New to Ubuntu (using 15.04) and trying to install a couple of Network printers I have. Brother HL-2280dw, and Brother MFC-J870dw. I downloaded the Linux deb drivers from Brother's support site and the application installer will give a message to not install.  Why?  Also tried directly by connecting to the network printer, and letting Ubuntu look for drivers and install from a free site it recommended, but during the installation it froze, and I had to reboot.
Guides on how to install drivers for a Brother all-in-one network printer are either for older devices or are rather complicated as they often involve editing or copying system files in my root directory.  Is there a newer, more convenient way to install the drivers needed?
How do I find out which process is eating up my bandwidth? I think I'm being the victim of a bug here. Sometimes while I'm working (I still don't know why), my network traffic goes up to 200 KB/s and stays that way, even tough I'm not doing anything internet-related.   This sometimes happens to me with the CPU usage. When it does, I just run a top command to find out which process is responsible and then kill it. Problem is: I have no way of knowing which process is responsible for my high network usage. Both the resource monitor and the top command only tell me my total network usage, neither of them tells me process specific network info.  I've found questions here about monitoring total bandwidth usage, but, as I mentioned, that's not what I need. Is there another command I can use to find out which process is getting out of hand?  The command iftop gives results that disagree entirely with the information reported by System Monitor. While the latter claims there's high network traffic, the former claims there's barely 1 KB/s.  I've already tried killing all the obvious ones (Firefox, update-manager, Pidgin, etc) with no luck. So far, restarting the machine is the only way I found of getting rid of the issue. Using Integration By Parts results in 0 = 1 I've run into a strange situation while trying to apply Integration By Parts, and I can't seem to come up with an explanation.  I start with the following equation:  $$\int \frac{1}{f} \frac{df}{dx} dx$$  I let:  $$u = \frac{1}{f} \text{ and } dv = \frac{df}{dx} dx$$  Then I find:  $$du = -\frac{1}{f^2} \frac{df}{dx} dx \text{ and } v = f$$  I can then substitute into the usual IBP formula:  $$\int udv = uv - \int v du$$  $$\int \frac{1}{f} \frac{df}{dx} dx = \frac{1}{f} f - \int f \left(-\frac{1}{f^2} \frac{df}{dx}\right) dx$$  $$\int \frac{1}{f} \frac{df}{dx} dx = 1 + \int \frac{1}{f} \frac{df}{dx} dx$$  Then subtracting the integral from both sides, I've now shown that:  $$0 = 1$$  Obviously there must be a problem in my derivation here...  What wrong assumption have I made, or what error have I made?  I'm baffled. Q) Prove that if $0\rightarrow V_1\rightarrow V_2\rightarrow...\rightarrow V_k\rightarrow 0$ is an exact sequence of finite dimensional vector spaces, then $\sum_{i=1}^{k}(-1)^idim (V_i) = 0$.  For $k=3$, I see that the statement is true. For $k&gt;3$, am I correct in understanding that $V_2 = V_1\oplus V_3 (\implies dim(V_2)=dim(V_1)+dim(V_3))$ and $V_3 = V_2\oplus V_4(\implies dim(V_3)=dim(V_2)+dim(V_4))$ and so on?   If it is true, then for $k=4$, $\sum_{i=1}^{4}(-1)^idim (V_i) = -m_1+m_1+m_3-m_2-m_4+m_4$ and am not sure why $m_2=m_3$ for the sum to be $0$? where $m_i = dim(V_i)$. Thanks. Non-perpendicular hyperplane decision trees The usual implementation of decision trees, and estimators based on ensembles of decision trees, use decision trees that threshold on a single predictor at each split. So they divide the feature space into perpendicular hyperlanes.  Is there any benefit to generalizing this a tiny bit. So at each split, the decision tree uses 2 features, and thresholds based on whether $ax_1 + bx_2 \ge T$. In other words, the feature space is now getting divided into hyperplanes that are not necessarily perpendicular.   I think the main disadvantages would be:    increased computation, and  overfitting since now using more degrees of freedom   I guess to help both of these issues, randomization could be used analogous to extra trees. The pair of features could be randomly selected, instead of exhaustively trying every pair, and the threshold T and parameters a and b could be chosen randomly as well (perhaps from a set of few discrete values such that the hyperplanes could have a few orientations).  Are there some cases where you think these non-perpendicular decision trees would be an improvement over the regular implementation of various decision tree algorithms? Is it supposed to be a HTML or an HTML Var[A+B+C] = Var [A] + Var [B] + Var [C] I know that this relationship is correct when A,B and C are pairwise independent. However, I'm finding it difficult to calculate it.  Let's say that: Var [A] = Var [B] = Var [C] = 0.25. How can I prove that Var[A+B+C] = 0.75?  I started to work with the formula: Var[A+B+C] =E[(A+B+C)^2]−(E[A+B+C])^2 but I got stuck.  Thanks! Show $\text{Hom}_{Ab}(A,\mathbb{Q}) = \{0\}$ for finite $A$ I have a finite abelian group $A$ and I have shown that $A\otimes\mathbb{Q} \cong \{0\}$, the trivial group.  Now I want to show that $\text{Hom}_{\textbf{Ab}}(A,\mathbb{Q}) = \{0\}$, i.e we only have the trivial zero group homomorphism from $A$ to $\mathbb{Q}$.  The question I'm trying to solve is phrased in such a way that $\text{Hom}_{\textbf{Ab}}(A,\mathbb{Q}) = \{0\}$ should fall out almost automagically once you have that the tensor product collapses but I'm really stuck on this even though some people I talked to said it was really easy.  