Evaluate:   $$0^0$$   Would you use this law of indices?     $$x^0=1$$   Or would you use that:     $$\frac{x^n}{x^n}=x^0$$ which would mean $$\frac{0}{0}=undefined$$
I have been a witness to many a discussion about numbers to the power of zero, but I have never really been sold on any claims or explanations.  This is a three part question, the parts are as follows...    1) Why does $n^{0}=1$ when $n\neq 0$?  How does that get defined?  2) What is $0^{0}$?  Is it undefined?  If so, why does it not equal 1?  3) What is the equation that defines exponents?  I can easily write a small program to do it (see below), but what about in equation format?    I just want a little discussion about numbers to the power of zero, for some clarification.    Code for Exponents: (pseudo-code/Ruby)  def int find_exp (int x, int n){     int total = 1;     n.times{total*=x}     return total; }
How do I change the launcher commands? Orbit followed by a particle around Schwarzschild Black Hole How to set systemd for automatically force fsck disks after crash (hard poweroff)?  When I used sysvinit (in Arch Linux) as /sbin/init I used the hack: in the rc.local I create /forcefsck file; in the rc.local.shutdown I remove it. At boot-up rc.sysinit enable force fsck if /forcefsck exists.  How to do same in systemd? Maybe it has built-in facilities for automatic fsck after crush? Why does DNS work the way it does? This is a  about DNS (Domain Name Service).   If my understanding of the DNS system is correct, the .com registry holds a table that maps domains (www.example.com) to DNS servers.   What is the advantage? Why not map directly to an IP address? If the only record that needs to change when I am configuring a DNS server to point to a different IP address, is located at the DNS server, why isn't the process instant? If the only reason for the delay are DNS caches, is it possible to bypass them, so I can see what is happening in real time? Primary decomposition of a monomial ideal Can anyone give me an idea about the primary decomposition of the ideal $I=(x^3y,xy^4)$ of the ring $R=k[x,y]$?    I am trying to connect the primary decomposition with the set Ass(R/I) which i have find before.. How can I force my browser to search Google in English? I'm too bored of seeing sites like Google and such show up in my native language, I would rather like them to be in English. Yet, I have to explicitly change the URL to .com and en and that kind of parameters in order for them to show up in English. Can I somehow force this?  So, how is Google configured?  However, it is set to English on the site itself so it has to be my browser:    Then, how does my browser land up on non-english pages, like Google?  It usually shows up in non-English when I'm performing a search, which uses:  {google:baseURL}search?{google:RLZ}{google:acceptedSuggestion}{google:originalQueryForSuggestion}{google:searchFieldtrialParameter}{google:instantFieldTrialGroupParameter}sourceid=chrome&amp;ie={inputEncoding}&amp;q=%s   When performing a search, it fills these variables in with non-english values.  How can I tell my browser to fill these in with the English values?  My Google Chrome options give preference to English: Evaluate $\int_0^\infty \frac{(\log x)^2}{1+x^2} dx$ using complex analysis How do I compute      $$\int_0^\infty \frac{(\log x)^2}{1+x^2} dx$$   What I am doing is take   $$f(z)=\frac{(\log z)^2}{1+z^2}$$  and calculating   $\text{Res}(f,z=i) = \dfrac{d}{dz} \dfrac{(\log z)^2}{1+z^2}$ which came out to be $\dfrac{\pi}{2}-\dfrac{i\pi^2}{8}+\dfrac{i\pi}{2}$  Im not too sure how to move on from here. the given answer is $\dfrac{\pi^3}{8}$  Any help will be appreciated. thank you in advanced. How would I prove that if an integer $n>2$, then $\bar{z}=z^{n-1}$ has $n+1$ solutions If $n$ is an integer and $n&gt;2$, then $$\bar{z}=z^{n-1}$$ has $n+1$ solutions. Does this have something to do with the rational root theorem? Is there a way to determine which features / variables of the dataset are the most important / dominant within a k-means cluster solution? $\ker \phi = (x_1-a_1, ..., x_n-a_n)$ for a ring homomorphism $\phi: R[x_1, ..., x_n] \to R$ Let $R$ be a commutative ring, $a_1, ..., a_n$ its elements and  $\phi: R[x_1, ..., x_n] \to R$  defined by  $ \phi(f(x_1, ..., x_n)) = f(a_1, ... ,a_n)$  a ring homomorphism.  Prove: $\ker \phi = (x_1-a_1, ..., x_n-a_n)$  It is obvious that $(x_1 - a_1, ..., x_n -a_n) \subseteq \ker \phi$. I'm not sure how to prove the converse.  At this point I don't know any division algorithms for multivariable polynomials, only for the ones in $R[x]$(and the book from where I taken the exercise doesn't assume the reader to know something beyond basics of rings and ideals, and the division algorithm for $R[x]$). Though I know this could be solved for $i = 1$ by dividing by $x-a$:   Let $f(x) \in \ker \phi$, divide by $x-a: f(x) = q(x)(x-a) + r, f(a) = r = 0$, so $f(x) = q(x)(x-a) \in (x-a)$. Should I activate the additional driver: Processor microcode firmware for Intel CPUs for intel-microcode I have recently noticed that there is an additional driver called Unknown: Unknown here which was not there before:    I have already read , and understand why it is there, but my question is, should I activate it? Will this make a difference? Or should I just leave it as it is as it sounds rather useful?    