what function that is differentiable 2019 times, but is not differentiable 2020 times? Please give a example of function that is differentiate 2019 times, but is not differentiate 2020 times
For every $n \in \mathbb{N}$, find a function $f$ which is differentiable $n$ times at $0$, but not $n+1$ times. For every $n \in \mathbb{N}$, (in this problem $\mathbb{N}$ starts at $n = 1$) find a function $f$ which is differentiable $n$ times at $0$, but not $n+1$ times.  The function I chose is the following:  $$ f_n(x) = \begin{cases} x^{n+1} &amp;\mbox{if } x \le 0\\ x^{n} &amp;\mbox{if } x &gt; 0 \end{cases} $$  Now I can't think of the best way to prove that $f$ is differentiable  $n$ times at $0$ but not $n+1$ times.
I am having trouble getting a basic example of a matrix of notes to show up in my beamer presentation. I am using the code copied from the pgf manual. Here is a complete working example. The error I get upon compilation is:  "Package pgfbasematrix Error: Single ampersand used with wrong catcode."  \documentclass{beamer} \usepackage{tikz} \usetikzlibrary{arrows,decorations.pathmorphing,backgrounds,fit,petri,positioning,matrix}  \begin{document}  \begin{frame}  \begin{tikzpicture} \matrix (magic) [matrix of nodes] { 8 &amp; 1 &amp; 6 \\ 3 &amp; 5 &amp; 7 \\ 4 &amp; 9 &amp; 2 \\ }; \draw[thick,red,-&gt;] (magic-1-1) |- (magic-2-3); \end{tikzpicture}  \end{frame} \end{document} Interval preserving transformations are linear in special relativity Is it possible to choose which OS will be loaded using a hardware switch? Tiny lifeforms on planetoid that live at accelerated time rate A book I once owned and read back in late-80's or early-90's that I have not been able to identify title &amp; author for:  It was a SF story concerning tiny native lifeforms, living on either a planetoid or possibly a dead/neutron star, that evolve at a seemingly accelerated time rate. Most of their story is told in the few days of observation (from orbit) by a human expedition to the planet, during which time-frame the race below progresses from stone-age tech level to being more advanced than the human observers, and eventually flying up to meet the humans in small advanced craft powered by micro-singularities... one of which accidentally 'disintegrates' the tip of a human scientist's nose in the process. arithmetic progression of primes Prove that there is no arithmetic progression that consists only of primes.  A question that I've been set; I'm guessing it makes use of primes being written in the form 4k+1 and 4k+3? Not sure where to start.  Thanks. I just came over this syntax in some of the questions in this forum, but Google and any other searchengine tends to block out anything but letters and number in the search so it is impossible to search out "=>".  So can anyone tell me what it is and how it is used? Use of comma separating two adjectives If I had the phrase       two blond haired, blue eyed people    would the comma be out of place?    should I hyphenate "blond haired" and "blue eyed"? The ^^ notation in various engines Basically, how does the ^^^... notation work in LuaTeX and XeTeX, exactly?  In 8-bit TeX engines (recent TeX, eTeX, pdfTeX, at least), two consecutive identical catcode 7 characters (typically ^), followed by two lowercase hexadecimal digits, are converted before the tokenization step to the corresponding byte.  Namely, ^^6f is exactly equivalent to o: for instance, \sh^^6fw ^^6f will cause TeX to show the letter o.  There is also the notation with two ^ (identical catcode 7 characters), followed by any ascii character (but not two lowercase hexadecimal digits), which is replaced by the character obtained by either subtracting or adding 64 to the character code, remaining among ASCII characters (range 0 to 128).  Unicode-aware engines (I'm thinking of LuaTeX and XeTeX, there are perhaps other less known ones around) also provide ^^^^xxxx and ^^^^^xxxxx for characters whose hexadecimal representation has 4, or 5 digits.  But this does not seem to be done in the same way across engines.  