I upgraded from 17.10 to 18.04 today and after a reboot, I noticed my password no longer works. I'm certain the password is correct, but I keep getting "bad passphrase or options". How can I fix this?
A few months ago I set full disk encryption during installation of Ubuntu 17.10. Now, I've decided to upgrade. The upgrade ran to the end without any problems. However, after reboot I am not able to log into my encrypted disk.  Where can be the problem? I am 100% sure I press the "right keys" on the keyboard, but technically I don't know what I am writing because of "*" characters and maybe after upgrade my keyboard layout changed. I use some characters which might be somewhere else on the keyboard. What language is default after upgrade?   By the way, I've already tried Caps-lock, but still no luck.  Help will be really appreciated. I am not a fan of installing my system and all backups over and over again.
Newton's cradle How do I display the nth result of an ls command? I used this command to display the first result of files in my directory.    ls | head -n 1   My simple question is, how can I modify this command to display say the nth result?  Thanks! When you watch a video in YouTube, you can seek it with the thumbnail preview. Many Windows' and MacOS' video players have this feature, but what about Linux's? I've already read , but the information in it seems out-of-date.  is still alive but the project is neglected (see ), and actually the player doesn't work on the latest Ubuntu or Linux Mint.  Is there any video player which has the feature and works on the latest linux distributions? Or must I use Windows' one with VirtualBox? If only VLC would implement this feature, but it seems that won't happen as far as I read .  YouTube's seek preview: What is the difference between "retired out" and "retired not out" in cricket? Learning Python programming with generic GIS goals in mind? In your opinion, what is the best book/site to learn Python with GIS goals in mind?  By "best", I mean:   not very long (book) easy to understand (book/site) good practical examples (book/site)   For answers specific to learning how to use the ArcPy site-package for Python to customize ArcGIS for Desktop there is a separate Q&amp;A: Nested Radical of Ramanujan I think I have sort of a proof of the following nested radical expression due to Ramanujan for $x\ge 0$.     $$\large x+1=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+\cdots}}}}$$ for $ x\ge -1$   I just want to know if my proof is okay or there is a flaw, and if there is one I request to give some suggestions to eliminate them. Thank you. The proof is the following:  Proof: Let us define $$ a_n(x)=\underbrace{\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+\cdots}}}}}_{n \ \mbox{terms}}$$ for $x\ge 0$ so that $$ a_1(x)=\sqrt{1+x},\ a_2(x)=\sqrt{1+x\sqrt{1+(x+1)}},\ a_3(x)=\cdots$$    and so on. Since $x\ge 0$ each one of the $a_n$ is defined (I am taking only the positive square root). Also, we note that $$a_{n+1}^2(x)=1+xa_{n}(x+1)$$  Now, we note that $\{a_n(x)\}$ is an increasing sequence and that $$a_n(x)&lt;x+1$$ $\forall n\ge 1$, this is easy to prove by induction as below:  For $n=1$, $a_1(x)=\sqrt{1+x}&lt;1+x$ since $x\ge 0\Rightarrow 1+x\ge 1$. SO it is true for $n=1$. Similarly, the truth can be proved for $n&gt;1$.  