In the sentence below, is there an adverb telling where (giving the location), or a noun as the subject?     There are many people on the beach.
According to      there         adverb          in or at that place (opposed to here ):      She is there now.         pronoun          (used to introduce a sentence or clause in which the verb comes before its subject or has no complement):      There is no hope.   Then, what element is there in:     There is the man.   I'm confused because "There is the man" is an adverb inversion, in which there is an adverb and "the man is" is inverted. But according to the definition above, there suddenly becomes a pronoun.
Show that if $n>2$, then $(n!)^2>n^n$. Huawei E3272 Ubuntu I have a problem with Huawei E3272 modem. Ubuntu 14.04 does not detect it as a mobile broadband modem. If I put my SIM card into my old modem(E3131) everything works, E3272 is just not detected. I have checked lsusb, and here I can see the system detects it as  12d1:14fe Huawei Technologies Co., Ltd.  but I cannot choose it in my Network Connections.  I tried adding usbserial vendor=0x12d1 product=0x14fe in /etc/modules, but it did not change anything. The same goes for adding   # Huawei E3272  ATTR{idVendor}=="12d1", ATTR{idProduct}=="157c", RUN +="usb_modeswitch '%b/%k'"  in /lib/udev/rules.d/40-usb_modeswitch.rules. I also tried installing Modem Manager, but it did not help.  Have you got any idea what could I do, so my system would detect my new modem? I would like to be able to use LTE. :) What is the definition of textbook or "raw" RSA?  What are some of the properties of textbook RSA?  How does it differ from other schemes based on RSA? "Any" followed by singular or plural countable nouns? This question has troubled me for ages despite my several attempts of looking it up in dictionaries or usage books. Do we say, "Do you have any ideas" or "Do you have any idea"? I do see an example where "any" means "it doesn't matter who/which/what", therefore "You can borrow any book you like." Also, does it matter between using it in questions and negatives? The Equation $11^x + 13^x + 17^x =19^x  $  Has    No Real Roots Only One Real Roots Exactly Two Real Roots  More than Two Real Roots    What I have done is   The function $f(x)=11^x + 13^x + 17^x -19^x  $ is strictly increasing and being always positive it have no roots A sequence $\{x_j\}$ is said to be fast-Cauchy if $\sum_1^\infty d(x_j,x_{j+1})&lt;+\infty$. Show that every Cauchy sequence has a fast-Cauchy subsequence.   **My attempt:**Argue by contradiction, suppose there is no such subsequence. Then for any subsequence $\{{x_j}_k\}$,  $\sum_1^\infty d({x_j}_k,{x_j}_{k+1})$ diverges. Vaguely recall from an analysis course, if a series diverges, it doesn't not satisfy the Cauchy criterion. That is to say, there exists $\epsilon&gt;0$ such that for all $k$ and all $p\ge1$, $d({x_j}_{k+1},{x_j}_{k+2})+...+d({x_j}_{k+p}+{x_j}_{k+p+1})\ge\epsilon$.  Then I intuitively think that this is a contradiction, since if the original series is Cauchy, then after a large $N$, since $d(x_m,x_n)$ has to be very small, so that their sum should be smaller than given $\epsilon$, which is a contradiction. But I don't know how I can write down the concrete proof? Inspired exercise Sheet from Indesign I would like to create this sheet which was made it by adobe indesign   Could someone create it with latex ?     Here's my code:  \documentclass[12pt,a4paper,twocolumn]{report} \usepackage[margin=0.5in]{geometry} \usepackage[french]{babel} \usepackage{fontspec} \usepackage{graphicx} \usepackage{amsthm,amssymb,amsfonts,mathtools} \usepackage[utf8]{inputenc} \usepackage{multicol} \usepackage{multirow} \usepackage{amssymb} \usepackage{array} \usepackage{graphicx} \setlength{\columnseprule}{0.5pt} \thispagestyle{empty} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{document} %%%%%%%%%%head%%%%%%%%%%%%%%%%%%%%%%% \twocolumn[    \begin{@twocolumnfalse}  \begin{center}  \fbox{\Large Exercise Sheets} \end{center}     \hrulefill \end{@twocolumnfalse}   ]  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  \fbox{ Exercise $1$:}\\  Compute the following limits \begin{multicols}{2} \begin{enumerate} \item $\lim_{x\to 2}\dfrac{4x^{3}-5x-22}{x^{2}-x-2}$ \item $\lim_{x\to 0^{+}}\dfrac{x-\sqrt{x}}{x+\sqrt{x}}$ \item $\lim_{x\to 1}\dfrac{\sqrt{x+3}+\sqrt{x}-3}{x-1}$ \item $\lim_{x\to 2}\dfrac{x^{2}\sqrt{x+2}-8}{4-x^{2}}$ \end{enumerate} \end{multicols}  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \fbox{ Exercise $2$:}\\ Let $n\in\mathbb{N}^{*}$ \begin{enumerate}     \item calculate : ${\displaystyle \lim_{x\to 0}\dfrac{1-\cos(x)\cos(2x)\ldots+\cos(nx)}{x^2} }$     \item b     \item c     \item d     \item e     \item f \end{enumerate} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \end{document} Massless charged particles Are there any massless (zero invariant mass) particles carrying electric charge?  