudo] password for niclas:  0% [Connecting to archive.ubuntu.com (2001:67c:1560:8001::11)] [Connecting to archive.canonical.com (2001:67c:1360:8c01::1b)]   I have no idea how to solve the problem. I already tried to ping6 archive.ubuntu.com with 100% packet loss while just ping wont have any loss.  I changed the DNS-Server (IPv4 &amp; IPv6) to Google's one no solution either. A post here on DBA.StackExchange () has spawned an interesting question (at least, interesting to me) regarding performance in MySQL.  The context is that we want to insert a record in a table for each row that is updated.  Before the row is updated, we want to store off a previous value and then increment one of the columns (a "version" column).  If we do this inside a trigger, it works out nicely.  For MySQL, the triggers are , so it would be an easy solution.  Select the data currently in the table, insert it into the logging table, and the update the "version" column in the new data.  However, it's possible to move this logic to a stored procedure. If you do that, you're performing the insert, then incrementing the "version" column in the table.  The entire thing would be set based.   So, when it comes to performing this insert, would this be more performant to use the set-based stored procedure approach or a trigger based approach?  This question is for MySQL (since it has row-by-row triggers), although it could apply to other row-by-row trigger DBMSs. Automated mapping of cost paths from multiple origin points I am currently mapping Lahar flows on Montserrat Island (as many have done before!). I have the Cost Distance analysis below, the circle is the origin and the triangle the destination. With a known destination a Cost Path analysis can easily be performed, however I would like to use the Cost Distance map below to automatically find the destination point, based purely on the the direction of flow from the origin/circle.  I was thinking of using a filter to include only edge cells of the raster, and then find the one with the lowest value, is there a way to do this?  Software: Arc GIS Desktop 10.5