Some thoughts:   I've tried to assume the existence of a nontrivial homorphism $A \to    \mathbb{Q}$ to generate a contradiction but with no success. We have the Tensor-hom adjunction theorem stating that for abelian groups $A,B,C$ we have a natural bijection  $$\text{Hom}_{\textbf{Ab}}(A\otimes B,C) \cong \text{Hom}_{\textbf{Ab}}(B ,\text{Hom}_{Ab}(A,C))$$ and by plugging in our particular groups we get that:   $$\text{Hom}_{\textbf{Ab}}(A\otimes \mathbb{Q},\mathbb{Q})\cong\text{Hom}_{\textbf{Ab}}(\mathbb{Q} ,\text{Hom}_{Ab}(A,\mathbb{Q}))$$  Since $A\otimes\mathbb{Q} \cong \{0\} \implies \text{Hom}_{\textbf{Ab}}(\mathbb{Q} ,\text{Hom}_{Ab}(A,\mathbb{Q})) = \{0\}$ so we only have the trivial morphism from $\mathbb{Q} \to \text{Hom}_{Ab}(A,\mathbb{Q})$ but I don't think this implies that $\text{Hom}_{Ab}(A,\mathbb{Q}) = \{0\}$ and I'm not sure if this reasoning helps me at all. Reference - What does this regex mean? I have functionality on a cloud page that depends on the cloud page url. I know how to Requestparameter(), but how do I get the entire url using ampscript or SSJS? Product of path connected spaces is path connected Statistical interpretation of Maximum Entropy Distribution I have used the principle of maximum entropy to justify the use of several distributions in various settings; however, I have yet to be able to formulate a statistical, as opposed to information-theoretic, interpretation of maximum entropy. In other words, what does maximizing the entropy imply about the statistical properties of the distribution?   Has anyone run across or perhaps discovered yourself a statistical interpretation of max. entropy distributions that does not appeal to information, but only to probabilistic concepts?  As an example of such an interpretation (not necessarily true): "For an interval of arbitrary length L on the domain of the RV (assuming its 1-d continuous for simplicity), the maximum probability that can be contained in this interval is minimized by the maximum entropy distribution."  So, you see there is no talk about "informativeness" or other more philosophical ideas, just probabilistic implications. I have a Godox X2Tc and a Godox X1Rc. I want to control my Canon Speedlite 430 EXII. (I also have a Godox V1, which is a beautiful thing!) I have the Canon Speedlite attached to the Godox receiver, so it should fire. It just wont! How do I set it up correctly so the receiver makes the Canon Speedlite fire?  Update: Michael C answered this question. I had the ID numbers set differently on the transmitter and receiver. I made them the same and now it all works. Perfect! Thanks Michael!! Possible Duplicate:          I'm looking for a good book on commutative algebra covering most of (but not limited to) :   Basic Galois theory and Module algebra Primary decomposition of ideals Zariski topology Nullstellensatz, Hauptidealsatz Noether's normalization Ring extensions "Going up" and "Going down"   The emphasis is on the approach, as I would like a book giving a good geometric intuition of ring theory that I could use as a solid basis to start learning algebraic geometry.  All in all, do you remember a book that gave you a deeper geometric insight of commutative algebra ? Swapping right shift and delete key using xmodmap I'm unsure how to swap these 2, this is the thought process I had but am unsure how to get it working:  remove sh = Shift_R remove dl = Delete  keysym Delete = Shift_R keysym Shift_R = Delete  add sh = Delete add dl = Shift_R   This is based on the code I use to swap caps lock and control.  Any thoughts?  xmodmap output:  dan@lantea:~$ xmodmap -pke | grep Shift_R keycode  91 = Shift_R NoSymbol Shift_R keycode 119 = Shift_R NoSymbol Shift_R dan@lantea:~$ xmodmap -pke | grep Delete keycode  62 = Delete NoSymbol Delete Short description of the scoping rules? What exactly are the Python scoping rules?  If I have some code:  code1 class Foo:    code2    def spam.....       code3       for code4..:        code5        x()   Where is x found?  Some possible choices include the list below:   In the enclosing source file In the class namespace In the function definition In the for loop index variable Inside the for loop   Also there is the context during execution, when the function spam is passed somewhere else. And maybe  pass a bit differently?  There must be a simple reference or algorithm somewhere.  It's a confusing world for intermediate Python programmers. Why can't you use the keyword 'this' in a static method in .Net? I'm trying to use the this keyword in a static method, but the compiler won't allow me to use it.  Why not? What is the difference between a class and a set? I know what a set is. I have no idea what a class is.   As best as I can make out, every set is also a class, but a class can be "larger" than any set (a so-called "proper class").  This obviously makes no sense whatsoever, since sets are of unlimited size. It's not like you get up to 25 elements and go "oh, hey, sets are limited to a maximum of 20 elements, so we'll have to call this 25-element thing something else".  Does anybody know what the actual difference between a set and a class is?  Several people have pointed out that you can write paradoxes which involve sets. I get that. But where do classes come into the picture? How to check type of variable in Java? Are the datum and the ellipsoid the same? What is a Datum?     The Earth is shaped like a flattened sphere. This shape is called an   ellipsoid. A datum is a model of the earth that is used in mapping.   The datum consists of a series of numbers that define the shape and   size of the ellipsoid and it's orientation in space. A datum is chosen   to give the best possible fit to the true shape of the Earth.  in this   article.... Is it possible to reset all folder permissions? I was wondering if it was possible to reset all the folders' rights. I was having trouble when trying to install LAMP on my Ubuntu 11.04 installation, and I modified many permissions like /var /etc and I want to cancel everything I did and have permissions set to how they would be after a new install.