Trying to do the upgrade of kubuntu from 19.04 to 19.10 Non-LTS versions are allowed (Prompt=normal in /etc/update-manager/release-upgrades)  update-manager says my system is up to date. Running do-release-upgrade produces "No new release found"  do-release-upgrade -d shows: "Upgrades to the development release are only available from the latest supported release."  Any ideas why I can't upgrade?  Thanks, Rudolf
Recently I have installed an older version of Ubuntu on my old machine. Whenever I try to install any software, I get an error saying it couldn't be found:  $ sudo apt-get install vlc Reading package lists... Done                Building dependency tree        Reading state information... Done     E: Couldn't find package vlc
If $f$ is measurable and $fg$ is in $L^1$ for all $g \in L^q$, must $f \in L^p$? How to prove that the sum and product of two algebraic numbers is algebraic? Suppose $E/F$ is a field extension and $\alpha, \beta \in E$ are algebraic over $F$. Then it is not too hard to see that when $\alpha$ is nonzero, $1/\alpha$ is also algebraic. If $a_0 + a_1\alpha + \cdots + a_n \alpha^n = 0$, then dividing by $\alpha^{n}$ gives $$a_0\frac{1}{\alpha^n} + a_1\frac{1}{\alpha^{n-1}} + \cdots + a_n = 0.$$     Is there a similar elementary way to show that $\alpha + \beta$ and $\alpha \beta$ are also algebraic (i.e. finding an explicit formula for a polynomial that has $\alpha + \beta$ or $\alpha\beta$ as its root)?    The only proof I know for this fact is the one where you show that $F(\alpha, \beta) / F$ is a finite field extension and thus an algebraic extension. Is "laic" an alternative word to the word "layperson"? 100 apples are/is considered as a large number of apples Rotation and Scale have suddenly stopped working? I was working quickly, (making mistakes), and suddenly rotation and scale transforms stopped working. I think I hit some shortcut or button by accident. Translate still works though. I checked that the transforms weren't locked and such, to no avail. What could be the cause? Why is MixColumns omitted from the last round of AES? How to open terminal, split to 9 terminals and switch between them using one script? In a triangle $\angle A = 2\angle B$ iff $a^2 = b(b+c)$ Prove that in a triangle $ABC$, $\angle A = \angle 2B$, if and only if:  $$a^2 = b(b+c)$$  where $a, b, c$ are the sides opposite to $A, B, C$ respectively.   I attacked the problem using the Law of Sines, and tried to prove that if $\angle A$ was indeed equal to $2\angle B$ then the above equation would hold true. Then we can prove the converse of this to complete the proof.   From the Law of Sines,   $$a = 2R\sin A = 2R\sin (2B) = 4R\sin B\cos B$$  $$b = 2R\sin B$$  $$c = 2R\sin C = 2R\sin(180 - 3B) = 2R\sin(3B) = 2R(\sin B\cos(2B) + \sin(2B)\cos B)$$  $$=2R(\sin B(1 - 2\sin^2 B) +2\sin B\cos^2 B) = 2R(\sin B -2\sin^3 B + 2\sin B(1 - \sin^2B))$$  $$=\boxed{2R(3\sin B - 4\sin^3 B)}$$  Now,   $$\implies b(b+c) = 2R\sin B[2R\sin B + 2R(3\sin B - 4\sin^3 B)]$$  $$=4R^2\sin^2 B(1 + 3 - 4\sin^2 B)$$  $$=16R^2\sin^2 B\cos^2 B = a^2$$  Now, to prove the converse:  $$c = 2R\sin C = 2R\sin (180 - (A + B)) = 2R\sin(A+B) = 2R\sin A\cos B + 2R\sin B\cos A$$  $$a^2 = b(b+c)$$  $$\implies 4R^2\sin^2 A = 2R\sin B(2R\sin B + 2R\sin A\cos B + 2R\sin B\cos) $$  $$ = 4R^2\sin B(\sin B + \sin A\cos B + \sin B\cos A)$$  I have no idea how to proceed from here. I tried replacing $\sin A$ with $\sqrt{1 - \cos^2 B}$, but that doesn't yield any useful results. A sine integral $\int_0^{\infty} \left(\frac{\sin x }{x }\right)^n\,\mathrm{d}x$ The following question comes from   post $$\int_0^{\infty} \left(\frac{\sin x }{x }\right)^n\,\mathrm{d}x$$ but now I'd be curious to know how to deal with it by methods of complex analysis. Some suggestions, hints? Thanks!!!  Sis. I have an unusual problem with Chrome when I try to open PDFs. If I have already downloaded a PDF and try to download it again, I can open it by tapping its name; in the screenshot below, I could tap "pdf.pdf" to open the file.   The issue is that tapping this link actually opens RollerCoaster Tycoon Classic! If I choose to download again, Chrome redownloads and opens the PDF correctly, so it's not a huge issue, but still an inconvenience.  In Settings, it says that RollerCoaster Tycoon Classic has no defaults, and I am unable to change the "open supported links" setting.    