OS Information:  Description:    Ubuntu 15.04 Release:    15.04 Looking for title/author of SF novel about malevolent creatures/gods residing inside stars I was trying to give the wording "isomorphic groups have the same structure" a precise interpretation, at least for the finite order case. A positive answer to my final question would reach the goal, as it would translate a "$\cong$" between groups into a "$=$" between sets.    For $n$ positive integer, let be:   $I_n:=\{1,\dots,n\}$; $G$, $\overline G$ groups of order $n$; $\psi\colon G \rightarrow \overline G$ isomorphism; $f$, $\bar f$ bijections; $\theta$, $\bar \theta$ embeddings; in general, $\varphi^{(\alpha)}$ the isomorphism between symmetric groups on sets of the same cardinality, defined by $\sigma \mapsto (g \mapsto (\alpha\sigma\alpha^{-1})(g))$, where $\alpha$ is a bijection between the sets; $S_n$ the symmetric group of degree $n$.   Visually:  $$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c}    I_n     &amp;   &amp;    I_n      \\     \da{f}         &amp;           &amp;     \da{\bar f}          \\    G &amp; \ras{\psi} &amp;    \overline G  \\     \da{\theta}         &amp;           &amp;     \da{\bar \theta}          \\    \operatorname{Sym}(G) &amp; \ras{\varphi^{(\psi)}} &amp;    \operatorname{Sym}(\overline G)  \\     \da{\varphi^{(f)}}         &amp;           &amp;     \da{\varphi^{(\bar f)}}          \\    S_n &amp;  &amp;    S_n  \\ \end{array} $$     Is it $(\varphi^{(f)}\circ\theta)(G)=(\varphi^{(\bar f)}\circ\bar \theta)(\overline G)$? I want to delete all *.o files in a directory and its sub-directories. However, I get an error: sashoalm@aspire:~/.Workspace.OLD$ rm -r *.o rm: cannot remove `*.o': No such file or directory  On the other hand, rm *.o works, but it's not recursive. Python Ctypes - loading dll throws OSError: [WinError 193] %1 is not a valid Win32 application How can I show that if some group $G$ has only one subgroup $K$ of order $n$, then $K$ is a normal subgroup? Would that mean that it only has one subgroup total? If so then I guess that makes sense. How to get around a problem with \usetikzlibrary{babel}? I am currently working on an update of tkz-euclide and new documentation but I just met the following problem. If I compile this code with pdflatex or lualatex and pgf 2.1 (The library babel does not exist with pgf 2.1) then the code compiles fine but with 3.0 pgf I must comment on the call \usetikzlibrary{babel} (pgf doc gives so I recommend to always load this library by saying \usetikzlibrary{babel}).  I'd like to know where the error occurred. Especially I would like to work around this problem.   With \usetikzlibrary{babel} I get the error Latex Error: ./untitled 2.tex:46 Undefined control sequence. l.1 \tkz@line@start  Bug in babel, bug in tikzlibrary{babel} bug in my code. I would like to keep  \usetikzlibrary{babel} but to avoid the error.  %!TEX TS-program = pdflatex \documentclass[a4paper]{article} \usepackage[french]{babel} \usepackage{tikz}  \usetikzlibrary{calc} %\usetikzlibrary{babel}  \makeatletter \pgfkeys{%     tkzdrawline/.cd,     start/.code         =  {\gdef\tkz@line@start{#1}},%     end/.code           =  {\gdef\tkz@line@end{#1}},%     start style/.code   =  {\tikzset{tkzstartstyle/.style={#1}}},%     end style/.code     =  {\tikzset{tkzendstyle/.style={#1}}},%    /tkzdrawline/.unknown/.code ={\let\searchname=\pgfkeyscurrentname\pgfkeysalso{%                          \searchname/.try=#1,/tikz/\searchname/.retry=#1}}}  %&lt;--------------------------------------------------------------–&gt; %                              Draw line %&lt;--------------------------------------------------------------–&gt; \pgfkeys{tkzdrawline/.cd, start/.code       =  {\def\tkz@line@start{#1}},% end/.code         =  {\def\tkz@line@end{#1}}, start style/.code =  {\tikzset{tkzstartstyle/.style={#1}}}, end style/.code   =  {\tikzset{tkzendstyle/.style={#1}}},   /tkzdrawline/.unknown/.code   = {\let\searchname=\pgfkeyscurrentname                                  \pgfkeysalso{\searchname/.try=#1,                                 /tikz/\searchname/.retry=#1}}}%         %&lt;--------------------------------------------------------------–&gt;  \def\tkzDrawLine{\pgfutil@ifnextchar[{\tkz@DrawLine}{%                                   \tkz@DrawLine[]}}%   \def\tkz@DrawLine[#1](#2,#3){%  \begingroup   \pgfkeys{tkzdrawline/.cd,start = {}, end = {}, start style={}, end style={},}   \pgfqkeys{/tkzdrawline}{#1}   \draw[ /tkzdrawline/.cd,#1] (#2) to%     node [at start,left,/tkzdrawline/.cd,tkzstartstyle] {\tkz@line@start}%     node [at end,right,/tkzdrawline/.cd,tkzendstyle] {\tkz@line@end} (#3);  \endgroup  }%   \makeatother   \begin{document}    \begin{tikzpicture}      \node (A) at (0,1) {};       \node (B) at (5,2) {};       \tkzDrawLine[end = {$\alpha$}](A,B)       \end{tikzpicture}  \end{document} Is a Schengen visa/transit visa required for a transit flight through two Schengen airports? EJB 3.1 Dependency Injection Failed Does there exist a magical shell piping which would allow easily to grep through bunch of .gz log files without needing to extract them somewhere?  .gz files are Apache logs, result of log rotation. I'd like to quickly check how often certain URIs are accessed in the past.