Here I am, poking around a remote server via SSH, and I want to find out about/read/edit/whatever a file. I don't like vi or other CMI stuff, I want to read it locally.  Is there any way I can download it without exiting my interactive SSH session? (Even better, would there be any way to run local commands without exiting it?)
If I'm logged in to a system via SSH, is there a way to copy a file back to my local system without firing up another terminal or screen session and doing scp or something similar or without doing SSH from the remote system back to the local system?
Perpendicular Chord parabola If $r_1,r_2$ be the length of the perpendicular chords drawn through the vertex of a parabola $y^2=4ax$, then show that $$(r_1r_2)^{4/3}=16a^2(r_1^{2/3}+r_2^{2/3})$$ I rooted my Dream/G1 last night and finally gave up on modding it (not for good mind you, but for the moment) and opted instead to just restore it to factory settings via Settings &gt; SD card &amp; phone storage &gt; Factory data reset, which did solve all the problems I was having with multitasking (and not being able to answer it, etc.)  Curiously, I still seem to have root on the phone, despite having reset it. I'd love to understand how that is possible! Superuser Permissions is the only non-stock app still on the phone, and I'm still able to use su to gain root privileges from adb shell. Is this because BusyBox installs to my SD Card?  My phone works just fine, so this is hardly critical, but the whole exercise was/is about knowing my phone better and this is most def. a mystery. Thanks! Possessive form for a surname ending with "z" What is the proper possessive form for a surname that ends with “z”? Is it z’ or z’s? How can I replace a string in a file(s)? In linear regression, when is it appropriate to use the log of an independent variable instead of the actual values? Am I looking for a better behaved distribution for the independent variable in question, or to reduce the effect of outliers, or something else? What is the difference between "owing to" and "due to"? every permutation is either even or odd,but not both How we can show every permutation is either even or odd,but not both......I can't arrive at a proof for this ..... Can anybody give me the proof...  Thanks in advance... Isomorphisms preserving integral domains and fields Let $R \simeq S$ be isomorphic commutative rings with unity. Prove the following:  a). If $R$ is an integral domain then $S$ is an integral domain.  For this, I said that If I let $f: R \to S$ be an isomorphism and $(R, + , \circ), (S, \oplus, \ast)$, then let $x, y\in R$,  such that $x \circ y = 0_R$, because $f$ is an isomorphism, $f(x \circ y) = f(x) \ast f(y) = f(0_R)$. If I let $R$ be an integral domain then, as per the definition $R$ is commutative, has unity, and no zero divisors.   So $f(x \circ y) = f(x) \ast f(y) = f(0_R)$ where we have either $f(x \circ y) = f(0_R) \ast f(y) = f(0_R)$ or $f(x \circ y) = f(x) \ast f(0_R) = f(0_R)$.   Since $f$ is an isomomorphism, let $s = f(x), t = f(y)| s,t \in S$. Then $s \ast t = f(0_R)$. By definition, $S$ also has unity so $0_S$ exists. If we map $0_R \to 0_S$. Then either $s = 0_S$ or $t = 0_S$.   So then, since $S$ has unity, is commutative, and has a no zero divisors as we showed above, it is also an Integral domain.  I'm not sure I have this correctly.  b). If $R$ is a field then $S$ is a field. I don't really understand how to do this one. Since $S$ is an integral domain, it has a $0_S$. So I need to show that $S$ distributes $\ast$ over $\oplus$.  Any help with these is appreciated Clearly if $\lambda$ is an eigenvalue of $A$ then $p(\lambda)$ is an eigenvalue of $p(A)$ where $p$ is a polynomial. And there are cases where $A$ may have eigenvalues other than these.   Is there a general rule for finding all eigenvalues of $p(A)$? Does it depend on the field from which the elements of $A$ are taken? Are their any partial solutions to this problem? For a given hardware configuration, how do I find out if Ubuntu will run on it? What considerations should I take into account when choosing an Ubuntu version and  such as:    with a lighter desktop than the usual Gnome and Unity  with the even lighter LXDE desktop   Obviously Ubuntu does not run on some processor architectures. So how do I go about choosing the right version and derivate. How can I find out the minmal system requirements? If I'm on a page that has web notifications (like Slack or others), Firefox asks me if I want to enable them. However, my choices are "Yes" or "Ignore for now". The problem is, if I select "Ignore", every time there is a notification it asks me again. This is incredibly annoying.   