Cloning an Ubuntu 18.04 physical server
Ubuntu 18.04 bond with netplan duplicate MAC address after cloning
Is the Formula Logically Valid? I have a question for my exam and I find it hard to understand.  I have to prove that the following formula is logically valid:    The professor told me to "push" all the symbols inside the brackets, and use the deduction theorem.  But I don't know how to do it, because I can't find the identities to push the "exist" symbol inside the brackets.  Your help is appriciated, thank you.  Alan Some sites are . What additional sites should the chat support?  One suggestion per answer. Let me start this with : Very long "About me" text squeezes both the left side (flair) and the right side (the section with location, GitHub link, etc.)  Now, the reverse thing has happened:         put a long text in their "location", and it causes the right section to be overly extended, effectively squeezing the About Me section.  When will the squeezing game end? When speaking with my mother a couple of days ago, I read to her a message I was sending to my cousin on her behalf ending with: "... the birthday of your youngest." [implying her child]  She immediately leapt on this and said that as my cousin only had two children, the use of the superlative was not permitted and the comparative must be used instead: "... the birthday of your younger child."  She was adamant that this was a solid grammatical rule that she was taught throughout her education.  As a younger Briton, I have relatively little formal grammatical education to older people who were better taught in this regard, so I usually defer to her on grammatical rule knowledge.  Is she right?  Are superlatives not permitted when the domain of the object is only two?  To me it seems bizarre.  The minimum or maximum of any set does not only exist when there are different minima and maxima, or indeed something which is neither. How can I configure the Ubuntu Dock to show windows only from the current workspace? I would like my Dock to show only the windows of the current workspace. Especially in the little preview window. I find it very confusing to jump around workspaces when trying to find the right window. There's no need to  the  that have led to a lack of trust between  of  and . Suffice it to say that astoundingly they continue to assert that  and, instead of apologizing and reinstating Monica have demanded that she go through a process with no transparency, due process, or involvement from the community.   Has the time has come for Stack Exchange, Inc to spin off the community and all the data needed to keep the sites running to a separate non-profit, who would then in turn license the software from the company?  What's in it for the community?   The community can practice self determination for issues from as basic as the number of moderators they have to as fundamental as the policies they enforce. If the company development continues to be unresponsive to the community's needs, the community would be able to develop its own open source software that would meet its needs. They can find their own funding model instead of having ads imposed on them.   What's in it for the company?   Community actions will no longer reflect on the company. Interaction with the community as a customer may come more naturally to current leadership than the present paradigm. They don't have to worry about paying for the community's operating expenses. They can focus on doing one thing well. Approximating $\arctan x$ for large $|x|$ The multi-part identifier could not be bound What is the value of $1^i$? $\,$ Prove that there is no continuous functions $f:[0,1]\to R$, such that   $$ f(x)+f(x^2)=x. $$   My try. Assume that there is a continuous function with this property. Thus, for any $n\ge 1$ and all $x\in [0,1]$, \begin{align*} f(x)&amp;=x-f(x^2)=x-\big(x^2-f(x^4)\big)=x-x^2+\big(x^4-f(x^8)\big)=\cdots\\ &amp;=x-x^2+x^4-\cdots+(-1)^n\left(x^{2^n}-f\big(x^{2^{n+1}}\big)\right) \end{align*}  since $f(0)=0$ and $\displaystyle\lim_{n\to \infty}x^{2^{n+1}}=0$ for any $x\in(0,1)$,it follows by the continuity of $f$ that $\displaystyle\lim_{n\to \infty}f\big(x^{2^{n+1}}\big)=0$, hence $$f(x)=x-x^2+x^4-x^8+\cdots+(-1)^nx^{2^n}+\cdots$$ for any $x\in (0,1)$  Why do I have prove that there is exists such a continuous functions $f$? Maybe  my example is wrong? Why?  if my   method is wrong,then How prove this problem ? Thank you. Direct Proof that $1 + 3 + 5 + \cdots+ (2n - 1) = n\cdot n$ Show that if $\sqrt{a}+\sqrt{b}\neq 0$, then $\mathbb{Q}(\sqrt{a}+\sqrt{b}) = \mathbb{Q}(\sqrt{a},\sqrt{b}), \forall a,b\in\mathbb{Q}$ Strictly Upper Triangular $n\times n$ Matrices form a Nilpotent Lie Algebra over $\mathbb{C}$ for $n\geq 2$ I have been attempting this problem for a while, it is an assignment problem so I don't want somebody to just post the answer, I'm just looking for hints.  