Proving that $\binom{r}{r} + \binom{r + 1}{r} + \cdots + \binom{n}{r} = \binom{n + 1}{r + 1}$
Proof of the Hockey-Stick Identity: $\sum\limits_{t=0}^n \binom tk = \binom{n+1}{k+1}$
Negative values for AICc (corrected Akaike Information Criterion) I have calculated AIC and AICc to compare two general linear mixed models; The AICs are positive with model 1 having a lower AIC than model 2.  However, the values for AICc are both negative (model 1 is still &lt; model 2).  Is it valid to use and compare negative AICc values? A closed ball in a metric space is a closed set Prove that a closed ball in a metric space is a closed set   My attempt: Suppose $D(x_0, r)$ is a closed ball. We show that $X \setminus D $ is open. In other words, we need to find an open ball contained in $X \setminus D$.   Pick $$t \in X-D \implies d(t,x_0) &gt; r \implies d(t,x_0) - r &gt; 0 $$ Let $B(y, r_1)$ be an open ball, and pick $z \in B(y,r_1)$. Then, we must have $d(y,z) &lt; r_1 $. We need to choose $r_1$ so that $d(z,x_0) &gt; r$. Notice by the triangle inequality  $$ d(x_0,t) \leq d(x_0,z) + d(z,t) \implies d(z,x_0) \geq d(x_0,t) - d(z,t) &gt; d(x_0,t) - r_1.$$  Notice, if we pick $r_1 = d(t,x_0)-r$ then we are done.  Is this correct? Are +1/+1 counters different from +1/+1 effects? SSMS wont give up file associations I want to calculate the posterior distribution given a gamma-prior and a poisson likelihood. The task is from a textbook and I just have the solutions (without a walkthrough). Please find all given information in the following:  Given information:  $m$ countries each with a population of $n_i$; $y_i$ is the number of people having a disease; $y_i$ is a poisson random variable with mean $\frac{n_i*\lambda}{100}$  The prior distribution for $\lambda$: $\lambda \sim Gamma(a_0,b_0)$  Task:  Show that the posterior distribution is also a gamma distribution with $a_m = a_0 + \sum_{i = 1}^{m}y_i$ and $b_m = b_0 + \sum_{i = 1}^{m} \frac{n_i}{100}$  My solution:  $Posterior \propto Likelihood \times Prior$  $p(\theta|y) \propto \prod_{i=1}^{m}(\theta^{y_i}e^{-\theta})\times \theta^{a_0-1}e^{-b_0\theta}$  $p(\theta|y) \propto \theta^{\sum_{i = 1}^{m}y_i}e^{-m\theta}\times \theta^{a_0-1}e^{-b_0\theta}$  $p(\theta|y) \propto \theta^{\sum_{i = 1}^{m}y_i+a_0-1}e^{-\theta(m+b_0)}$  $a_m = a_0 + \sum_{i = 1}^{m}y_i$  $b_m = b_0 + m \neq b_0 + \sum_{i = 1}^{m} \frac{n_i}{100}$  I think I am doing a mistake, because I don't know how to incorporate the information that the mean of $y_i$ is given as $\frac{n_i*\lambda}{100}$. Or do I just have to transform $b_m$ somehow? Can someone may give me a hint?   Thank you already in advance!  Edit 1: Considering first advices  $Likelihood: p(y | \theta) \propto \prod_{i=1}^{m}(\frac{n_i\lambda}{100})^{y_i}e^{-\frac{n_i\lambda}{100}}$  $\propto \prod_{i = 1}^{m} ((\frac{n_i}{100})^{y_i})\lambda^{\sum_{i=1}^{m}y_i}e^{-\lambda\sum_{i=1}^{m}\frac{n_i}{100}}$  $\propto \lambda^{\sum_{i=1}^{m}y_i}e^{-\lambda\sum_{i=1}^{m}\frac{n_i}{100}}$  Is the transformation of $\prod_{i=1}^{m}$ I did from the first to the second step correct? Nonlinear function continuous but not bounded I would like an example of a map $f:H\rightarrow R$, where $H$ is a (infinite dimensional) Hilbert space, and $R$ is the real numbers, such that $f$ is continuous, but $f$ is not bounded on the close unit ball $\{ x\in H : \|x\| \leq 1\}$.  Actually, $H$ could be replaced by any Banach space (but not just a normed space-- that's too easy).  My motivation is that if $f$ is linear, this is impossible; but I have next to no intuition about non-linear functions.  Edit: Here's an example for $c_0$ which is even differentiable (disclaimer: I found it here: ).  Define $f:c_0\rightarrow F$ (where F is your field, real or complex) by $$ f(x) = \sum_{n=1}^\infty x_n^n \qquad (x=(x_n)). $$  You can estimate the sum by a geometric progression, so it does converge.  A bit of checking shows that f is Frechet differentible (so certainly continuous).  But $f(1,1,\cdots,1,0,\cdots)=n$ (if there are $n$ ones) so $f$ is not bounded on the closed unit ball.  What I don't immediately see is how to adapt this to $\ell^2$, say. Can I deploy Juju on Eucalyptus? Please let me know if I can deploy juju on Eucalyptus, if yes please provide me the configuration details which i need to edit on environments.yaml file. Get name of caller function in PHP? How can I install Sun/Oracle's proprietary Java JDK 6/7/8 or JRE? HTML Element Array, name="something[]" or name="something"? Missing abstract with amsart class I'm not very good at LaTeX but I've been using it to write my dissertation. Have just come to the end and trying to format everything correctly. Whenever I try to put my Abstract paragraph into the form \begin{abstract} (writing here) \end{abstract} and click play the whole abstract section just disappears. What am I doing wrong?  \documentclass[12pt]{amsart}  \begin{document}  \begin{abstract} (my abstract here) \end{abstract} \end{document} I can't seem to find a satisfactory answer to this anywhere on the web. While  poses a related question, I don't think it sufficiently answers mine...  