If not, why not? Do we expect to see any or are they a theoretical impossibility? I am working on binary classification problem, I try to evaluate the performance of some classification algorithms (LR,Decission Tree , Random forest ...). I am using  a cross validation technique (to avoid over-fitting) with AUC ROC as scoring function to compare the performance of the algorithms, but I am getting a weird results with Random forest and AdbBoost, I have a perfect AUC_ROC score (i.e. =1) despite the fact that the recall(TPR) and FPR of this algorithms are different from 1 and 0 respectively .  def FPR(y_true, y_pred):     tn, fp, fn, tp = confusion_matrix(y_true, y_pred).ravel()     result = fp / (fp+tn)     return result def FNR(y_true, y_pred):     tn, fp, fn, tp = confusion_matrix(y_true, y_pred).ravel()     result = fn / (tp+fn)     return result FPR_scorer = make_scorer(FPR) FNR_scorer = make_scorer(FNR)     def get_CrossValResults2(model,cv_rst,bestIndx):     best=pd.DataFrame.from_dict(cv_rst).iloc[bestIndx]     roc=&quot;{:.12f}&quot;.format(best['mean_test_roc_auc'])     acc =&quot;{:.0%}&quot;.format(best['mean_test_accuracy'])     prec =&quot;{:.0%}&quot;.format(best['mean_test_precision'])     rec =&quot;{:.0%}&quot;.format( best['mean_test_recall'])     f1 =&quot;{:.0%}&quot;.format(best['mean_test_f1'])     r2=&quot;{:.2f}&quot;.format(best['mean_test_r2'])     g_mean=&quot;{:.2f}&quot;.format(best['mean_test_gmean'])     pr_auc=&quot;{:.8f}&quot;.format(best['mean_test_pr'])     fnr=&quot;{:.0%}&quot;.format(best['mean_test_fnr'])     fpr=&quot;{:.0%}&quot;.format(best['mean_test_fpr'])     rst = pd.DataFrame([[ model, acc,prec,rec,fpr,fnr,f1,roc,pr_auc,g_mean,r2]],columns = ['Model', 'Accuracy', 'Precision', 'Recall','FPR','FNR', 'F1-Score','ROC_auc','PR_auc','gmean','r2'])     return rst        cross_val_rst = pd.DataFrame(columns = ['Model', 'Accuracy', 'Precision', 'Recall','FPR','FNR', 'F1-Score','ROC_auc','PR_auc','gmean','r2'])              scoring = {'accuracy':'accuracy','recall':'recall','precision':'precision','fpr':FPR_scorer,'fnr':FNR_scorer,'f1':'f1' ,'roc_auc':'roc_auc','pr':'average_precision','gmean':Gmean_scorer,'r2':'r2'}     param_grid = {'n_estimators': [200],               'max_depth': [80,90],               'min_samples_leaf': [2,3, 4],             'min_samples_split': [2,5,12],             'criterion': [ 'gini'],              'class_weight' : [class_weights], 'n_jobs' : [-1]}  clf = GridSearchCV(RandomForestClassifier(class_weight=class_weights), param_grid, cv=kfold,scoring=scoring,refit=refit)#Fit the model bestmodel = clf.fit(X,Y)       cross_val_rst = cross_val_rst.append(get_CrossValResults2(model='Random Forrest',bestIndx=bestmodel.best_index_,cv_rst=bestmodel.cv_results_),ignore_index=True) Prove that f is one-to-one on D How do we show that equality holds in the triangle inequality $|a+b|=|a|+|b|$ iff both numbers are positive, both are negative or one is zero? I already showed that equality holds when one of the three conditions happens. I've been trying to understand $\sigma$-algebras and how it encodes information in context of filtration. While certain parts seem clear and logical, I can't say I get the whole picture.  I'll try to explain the counter-intuition I get with the classical example of : the probability space $\Omega = \{ HH, HT, TH, TT \}$ and a r.v. $X(\omega)$ equal to the number of heads.  At times $0$, $1$ and $2$ the available information is represented using $\sigma$-algebras $\mathcal{F}_0=\{\emptyset,\Omega\}$, $\mathcal{F}_1=\{\emptyset, \Omega, \{HH,HT\},\{TH,TT\}\}$ and $\mathcal{F}_2=\{\emptyset, \Omega,\{HH,HT\},\{TH,TT\},\{HH\},\{HT\},\{TH\},\{TT\}\}$.  