Let $\mathfrak{u}(n,\mathbb{C})$ be the Lie algebra of strictly upper triangular $n\times n$ matrices over $\mathbb{C}$.  I am trying to show that $\mathfrak{u}(n,\mathbb{C})$ is a nilpotent Lie algebra for $n\geq 2$.  This is what I have looked at so far.  I have done some explicit calculations (I shall omit the bulk of them and just summarise below) for $n=2,3,4$ and have come up with some ideas.  Let $L=\mathfrak{u}(n, \mathbb{C})$.  If $n=2$,  $L^1 = [L, L] = &lt;0&gt;_{\mathbb{C}}$  So $L$ is nilpotent.  If $n=3$,  $L^1 = [L, L] = &lt;e_{13}&gt;_{\mathbb{C}}$  $L^2 = [L, L^1] = &lt;0&gt;_{\mathbb{C}} = \left\{ 0 \right\}$  So, $L$ is nilpotent.  If $n=3$  $L^1 = [L,L] = &lt;e_{13},e_{14},e_{24}&gt;_\mathbb{C}$  $L^2 = [L,L^1] = &lt;e_{14}&gt;_{\mathbb{C}}$  $L^3 = [L, L^2] = &lt;0&gt;_{\mathbb{C}} = \left\{ 0 \right\}$  So, $L$ is nilpotent.  So my hypothesis so far is that for an arbitrary $m\in\mathbb{N}$, $L^{m-1}=\{0\}$  My instinct is to do this by induction. My base case, $n=2$ has already been shown. Now I assume that $\mathfrak{u}(n,\mathbb{C})$ is nilpotent for $n=k$ ($k\in\mathbb{N}$) and consider the case for n=k+1.  However, I have no idea how to link these two cases together. I have read in Introduction to Lie Algebras and Representation Theory - J.E. Humphreys that $L^k$ should be spanned by basis vectors $e_{ij}$ where $j-i=k+1$ - however in my calculations for $L^1$, I obtain an $e_{14}$ which does not satisfy 4-1=2.  Am I missing something completely obvious? Or have I made a mistake somewhere? (I have checked my calculations numerous times, but cannot find any errors!)  Thanks,  Andy. Why is the notation $\frac{\partial f}{\partial x}$ used? Say a security has been trading at under $20.10, and there are limit orders selling at $20.10 and $20.21. Then Simon issues a limit order to buy at $20.21. According to , Simon would end up buying shares at $20.10.  Why does the exchange favor the buyer here?  I.e., why does the transaction execute at the seller's limit of $20.10 instead of the buyer's limit of $20.21?  (I.e., why don't Inara/River get to sell at $20.21 / above their limit, but Simon gets to buy at $20.10 / below his limit?)  Is the example flawed?  If it makes a difference, I'm mainly curious about NASDAQ and NYSE.  I tried Googling for specs (e.g. ), but couldn't find anything.  Only found that some  implementation executes the average of the buy/ask.  Just curious how the system works. Failed to load map. Error contacting Google servers. This is probably an authentication issue $(function(){   $('.slider').mobilyslider({     content: '.sliderContent',     children: 'div',     transition: 'horizontal',     animationSpeed: 500,     autoplay: true,     autoplaySpeed: 9000,     pauseOnHover: true,     bullets: true,     arrows: true,     arrowsHide: true,     prev: 'prev',     next: 'next',     animationStart: function(){},     animationComplete: function(){}   });    });   I am integrating the above JavaScript file in a Drupal theme, but it is giving me the "$ is not function" error. I googled it, and changed the code bro the below, but it doesn't work either.  (function($) {   $('.slider').mobilyslider({     content: '.sliderContent',     children: 'div',     transition: 'horizontal',     animationSpeed: 500,     autoplay: true,     autoplaySpeed: 9000,     pauseOnHover: true,     bullets: true,     arrows: true,     arrowsHide: true,     prev: 'prev',     next: 'next',     animationStart: function(){},     animationComplete: function(){}   });    })(jQuery);   The above code gives me following error:     $('.slider').mobilyslider is not a function The problem statement is: for $a,b$ coprime, prove that there exists positive integers $x,y$ such that $ax-by=1$.  The question is form Arthur Egnels' problem solving text, and from the chapter on the pigeonhole principle.    My question: His proof begins by generating a list $a,\ldots,a(b-1)$, where he then points out that each element in this list(which proceeds sequentially) doesn't have remainder $0\pmod{b}$.  He then shows that we arrive at a contradiction if we assume that we also don't get remainder $1$ in this list $\bmod b$.  To show the contradiction in this last statement, he first states that we would have positive integers $p,q $ where $ 0&lt;p&lt;q&lt;b$ so that $pa\equiv qa \pmod{b}$.  He then goes on to point out that since we $a$ and $b$ are coprime, we then have that $b| q-p$.  I understand the last part here, but don't see why we get $pa\equiv qa \pmod{b}$.  Thanks in advance Applied modifiers are not visible A triangle with vertices on the sides of a square, with one at a midpoint, cannot be equilateral