I try to change ip_forward from 0 to 1 but fail, even with root permission in Xubuntu 11.10. I don't have such similar problem while using Ubuntu 11.10  chiaki@chiaki:~$ sudo echo 1 &gt; /proc/sys/net/ipv4/ip_forward  bash: /proc/sys/net/ipv4/ip_forward: Permission denied   any idea?
I have a small question regarding using sudo with output redirect &gt;. To enable IP forwrding, someone can use the command:  echo 1 &gt; /proc/sys/net/ipv4/ip_forward   Executing this command will give permission denied as it requires root privileges. However, executing the same command with sudo gives also permission denied error! It seems that output redirect &gt; does not inherit the permissions of the preceding command echo. Is this right?  As a workaround I do:  echo 1 | sudo tee /proc/sys/net/ipv4/ip_forward   Is this the best way to do it? Am I missing something?  Please, note that this is an example and it applies to all commands that use output redirect.
Live CD asks for a username and password In Windows 8, when Alt+Tab, the desktop would appear as an option to switch to at the end of the list of open programs (thus Alt-Shift-Tab helpfully took you right to the desktop).  In Windows 10 that desktop option is gone. Is there any way to get it back?  I know about Windows+D but Alt+Tab is just more convenient in my opinion. And  also has no desktop and is uglier. Examples proving why the tensor product does not distribute over direct products. I recently read about the result that the tensor product distributes over direct sums. I was curious if it also distributes over direct products, but google tells me it doesn't.   What are some simple counterexamples to why this property isn't true? I know that there is a natural homomorphism  $$ \left(\prod M_i\right)\otimes N\to \prod (M_i\otimes N) $$ given by $(\prod m_i)\otimes n\mapsto \prod (m_i\otimes n)$ when $M$ and $N$ are modules over some commutative ring $R$. Are there standard examples where this homomorphism is not injective/surjective and hence not an isomorphism? I'm trying to execute a string into my bash script. String contains variables (so I could substitute some CLI arguments depending on circumstances).  The problem is when I use variable containing --argument=value, everything breaks up, giving error:  Missing trailing-' in remote-shell command.   Example code:  #!/bin/bash  rsync_path="/usr/bin/rsync -azhrP" rsync_ssh_cmd="--rsh='ssh -o Compression=no'"  `$rsync_path /path/to/file $rsync_ssh_cmd username@localhost:/tmp`   If I just echo the last string from this code, it looks fine. But if i try to execute this string from script, it doesn't executes as expected.  I found a workaround - everything became OK if I try  eval "my-command-string"   instead of  `my-command-string`   or  (my-command-string)   Why does it so? How could I (should I really?) avoid to use eval?  I use bash 4.3.11(1)-release (x86_64-pc-linux-gnu) (Kubuntu 14.04.3). Is it possible to use NBT tags in commands in Minecraft: Bedrock Edition? It is not unusual for an Ubuntu (or other distro with apt package management) user to encounter the error: user@box ~ $ sudo apt install x [sudo] password for user:  Reading package lists... Done Building dependency tree        Reading state information... Done Package x is not available, but is referred to by another package. This may mean that the package is missing, has been obsoleted, or is only available from another source  E: Package 'x' has no installation candidate  There have been several questions asked concerning this on various StackExchange sites,for example  on AskUbuntu, but these are generally about how to obtain the specific package despite the error. I have found no package that is not found simply (though not always easily) by determining if it is still distributed, and under which repository if so, and ensuring that repository is in apt.sources, or the worst comes to the worst downloading a .deb or source to compile. So my question is more general, and although I am guessing it has a rather simple and therefore non-exciting answer I cannot think how to find it by conventional search engine methods, so I turn to AskUbuntu: Is there a command or option I can use to determine what package / which packages reference the missing package? I want to draw a perfect rectangle. How can I do that with QGIS v.3+?  .  Is there a similar solution for the latest QGIS version? Find all functions $f$ such that $(x+y)f(x)+f(y^2)=(x+y)f(y)+f(x^2)$ Find all functions $f$ that assign a real number $f(x)$ to every real number $x$ such that $$(x+y)f(x)+f(y^2)=(x+y)f(y)+f(x^2)$$  I've tried subbing in heaps of values but I keep getting things like $f(0)=f(0)$ and other such useless results.  Any help would be hugely appreciated. Will $\sum^{\infty}_{n=1}|a_n|^p$ converge? Given that $\sum^{\infty}_{n=1} a_n^2$ converges,then, will $$\sum^{\infty}_{n=1}|a_n|^p$$ converge for $p&gt;2$?  I couldn't find a counterexample neither a proof but it looks to me that this is true. For $p=3/2$, I found the counter example $\sum a_n=\displaystyle \sum^{\infty}_{n=1} \frac{1}{n^{2/3}}$ Understanding hard drive space Can there be a function that's even and odd at the same time? How can I deactivate Bluetooth on system startup? How can I deactivate Bluetooth on system startup? How do shock waves form in space? Today my aerodynamics professor mentioned that the equations we were learning for atmospheric shock waves can also be applied by astrophysicists to study black hole-related shock waves in space.  