I have two machines A and B which are in different subnets, both behind separate firewalls. Machine A can see B, but B cannot see A. I have a user account (non-root) on both machines, I can SSH B from A, and I would like to be able to SSH A from B instead, but that cannot be done directly.   I have used tunnelling to hop through intermediate servers with SSH, but what I am asking is different here, and I don't know what it would be called. Is there a way to open a connection from A to B, which could in turn be used "in reverse" from machine B to run commands on A?
As I understand this, firewalls (assuming default settings) deny all incoming traffic that has no prior corresponding outgoing traffic.  Based on  and , reverse SSH tunneling can be used to get around pesky firewall restrictions.  I would like to execute shell commands on a remote machine. The remote machine has its own firewall and is behind an additional firewall (router). It has an IP address like 192.168.1.126 (or something similar). I am not behind a firewall and I know the remote machine's IP address as seen from the Internet (not the 192.168.1.126 address). Additionally, I can ask someone to execute ssh (something) as root on the remote machine first.  Could anyone explain me, step by step, how reverse SSH tunneling works to get around the firewalls (local and remote machines' firewalls and the additional firewall between them)?  What is the role of the switches (-R, -f, -L, -N)?
Selecting ARIMA orders based on ACF-PACF vs. auto.arima I'm working on a 2nd order passive low pass filter, consisting of two passive low pass filters chained together.         &ndash; Schematic created using   Let \$ H(s) = H_1(s)H_2(s) \$ where \$ H_1(s) \$ and \$ H_2(s) \$ are the transfer functions for each separate filter stage.  Then \$ |H(s)| = |H_1(s)||H_2(s)| \$  Knowing the magnitude of a passive low pass filter,   $$|H(s)| = \dfrac1{\sqrt{ (\omega R_1C_1)^2 + 1} } \times \dfrac1{\sqrt{ (\omega R_2C_2)^2 + 1} } = \dfrac1{\sqrt{((\omega R_1C_1)^2 + 1)((\omega R_2C_2)^2 + 1)}} $$  Then trying to find the cutoff frequency:  $$ \left(\dfrac1{\sqrt{2}}\right)^2  = \dfrac1{\sqrt{((\omega R_1C_1)^2 + 1)((\omega R_2C_2)^2 + 1)}} $$ $$ 2  = \sqrt{((\omega R_1C_1)^2 + 1)((\omega R_2C_2)^2 + 1)} $$ $$ 4  = ((\omega R_1C_1)^2 + 1)((\omega R_2C_2)^2 + 1) $$ $$ 4  = (\omega R_1C_1)^2(\omega R_2C_2)^2 + (\omega R_1C_1)^2 + (\omega R_2C_2)^2 + 1 $$  And I'm stuck. Research on the web tells me \$ \omega_c = \dfrac1{\sqrt{R_1C_1R_2C_2}} \$, but I can't find why? Can anybody show my the derivation to find this? I have read about copying files with terminal but these examples will help me a lot. So here is what I want to do:  Examples:   I have a file in /home/levan/kdenlive untitelds.mpg and I want to copy this file to /media/sda3/SkyDrive and do not want to delete any thing in SkyDrive directory. I have a file in /media/sda3/SkyDrive untitelds.mpg and I want to copy this file to /home/levan/kdenlive  and do not want to delete any thing in kdenlive directory  I want to copy a folder from home directory to sda3 and do not want to delete any thing on sda3 directory and opposite I want to cut a folder/file and copy to other place without deleting files in that directory I cut it into. What is the difference between a[bc]d and a{b,c}d? Why do people use a{b,c}d when there is already a[bc]d? Today I lost 25 reputation points and I'm just curious why? Apparently it was an upvoted and accepted answer and I thought that questions with answers can't be deleted easily. Talking about the general case, when a question gets removed by a Moderator, I'd think it'd be unfair to lose the points, because one still took the time to answer the question and probably invested a significant amount of work.   Unfortunately I don't have a link to the question. All I have is this:        but the "link" to the question is not clickable.  A syntax error question is a bad example for "probably invested a significant amount of work", I know, but again, I'm just curious what happened and want to know, why one does loose the reputation points ? Show that the set $\mathbb{Q}[\sqrt{2}] = \{a + b \sqrt{2} \mid a, b \in \mathbb{Q}\}$ is a field with the usual multiplication and addition. Show that the set $\mathbb{Q}[\sqrt{2}] = \{a + b \sqrt{2} \mid a, b \in \mathbb{Q}\}$ is a field with the usual multiplication and addition.  It is easy enough to show that it is closed under addition and multiplication ( $\forall a,b \space)$ we have $a+b \in \mathbb{Q}\sqrt{2}$ and $a * b = ab \sqrt{2} \in \mathbb{Q}\sqrt{2}$.  However, I had trouble proving the other axioms (associativity, commutativity, unique neutral element, unique inverse, and distributivity of multiplication over addition). I would really appreciate it if someone could help me with those. Is it the case that for all sets $A, B, C,$ and $D$, $(A \times B) \cup (C \times D) = (A \cup B) \times (C \cup D)$? I think I have managed to work this out, however there is no solution provided with the question so I thought it best to check.   