I'm working on Project Euler, :     The series, 11 + 22 + 33 + ... + 1010 = 10405071317.      Find the last ten digits of the series, 11 + 22 + 33 + ... + 10001000.   This would be the equivalent of: $$ f(n) = \sum_{i = 1} ^ n i^i $$  The original question aside, it got me thinking. The rate of change from n to n+1 would be i^i, so if I could take the integral of ii, I could get a generalized formula for any given n. Upon plugging it into Wolfram Alpha, I got a monstrous expansion, but it said, "no result found in terms of standard mathematical functions". It's been years since I've taken calculus and I'm mostly a computer science guy to begin with, but is this premise sound? And if not, where did I go wrong?
I'm trying to find $\int x^x \, dx$, but the only thing I know how to do is this:  Let $u=x^x$.  $$\begin{align} \int x^x \, dx&amp;=\int u \, du\\[6pt] &amp;=\frac{u^2}{2}\\[6pt] &amp;=\dfrac{\left(x^x\right)^2}{2}\\[6pt] &amp;=\frac{x^{2x}}{2} \end{align}$$  But it's certain that this isn't the correct way to evaluate that, and the answer must be wrong.
I am trying to prove that if I have $f:\mathbb{R} \to \mathbb{R}$ satisfying $\forall x,y\in\mathbb{R},f(x+y) = f(x) + f(y)$. Which is assumed continuous at $0$, that $f$ is continuous on $\mathbb{R}$  I am fairly sure it is, for that property seems to be a property of polynomials, and we know polynomials are continuous where defined(all reals)  Any ideas for rigor?  $|x-a| \lt \delta \implies |f(x+a)-f(x)-f(a)| \lt \epsilon$  No idea where to go from here, this is my first time doing $\epsilon-\delta$ stuff. Vanishing of Taylor series coefficient Comment flags are broken If I flag a comment now, it puts the dialog off the page:        It is only fixed by performing one of these actions:   Maximize window (if not already maximized) Restore window (if maximized)         I am running Windows and Firefox; also I tried Firefox in safe mode. I would be interested to see what testing was done on this beforehand - as this seems like a pretty glaring error. Please and thank you.    Inspired by the answer, the following CSS worked for me. It makes me sad to have to fix Stack Exchange CSS; I would figure this community over all others would get it right. Oh well.  aside {   transform: none !important; } I masked an image in the UV/Image editor masking section. I then used that mask in the compositor for background replacement. However, the mask size seems to be different. Why is that? Here's a screenshot: I read on the :     Focal-length determines the angle-of-view seen through a lens for a   given sensor-size. With a full-frame sensor a lens gives the same   angle-of-view as it would on a 35mm film camera. With a smaller   sensor, the angle-of-view becomes smaller. The crop-factor, also   called FLM, is the ratio representing the difference in equivalent   focal-lengths. So a 150mm on a full-frame DSLR such as the Nikon D700   gives the same angle-of-view as a 100mm on a D7000 since its FLM is   1.5X.      Short focal-lengths show a greater angle-of-view compared to longer   ones. [...]   In layman's terms, what is angle-of-view? Is it the same as focal length? If not, how is it different? How is it used? Why do I need to know about it? I get this message while trying to install Ubuntu 12.04 as a guest in VirutalBox 4.1.18, on an Ubuntu 10.04 host.         This kernel requires the following feature not present on the CPU: pae   Some host specs:     The host's kernel is: Linux 2.6.32-41-generic-pae GNU/Linux   lscpu (host): Architecture: i686, CPU op-mode(s): 32-bit, 64-bit   grep --color=always -i PAE /proc/cpuinfo   does show pae in its output.   The 12.04 iso used is: ubuntu-12.04.0-desktop-i386.iso      As a comparison/check, I downloaded and installed Linux Mint 13 Cinnamon to the same host on the exact same VM (I just changed the .iso image). It worked fine. Its iso is: linuxmint-13-cinnamon-dvd-32bit.iso   It seems (to me) that I have pae.. what is going on here?   Update: I had assumed that Linux Mint also required pae (being Ubuntu based), but I've just run;   grep --color=always -i PAE /proc/cpuinfo   in the Mint VM.   It showed no output.   So it seems the issue may lie with VirtualBox.   If that is the case, how can I get Virtualbox into pae mode? How fast does electricity flow? I've been studying the Huddleston and Pullum book for four months now. So far only one thing confuses me: the identity of gerund. Is it a noun or a verb?        His constant smoking upset me.  smoking seems  noun because of adjective constant.   Him/His constantly smoking upset me. smoking seems to be verb because of adverb constantly.   Him/His smoking cigars upset me. smoking seems to be a verb because of object cigars.      "His smoking upset me." So is this smoking gerund a noun or verb? Because there is no differentiation, is it both verb and noun? Maybe it's new word category?  Also his seems to be both  sometimes a subject and at other times a possessive determiner. Is that correct? I'd really like your help proving:  $(A\rightarrow B),(A\rightarrow C)\rightarrow B,  \mapsto_{HPC} B $  Where $HPC$ is the Hilbert's system proof which contains the following relevant axioms:   $A\rightarrow(B \rightarrow A)$ $(A\rightarrow(B\rightarrow C)) \to ((A\rightarrow B)\rightarrow(A\rightarrow C))$ $(A\rightarrow B)\rightarrow ((A\rightarrow\bar{B})\rightarrow \bar{A})$ $\bar{\bar{A}} \rightarrow A$   In addition tried to use these following lemmas: $\bar{A} \rightarrow (A \rightarrow C) $ and $(A\rightarrow B)\rightarrow (\bar{B} \rightarrow \bar{A})$.  