Is there an option to apply a password with the command in the terminal? Somethhing like:  sudo some-command -pass=mypass   I'm just curious, I know that this is not safe.
I want to turn my system on automatically every day. So I use the below code in my Python script, but sudo asks me for a password every time:  os.system('sudo sh -c "echo date \'+%s\' -d \'+ \        24 hours\' &gt; /sys/class/rtc/rtc0/wakealarm"')   How can I run this script without sudo asking for the password every time?
$f(x+1)-f(x)$ converges $\Rightarrow\frac{f(x)}x$ converges Let $f:\mathbb{R}\to\mathbb{R}$ continuous such that $\lim_{x\to\infty}f(x+1)-f(x)=l$.  How can I prove that $\dfrac{f(x)}x$ converges for $x\to+\infty$ ?  I am just trying to prove the convergence, not that it converges to $l$ (that part be deduced immediately by Stolz-Cesaro and the sequential characterization of limits) When should N-Gons be used, and when shouldn't they? Multiple Author and email with location? I am trying to add multiple Author with their email and location but not working.  I am trying to get something like this     first_person^1 , second person^2, third_person^3            {1^mail, 2^mail, 3^mail}@gmail.com                   institution name                      location The line numbers for the text of a paragraph with an equation defined by \begin{align} command do not show as expected. How to fix this?    The LaTeX codes are attached here.   \documentclass[a4paper, 12pt]{article} \usepackage[utf8]{inputenc}    % Use the UTF-8 encoding \usepackage{amsmath}           % For the use of math \usepackage{amsfonts} \usepackage{amssymb}           % For the use of AMS symbols \usepackage{indentfirst}       % For indentation \usepackage[top=2cm, bottom=2cm, left=2cm, right=2cm]{geometry}  \usepackage{lineno} \linenumbers  \usepackage{titlesec} \titleformat*{\section}{\large\bfseries} \titleformat*{\subsection}{\normalsize\bfseries} \titleformat*{\subsubsection}{\normalsize}   \renewcommand{\baselinestretch}{1.5}    \usepackage{authblk}           % Use author package \renewcommand*{\Authfont}{\large} \renewcommand*{\Affilfont}{\normalsize\normalfont}  \makeatletter \g@addto@macro\normalsize{%   \setlength\abovedisplayskip{0pt}   \setlength\belowdisplayskip{0pt}   \setlength\abovedisplayshortskip{0pt}   \setlength\belowdisplayshortskip{0pt} } \makeatother  \title{\textbf{Globally cumstomize the spacing between the text and equation}}    \date{}  \begin{document}  \maketitle  \section{Air filter} The air filter is simulated by the Control Valve module (AFT in Fig. 2(a)). The pressure drop through the air filter ($ \Delta P_{af} $) is given by the following equation [11,16]. \begin{align} \Delta P = \Delta P_{d} \left( \frac{m}{m_{d}}\right)^{1.84}  \left( \frac{T}{T_{d}}\right)  \left( \frac{P}{P_{d}}\right)^{-1}  \end{align} where, $ \Delta P_{af} $ is the pressure drop, $ m $ is the mass flow rate, $ T $ is the temperature, and $ P $ is the pressure, and subscript $ d $ denotes the design condition.  \newpage \section{Air compressor (AC)} The AC operating characteristics can be described by its performance map, which relates the following dimensionless operational variables. The AC operating characteristics can be described by its performance map, which relates the following dimensionless operational variables. \begin{align} &amp; \text{Relative pressure ratio:} &amp;&amp; PR_{r} = \frac{PR-1}{PR_{d}-1}\\ &amp; \text{Relative isentropic efficiency:} &amp;&amp; \eta_{r} = \frac{\eta}{\eta_{d}} \\ &amp; \text{Relative corrected mass flow:} &amp;&amp; m_{cor,r} = \frac{\left(\frac{m_{in}}{P_{in}}\sqrt{T_{in}}\right)}{\left(\frac{m_{in,d}}{P_{in,d}}\sqrt{T_{in,d}}\right)} \\ &amp; \text{Relative corrected speed:} &amp;&amp; N_{cor,r} = \frac{\left(\frac{N}{\sqrt{T_{in}}}\right)}{\left(\frac{N_{d}}{\sqrt{T_{in,d}}}\right)} \end{align} where, $ \eta $ is the efficiency, $ N $ is the shaft speed, and $ PR = P_{in}/P_{out} $. Subscript $ cor $ denotes the corrected value, $ r $ denotes the relative value, $ in $ denotes the inlet, and $ out $ denotes the outlet.  \end{document} How to get over "device or resource busy"? Is either "can" or "could" more polite? Is it possible to make a sequence of renders and give the user the option to cancel the process through the UI at any given time? I'm trying to make a script where I have to render several images of the same scene at the same frame, and write to disk with appropriate names. The problem is that although this works fine with a non-responsive UI, it doesn't work well when the render progress appears in the Blender UI.  