Why are FETs preferred over BJTs?
When is a MOSFET more appropriate as a switch than a BJT?
I am trying to make a histogram with the fitted curve plotted on it. Here's the mwe:   \documentclass[a4paper]{scrartcl}     \usepackage{tikz}     \usepackage{pgfplots,pgfplotstable}     \usepackage{filecontents}     \pgfplotsset{compat=1.10}     \usepackage[active,tightpage]{preview}     \setlength\PreviewBorder{2pt}     \newcommand{\plots}{0.611201}     \newcommand{\plotm}{2.19882}     \begin{filecontents}{data.dat}     2 0.0629921259843     3 0.0236220472441     4 0.0314960629921     5 0.125984251969     6 0.0629921259843     7 0.102362204724     8 0.110236220472     9 0.0551181102362     10 0.0629921259843     11 0.0314960629921     12 0.0236220472441     13 0.0314960629921     14 0.0629921259843     15 0.0551181102362     16 0.0393700787402     17 0.0472440944882     18 0.00787401574803     19 0.0393700787402     24 0.00787401574803     27 0.00787401574803     33 0.00787401574803     \end{filecontents}     \begin{document}     \begin{preview}      \begin{tikzpicture}     \begin{axis}[     yticklabel style={/pgf/number format/fixed},     scaled y ticks = false,     minor y tick num={1},     xtick pos=left,     legend cell align = left,     legend style={draw=none},     xlabel = {group size},     ylabel = {ratio}     ]     \addplot[blue,ybar,fill, fill opacity=0.3, bar width = 0.8,] table {data.dat};     \addplot[red, line width = 1,domain=1:40,samples=100] {1/(x*sqrt(2*pi)*\plots)*exp(-(ln(x)-\plotm)^2/(2*\plots^2))};     \legend{empirical,lognormal fit}     \end{axis}     \end{tikzpicture}     \end{preview}     \end{document}     As you can see, the legend is not really good. How can I change the legend so it looks more like this? Integral $\int\csc^3{x} \ dx$ I found these step which explain how to integrate $\csc^3{x} \ dx$. I understand everything, except the step I highlighted below.  How did we go from: $$\int\frac{\csc^2 x - \csc x \cot x}{\csc x - \cot x}\,dx%$$ to  $$\int \frac{d(-\cot x + \csc x)}{-\cot x + \csc x} \quad?$$  Thank you for your time! $$ \int \csc^3 x\,dx = \int\csc^2x \csc x\,dx$$ To integrate by parts, let $dv = \csc^2x$ and $u=\csc x$. Then $v=-\cot x$ and $du = -\cot x \csc x \,dx$. Integrating by parts, we have: $$\begin{align*} \int\csc^2 x \csc x \,dx &amp;= -\cot x \csc x - \int(-\cot x)(-\cot x\csc x\,dx)\\ &amp;= -\cot x \csc x - \int \cot^2 x \csc x\,dx\\ &amp;= -\cot x\csc x - \int(\csc^2x - 1)\csc x\,dx &amp;\text{(since }\cot^2 x = \csc^2-1\text{)}\\ &amp;= -\cot x \csc x - \int(\csc^3 x - \csc x)\,dx\\ &amp;= -\cot x\csc x - \int\csc^3 x\,dx + \int \csc x\,dx \end{align*}$$ From $$\int \csc^3 x\,dx = -\cot x\csc x - \int\csc^3 x\,dx + \int \csc x\,dx$$ we obtain $$\begin{align*} \int\csc^3x\,dx + \int\csc^3 x\,dx &amp;= -\cot x \csc x + \int\csc x\,dx\\ 2\int\csc^3 x\,dx &amp;= -\cot x\csc x + \int\csc x\,dx\\ \int\csc^3x\,dx &amp;= -\frac{1}{2}\cot x\csc x + \frac{1}{2}\int\csc x\,dx\\ &amp;=-\frac{1}{2}\cot x\csc x + \frac{1}{2}\int\frac{\csc x(\csc x - \cot x)}{\csc x - \cot x}\,dx\\ &amp;= -\frac{1}{2}\cot x \csc x + \frac{1}{2}\int\frac{\csc^2 x - \csc x\cot x}{\csc x - \cot x}\,dx\\ &amp;= -\frac{1}{2}\cot x \csc x + \frac{1}{2}\int\frac{d(-\cot x+\csc x)}{-\cot x +\csc x}\\ &amp;= -\frac{1}{2}\cot x\csc x + \frac{1}{2}\ln|\csc x - \cot x|+ C \end{align*}$$ How do I convert materials from Blender Internal to Cycles? I know this may be a tough or confusing question, but I have numerous existing materials I created and reuse for a project, which were all prepared for Blender Internal.  