How is this possible? Isn't space too rarefied for shock waves? What is an example application? He mentioned something about the density limit before/after a relativistic shock approaching 6. Sharing a Bluetooth Keyboard and Mouse sequentially between two Macs I work out of my house where I have two computers: a mid-2010 Macbook Pro (for my day job) and a mid-2010 iMac (personal). During the workday, my iMac acts as a second display for my Macbook Pro via Mini Displayport.  Currently, I use a USB keyboard and mouse for my work computer and a Bluetooth keyboard and Magic Mouse + Magic Trackpad for use with my personal computer. I'd like to get rid of the wired keyboard and mouse and just use the Bluetooth stuff for both machines (but never simultaneously).   Is it possible to easily share the single set of input devices without having to repeatedly break and reestablish the pairing of the input device to the correct computer? I'm not opposed to buying some type of additional hardware to accomplish this, provided it's not prohibitively costly or cumbersome to use or configure.  It's not clear to me how I would tell the MacBook Pro to leave the devices alone so the iMac can have them for the evening. I want to avoid de-pairing and re-pairing the devices each time I want to shift them back or forth.  Basically is there software or hardware that works as a Bluetooth KVM switch? End of javascript sidebar workarounds? (Salesforce Classic / Aloha) Reading through this doc:      We replaced the rich text editor in HTML Area home page components   with a new version that supports more markup but doesn’t allow HTML to   be manually entered.      In Summer ’15 we will start removing unsupported code from HTML Area   home page components. As a result, components that contain JavaScript,   CSS, iframes, or other unsupported markup might stop working properly.   To use JavaScript or other advanced HTML elements in your home page   component, we recommend that you use a Visualforce Area component   instead.   With Visualforce homepage component the best we could do is design fancy home page components, which has not been the big use of sidebar amongst developers. How do we do the fancy hack we do today using html components since visualforce is hosted in a different domain will the sidebar hacks even work after summer 15?     You can include JavaScript in your Visualforce page, but because the   component is rendered in an iframe on the home page layout, the   JavaScript can’t interact with the page that contains the component.   Finally why this move :(? $T(av_1 + bv_2) = aT(v_1) + bT(v_2)$  Why is this called linear? $f(x) =ax + b$, the simplest linear equation does not satisfy $f(x_1 + x_2) = f(x_1) + f(x_2)$.  Thank you. In-Page Editor Line Distortion The SFDC In-page editor has always had distortion for me, and I'm just wondering if it's just a lousy browser plugin they're using or if someone has found a good browser / operating system combination that stops the obnoxious distortion.   Obviously Salesforce doesn't have much interest in fixing it: as it's been this way on multiple operating systems and browsers I've used since I've used Salesforce.  However, my frustration with their other developer tools has lead me to use it occasionally and it's always the same: ST_Length (GEOM, true) for calculating distances in Geographic Reference System I'm stuck on the following phrases. Is it correct to use preposition "of" in the sentences bellow?    It's not a little unpleasant of being her so fussy. It's awfull of him having done this.   Thank you. Prove $\int\cos^n x \ dx = \frac{1}n \cos^{n-1}x \sin x + \frac{n-1}{n}\int\cos^{n-2} x \ dx$ I am trying to prove  $$\int\cos^n x \ dx = \frac{1}n \cos^{n-1}x \sin x + \frac{n-1}{n}\int\cos^{n-2} x \ dx$$  This problem is a classic, but I seem to be missing one step or the understanding of two steps which I will outline below.   $$I_n := \int\cos^n x \ dx = \int\cos^{n-1} x \cos x \ dx \tag{1}$$  First question: why rewrite the original instead of immediately integrating by parts of $\int \cos^n x \ dx$?  Integrate by parts with $$u = \cos^{n-1} x, dv = \cos x \ dx \implies du = (n-1)\cos^{n-2} x \cdot -\sin x, v = \sin x$$  which leads to  $$I_n = \sin x \ \cos^{n-1} x +\int\sin^2 x (n-1) \ \cos^{n-2} x \ dx \tag{2}$$  Since $(n-1)$ is a constant, we can throw it out front of the integral:  $$I_n = \sin x \ \cos^{n-1} x +(n-1)\int\sin^2 x \ \cos^{n-2} x \ dx\tag{3}$$  I can transform the integral a bit because $\sin^2 x + cos^2 x = 1 \implies \sin^2 x = 1-\cos^2 x$  $$I_n = \sin x \ \cos^{n-1} x + (n-1)\int(1-\cos^2 x) \ \cos^{n-2} x \ dx \tag{4}$$  According to Wikipedia as noted , this simplifies to:  $$I_n = \sin x \ \cos^{n-1} x + (n-1) \int \cos^{n-2} x \ dx - (n-1)\int(\cos^n x) \ dx \tag{5}$$  Question 2: How did they simplify the integral of $\int(1-\cos^2 x) \ dx$ to $\int(\cos^n x) \ dx$?  Assuming knowledge of equation 5, I see how to rewrite it as  $$I_n = \sin x \ \cos^{n-1} x + (n-1) I_{n-2} x  - (n-1) I_{n} \tag{6}$$  and solve for $I_n$. I had tried exploiting the fact that  $$\cos^2 x = \frac{1}{2} \cos(2x) + \frac{1}{2} $$  and trying to deal with $\int 1 \ dx - \int \frac{1}{2} \cos (2x) + \frac{1}{2} \ dx$  which left me with $\frac{x}{2} - \frac{1}{4} \sin(2x)$ after integrating those pieces. Putting it all together I have:  $$I_n = \sin x \ \cos^{n-1} x + (n-1) I_{n-2} x \left(-(n-1) (\frac{x}{2} - \frac{1}{4} \sin 2x) \right) \tag{7}$$  but I'm unsure how to write the last few terms as an expression of $I_{something}$ to get it to match the usual reduction formula of   $$\int\cos^n x \ dx = \frac{1}n \cos^{n-1}x \sin x + \frac{n-1}{n}\int\cos^{n-2} x \ dx$$