Then $a_n(x)$ converges to $$l(x)=\sup_{n}a_n(x)\le x+1$$ Now I make the following claim:  Claim: $l(x)=x+1\quad \forall i\ge 0$  Proof: Fix $x$. Let $l(x)&lt;x+1$. Then, $l(x)=x+1-\epsilon$ for some $\epsilon&gt;0$. Now, I claim that there must be a $n$ such that $$x+1-a_n(x)&lt;\epsilon$$, and if that is true then $$a_n(x)&gt;x+1-\epsilon=l(x)$$ which is a contradiction since $$l(x)=\sup_{n}a_n(x)$$ and then it implies that $$l(x)=x+1$$ To prove my claim it requires $$x+1-a_n(x) &lt;  \epsilon$$ Now, \begin{align} x+1-a_n(x) = &amp; x+1-\sqrt{1+xa_{n-1}(x+1)} \\ \ =&amp; \frac{(x+1)^2-({1+xa_{n-1}(x+1)})}{x+1+\sqrt{1+xa_{n-1}(x+1)} } \\ \ =&amp; x\frac{(x+1)+1-a_{n-1}(x+1)}{x+1+\sqrt{1+xa_{n-1}(x+1)} } \\ \ &lt;&amp; \frac{x}{x+2}((x+1)+1-a_{n-1}(x+1))\\ \ &lt;&amp; \frac{x}{x+2}\cdot\frac{x+1}{x+3}((x+2)+1-a_{n-2}(x+2))\\ \ &lt;&amp; \frac{x}{x+n-1}\cdot \frac{x+1}{x+n} ((x+n-1)+1-a_{1}(x+n-1))\\ \ &lt;&amp; \frac{x}{x+n-1}\cdot \frac{x+1}{x+n}(x+n-\sqrt{x+n})\\ \ &lt;&amp; \frac{x(x+1)}{x+n-1} \end{align}  Now, if one is able to find $n$ such that $$\frac{x(x+1)}{x+n-1}&lt;\epsilon \Rightarrow x&lt; \frac{-(1-\epsilon)+\sqrt{(1-\epsilon)^2+4\epsilon(n-1)}}{2}$$ then we're done.   Now from the upper bound it seems that there always exists some $n$ that satisfies this requirement. Hence the claim is proved. How to read off coefficients of tensor-like expression in a speedy way? Investment options for f1 visa students in USA formal proof from calulus Given $f:R \to R$, $f$ is differentiable on $R$ and $\lim_{x \to \infty}(f(x)-f(-x))=0$. I need to show that there is $x_0 \in R$ such that $f'(x_0)=0$  I am trying to prove it by contradiction .... so i assume there is no $x_0 \in R$ such that  $f'(x_0)=0$ this means this function is strictly monotonic because the derivative "respects" the mean value theorem so it cannot be negative and positive without passing a zero value ....  (from what i understand the derivative function of a continuous/differentiable functions isn't necessarily continuous but it still respect the mean value theorem i think it's called darboux's theorem )  now from $\lim_{x \to \infty}(f(x)-f(-x))=0$ ,we get $\lim_{x \to \infty}f(x)=\lim_{x \to \infty}f(-x)$  now the last step is the contradiction its very intuitive and even maybe obvious that we can't get $\lim_{x \to \infty}f(x)=\lim_{x \to \infty}f(-x)$ for strictly monotonic function but i cant think of a formal way to prove it . Sphere inside a conical vessel touches the lid A conical vessel with semi-vertical angle 30 degrees and height 10.5 cm has a thin lid. A sphere kept in the vessel touches the lid. The radius of the sphere in cm is ... Typesetting UML class diagrams What's the best way to draw UML class diagrams in LaTeX? We have the following points: $$ (0,0)(1,51.8)(1.9,101.3)(2.8,148.4)(3.7,201.5)(4.7,251.1)  \\ (5.6,302.3)(6.6,350.9)(7.5,397.1)(8.5,452.5)(9.3,496.3) $$ How can we find the best fitting line $y=ax$ through the points? My calculator has the option to find the best fitting line $y=ax+b$ through these points, which is:  $$y = 53.28x + 0.37$$  How can I find the best fitting $y=ax$? It seems to me we can't just remove the $0.37$ without compensating in the $a$? Proving that the triangle inequality holds for a metric on $\mathbb{C}$ Show that $(X,d)$ is a metric space where $X =\Bbb C $ and the distance function is defined as:   $$d(x,y) = \frac {2|x-y|}{\sqrt {1+|x|^2} +  \sqrt {1 + |y|^2}}, \text{ for } x,y \in \Bbb C.$$   I have done the proof of the first two propositions for being a metric, but I'm having a problem in proving the triangle inequality. Memory Usage of SQL Server SQL Server instance on my server is using almost full memory available in my Physical Server. Say if i am having 8GB of RAM than SQL Server is using 7.8 GB of RAM from system. I also have read articles and also read many similar questions regarding same on this forum and i understand that memory is reserved and it is using memory.  