One can notice that $X(\omega)$ is not measurable with respect to $\mathcal{F}_0$ and $\mathcal{F}_1$, because $X^{-1}((\frac{3}{2}; +\infty))=\{HH\}$. To me it is quite surprising: intuitively $X$ makes perfect sense at all times. In particular it has an expected value at time $0$, which I interpret as that the probability and value of all outcomes $\{\omega\}$ can be computed. How do you think of a non-measurable function?  Here's another way of expressing the same confusion. The most natural choice of $\sigma$-algebra in a finite discrete case is $\mathcal{F}=2^\Omega$, and it is implicitly used in all elementary probability problems. However, this choice of $\mathcal{F}$ does not reflect the fact that some information is known or unknown, conditional probability does. Does it mean that the statement "$\sigma$-algebra is known information" make sense only in conditioning? Why is it convenient then? Please describe direct products in the category Rel. LaTeX figures side by side Monic $f(x)\in\Bbb Z[x]$ has no rational root if $f(0)\ \&\ f(1)$ odd [Parity Root Test, Modular Root Test] A polynomial problem from my old algebra textbook:     $f(x)\in\Bbb Z[x]$ with leading coefficient $1$, $\deg f(x)\ge 1$, and both $f(0)$ and $f(1)$ are odd numbers, prove:   $f(x)$ has no root within $\Bbb Q$   Eisenstein's criterion seems to be of little help here because we know virtually nothing about the coefficients. So I tried some other approach. Let $$f(x)=a_0+a_1x+a_2x^2+\cdots+a_{n-1}x^{n-1}+x^n\in \Bbb{Z}[x]$$ What we have known so far is $$f(0)=a_0\text{ is an odd number}$$ and $$f(1)=a_0+a_1+a_2+\cdots+a_{n-1}+1 \text{ is an odd number}$$ from which we can also conclude that $$f(1)-f(0)=a_1+a_2+\cdots+a_{n-1}+1\text{ is an even number}$$ To be honest I don't know how to proceed now. It isn't until just now that I realized "irreduicible" is not equivalent to "has no roots".. So all my previous attempt is fundamentally wrong.. Can anybody help or drop a hint? Best regards! I am designing a hobby project that will run on a battery so I am trying to reduce power consumption. The SoC I am using has the ability to go into a deep sleep mode where it consumes very little power, and then can be interrupted by a falling edge on a pin.  I want to generate a high-low-high pulse when the state of a reed switch changes in either direction. I have been reading a lot and it seems that a one-shot monostable circuit can be built from a couple of transistors and an RC circuit to generate an output pulse when an edge is detected. I understand how that works. However, I can't figure out the best way to also pulse on the opposite edge.  One thought, was to just combine two one-shots that are configured to trigger on rising and falling edges respectively, and use the transistors at the output to pull the common pull-up low (kind of like a NOR gate).  Since I am a hobbyist, I am mostly using passive components. What is the "real" way of doing this in a low-power configuration using logic devices?  Edit: I figure I should mention something about the timing requirements. Normally the reed switch will be either open or closed for many seconds at minimum before changing states.  Edit: Here is an example of the circuit I was using for detecting only the falling edge:    And here is the ~20ms pulse it was producing for ~200ms button press simulating the reed switch (Channel 1 is Vbe for T1 and Channel 2 is measured at RST): In Triangle, $\sin\frac A2\!+\!\sin\frac B2\!+\!\sin\frac C2\!-\!1\!=\!4\sin\frac{\pi -A}4\sin\frac{\pi -B}4\sin\frac{\pi-C}4$ To prove  $$\sin\frac A 2+\sin\frac B 2+\sin\frac C 2-1=4\sin\frac{\pi -A}4\sin\frac{\pi -B}4\sin\frac{\pi-C}4$$  My approach :   $$ \begin{align} \text{L.H.S.} &amp; = \sin\frac{A}{2}+\sin \frac{B}{2}+\sin\frac{C}{2} -1 \\[8pt] &amp; = 2\sin\frac{A+B}{4}\cos\frac{A-B}{4}+\cos\left(\frac{\pi}{2}-\frac{C}{2}\right)-1 \\[8pt] &amp; = 2\sin\frac{\pi -C }{4}\cos\frac{A-B}{4} - 2\sin^2\left(\frac{\pi -C}{4} \right) \\[8pt] &amp; =2\sin\frac{\pi -C }{4}\left\{ \cos\frac{A-B}{4} - \sin\left(\frac{\pi -C}{4} \right)\right\} \end{align} $$  Unable to move further please help. thanks. Better than premium Is there something higher (better) than the word premium, or does it mean the best? Also, is there something in between premium and deluxe?  Graphically: Does any word fit here? > Premium > Does any word fit here? > Deluxe How can I get the DateTime for the start of the week? 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 $(\varphi^{(f)}\circ\theta)(G)=(\varphi^{(\bar f)}\circ\bar \theta)(\overline G)$?