I mean,  bpy.ops.render.render()   uses "EXEC_DEFAULT" when called from a script. This leaves a blocked User Interface until the render finishes, meaning that you can't cancel the process, except by killing Blender. The problem is that the render may be too big in resolution, and the user could want to stop the process when realizing that the render could take hours, so the idea is to cancel this process without losing the currently opened project.  If I call,  bpy.ops.render.render("INVOKE_DEFAULT")   I get a nice preview bar in the Info Window with a clickable "X" button. I want the user to be able to press that button to cancel the operation if needed, but for a series of renders (not animation).  The problem here seems to be when trying to loop this command with the INVOKE_DEFAULT context to write several renders. I made a lot of testing, and this way of calling the render operator seems to continue with whatever is next in the script, so it loops very fast and ends before the rendering actually finishes.  I think that I need some way to stop the script and make it wait until render finishes to continue looping with the next ones. I tried the "render_complete" and "render_cancel" handlers, but I can't use them to start a new render or even make the rest of the script wait until the loop of renders is complete. I tried threading and events without success either. Maybe I'm doing something wrong.  import bpy  render = bpy.context.scene.render path = render.filepath = "/tmp/" shots = ["one", "two", "three"]  for x in shots:     render.filepath = path + x     bpy.ops.render.render("INVOKE_DEFAULT", write_still=True) # this doesn't work      # but if I use:     # bpy.ops.render.render(write_still=True)     # without the INVOKE thing, images are written as expected,     # but losing the possibility to see the progress and cancel rendering When does $\lim_{n\rightarrow \infty} a_n = 0$ imply $\sum_{n = 0}^{\infty} a_n$ converges? This is just a general question. I know the opposite statement is always true. I'm just asking when the converse holds. I know that the converse is not true in some cases, such as for harmonic series. What additional restrictions are needed to make this statement true? Where to set proxy ignore list? I upgraded to ubuntu 11.10 from 11.04, shocked to see few things...    My system is behind a proxy, so I did set a proxy setting. Being a web developer, I want to remove a lot of addresses that I type in browser address bar to not hit proxy. There's no such ignore list in ubuntu 11.10, what/where do I do it? What was the requirement to change the way network gui worked? Are all limits solvable without L'Hôpital Rule or Series Expansion Is it always possible to find the limit of a function without using  or ?  For example,   $$\lim_{x\to0}\frac{\tan x-x}{x^3}$$  $$\lim_{x\to0}\frac{\sin x-x}{x^3}$$  $$\lim_{x\to0}\frac{\ln(1+x)-x}{x^2}$$  $$\lim_{x\to0}\frac{e^x-x-1}{x^2}$$  $$\lim_{x\to0}\frac{\sin^{-1}x-x}{x^3}$$  $$\lim_{x\to0}\frac{\tan^{-1}x-x}{x^3}$$ I'm relatively new to ring theory so this is probably a simple question.  Its easy to see that a finite ring that is commutative and has no-zero divisors (i.e. an integral domain) must have multiplicative inverses.  I am wondering if we can rearrange these properties and always get implication or produce finite rings with only 2 of the above properties.  Explicitly my questions are:  1) Does a finite ring that is commutative with multiplicative inverses always have no-zero divisors? (EDIT: this one is pretty easy too, let $xy=0$ and assume $x\neq 0$. then $x^{-1}xy=x^{-1}0$. Hence, $y=0$ as desired.)  2) Is a finite ring that has multiplicative inverses and no zero divisors always commutative?  If these are actually simple exercises, I'd just like a hint to get started with the proof or any counterexamples.  Thanks :) Complexity of $T(n) = 2T(n/2) + n$ Drop shadow for text in PGF/Beamer I want to really put an emphasis on some central piece of text I'm using in a Beamer slide.  I'm using tikz and overlay to put it, enlarged, on the center of the page, using the following code:  \begin{tikzpicture}[remember picture,overlay]   \node[at=(current page.center)] {     \scalebox{2}{\Huge\texttt{{This is important!