Let n and r be positive integers with n ≥ r.
How to prove this statement: $\binom{r}{r}+\binom{r+1}{r}+\cdots+\binom{n}{r}=\binom{n+1}{r+1}$
Newcommand + Stepcounter I'm trying to come up with a command which should increase a counter and return a string including the current value of the counter.  Have a look at the following minimal example:  \documentclass{article}  \newcounter{mycounter} \newcommand{\mystepFAILS}{\stepcounter{mycounter}foo\themycounter} \newcommand{\mystepWORKS}{foo\themycounter}  \begin{document}  \section{\mystepFAILS}  \section{\mystepWORKS}  \end{document}   Why does the command with \mystepFAILS throw an error? What is so special about \stepcounter? Any ideas? Should the site prompt new users to include an @name in comments? Examples:   In a car, van, etc. On a bus, boat, motorcycle, etc.   How can one decide which preposition to use? Is memorization the only way or is there a better way?  Note: People generally explain this by either distinguishing between open and closed vehicles or between large and small vehicles. However, the examples I've given defeat both explanations. Do database engines update when the updated value is the same? What is an unsigned char? In C/C++, what an unsigned char is used for? How is it different from a regular char? Let's say I am missing my novel and I think that my friend might have taken it after I left the place.  1: By the time I left my apartment, my friend was still there, so he might have taken it. 2:When I left the apartment, my friend was still there, so he might have taken it.  I know what 2nd expression means, but not so sure about the first expression. Does it convey the same meaning as the 2nd expression? Trouble understanding Landau & Lifshitz writing about Lagrangians and Galilean Relativity We have two inertial coordinate systems, $K'$ and $K$. $K$ is moving with infinitesimal velocity ${\epsilon}$ relative to $K'$. Using Galilean relativity we can transform this into $v'=v+{\epsilon}$. Here Lagrangian is only a function of speed $L=L(|v|^2)$. $L'$ may differ from $L$ only by a total derivative of function of coordinates and time, i.e. $L'=L+\frac{d}{dt}f(q,t)$  $"$We have $L'=L(|v|^2+2{v}·{\epsilon}+{\epsilon}^2)$. [1. ] Expanding this expression in powers of ${\epsilon}$ and neglecting terms above the first order, we obtain $L(|v'|^2)=L(|v|^2)+\frac{\partial{L}}{\partial{|v|^2}}2v·{\epsilon}$ [2.] The second term on the right of this equation is a total time derivative only if it is a linear function of the velocity $v$ $"$  I understand nothing in the quoted part. How do we get [2.]? Why does $\frac{\partial{L}}{\partial{|v|^2}}2v·{\epsilon}$ being a linear function of velocity imply that it's a total time derivative? Is fractional scaling ( ) available in 18.04 LTS yet? I am waiting until this is available before I install. Inaccurate shading on object I am relatively new to Blender so please excuse me if this is rather amateur!   The bottom of my torus has poor surface quality, I have scaled the vertices down for a smaller (variable) tubular radius all the way around the torus and at the bottom the surface quality is awful, with inaccurate shading. Any ideas as to what is wrong and/or how I could fix this?  Thanks, itsanate I am on Ubuntu and I did this command:  $ uname -a Linux slabrams-desktop 2.6.32-29-generic #58-Ubuntu SMP Fri Feb 11 19:00:09 UTC 2011 i686 GNU/Linux   Does it mean I am on 32bit or 64 bit processor?  The reason I am trying to figure this out is that I was getting errors which looked like this:  cannot execute binary file   and from Googling, I thought it was a processor issue.  Any ideas? Including the interaction but not the main effects in a model Is there an easy way to remove overlapping faces? In the example remove doubles doesn't work  You might be able to do the following: A to select all faces. X &gt; Only Faces. F to create faces, but after deleting the faces I'm left with something like this.  