For instance, both LuaTeX and XeTeX appear to accept the notation with 4, 5, or 6 carets followed by the same number of hexadecimal digits, but XeTeX also accepts it for 3, while LuaTeX doesn't. Compiling the following with pdfTeX, LuaTeX, and XeTeX gives different results.  \catcode0=12 \newlinechar=10 \def\loopshow#1{\message{\meaning #1^^J}\loopshow} \loopshow \/   ^^56   ^^^056   ^^^^0056   ^^^^^00056   ^^^^^^000056   ^^^^^^^0000056 {\end\iffalse} \fi} \bye   One weird fact about XeTeX (a bug?) is that  \show ^^^^^^010101   shows the character displaywidth.  My goal (there may be a better way to do this) is to provide a way to test whether passing a given list of tokens through \scantokens is safe.  For that, my plan is to go through the \detokenized token list one character at a time, applying TeX's rule for tokenizing (but no need to fully tokenize), and detecting begin-group and end-group tokens, as well as invalid characters. I'm trying to  working on Linux. I have PAINSTAKINGLY installed NVidia drivers, Cuda 5.0 and incidentally I'm failing on the last step when I run make:  make  all-recursive make[1]: Entering directory `/var/progs/CudaMiner' Making all in compat make[2]: Entering directory `/var/progs/CudaMiner/compat' Making all in jansson make[3]: Entering directory `/var/progs/CudaMiner/compat/jansson' gcc -std=gnu99 -DHAVE_CONFIG_H -I. -I../..     -g -O2 -MT dump.o -MD -MP -MF .deps/dump.Tpo -c -o dump.o dump.c /bin/bash: gcc: command not found make[3]: *** [dump.o] Error 127 make[3]: Leaving directory `/var/progs/CudaMiner/compat/jansson' make[2]: *** [all-recursive] Error 1 make[2]: Leaving directory `/var/progs/CudaMiner/compat' make[1]: *** [all-recursive] Error 1 make[1]: Leaving directory `/var/progs/CudaMiner' make: *** [all] Error 2   I'm pretty sure I have gcc, in fact I have multiple versions of it. I installed version 4.6 (which is what cuda requires) but it doesn't seem to find it. I'm so close!  Thanks in advance. Many of the UK visa refusals we see here share a common pattern and the prevailing reasons refer to V 4.2 (a) and (c).    I do understand that, while the applicants may describe very different individual circumstances, there's a consistent pattern and, broadly, fall into specific categories:    Credibility (lifestyle, lack of ties, visit history, lies &amp; omissions) Funding (insufficient funds, provenance of funds, funds parking) Sponsorship (family, friends, employer)   Question: What most commonly triggers a UK visa refusal where V 4.2 (a) and (c) are given as the grounds?  Secondarily: Given that there is a clear pattern, to what extent are these refusals predictable? Does sponsorship make a difference? Is there a set of personal circumstances, however abstract, where a refusal is all but guaranteed? For example, why do those we see here on TSE who sought entry for the PLAB or British Army appear to be refused with a common theme? Is there a uniform shortcoming or is it just discrimination?   PLAB: What is about applications for a visa, for the purpose of sitting the Professional and Linguistic Assessments Board (PLAB) exam, that seem to invite refusals. After all, the test is given so that international medical graduates can show that they qualify to practise medicine in the UK. The first part, PLAB 1, can be taken in centres outside of the UK. However, PLAB 2 can be taken only in the UK. Why can't you can't get a visa just to sit the exam, promising that you'll leave immediately after?    British Army: Since Commonwealth citizens are eligible to apply online to join, even those who don't reside in the UK, why is it so difficult to get a visa just to attend an interview to see whether you are suitable? Even with an invitation from the Army, such visa applications seem to be unsuccessful. Isn't a career in the British Army a valid reason?     