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Word for low level noise among restless crowd, as sign of confusion or complaint? Dictionary gives me "drone" and "murmur" but I'm not sure if that's the word I am thinking of.  I want to use it to describe a group of people on strike who hear that their demands are not met and the CEO has resigned.  They're not shouting, but they're not whispering either.  They're restless and confused.  I want to say that a certain something (murmur?) was heard among the crowds, to indicate their state. Object not visible in render I have a model which you can see here (camera perspective).    In the second image, you can see how the render turned out.    I'm at my wit's end trying to figure our why it's not showing up in the render. If anyone needs more information, just ask and I'll get it to you. I've already checked the normals and they're not the cause. When to use RSA and when ElGamal asymmetric encryption Cauchy condensation test The following is Cauchy condensation test:  If $b_n \ge 0$ and $b_n \ge b_{n+1}$ then $\sum b_n$ converges if and only if $\sum 2^n b_{2^n}$ converges.  Is the following modified Cauchy condensation test possible:  Let $N \in \mathbb N$. Then if $b_n \ge 0$ and $b_n \ge b_{n+1}$ then $\sum b_n$ converges if and only if $\sum N^n b_{N^n}$ converges? I have a 32GB iPhone 4 version 4.0.1. I have many photos on my phone, but I have deleted an important photo from my iPhone. How can I get this photo back from my iPhone? How to justify the definition of summation $s_n=\sum_{i=1}^n a_i=a_1+\cdots+a_n$? I read  and found that they define summation via recursion, so I decided to formalize the proof that that this definition is actually valid. I've two questions:   Does my proof contain any error? Are there other simple ways to define summation?   Thank you so much!     Suppose that $(a_1,\cdots,a_n)$ is a finite sequence in $\mathbb N$. Show that there is a sequence $(s_1,\cdots,s_n)$ such that $s_1=a_1$ and $s_{i+1}=s_i+a_{i+1}$ for all $1\leq i&lt;n$.   My attempt:  We define mapping $f$ as follows: $f: \mathbb N\times\mathbb N\to\mathbb N\times\mathbb N: (i,a)\mapsto\begin{cases} (i+1,a+a_{i+1})&amp;\text{if }i&lt;n\\ (i+1,a)&amp;\text{if }i\geq n \end{cases}$  By recursion theorem, there is a unique sequence $(p_i\mid i\in\mathbb N)$ such that $p_0=(1,a_1)$ and $p_{i+1}=f(p_i)$. Let $\pi:\mathbb N\times\mathbb N\to\mathbb N$ be the projection to the second co-ordinate i.e. $\pi(i,a)=a$. Let $s_i=\pi(p_i)$ for all $1\leq i\leq n$, then $(s_i\mid 1\leq i\leq n)$ is the required sequence. It's clear from the definition of $s_i$ that $s_{i+1}=s_i+a_{i+1}$ for all $1\leq i&lt;n$. A little bird told me that if $2^n+1$ is prime, then $n$ is a power of $2$. I tend not to trust talking birds, so I'm trying to verify that statement independently.  Suppose $n$ is not a power of $2$. Then $n = a \cdot 2^m$ for some $a$ not a power of $2$ and some integer $m$. This gives $2^n+1 = 2^{a \cdot 2^m}+1$. Now I suspect there's a way to factor that, but I don't see how. Can someone give me a hint? Definition of complex number In many situations (problems as well as solutions) I encounter the complex number $i$ which many times is defined as $i^2=-1$ instead of $i=\sqrt{-1}$, since it is "more preferred". Does anyone know why? Is it because there are two solutions to the equation $x^2+a=0, \ \ a&gt;0$? If $M$ isn't Noetherian, $M$ has a submodule maximal with respect to being not finitely generated. I'm answering this question:     Let $M$  be a module. Show that if  $M$ is not Noetherian then $M$ has a submodule $N$ such that $N$ is not finitely generated but $A$ is finitely generated whenever $N &lt; A ≤ M$.   The solution starts as follows (I'm following a given solution): Let $Y$ be the set of non-finitely-generated submodules of $M$. If $(N_α)$ is a chain in $Y$ then $Q= \bigcup_αN_α $ is in $Y$, else $Q$ is f.g., which implies $Q=N_α$ for some $α$, whence $N_α$  is f.g., contradiction.  I don't understand why if $Q$ is f.g., then $Q=N_α$ for some $α$? Am I missing some fact?  Thanks! Box Cox Transforms for regression I'm trying to fit a linear model on some data with just one predictor (say (x,y)). The data is such that for small values of x, the y values give a tight fit to a straight line, however as x values increase, the y values become more volatile. Here is an example of such data (R code)  y = c(3.2,3.4,3.5,3.8,4.2,5.5,4.5,6.8,7.4,5.9) x = seq(1,10,1)   I'm curious to know if there exists any power transform (Box cox perhaps?) that lets me get a better fit for the data than simply doing a linear fit as shown below.  