During the opening chapters of Jedi Search, Luke addresses the new Senate on the need to find new Jedi talent, or Force sensitive individuals. During this proposal, he cites that the most fruitful leads of Jedi potential are in the descendants of Jedi Knights: as Force sensitivity is hereditary. The book even discusses the notion that Leia should be comparatively well attuned to the Force, due to her being the child of Anakin Skywalker.  I understand that the concept of the Jedi as a totally celibate and monastic order was something that was retconned into the lore after Episode II: Attack of the Clones, and that there are examples of entire nuclear Jedi families exist (such as Andur Sunrider and Plo Koon)...  But if Force-manipulative ability is passed down genetically (as a potentially mutative property of their genetic makeup), why would the Jedi order be so numerous if they practice Chastity?  Even if procreation is not expressly forbidden (as in the pre-retcon Expanded Universe): why would Jedi need special permission to produce more Jedi potential children?  How are more Jedi created, if not through procreation? Has their profusion been deliberately retarded, and if so why? is there another explanation for their (pre-empirical) plenitude? I have this directory /foo/bar/. Inside, I have a python file named myfile.py. I made it executable by adding (at the top):  #!/usr/bin/env python   Then, I made it executable by issuing   chmod +x myfile.py   I edited my path by issuing  sudo nano /etc/paths   I added /foo/bar/ to it.  The problem is that although I can run myfile.py by typing ./myfile.py while I am inside /foo/bar/, it doesn't work from any other location. What am I doing wrong?  Thank you. Does voltage drop occur with zero load and zero resistance Is there an intelligent way to left-align the parts right of &amp; of the two cases environments shown below (if x >= 0, otherwise, if y &lt; 0, otherwise). I wonder if there is a better way than using manual spaces:  \begin{align*}  u(x) &amp;=   \begin{cases}    0        &amp; \text{if } x \geq 0 \\    1        &amp; \text{otherwise}   \end{cases}   \\  \tau &amp;=   \begin{cases}    a+b+c+d  &amp; \text{if } y &lt; 0 \\    1        &amp; \text{otherwise}   \end{cases} \end{align*} Applications of complex numbers to solve non-complex problems Recently I asked a question regarding the diophantine equation $x^2+y^2=z^n$ for $x, y, z, n \in \mathbb{N}$, which to my surprise was answered with the help complex numbers. I find it fascinating that for a question which only concerns integers, and whose answers can only be integers, such an elegant solution comes from the seemingly unrelated complex numbers - looking only at the question and solution one would never suspect that complex numbers were lurking behind the curtain!  Can anyone give some more examples where a problem which seems to deal entirely with real numbers can be solved using complex numbers behind the scenes? One other example which springs to mind for me is solving a homogeneous second order differential equation whose coefficients form a quadratic with complex roots, which in some cases gives real solutions for real coefficients but requires complex arithmetic to calculate.  (If anyone is interested, the original question I asked can be found here: )  EDIT:  I just wanted to thank everyone for all the great answers! I'm working my way through all of them, although some are beyond me for now! US and Canada border crossing and computer privacy: What do I need to know? I've read stories about border officials searching laptops when crossing the border in one direction or the other. Does this really happen, typically speaking?  As a US citizen, can I be forced to give them my laptop's password for any/no reason? I don't believe they are allowed to deny entry to the US to a US citizen? ....but who knows these days?  Canada can deny me entry for any reason, so that's fine. But in practice, does this even happen in either crossing direction?  EDIT - I've now made the trip. They didn't so much as glance in the direction of my laptop. My biggest issue turned out to be lack of international wireless data service! Why does the Community user approve and reject edits? Why does the Community user sometimes approve or reject edits? How does a machine know if my edit is valid? What does it mean if my edit was rejected by Community?     See also: Etiquette for revision comments I have two users, user1 and user2, that are both members of groupA. user2 has a folder in their home directory called folderA. If they wish to allow read-write-execute permissions for all members of groupA, how would they do this?  What if folderA contains many files and additional folders that also need to have read-write-execute permission?  Information regarding groups is a little 'spotty' across the web, so I am putting my question here in the hope someone posts a clear answer that might help others out too.  Thanks!