Any suggestions? su command + authentication failure Simple example of non-arithmetic ring (non-distributive ideal lattice) Can anyone provide a simple concrete example of a non-arithmetic commutative and unitary ring (i.e., a commutative and unitary ring in which the lattice of ideals is non-distributive)? Should "each" be followed by a singular or plural possessive? If a possessive noun, which is plural, is preceded by "each", then should it use the singular or plural possessive form?  For example, which of the following is correct?    spend time in each other's presence spend time in each others' presence Proving the limit of a function of a sequence is equal to the function of the limit of that sequence Suppose $f$ is a continuous function at $x = c$ in $[a,b]$.  Prove that for any sequence ${x_n}$ in $[a,b]$ converting to $c$, the sequence $\{f(x_n)\}$ converges to $f(c)$.  That is, $$ \lim_{n\to\infty}f(x_n)= f\left(\lim_{n\to\infty}x_n\right)$$    This proof seems simple but there are a few things that I need to know first. If $\{x_n\}$ converges to $c$, is it sufficient to substitute $c$ in for $\lim_{n\to\infty}x_n$?  Also needing some guidance on the structure of this proof.  Thanks! I tried to use the usb-creator.exe from Ubuntu 10.10 desktop i386 ISO image and it does create an USB stick that doesn't boot. When I try to boot from it I get:     Operating System not found.   I try to do this from Windows 7 x64. I'm not looking to create a live Ubuntu media, only an installation USB disk.  I repeated the process several times and I even tried to reformat the USB drive (FAT32), no success yet.  The USB stick is a 8GB one. If I were to attack with a 4/4 indestructible creature and it's blocked by a 4/2 creature with first strike, would my indestructible creature still attack during the second combat damage step, or would it just tap, not assigning any damage? Finite abelian group contains an element with order equal to the lcm of the orders of its elements I will quote a question from my textbook, to prevent misinterpretation:     Let $G$ be a finite abelian group and let $m$ be the least common multiple of the orders of its elements. Prove that $G$ contains an element of order $m$.   I figured that, if $|G|=n$, then I should interpret the part with the least common multiple as $lcm(|x_1|,\dots,|x_n|)=m$, where $x_i\in G$ for $0\leq i\leq n$, thus, for all such $x_i$, $\exists a_i\in\mathbb{N}$ such that $m=|x_i|a_i$. I guess I should use the fact that $|x_i|$ divides $|G|$, so $\exists k\in \mathbb{N}$ such that $|G|=k|x_i|$ for all $x_i\in G$. I'm not really sure how to go from here, in particular how I should use the fact that $G$ is abelian. We are doing research on monitor human activities to avoid adverse events. In part of it, we did an experiment in our MoCap lap to detect the human motion to monitor his activities. We have f-curves from that motion. But we need all the xyz coordinates values for that motion for that given time (just say about 2-3 minutes) to do a scientific analysis of the data.  Do anyone have an idea how to get xyz values from f-curves? I am trying to solve the following exercise:     Prove that complex multiplication does not extend to a multiplication   on $\mathbb R^3$ so as to make $\mathbb R^3$ into a real division   algebra.   I am aware of Frobenius' theorem that there are only three finite dimensional real associative division algebras. But the exercise would be pointless if the theorem was assumed.   Let $x,y \in \mathbb R^3$. Since $x y$ has to extend the complex multiplication:  $$xy = (x_1y_1-x_2y_2, x_1y_2 + x_2 y_1, ?)$$  I have no idea how I can proceed from here. Could someone tell me how to do this? It should be easy as the other exercises I did so far were also easy. Two column float with [t] using the cas-dc template I am using the cas-dc template for Elsevier. When trying to use \begin{figure*}[t], except I get the error 'The module 'cas/fig' does not have a key called 'cas/fig/t'. Other templates I use work fine. I have included \usepackage{stfloats}. I don't see anything mentioned that strictly forbids it, unless I missed this. Anyone know why? I have made a fancy title page using titleformat and tikz, but I now face the following problem. The \chaptertitlename command returns \chaptername or \appendixname. I would like to spread the letters of the text returned by \chaptertitlename to a given length.  For example I would like to be able to call: \mycommand{\chaptertitlename}{3cm} in order to space the letters of \chaptertitlename on a total length of 3cm. How to do that ?  EDIT: This example illustrates a nearly working thing:  \documentclass[11pt]{article} \usepackage{fontspec} \newcommand{\spreadletters}[2]{\makebox[#2][s]{#1}}  \begin{document} \spreadletters{m y w o r d}{3cm} \end{document}   How to convert myword to m y w o r d ?  EDIT 2: Can someone explain me why the first work and not the second one (and how to make it work) ?  \documentclass[11pt]{book} \usepackage{fontspec} \usepackage{seqsplit} \usepackage[explicit]{titlesec}  \newcommand{\spreadletters}[2]{\makebox[#2][s]{\seqsplit{#1}}}  \begin{document} \chapter{First} \spreadletters{myword}{10cm} \linebreak % This is working \spreadletters{\chaptertitlename}{10cm} % This is not working \end{document}