Now, I want to start the transition to Cycles, but I've hit a roadblock. How can I transition my materials without completely starting over? Or should I just forget it and start with fresh materials? How do I enable the "multiverse" repository? I keep seeing places refer to the "multiverse" repository as a place I can get software, how can I enable this repository? Please specify how to do this graphically and via command line. pgfplots from file: search path, looking for \graphicspath equivalent How to execute hostapd command in cli? I always got error. When i execute hostapd command. I was changed everything in my hostapd.conf file but i got error.  hostapd_cli wps_pbc Failed to connect to hostapd - wpa_ctrl_open: No such file or directory  Can anyone help to me.? How do I effectively defeat the Wall of Flesh? TAR 6 GB worth of files into separate 1.9 GB TAR files How find this limit $\lim\limits_{x\to 0^{+}}\frac{\sin{(\tan{x})}-\tan{(\sin{x})}}{x^7}$ Why can't the end code of an environment contain an argument? Can I rephrase the sentence "Can you explain this word to me?" into "Can you explain me this word." When you pick up items in Skyrim it tells you the value of each item. However when I go and sell the items the merchants are only paying 30% of this value, at best.  Discounting any perks or other effects, how high does Speechcraft have to be before merchants start paying full value for items? Commutator of a group Finding the Transform matrix from 4 projected points (with Javascript) Books for self-studying time series analysis? Stop iPhone alerting me when I'm in a phone call When I'm in a phone call, I still get alerts for emails and texts. The noise will play and, if the phone is on Speaker, it will vibrate -- often leading the person I'm talking to wonder what that weird noise was.  Surely there must be a way to silence the phone while you're talking on it. It's very distracting to have beepings going off right next to your ear! I would like to have \alph{counter} for values larger than 26. The following MnWE gives me the error ! LaTeX Error: Counter too large.  \documentclass{article} \begin{document}  \begin{enumerate} \def\theenumi{\alph{enumi}} \setcounter{enumi}{24} \item twenty-five \item twenty-six \item twenty-seven \end{enumerate}  \end{document}   I would like it to behave like one of these:  ..., x, y, z, aa, bb, cc, ..., xx, yy, zz, aaa, bbb, ccc, ... (preferred) ..., x, y, z, aa, ab, ac, ..., ax, ay, az, ba,  bb,  bc,  ... (can be if nothing else works)   A solution without LaTeX3 extensions would be awesome as I cannot run LaTeX3 extensions. How to get the current date/time in Java Limit of $\lim_{(x,y)\rightarrow(0,0)}\frac{x^2y^2}{x^3+y^3}$ I want to know whether $\lim_{(x,y)\to (0,0)}\dfrac{x^2y^2}{x^3+y^3}$ exists or not. I tried to approximate to (0,0) from different "paths" and the result was always 0. For example,  $f(x,mx^2) = \dfrac{m^2x^3}{1+m^3x^3}$  But that doesn't show that the limit is 0. I'm having trouble understanding Cantor's diagonal argument.  Specifically, I do not understand how it proves that something is "uncountable".  