Lastly:  After such a refusal, what approaches would increase the chance of a successful application? Automatically run commands over SSH on many servers There is a list of IP addresses in a .txt file, ex.:  1.1.1.1 2.2.2.2 3.3.3.3   Behind every IP address there is a server, and on every server there is an sshd running on port 22. Not every server is in the known_hosts list (on my PC, Ubuntu 10.04 LTS/bash).  How can I run commands on these servers, and collect the output?  Ideally, I'd like to run the commands in parallel on all the servers.  I'll be using public key authentication on all the servers.  Here are some potential pitfalls:   The ssh prompts me to put the given servers ssh key to my known_hosts file. The given commands might return a nonzero exit code, indicating that the output is potentially invalid. I need to recognize that. A connection might fail to be established to a given server, for example because of a network error. There should be a timeout, in case the command runs for longer than expected or the server goes down while running the command.   The servers are AIX/ksh (but I think that doesn't really matter. The probability of two dice summing to k is simple enough, make a diagram of the throws,  $$ \begin{array}{c}  &amp;|&amp;1&amp;2&amp;3&amp;4&amp;5&amp;6\\\hline 1&amp;|&amp;2&amp;3&amp;4&amp;5&amp;6&amp;7\\ 2&amp;|&amp;3&amp;4&amp;5&amp;6&amp;7&amp;8\\ 3&amp;|&amp;4&amp;5&amp;6&amp;7&amp;8&amp;9\\ 4&amp;|&amp;5&amp;6&amp;7&amp;8&amp;9&amp;\color{green}{10}\\ 5&amp;|&amp;6&amp;7&amp;8&amp;9&amp;\color{green}{10}&amp;11\\ 6&amp;|&amp;7&amp;8&amp;9&amp;\color{green}{10}&amp;11&amp;12\\ \end{array} $$  and the probability of a sum is just the length of the corresponding diagonal divided by $36$. For instance, $\mathrm{P}(\mathrm{sum} = 10) = \frac{3}{36} = \frac{1}{12}$. However, it's not as easy to draw a diagram for three dice or more, so how would a general formula look? Evaluate Current User for Flow - 15 - 18 digit User ID I need to evaluate the current user running a flow. Someone suggested I use the Global Variable {!User.Id}, but it's returning a 15 digit Id. I need to compare the current user with the Account owner. The Account Owner field is returning an 18 digit id. Is there another way to evaluate the current user in my flow and compare it to an Account owner?   The above resolution doesn't really work for my purposes for evaluating a user if I can't compare the user to the Account Owner ID. In the above example, the Account Owner will never = the current user as it is comparing a 15 digit id to a 18 digit id.  Is there a way turn the User ID from the Global Variables into 18 digits? Or vice versa, make the Account Owner ID return as 15 digits? Thanks so much! Quantum computers are known to be able to crack  a broad range of cryptographic algorithms which were previously thought to be solvable only by resources increasing exponentially with the bit size of the key. An example for that is .  But, as far I know, not all problems fall into this category. On , we can read     Researchers have developed a computer algorithm that doesn’t solve problems but instead creates them for the purpose of evaluating quantum computers.   Can we still expect a new cryptographic algorithm which will be hard to crack using even a quantum computer? For clarity: the question refers to specifically to the design of new algorithms. Prove that $\displaystyle \sum^{n}_{r=0}(-1)^r\cdot \large\frac{\binom{n}{r}}{\binom{r+3}{r}} = \frac{3!}{2(n+3)}$   $\bf{My\; Try::}$ We can write $$\frac{\binom{n}{r}}{\binom{r+3}{r}} = \frac{n!}{r!\times (n-r)!}\times \frac{r!\times 3!}{(r+3)!} = \frac{3!\times n!}{(n-r)!\times (r+3)!}$$  So our sum is $$ 3!\sum^{n}_{r=0}(-1)^r\frac{n!}{(n-r)!\times (r+3)!} = \frac{3!}{(n+1)(n+2)(n+3)}\sum^{n}_{r=0}(-1)^r\frac{(n+3)!}{(r+3)!\times (n-r)!}$$  So we get Sum $$ \frac{3!}{(n+1)(n+2)(n+3)}\sum^{n}_{r=0}(-1)^r\binom{n+3}{r+3} = -\frac{3!}{(n+1)(n+2)}\sum^{n}_{r=0}(-1)^{r+3}\binom{n+3}{r+3}$$  $$  = -\frac{3!}{(n+1)(n+2)(n+3)}\left[-\binom{n+3}{3}+\binom{n+3}{4}-\binom{n+3}{5}--.......+(-1)^{n+3}\binom{n+3}{n+3}\right]$$  $$=-\frac{3!}{(n+1)(n+2)(n+3)}\left[\binom{n+3}{0}-\binom{n+3}{1}+\binom{n+3}{2}-\binom{n+3}{3}+\binom{n+3}{4}-\binom{n+3}{5}--.......