How do I disable this completely? (I would accept disabling this feature entirely if that's easiest.) When & How often do 4e Monster powers recharge? Just installed Ubuntu for the first time... Is there any simple way to browse installed programs?  On Mac there is the Applications top-level folder... here I see Desktop, Documents, Downloads etc but no Apps? That'd be a good idea, no? (is there some way to create this myself?)  On Windows there is the Start menu...  Here it seems like I have to click the 'Dash Home' and then I can see 'Recent Apps' ... I can search for them (I'd rather not)  ...or I can click the ruler/pencil icon at the bottom and see 'Recent Apps' and click to expand 'Installed Apps' ...ok that's all of them - is it possible to get a straight list view instead of the hard-to-scan wrapped rows?  Also, is there any way to get rid of the pointless random selection of 'Apps available for download' at the bottom? Is there a way to calculate the improper integral $\int_0^\infty \big(\frac{\ln x}{x - 1}\big)^2 dx$? Is there a way to calculate the improper integral $\int_0^\infty \big(\frac{\ln x}{x - 1}\big)^2 dx$?  What have I tried:  $$\int_0^\infty \Big(\frac{\ln x}{x - 1}\Big)^2 dx = \int_0^1 \Big(\frac{\ln x}{x - 1}\Big)^2 dx + \int_1^\infty \Big(\frac{\ln x}{x - 1}\Big)^2 dx = \int_1^\infty \Big(\frac{\ln x}{x - 1}\Big)^2dx + \int_1^\infty\Big(\frac{\ln t}{\big(\frac{1}{t} - 1\big)t}\Big)^2dt = 2\int_1^\infty \Big(\frac{\ln x}{x - 1}\Big)^2dx$$  From the above statement we can conclude, that the integral does indeed converge, as $0 \leq (\frac{\ln x}{x - 1})^2 \leq \frac{C}{(x - 1)^{\frac{3}{2}}}$, for some constant $C$. However, this new form of the integral is not much helpful in finding its exact value.  I also tried substituting $z = \ln x$ , then $dz = \frac{dx}{x}$. Then the integral becomes reduced to $2\int_{0}^{\infty} \frac{z^2}{e^z(e^z - 1)^2}dz$, which, unfortunately, also does not seem to be any easier. I have a mathematical problem with the following formula. The goal is to transform this equation to have a seperated real and imaginary part. I have already done this with several other equasions, but for this particualar one I don't know where to start.  I dont expect you to do the complete work of transforming for me, it would already be very nice if you could give me a hind on how you would start.  Thank you in advance!  C||RLC      This is my current progress:    My question is:  How can I get rid of the imaginary part jωC2*R?  When the denominator is completely real I can just split up the formula and have an imaginary and a real part. Is it possible to decompose a hermitian and unitrary matrix $A$ into the sum of the Pauli matrix Kronecker products?   For example, I have a matrix 16x16 and want it to be decomposed into something like $$A =\sum_{i,j,k,l}h_{ijkl}\sigma_i\otimes\sigma_j\otimes\sigma_k\otimes\sigma_l$$ I can't find any clear explanation how it can be done. I found only this for two-dimentional case: How can I randomize order of elements in a list? Font characters displayed as squares in Ubuntu 18.04 I'm trying to get a Schengen Visa from my first port of entry, France. I am trying to book an appointment with the French Consulate in SF, and the URL I am using is the following:   I've been checking this page periodically for the past few weeks, and I always see a dialog that that says "We are currently at full capacity. Please try again later." Their calendar provides no direct option for going to the next month. (Using the in-browser developer tools, I noticed that the button for next month is hidden. Un-hiding it and clicking on the Next Month button displays no available appointments either. I checked all the way until September 2016).  My guess is that there are absolutely no appointments for the next 6 months. Or, more likely, their page isn't functioning properly. I'll be traveling in July. Does anybody have any alternate suggestions to book an appointment? Value of Vandermonde type determinant