I want to install Kali Linux tools in Ubuntu 16.04 LTS. But I'm getting this error message. How do I solve this issue?
I get this error whenever I try to install programs using the terminal:  home@ubuntu:~$ apt-get install myunity E: Could not open lock file /var/lib/dpkg/lock - open (13: Permission denied) E: Unable to lock the administration directory (/var/lib/dpkg/), are you root?   Also I'm unable to install updates using the terminal.
Is there any way to view saved wifi passwords on the iPhone? Getting a FILE* from a std::fstream Is there a (cross-platform) way to get a C FILE* handle from a C++ std::fstream ?  The reason I ask is because my C++ library accepts fstreams and in one particular function I'd like to use a C library that accepts a FILE*. LaTeX Classes or Styles for Schedules and/or Calendars? Maybe TeX isn't the best for this particular problem, but any teachers out there use a particular class or package to plan lessons?  Right now I'm just using a spreadsheet, but I always find myself wishing there was a 'lessonplan.sty' or 'classschedule.sty' file out there somewhere so I could still use the LaTeX for equations, tables, lists etc.    A bit of searching yielded a few results, but there doesn't seem to be a consensus.  Anyone used any of these:        Any recommendations from anyone on best way to make calendar-based class schedules and/or lesson plans? $$f(x,y) = \left\{\begin{array}{cc} \frac{xy}{x^2+y^2} &amp; (x,y)\neq(0,0) \\ f(x,y) = 0 &amp; (x,y)=(0,0) \end{array}\right.$$  In order to verify if this function is differentiable, I tried to prove it by the theorem that says that if $\frac{∂f}{∂x}$ and $\frac{∂f}{∂y}$ exist and are continuous at the point $(x_0,y_0)$, then the function is differentiable at this point. So I did:  $$\frac{\partial f}{\partial x}(0,0) = \lim_{h\to 0}\frac{f(0+h,0)-f(0,0)}{h} = 0$$ $$\frac{\partial f}{\partial y}(0,0) = \lim_{h\to 0}\frac{f(0,0+h)-f(0,0)}{h} = 0$$  so we have that the partial derivatives at point $(0,0)$ is $0$. Now, if we take the derivative at $(x,y)\neq (0,0)$ and then take the limit of it as $(x,y)\to(0,0)$, we can see if the derivatives are continuous or not. So here it is:  $$\frac{\partial f}{\partial x}(x,y) = \frac{y(y^2-x^2)}{(x^2+y^2)}$$  but   $$\lim_{(x,y)\to(0,0)} \frac{y(y^2-x^2)}{(x^2+y^2)} $$ does not exist (by wolfram alpha... but can anybody tell me an easy way to prove this limit does not exist? easier than taking the limit in different directions?), therefore the derivative is not continuous at $(0,0)$, so we can't say $f$ is differentiable at $(0,0)$, but for $(x,y)\neq (0,0)$ the function is continuous, as it is a quotient of continuous functions. So $f$ is at least differentiable at $(x,y)\neq (0,0)$.   Now, to verify differentiability at $(0,0)$ I think we must use the limit definition of differentiablity: A function is differentiable at $(0,0)$ iff: $$\lim_{(h,k)\to (0,0)} \frac{f(0+h,0+k)-f(0,0)-\frac{\partial f}{\partial x}(0,0)-\frac{\partial f}{\partial y}(0,0)}{\|(h,k)\|} = 0$$  Let's calculate this limit:  $$\lim_{(h,k)\to (0,0)} \frac{f(0+h,0+k)-f(0,0)-\frac{\partial f}{\partial x}(0,0)-\frac{\partial f}{\partial y}(0,0)}{\|(h,k)\|} = \\ \lim_{(h,k)\to (0,0)} \frac{\frac{hk}{h^2+k^2}}{\sqrt{h^2+k^2}} = \\ \lim_{(h,k)\to (0,0)} \frac{hk}{(h^2+k^2)\sqrt{h^2+k^2}}$$  which I think, it's a limit that does not exist, therefore the function isn't differentiable at $(0,0)$ I'm using Terminator version 0.97. Is there a way that I can scroll unlimited? In the current version I only can scroll back limited history Encouraging users to create an account (and keep it) Initial connection to SQL Server Connection Is Slow. Why? How to get my own IP address and save it to a variable in a shell script? Is there a way to show that $\sqrt{p_{n}} < n$? Is there a way to show that $\sqrt{p_{n}} &lt; n$?  