+(-1)^{n+3}\binom{n+3}{n+3}\right]+\frac{3!}{(n+1)(n+2)(n+3)}\left[\binom{n+3}{0}-\binom{n+3}{1}+\binom{n+3}{2}\right]$$  So we get $$ = \frac{3!}{(n+1)(n+2)(n+3)}\left[1-(n+3)+\frac{(n+3)(n+2)}{2}\right] = \frac{3!}{(n+1)(n+2)}\times \frac{n^2+3n+2}{2} = \frac{3!}{(n+1)(n+2)(n+3)}\times \frac{(n+1)(n+2)}{2} = \frac{3!}{2(n+3)}$$  Can we prove it Using Combinatorial way,bcz above method is very lengthy  If yes then plz explain here  Thanks Do you have to want to hurt someone for the cruciatus curse to work? OK so the scene where Harry sees Voldemort making Draco tutur Rowel brings a few questions to my mind.      he felt the rage that did not belong to him posses his soul, saw a long room, only light by fire light, and the grate blond Death Eater on the floor,screaming and writhing, and a slighter figure standing over him, wand outstretched, while Harry spoke in a high, cold, merciless voice.   "More, Rowel, or shall we end it and feed you to Nagini? Lord Voldemort is not sure he will forgive you this time...You called me back for this, to tell me Harry Potter has escaped again? Draco, give Rowel another taste of our displeasure... do it, or feel, my wrath yourself!"   A log fell in the fire: flames reared, their light darting across a terrified, pointed white face-   I remember in the Order of the Phoenix Harry tries to do the cruciatus curse on Bellatrix, it doesn't work. She say this is because Harry doesn't mean it, he doesn't want to hurt he enough. If she's right about that (I'm sure she is) I don't see how Draco is managing to make to make Rowel writhe and scream, because it is clear he doesn't want to be doing it.   Also Harry was actually angry at Bellatrix and if his anger at her killing his godfather won't work... well you get my drift.   I know Draco is probably better at magic, but I feel Bellatrix is right, you really have to want them to hurt for it to work.  Next thing, Voldemort's having Draco do it why? Is he doing to upset Lucius, because that and lot of effort to go to, seeing as Voldemort is angry and he'd probably do a better job himself. Or is he trying to train himself a new soldier, I mean the other ones are getting old. Is it possible to add subtitles of a figure which contains subfigures I have a single image which contains two sub-images. I want the following subtitles (a) and (b) to be shown correctly.   In which, left sides is my current image which contains two rectangular. And right sides is my expected result. I found a solution is that we can use subfig package. However, it requires separating the original figure into two subfigure. In that question, I am regarding that we do not need split original figure, I want to write (a),(b) directly below figure. Thanks Current solution is  \begin{figure}[H] \captionsetup{labelfont={bf}}   \centering     \subfloat[]{\centering\includegraphics[scale=0.5]{Fig1a.png}} \\     \subfloat[]{\centering\includegraphics[scale=0.5]{Fig1b.png}}   \caption{This is main figure}   \label{fig:9} \end{figure} How to create a bootable Ubuntu USB flash drive from terminal? Is there any way to create a bootable Ubuntu USB flash drive from the terminal without using any third-party applications like , , ,  etc.  I tried to create a bootable Ubuntu flash drive with dd method,  sudo umount /dev/sdb sudo dd if=/path/to/ubuntu.iso of=/dev/sdb bs=1M   It creates files on the USB disk, but when I try to boot the USB disk it shows an Operating System Not Found error. What do you call a response which does not address the question? When some one is asked a question, sometimes if they are trying to avoid answering the question, they respond with something unrelated. What is the word for that response?  Eg. A: Why were you late?     B: This bagel tastes good.  What I am looking for is the name of that response, not the action. Multiplicative order and powers Let's consider two number $a$ and $b$ such as $gcd(a,b)=1$. Can you explain me intuitively why there exist $n&gt;0$ such as $a^{n} \equiv 1 \space (mod \space b)$  ?