If I press A to select all and then press F to create faces then there are still overlapping. Blend file Can't reset WordPress password I haven't logged onto my blog for years, and have, of course, forgotten the password. I click the 'Forgot' button and get a mail with a link, but that mail already says     Sorry, that key does not appear to be valid.   And sends me round the click to reset, receive a mail, click the link again etc. As if the timeout for password resets has been set to 0.5s or something.  This is a shared hosted blog, so I have full access to cPanel and all of the WP files via FTP, if that helps anything. E: Could not get lock /var/lib/dpkg/lock-frontend - open (11: Resource temporarily unavailable) How can I execute a function on pressing the enter key in an  field? Proof that $\frac {1} {2\pi i} \oint \frac {{\rm d}z} {P(z)} $ over a closed curve is zero. Prove that $$\frac {1} {2\pi i} \oint \frac {{\rm d}z} {P(z)} $$ over a closed  smooth simple curve that contains all of the roots of the polynomial $P(z)$ is either zero if $n \geq 2$ or equals $\frac{1} {a_0} $ if it's degree is $1$.   Proof: $n\geq 2$ $$\frac {1} {P(z)=(z-z_0)(z-z_1)\dots(z-z_n)}$$ So after partial decomposition of the fraction I'll have something like $$\frac {A} {z-z_0} +\frac{B} {z-z_1}+\cdots$$ where $A,B \in \mathbb{C}$ (Is this fact true that $A,B$ wont be polynomials and just coefficients I can see it why in an intuitive way but cannot write a full proof of it for the decomposition trying to get a linear system or something) then use Cauchy's theorem and integrate all of these on  different circles with their center at each root  and argue that  the integral is the same as the sum of those circle integrals. Now since $\frac {1}{z-z_n}$ will be analytic inside the circle when I'm integrating on circles that do not contain $z_n$  some  of those integrals  will be zero. and now since my curves are of the form $\gamma(t)=z_n+e^{it}$ , $t \in [0,2\pi]$ the rest of the integrals  will look like $$\int_0^{2\pi}  \frac{Aie^{ti}} {e^{it}}\,{\rm d}t $$  and all of them will be $A_i$.   Is my proof correct? Only part I cannot show is that I won't have any problem with decomposing the fraction. And I'm also using the assumption that after the decomposition I'll have as numerators simple complex numbers not not functions of any kind. Windows says RAM ran out while there is still 4 GB of physical memory available Baby Rudin Theorem 1.20 (b) Proof I have a question about Rudin's proof of Theorem 1.20 (b) in his book Principles of Mathematical Analysis.  Theorem 1.20 is stated as follows:     (a) If $x\in R, y\in R$, and $x&gt;0$, then there is a positive integer $n$ such that $$nx&gt;y.$$   (b) If $x\in R, y\in R$, and $x&lt;y$, then there exists a $p\in Q$ such that $x&lt;p&lt;y$.     I understand Rudin's proof of (a).  The beginning of Rudin's proof of (b) is given below:     Since $x&lt;y$, we have $y-x&gt;0$, and (a) furnishes a positive integer $n$ such that $$n(y-x)&gt;1.$$  Apply (a) again, to obtain positive integers $m_1$ and $m_2$ such that $m_1&gt;nx$, $m_2&gt;-nx$.  Then $$-m_2&lt;nx&lt;m_1.$$  Hence there is an integer $m$ (with $-m_2\leq m\leq m_1$) such that $$m-1\leq nx&lt;m.$$   I don't understand the justification for this last sentence beginning "Hence...."  How is $m$ found, and why are $m_1$ and $m_2$ needed to find $m$? Angle Constructions Over the numerous math tests I've taken, I have yet to understand what angle can be constructed and what cannot. From my understanding, 30, 45, 60, and 90 degree angles can be constructed. You can then use these angles and subtract or add them to create other angles like 15 or 75. Also, any angle can be bisected. Are there any other basic angle measures that can be constructed like 30, 45, etc.? So 82.5° is an angle measure  that can be constructed because you can add 60+60+45 to get 165 and then bisect it. Just a quick question from an Objective-C beginner: what does the asterisk in the parenthesis mean in the following sample code? Is it a pointer? Thanks in advance.  #import &lt;Foundation/Foundation.h&gt;  @interface Tire : NSObject @end @implementation Tire  - (NSString *) description {    return (@"I am a tire."); }  @end