My working:  Let $(x,y) \in (A \times B) \cup (C \times D)$  $\Rightarrow x \in (A$ or $C)$ and $y \in (B$ or $D)$  $\Rightarrow (x, y) \in (A \cup C) \times (B \cup D)$  $\Rightarrow (A \times B) \cup (C \times D) \subseteq (A \cup C) \times (B \cup D)$  Now let $(x,y) \in (A \cup C) \times (B \cup D)$  $\Rightarrow x \in (A$ or $C) $ and $y \in (B$ or $D)$  $\Rightarrow (x, y ) \in (A \times B) \cup (A \times D) \cup (C \times B) \cup (C \times D)$  $\Rightarrow (A \cup C) \times (B \cup D) \nsubseteq (A \times B) \cup (C \times D)$  $\Rightarrow (A \times B) \cup (C \times D) \not= (A \cup C) \times (B \cup D)$  Is this correct?  Thank you. How do I get started with RAW photography? Having read about the benefits of shooting in RAW as opposed to JPEG (for example ), I'd like to have a go with it myself. However, I'm not really sure how to go about it.  I can turn RAW mode on in my camera (that's easy!), but what do I do with the files when I get them on to my computer? I presume I need to process them to get the RAW data into something that I can work with, but do I need a specific tool for my camera (a Finepix S5100), or will something like the GIMP do what I want? I'm also a little confused about how I then take advantage of the abilities that RAW gives me when I'm processing it on the computer — do RAW processing tools have more processing options than standard tools? Do we need a global test before post hoc tests? Pass an extra argument to a callback function Is it better to have an ErrorDocument 404 redirect back to the homepage or a standard 404 error page? How did this character come to be in the Nethersphere? How is it that       The Master   Came to be in the Nethersphere?  The last time we saw him      He died and refused to regenerate.  The Doctor even cremated his body.   So, given what previously happened to him, how was he now able to      Regenerate as Missy and take over the Nethersphere? How do I use superposition to solve a circuit? Yes, this is a pedagogical question. While answering another recent question, I wanted to refer the OP to concise instructions for using superposition to solve circuits. I found that all the easily found resources online were somewhat deficient. Typically they were unclear about what kinds of circuits superposition applies to, or about the actual method to apply the superposition theorem to a circuit problem. So,   What kinds of circuits can be solved by superposition?  How are different kinds of sources treated when solving by superposition?  What are the steps to solve a circuit using the superposition theorem? I have requested for a French long term (Type D) student visa from a consulate in India. It's a national Type D visa and not a Schengen visa. My question is that can we still go to other Schengen countries on this visa? If not then can we get a Schengen visa from a consulate in France? Cannot select vertices from different object while in edit mode Whenever I enter edit mode, with 2 separate meshes (cubes), I can only select vertices/edges/faces of 1 mesh.  Example (the screen shot did not include the mouse cursor):    Also, when I enter edit mode with both meshes selected, one of the meshes has a red border and cannot be selected (it also has this border in Object mode): How do I use Moq to mock an extension method? I am writing a test that depends on the results of an extension method but I don't want a future failure of that extension method to ever break this test. Mocking that result seemed the obvious choice but Moq doesn't seem to offer a way to override a static method (a requirement for an extension method). There is a similar idea with Moq.Protected and Moq.Stub, but they don't seem to offer anything for this scenario. Am I missing something or should I be going about this a different way?  Here is a trivial example that fails with the usual "Invalid expectation on a non-overridable member". This is a bad example of needing to mock an extension method, but it should do.  public class SomeType {     int Id { get; set; } }  var ListMock = new Mock&lt;List&lt;SomeType&gt;&gt;(); ListMock.Expect(l =&gt; l.FirstOrDefault(st =&gt; st.Id == 5))         .Returns(new SomeType { Id = 5 });   As for any TypeMock junkies that might suggest I use Isolator instead: I appreciate the effort since it looks like TypeMock could do the job blindfolded and inebriated, but our budget isn't increasing any time soon. Normal ordering of the commutator between annihilation and creation operator According to the commutation relation of annihilation and creation operators,   $$[a,a^{\dagger}]=1. \tag{1}$$   I would like to calculate the vacuum expectation value of the normal order of this commutator. We claim that since this commutator is just a c-number, then there is no effect from the normal ordering. We therefore get   $$\langle 0|(:[a,a ^\dagger]:)|0 \rangle=1. \tag{2}$$  However, if we expand the commutator first and do normal order later, we will get something like   $$\langle0|(:[a,a^\dagger]:)|0\rangle=\langle0|(:aa^\dagger-a^\dagger a:)|0\rangle=\langle0|(a^\dagger a-a^\dagger a)|0\rangle=0.\tag{3}$$ Which is contradicting with itself. Is "you are welcome" more polite than "no problem"?  What is the general proper reply to a thank-you? Why are relays so frequently driven by optocouplers? How do I get rid of OneDrive completely?