}}}   }; \end{tikzpicture}   Can I add a drop shadow to the letter themselves, so they would look even more emphasized?  Is there some other decoration that can be added to the text itself, like a glow, etc.?  Or does PGF ignores node text altogether?  The text should be in typewriter style, so using special fancy fonts is probably not the solution. restrict 1 word as case sensitive and other as case insensitive in python regex | (pipe) Seriously, I had not noticed e.g. Jeff's edit  had he not also retagged my question. I have an array of Foo objects. How do I remove the second element of the array?  I need something similar to RemoveAt() but for a regular array. Move item to folder when added/edited (SP2013 or O365) How to move item to folder automatically when it has been added or edited in a library? The folder is in the same library. The correct folder will be defined by metadata. This should happen automatically when the item has been added or edited. The listitem Id should be preserved. No server-side code should be deployed.  Can this be done using workflows (SP2010/SP2013)? If not then what would be the other options? When I analyze my variables in two separate (univariate) logistic regression models, I get the following:  Predictor 1:    B= 1.049,    SE=.352,    Exp(B)=2.85,    95% CI=(1.43, 5.69),    p=.003    Constant:    B=-0.434,    SE=.217,    Exp(B)=0.65,                            p=.046  Predictor 2:    B= 1.379,    SE=.386,    Exp(B)=3.97,    95% CI=(1.86, 8.47),    p&lt;.001    Constant:    B=-0.447,    SE=.205,    Exp(B)=0.64,                            p=.029   but when I enter them into a single multiple logistic regression model, I get:  Predictor 1:    B= 0.556,    SE=.406,    Exp(B)=1.74,    95% CI=(0.79, 3.86),    p=.171 Predictor 2:    B= 1.094,    SE=.436,    Exp(B)=2.99,    95% CI=(1.27, 7.02),    p=.012    Constant:    B=-0.574,    SE=.227,    Exp(B)=0.56,                            p=.012   Both predictors are dichotomous (categorical). I have checked for multicollinearity.  I am not sure if I have given enough info, but I cannot understand why predictor 1 has gone from being significant to non-significant and why the odds ratios are so different in the multiple regression model. Can anyone provide a basic explanation of what is going on? limsup of intersection of events as a subset of intersection of limsups Let $A_1, A_2, \ldots$ and $B_1, B_2, \ldots$ be two sequences of events in some probability triple $(\Omega, \mathcal{F}, \mathbf{P})$. Now, it is true that $\left(\limsup_n A_n\right) \cap \left(\limsup_n B_n\right) \supseteq \limsup_n (A_n \cap B_n)$, and I can think of many different cases where this relation holds true for equality (e.g., let $A_n = B_n$, or $A_n = B_n^c$, or $B_n = \emptyset$, or $B_n = \Omega$). However, I cannot think of any case where $\left(\limsup_n A_n\right) \cap \left(\limsup_n B_n\right) \supset \limsup_n (A_n \cap B_n)$ holds for strict inclusion.  My proof for the relation is simple. Since $A_n \cap B_n \subseteq A_n$ and $A_n \cap B_n \subseteq B_n$, then we can perform some set algebra as follows:  $$\bigcup_{k=n}^\infty (A_k \cap B_k) \subseteq \bigcup_{k=n}^\infty A_k$$ $$\bigcap_{n=1}^\infty \bigcup_{k=n}^\infty (A_k \cap B_k) \subseteq \bigcap_{n=1}^\infty \bigcup_{k=n}^\infty A_k$$  Then we can use the same logic for $B_n$, and come to the conclusion as stated in the beginning. Finding $\int_0^{\infty}xe^{-\lambda x} \, dx$ How to integrate:  $$\int_0^\infty x \, \lambda e^{-\lambda x} \, dx$$    I tried using integration by parts:  Let $u = x$ and $dv = \lambda e^{-\lambda x} \, dx$  Then $du = dx$  And $v = - \lambda e ^{-\lambda x}$  Correct so far?   Then   $$\begin{aligned} uv - \int v \, du &amp;= -\lambda x e^{-\lambda x} - \int (- \lambda e^{- \lambda x}) dx \\ &amp;= -\lambda x e^{-\lambda x} - \lambda e^{-\lambda x} \end{aligned}$$    The correct answer from lecture notes      UPDATE 2  Let $u=\lambda x$ and $dv = e^{-\lambda x} dx$  Then $du = \lambda \, dx$  $$dv = e^{-\lambda x} \, dx$$  Let $y = -\lambda x$  Then $dy = -\lambda \, dx$  So $dx = -\frac{1}{\lambda} dy$  $$\begin{aligned} v &amp;= \int e^u \cdot - \frac{1}{\lambda} du \\ &amp;= - \frac{1}{\lambda} e^{-\lambda x} \end{aligned}$$  But here I seem to have an extra $- \frac{1}{\lambda}$ in $v$?  If I continue using integration by parts, I get:   $$-x e^{-\lambda x} + \color{red}{\frac{1}{\lambda}} \int e^{-\lambda x} \, dx$$