But i have 2 same servers and 2 SQL Servers, why this is happening on a single SQL Instance not on other. Also when i run DBCC MemoryStatus than it is showing up...  VM Reserved 8282008 VM Committed 537936   so from this we know that SQL reserved whole 8GB memory, but why this VM Committed keeps increasing. What i understand is VM Committed is:      VM Committed: This value shows the overall amount of VAS that SQL Server has committed. VAS that is committed has been associated with physical memory.   So this is the memory SQL Server has committed (from this i understand that physical memory actually SQL Server is using at instance).  So like to know the reason behind this ever increasing VM Committed memory on my server and not on another. Thanks in Advance. Proof by induction that $2 + 4 + 6 + \cdots + 2n = n(n+1)$ Pre-image of a measurable set A is always measurable? Let $f$ be a continuous function on a set $E$.  Is it always true that $f^{-1}(A)$ is always measurable if $A$ is measurable?  I say no.  We know that $\bar{\psi(x)}:=\frac{\phi(x)+x}{2}$ where $\phi$ is the Cantor (or Cantor-Lebesgue) function and $\bar{\psi}:[0,1] \rightarrow [0,1]$.  We know this function maps a measurable subset of $C$ (the Cantor set) into a non-measurable set $W$.  A few observations about $\bar{\psi(x)}$.  It is strictly increasing and continuous, so it has a continuous inverse $\bar{\psi(x)}^{-1}$.  Thus $\bar{\psi(x)}^{-1}(W)=c \subset C$; that is, $\bar{\psi(x)}^{-1}(W)$ is non-measurable, but c is.  Is this correct? Quantum teleportation - the alternative to destroying atoms Can't you just disassemble, not destroy our atoms and transport them to another teleporter via networking or telecoms?  This is my thinking of how to keep the same person, not an exact copy of him/her, alive when teleporting.  Please answer to tell me if it's possible or not. In Stein's Fourier Analysis i'm having trouble Justifying the following calculations for the Fourier Coefficients in $(1.)$  $(1.)$  Let $f(\theta) = \theta$ for $ -\pi \leq \theta \leq \pi$ in the case when $n \neq 0$:  $$f(n)=\frac{1}{2 \pi}\int_{-\pi}^{\pi}\theta e^{-in \theta} d \theta = \frac{1}{2 \pi}[-\frac{\theta}{in}e^{-in \theta}]_{-\pi}^{\pi} + \frac{1}{2 \pi in} \int_{-\pi}^{\pi}e^{-in\theta}d \theta = \frac{(-1)^{n+1}}{in} $$  Then the Fourier Series of $f$ is given by:  $$ f(n) = \sum_{n \neq 0}  \frac{(-1)^{n+1}}{in}$$ My website used to have sitelinks and now it doesn't. It's very possible that it's due to changing the website to a sidebar design instead of having an "interstitial" type landing page which limited the number of choices, but I'm not sure.  Here is how sitelinks might look for a site:    What are some things that I can do to improve my chances of getting sitelinks? How do I detect unsigned integer multiply overflow? I was writing a program in C++ to find all solutions of ab = c, where a, b and c together use all the digits 0-9 exactly once. The program looped over values of a and b, and it ran a digit-counting routine each time on a, b and ab to check if the digits condition was satisfied.  However, spurious solutions can be generated when ab overflows the integer limit. I ended up checking for this using code like:  unsigned long b, c, c_test; ... c_test=c*b;         // Possible overflow if (c_test/b != c) {/* There has been an overflow*/} else c=c_test;      // No overflow   Is there a better way of testing for overflow? I know that some chips have an internal flag that is set when overflow occurs, but I've never seen it accessed through C or C++.    Beware that signed int overflow is undefined behaviour in C and C++, and thus you have to detect it without actually causing it. For signed int overflow before addition, see .