Inaccuracy of Piecewise function?
Plot showing discontinuity where it shouldn't
Is the following layout possible with the subfigure package? Search a specific word and copy line to another Sheet Orthogonal projections with $\sum P_i =I$, proving that $i\ne j \Rightarrow  P_{j}P_{i}=0$ Combining  with allowframebreaks in beamer If $n&gt;1$ is an integer, then $\sum \limits_{k=1}^n \frac1k$ is not an integer.  If you know , then you know there must be a prime $p$ between $n/2$ and $n$, so $\frac 1p$ appears in the sum, but $\frac{1}{2p}$ does not. Aside from $\frac 1p$, every other term $\frac 1k$ has $k$ divisible only by primes smaller than $p$. We can combine all those terms to get $\sum_{k=1}^n\frac 1k = \frac 1p + \frac ab$, where $b$ is not divisible by $p$. If this were an integer, then (multiplying by $b$) $\frac bp +a$ would also be an integer, which it isn't since $b$ isn't divisible by $p$.  Does anybody know an elementary proof of this which doesn't rely on Bertrand's Postulate? For a while, I was convinced I'd seen one, but now I'm starting to suspect whatever argument I saw was wrong. Easy way to concatenate two byte arrays Why can you see virtual images? Convergence of the sequence $\sqrt{1+2\sqrt{1}},\sqrt{1+2\sqrt{1+3\sqrt{1}}},\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1}}}},\cdots$ I recently came across this problem     Q1 Show that $\lim\limits_{n \rightarrow \infty} \underbrace{{\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots+n\sqrt{1}}}}}}}_{n \textrm{ times  }} = 3$   After trying it I looked at the solution from that book which was very ingenious but it was incomplete because it assumed that the limit already exists.  So my question is     Q2 Prove that the sequence$$\sqrt{1+2\sqrt{1}},\sqrt{1+2\sqrt{1+3\sqrt{1}}},\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1}}}},\cdots,\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots+n\sqrt{1}}}}}$$ converges.   Though I only need solution for Q2, if you happen to know any complete solution for Q1 it would be a great help .  If the solution from that book is required I can post it but it is not complete as I mentioned.   Edit: I see that a similar question was asked before on this site but it was not proved that limit should exist. How to set mount point of boot partition to /boot/efi? Is there a word to describe the virtue of "to do something and ask for nothing in return"? I am wondering if there is a positive, formal counterpart to the slang "No strings attached" to describe, for example, a parent's unquestioned devotion to their children, or the virtue of someone's "a friend in need is a friend in deed", help first, question later mentality. If $\Sigma \vdash \varphi$ then $\Sigma \vdash \forall x \varphi$. Why when $(\varphi \implies \forall x \varphi)$ is not true? In the book of first order logic I'm reading they say that if $(\varphi_1,...,\varphi_n)$ is a proof from a set of formulae $\Sigma$, then so is $(\varphi_1,...,\varphi_n,\forall x \varphi_j)$ for all $1\leq j \leq n$. They assume that as an axiom but it doesn't make sense to me. Why if $\varphi$ is true then so is $\forall x \varphi$? For example if $\varphi$ is $x=1$ that wouldn't mean that $\forall x : x=1$ , or would it?  Maybe what I don't understand is: why $(\Sigma \vdash \varphi \implies \Sigma \vdash \forall x \varphi)$ is true but $(\varphi \implies \forall x \varphi)$ is not true?  The book is this: I was answering a question on Stack Overflow and required to use a trailing space in one of my code snippets.  I formatted it like so:  `&lt;a `   And it renders without the trailing space: &lt;a  Using the &lt;code&gt; tag does account for trailing spaces, though in my case it busts because of the &lt;. Here's my formatting with the &lt;code&gt; tag:  &lt;code&gt;&lt;a &lt;/code&gt;   Here's how it renders:  (Yes, you should see nothing there).  This is a pretty annoying bug. Can we get it fixed? How can I display only the unique values in one column in Excel? Is there a way to display the unique values in a column ONLY?  Example:   AAPL AAPL AAPL AAPL AKAM AMZN AMZN AMZN AMZN AMZN AMZN  Expected result:   AAPL AKAM AMZN  I have already done a sort for multiple factors, and the first match for each is the best match.  I'm not sure if this is a vlookup or something else. How can I find a saved Wi-Fi password? I have an Android phone which is connected to a secured Wi-Fi hotspot.  How can I find out the password of the hotspot from the phone? Detect when browser receives file download Handle null value using Guava MapMaker/CacheBuilder Why is the determinant the volume of a parallelepiped in any dimensions? For $n = 2$, I can visualize that the determinant $n \times n$ matrix is the area of the parallelograms by actually calculating the area by coordinates.  But how can one easily realize that it is true for any dimensions? Auto-inserting Stack Overflow affiliate into all Amazon links We have officially discontinued the Amazon book affiliate remnant ads, since despite our best efforts they don't perform well for our audience and cannot be made to perform well.  We've often wondered,   I'm thinking this isn't a problem; it has gotten lots of support historically. So we're going to go ahead and do it, so the company Amazon account is still getting some use, and we benefit a little.   To be clear: from this point on, all Amazon links posted on all sites will automatically be rewritten to add our affiliate link. We have a clever affiliate redirect that geolocates the IP address and redirects to the appropriate regional Amazon store (amazon.uk, amazon.ca, et cetera).  If you have any feedback on this, or suggestions, as they say &hellip; speak now or forever hold your peace! Let $T=\{z:|z|=1 \mathbin{\text{and}} z \in \mathbb C\}$ be the unit circle in the complex plane, considered as a topological group under complex multiplication and the usual topology. Show that every subgroup of $T$ is either dense in $T$ or finite.  How to prove the denseness? How can I stop Lollipop ignoring my wifi network? Android 5.0 has a new "feature": If a wifi network has no connection to the internet, it will remain connected to it but route traffic over mobile data instead. A small exclamation mark shows in the status bar over the wifi symbol.  Unfortunately, for some reason Android has decided that my home wifi does not have a good internet connection - which is not true, as I'm using it to access Stackexchange at the moment - and is, instead, trying to send everything over GPRS.  With mobile data disabled, things work over wifi just fine, despite the exclamation mark - so it isn't a problem with the wifi not working, per se.  Is there any way to override this "smart" feature? The only way I have found to make things work at the moment is to disable mobile data while I am in my house - which is not a practical long-term solution.   EDITING to add that I have now solved this problem by buying a new router! I was using something very old... But, I'll leave the question here for the benefit of anybody else with the same difficulty.