My understanding of the argument is that it takes the following form (modified slightly from the  article, assuming base 2, where the numbers must be from the set $ \lbrace 0,1 \rbrace $):  \begin{align} s_1 &amp;= (\mathbf{0},1,0,\dots)\\ s_2 &amp;= (1,\mathbf{1},0,\dots)\\ s_3 &amp;= (0,0,\mathbf{1},\dots)\\ \vdots &amp;= (s_n \text{ continues}) \end{align}  In this case, the diagonal number is the bold diagonal numbers $(0, 1, 1)$, which when "flipped" is $(1,0,0)$, neither of which is $s_1$, $s_2$, or $s_3$.  My question, or misunderstanding, is: When there exists the possibility that more $s_n$ exist, as is the case in the example above, how does this "prove" anything?  For example:  \begin{align} s_0 &amp;= (1,0,0,\mathbf{0},\dots)\ \ \textrm{ (...the wikipedia flipped diagonal)}\\ s_1 &amp;= (\mathbf{0},1,0,\dots)\\ s_2 &amp;= (1,\mathbf{1},0,\dots)\\ s_3 &amp;= (0,0,\mathbf{1},\dots)\\ s_4 &amp;= (0,1,1,\mathbf{1},\dots)\\ s_4 &amp;= (1,0,0,\mathbf{1},\dots)\ \ \textrm{ (...alternate, flipped } s_4\textrm{)}\\ s_5 &amp;= (1,0,0,0,\dots)\\ s_6 &amp;= (1,0,0,1,\dots)\\ \vdots &amp;= (s_n \text{ continues}) \end{align}  In other words, as long as there is a $\dots \text{ continues}$ at the end, the very next number could be the "impossible diagonal number", with the caveat that it's not strictly identical to the "impossible diagonal number" as the wikipedia article defines it:     For each $m$ and $n$ let $s_{n,m}$ be the $m^{th}$ element of the $n^{th}$ sequence on the list; so for each $n$,      $$s_n = (s_{n,1}, s_{n,2}, s_{n,3}, s_{n,4}, \dots).$$      ...snip...      Otherwise, it would be possible by the above process to construct a sequence $s_0$ which would both be in $T$ (because it is a sequence of 0s and 1s which is by the definition of $T$ in $T$) and at the same time not in $T$ (because we can deliberately construct it not to be in the list). $T$, containing all such sequences, must contain $s_0$, which is just such a sequence. But since $s_0$ does not appear anywhere on the list, $T$ cannot contain $s_0$.      Therefore $T$ cannot be placed in one-to-one correspondence with the natural numbers. In other words, it is uncountable.   I'm not sure this definition is correct, because if we assume that $m = (1, \dots)$, then this definition says that "$s_n$ is equal to itself"&mdadsh;there is no "diagonalization" in this particular description of the argument, nor does it incorporate the "flipping" part of the argument, never mind the fact that we have very clearly constructed just such an impossible $T$ list above.  An attempt to correct the "diagonalization" and "flipping" problem:  $$s_n = (\lnot s_{m,m}, \lnot s_{m,m}, \dots) \quad \text{where $m$ is the element index and} \quad\begin{equation}\lnot s_{m,m} = \begin{cases}0 &amp; \mathrm{if\ } s_{m,m} = 1\\1 &amp; \mathrm{if\ } s_{m,m} = 0\end{cases}\end{equation}$$  This definition doesn't quite work either, as we immediately run in to problems with just $s_1 = (0),$ which is impossible because by definition $s_1$ must be $ = (1)$ if $s_1 = (0)$, which would also be impossible because...   Or more generally, with the revised definition there is a contradiction whenever $n = m$, which would seem to invalidate the revised formulation of the argument / proof.  Nothing about this argument / proof makes any sense to me, nor why it only applies to real numbers and makes them "uncountable". As near as I can tell it would seem to apply equal well to natural numbers, which are "countable".  What am I missing?