In , I show that $f_{2}(x)=\frac{x}{ln(x)} - \sqrt{x}$ is ascending, for $\forall x\geq e^{2}$. As a result, $\forall n \geq 3$ $$\frac{p_{n}}{ln(p_{n})} - \sqrt{p_{n}}\leq \frac{p_{n+1}}{ln(p_{n+1})} - \sqrt{p_{n+1}}$$ Also (and as a result), $\forall n \geq 3$ $$ \frac{p_{n}}{ln(p_{n})} - \sqrt{p_{n}} &gt; 0$$ Or $$ \frac{\pi (p_{n})}{p_{n}/ln(p_{n})} &lt; \frac{\pi (p_{n})}{\sqrt{p_{n}}}$$  According to  $$\displaystyle\smash{\lim_{n \to \infty }}\frac{\pi (p_{n})}{p_{n}/ln(p_{n})}=1$$ Or, $\forall \varepsilon &gt;0$, $\exists N(\varepsilon )$: $\forall n&gt;N(\varepsilon )$ $$1- \varepsilon &lt; \frac{\pi (p_{n})}{p_{n}/ln(p_{n})} &lt; 1+ \varepsilon$$ Or $$1- \varepsilon &lt; \frac{\pi (p_{n})}{p_{n}/ln(p_{n})} &lt; \frac{\pi (p_{n})}{\sqrt{p_{n}}}$$ As a result $\forall \varepsilon &gt;0$, $\exists N(\varepsilon )$: $\forall n&gt;N(\varepsilon )$ $$(1 - \varepsilon ) \cdot  \sqrt{p_{n}} &lt; \pi (p_{n}) = n$$  But this is not enough.  Interestingly,  is true iff function $f_{4}(x)=\pi (x) - \sqrt{x}$ is strictly ascending ($x &lt; y \Rightarrow f(x) &lt; f(y)$) for prime arguments.  If $f_{4}(p_{n}) &lt; f_{4}(p_{n+1})$ then $$\pi (p_{n}) - \sqrt{p_{n}} &lt; \pi (p_{n+1}) - \sqrt{p_{n+1}}$$ Or $$\sqrt{p_{n+1}} - \sqrt{p_{n}} &lt; \pi (p_{n+1}) - \pi (p_{n}) =1$$  And vice-versa, if $$\sqrt{p_{n+1}} - \sqrt{p_{n}} &lt; 1$$ Then $$-\sqrt{p_{n}} &lt; -\sqrt{p_{n+1}} + 1$$ Or $$\pi (p_{n})-\sqrt{p_{n}} &lt; \pi (p_{n}) + 1 -\sqrt{p_{n+1}} = \pi (p_{n+1}) -\sqrt{p_{n+1}}$$  So, if Andrica's conjecture is true then $\forall n \geq 3$ $$\pi (p_{n})-\sqrt{p_{n}} &gt; 0$$ Or $$\sqrt{p_{n}} &lt; \pi (p_{n})= n$$ Imagine your college buddy is now running a multi-billion dollar company, while you're still stuck at the same white-collar day job 5 years after graduation. You chat with him and want to express the idea that you're happy for him and fancy having his lifestyle--something like "I wish I could buy a yacht and cruise around the world like you do; I'm glad you're doing really well!"  Basically, it's the feeling of envy when you take out the resentment from it. Is there a word for it?   (I thought about "admire", but found it too strong and broad) translation invariant of integral on $\mathbb{R}$ Problem 1: Let $f\in L^1(\mathbb{R})$, show $\int_\mathbb{R}f(t)dt=\int_{\mathbb{R}}f(x+t)dt,\forall x\in (-\infty, \infty)$.  Problem 2: Let $f\in L^1(\mathbb{R})$,show $\displaystyle \lim_{t\rightarrow 0}\int_{\mathbb{R}}|f(x+t)-f(x)|dx=0$.  My try for Problem 1:  Consider $\displaystyle \int_{\mathbb{R}}f(t)dt=\lim_{n\rightarrow \infty}\int_{-n}^nf(t)dt=\lim_{n\rightarrow \infty}\int_{\mathbb{R}}f_n(t)dt$, where $f_n(t)=f(t)\chi_{[-n,n]}$.  And since $dt=d(x+t)$, we have $\displaystyle \int_{-n}^nf(x+t)dt=\int_{-n+x}^{n+x}f(t)dt$.  So for each $x\in (-\infty,\infty)$,  $$ \begin{split} \int_{\mathbb{R}}f(t)-f(t+x)dt &amp;=\lim_{n\rightarrow \infty}\int_{\mathbb{R}}f(t)(\chi_{[-n,n]}-\chi_{[-n+x,n+x]})dt\\ &amp;=\lim_{n\rightarrow \infty}\int_{\mathbb{R}}f(t)(\chi_{[-n,-n+x]}-\chi_{[n,n+x]})dt\\ &amp;=\lim_{n\rightarrow \infty}\lim_{k\rightarrow \infty}\int_\mathbb{R}\varphi_k(t)(\chi_{[-n,-n+x]}-\chi_{[n,n+x]})dt\\ &amp;=\lim_{k\rightarrow \infty}\lim_{n\rightarrow \infty}\int_\mathbb{R}\varphi_k(t)(\chi_{[-n,-n+x]}-\chi_{[n,n+x]})dt \end{split} $$ where $\varphi_k(t)$ is simple function pointwise converges to $f(t)$, simple function vanish out of set of finite measure, so $\varphi_k(t)(\chi_{[-n,-n+x]}-\chi_{[n,n+x]})$ vanishes when $n$ big enough(is this right?)    Another try:  Since $f\in L^1$, $f$ is bounder a.e. so can be approximated by simple functions, and $\chi_{A+t}(x)=\chi_A(x-t)$,thus $$\int f(t)=\lim_n\int\varphi_n=\lim_n \sum_{i=1}^nc_i\mu(E_i)=\lim_n \sum_{i=1}^nc_i\mu(E_i-x)=\int f(t+x)$$  For problem 2, I guess they share the same method.   Any help or correction about these two problems? Thanks. Change Linux Mint to Ubuntu What is the reason that the value "47" is so popular in electrical engineering? We often see component values of 4.7K Ohm, 470uF, or 0.47uH.  