fit = lm(y ~ x) Comparison of backup tools How to install software or upgrade from an old unsupported release? There is the next characterization for valuation domains:  Let $D$ be an integral domain and $K$ its field of fractions. The following are equivalent:   For every nonzero $x$ in $K$, either $x$ in $D$ or $x^{−1}$ in $D$. The ideals of $D$ are totally ordered by inclusion. The principal ideals of $D$ are totally ordered by inclusion (i.e., the elements in D are totally ordered by divisibility.) There is a totally ordered abelian group $Γ$ (called the value group) and a surjective group homomorphism (called the valuation) $v :$ $K^×$ $→ Γ$ with $D = \{ x ∈$ $K^×$ | $v(x)$ ≥ $0 \}$ $∪$ {0}.   The equivalence between 1, 2 and 3 are clear, and I understand that the fourth can be found in Krull's article (1936), but I don't speak german and I can't find a proof anywhere. Any link or hint will be much appreciated. Define a relation and find its equivalence classes. Show all the file up to the match grep --before-context 5 shows 5 lines before the match.  I want to show everything before the match. Doing grep --before-context 99999999 would work but it is not very... professional.  How to show all the file up to the match? I frequently find myself needing a word to express happiness that acknowledges a tragic or unfortunate circumstance underpinning that happiness. Preferably the word could have both adjective and verb forms. For example, "Gregory [verb-ed | felt adjective about] the balmy February day, suspecting it was yet another consequence of catastrophic and irreversible climate change."  "Bittersweet" is close, but not precisely right. It implies ambivalence rather than genuine gladness. "Schadenfreude" is also in the neighborhood, but I don't want to imply that the tragic part of the situation is what causes the pleasure. (Also, I recognize that the word may not quite exist in English, and I don't know if this is forbidden for this forum, but I'd welcome a loan word or phrase if there's a better one.) Enable Wireless Monitor mode for traffic sniffing I need to sniff all the wifi traffic that I can capture using wireshark in Ubuntu. My wireless interface is named as eth1 in the interfaces list. When I select that interface in Wireshark and start new sniff, I got following error;    As  thread says, I tried to installed the mentioned driver but I got an error in installation saying that that driver is not supported by my device. After that I tried the second answer in the same thread and run following command to enable monitor mode in my wireless card.  echo 1 &gt; /proc/brcm_monitor0   Then I saw a new Ethernet interface (not a wireless interface ) called prism0 in wireshark interface list. I couldn't start a sniff using that interface using monitor mode because in that interface settings, monitor mode check box has been disabled.  I have installed a broadcom wireless driver in Additional Drivers as follows:    Is there an error listing my Wifi interface as eth1 rather than wlan0 in the interfaces list? How can I start my wireless interface in monitor mode for sniffing?   Thank you. Is it allowed/desirable to post code snippets? Proof of the First Isomorphism Theorem for Groups For every twice differentiable function $f : \bf R \rightarrow [–2, 2]$ with $(f(0))^2 + (f'(0))^2 = 85$, which of the following statements are TRUE? For every twice differentiable function $f : \mathbf R \rightarrow [–2, 2]$ with $(f(0))^2 + (f'(0))^2 = 85$, which of the following statement(s) is (are) TRUE ?  (A) There exist $r, s\in \bf R$, where $r &lt; s$, such that $f$ is one-one on the open interval $(r, s)$  (B) There exists $x_0 \in (–4, 0)$ such that $|f'(x_0)| &lt; 1$   (C) $\lim\limits_{x\to\infty}f(x) = 1$   (D) There exists $\alpha\in(–4, 4)$ such that $f(\alpha) + f''(\alpha) = 0$ and $f '(\alpha)\ne0$  My attempt is as follows:-  $A)$ If function is continuous, then definitely in some part of the interval it will be increasing or decreasing hence will be one-one on the open interval $(r,s)$   $f(x)$ cannot be a constant function as in that case $f'(x)=0$ which means $f(0)=\sqrt{85}$, but it doesn't belong to co-domain of $f$  B) By mean value theorem we can say that there exists some $c\in(-4,0)$ for which $4f'(c)=f(0)-f(-4)$  Suppose if $f(0)=2$ and $f(-4)=-2$, then we can only say that there exists $c$ such that $f'(c)=1$. It will not be necessary that there exists a $c$ for which $|f'(c)|&lt;1$  C) $\lim\limits_{x\to\infty}f(x) = 1$, this is not at all necessary.  D) If function is not constant in $(-4,4)$, then we can safely say that at some point $\alpha$, $f(\alpha)\ne0$  But how to know $f(\alpha)+f''(\alpha)=0$  Actual answer is $A,B,D$ but I am able to ascertain only option $A)$. Please help me in this.