How do I stop RollerCoaster Tycoon  trying to be a PDF reader? Differences between astronomy, astrophysics and cosmology? How does the system determine which one close reason should be displayed on a closed question, given the close votes? When voting to close a question, it's possible for users to vote for any close reason, even if others have voted for different reasons. However, in the end, once the question reaches five close votes for any reason, only one of those reasons can be displayed.  How does the system determine which one close reason to display, given the votes made by users? Want to switch from software development to testing domain What is the number of strings at size $n$ that is constructed from ${a,b,c,d}$ and there is an even num of $a$ What is the number of strings at size $n$ that is constructed from ${a,b,c,d}$ and there is an even num of $a$?  I have tried to answer it as recursion formula in the following logic:  In order to build $ A(n) $, I will split into several cases:   if the string starts with $a$, then I will need another $a$ to make an even number of $a's$. so we get $aa[A(n-2)]$ meaning 2 $a's$ at the start + a valid string at the size of $n-2$. if it starts with $b,c,d$ we get $b[A(n-1)]$ + $c[A(n-1)]$ +$d[A(n-1)]$, meaning a one of the letters + a valid string at the size of $n-1$.   So we will get a recursion formula that looks like that:  $A(n) = A(n-2) + 3A(n-1)$  I'm not 100% on my answer, am I right to approach it in that direction? Do electrons move around a circuit? We could imagine a simple electronic circuit composed by a power source and a resistor.   It is usual find descriptions as "The moving charged particles in an electric current are called charge carriers. In metals, one or more electrons from each atom are loosely bound to the atom, and can move freely about within the metal. These conduction electrons are the charge carriers in metal conductors." (wikipedia ). That made me imagine electrons going from power -, running by the wires, crossing the resistor and returning to power +, where they suffer a new kick to power -. All that at near the speed of light.  In this model, it is easy to define "electrical current intensity" as proportional to the number of electrons per second that cross one specific point of the circuit.  However, looking for the physics of resistivity, I can read the phrase "The actual drift velocity of electrons is very small, in the order of magnitude of a meter per hour." (wikipedia ). This is a complete different point of view, where electrons are more or less fixed around a fixed position (or moving very slowly), an electricity is a wave of "collisions" or "jumps" of electrons, like in a Newton's cradle (or "do the wave" as public in a football match). Obviously, we do not need severals hours from the switch on to the light on.  So, the question is, do electrons move around a circuit ? References to band theory are welcome.  A related question is: if electrons doesn't moves along the circuit, what is "electrical current intensity"?  Note: I known there are similar questions already posted, but answers of the ones I've read are not very detailed (could be I've not found the best ones). Please, be a few patient before "close vote". How to boot from non-Ubuntu live ISO images like Fedora or CentOS? I have studied some of Einstein's Theory of General Relativity, and I understand that it states that gravity isn't a force but rather the effects of objects curving space-time. If this is true, then why are we instructed in middle school that it is a force? "invalid use of incomplete type" error with partial template specialization Extracting embedded images from a PDF An equality involving the mean value property I am trying to establish that  $$\frac2{\pi}\int_0^{2\pi} f(e^{i\theta})\cos^2\left(\frac{\theta}{2}\right)\;\mathrm d\theta=2f(0)+f^\prime(0)$$  for $f$ analytic on $|z| &lt; R,\; R &gt; 1$  I think I am making a mistake somewhere, please point it out to me.Here's what I've done unsing the mean value property and Cauchy's integral formula:  $$2f(0)+f&#39;(0)=2(\frac1{2\pi}\int_0^{2\pi} f(e^{i\theta})d\theta + \frac1{2\pi i}\int_0^{2\pi}\frac{f(e^{i\theta})ie^{i\theta}}{e^{i2\theta}}=\frac{1}{\pi}\int_0^{2\pi} f(e^{i\theta})d\theta\frac1{2\pi}\int_0^{2\pi}\frac{f(e^{i\theta})}{e^{i\theta}}\quad\text{(1)}$$  But when I develop the expression $\frac2{\pi}\int_0^{2\pi} f(e^{i\theta})\cos^2(\frac{\theta}{2}) d\theta$ I get:  $$\frac1{\pi}\int_0^{2\pi} f(e^{i\theta})d\theta+\frac1{2\pi}\int_0^{2\pi}f(e^{i\theta})[{e^{i\theta}}+e^{i-\theta}]d\theta \quad\text{(2)}$$  The second integral in (1) differs from the second integral in (2). Thanks.