For example, digikey has millions of 4.7uF ceramic capacitors, and not a single 4.8uF or 4.6uF and only 1 listed for 4.5uF (specialty product).  What's so special about the value 4.7 that sets so far apart from say 4.6 or 4.8 or even 4.4 since in the 3.. series we usually 3.3,33, etc.  How did these numbers come to be so entrenched?  Perhaps a historical reason? Moving files of specific type/extension from one directory to another? How do you check if a JavaScript Object is a DOM Object? I'm trying to get:  document.createElement('div')  //=&gt; true {tagName: 'foobar something'}  //=&gt; false   In my own scripts, I used to just use this since I never needed tagName as a property:  if (!object.tagName) throw ...;   So for the second object, I came up with the following as a quick solution -- which mostly works. ;)  The problem is, it depends on browsers enforcing read-only properties, which not all do.  function isDOM(obj) {   var tag = obj.tagName;   try {     obj.tagName = '';  // Read-only for DOM, should throw exception     obj.tagName = tag; // Restore for normal objects     return false;   } catch (e) {     return true;   } }   Is there a good substitute? How do I find the package that provides a file? Simple enough question: is there some shell command (or GUI method) I can use that, given the path to a file on my system, tells me what package put it there? Assuming the file did in fact come from a package, that is.  Bonus question: what if it's a file that isn't installed on my system? Is there, say, a website that will let me look up a file and see what packages, if any, provide it? Let G be a group, where $(ab)^3=(a^3)(b^3)$ and $(ab)^5=(a^5)(b^5)$. Prove that $G$ is abelian group.  Thank you in advance. Any help is appreciated. Let's say I have the following class X where I want to return access to an internal member:  class Z {     // details };  class X {     std::vector&lt;Z&gt; vecZ;  public:     Z&amp; Z(size_t index)     {         // massive amounts of code for validating index          Z&amp; ret = vecZ[index];          // even more code for determining that the Z instance         // at index is *exactly* the right sort of Z (a process         // which involves calculating leap years in which         // religious holidays fall on Tuesdays for         // the next thousand years or so)          return ret;     }     const Z&amp; Z(size_t index) const     {         // identical to non-const X::Z(), except printed in         // a lighter shade of gray since         // we're running low on toner by this point     } };   The two member functions X::Z() and X::Z() const have identical code inside the braces.  This is duplicate code and can cause maintenance problems for long functions with complex logic.    Is there a way to avoid this code duplication? How do I backup all files? (.dbus and dconf backup problems) What are /home/user/.cache/dconf and /home/user/.dbus? Why does their backup fail?  I tried following  (see code below) but while it allowed me to backup the above files, now I am told it couldn't backup /home/[user]/.dbus/session-bus.    $ man chown $ sudo chown user .cache/dconf   [sudo] password for user:  $ sudo chown user .dbus $ man chgrp $ sudo chgrp user .cache/dconf $ sudo chgrp user .dbus   How do I fix it so it will not take my time closing an error message after each backup? Should I exclude these folders ? Why are they not needed? Why would someone want to back them up? Analytic continuation of a power series 2 Another Qual question here,  For the function $$\sum_{n=0}^\infty z^{2^n}$$,  Prove the following:  i) $f$ converges to a function analytic in the open unit disk $D$,  ii) $f(z) =z+f(z^2)$ and  iii) $f(z)$ can not be analytically continued past any point on the unit circle.  I can even see (ii) very easily, but I can not see how can I prove rigorously (i) and  (iii). Help please.