I am trying to find quite a simple limit i.e.  for $x_n = \sqrt{2+x_{n-1}}$ and $x_1 = \sqrt{2}$, find:     $$\lim_{n \to \infty}{x_n}$$   So in order to use recurrence relations to find the limit, I must first prove that this limit exists. So i thought I would prove that this sequence is monotone sequence and bounded from above.   By solving $x_{n+1} &gt; x_n$ I got that this holds for $-1 &lt; x_n &lt;2$ . So i thought immediately that by proving $x_n &lt;2$ proves the existence of this limit.   This part seems to be causing me problems. Proving:     $$\sqrt{2+\sqrt{2+\sqrt{2+...}}} &lt;2$$   Now this is intuitive and very simple but I need a formal proof. I did this: $$\sqrt{2} &lt; 2 $$ Adding 2 to both sides gives: $$2+\sqrt{2} &lt; 4$$ Square rooting : $$\sqrt{2+\sqrt{2}} &lt; 2$$ Repeating the process $n$ times: $$\sqrt{2+\sqrt{2+\sqrt{2}...}} &lt; 2$$  But i am not sure of this proof is formal enough or if it is a valid proof. I believe this problem has been solved before but i couldn't find it on math.stackexchange.com .  Of course, calculating the limit gives: $$\lim_{n \to \infty } {x_n} = 2$$  Summary: Is the proof provided above formal enough, and if there are other proofs, I would like to see them?
I need to show that $\sqrt{2}$, $\sqrt{2+\sqrt{2}}$, $\sqrt{2+\sqrt{2+\sqrt{2}}},\ldots$ converges and find the limit.   I started by defining the sequence by $x_1=\sqrt{2}$ and then $x_{n+1}=\sqrt{2+x_n}$. Then I proved by induction that the sequence is increasing and that it is bounded. Then I use the Monotone Convergence Theorem to prove that it converges.  Now, I claim that the limit of the sequence is $2$. So I need to show that for all $\epsilon&gt;0$, there exists $N$ such that for all $n\geq N$ we have $|x_n-2| &lt; \epsilon$. This is kind of where I'm stuck. I don't know how to proceed to prove the convergence to $2$.
Ubuntu 20.04 brightness adjust not working I am unable to adjust brightness levels through Gnome. I had the same issue with 18.04 but it was resolved there (honestly, I can't remember how).  lspci|grep VGA 00:02.0 VGA compatible controller: Intel Corporation Core Processor Integrated Graphics Controller (rev 18)  xrandr --output LVDS-1 --brightness 0.95 &lt;-- this works through the terminal excerpt from /etc/default/grub GRUB_CMDLINE_LINUX_DEFAULT="quiet splash acpi_backlight=vendor"   Thank you in advance. Make specific author bold using biblatex Import Large Excel File How do I deal with a compromised server? In  when Ambassador Sarek became critically ill, why didn't they send him home or send for a Vulcan medical expert? Since they were hosting many prominent political figures from many races, it seems strange that they did not have much medical support for a race so common as Vulcans (who are part of the Federation). Since I want to protect my ssh connections, I set some global cipher suite options to restrict set of used algorithms. But recently I've encountered a server which doesn't support some of those algorithms. So, I need to selectively enable deprecated algorithms for a specific host record in client (my system) configuration.  I found out that the options override is not working as I expected. Let's take a minimal (not-)working example for the github:  HostKeyAlgorithms ssh-ed25519-cert-v01@openssh.com,ssh-ed25519,ecdsa-sha2-nistp521-cert-v01@openssh.com,ecdsa-sha2-nistp521,ecdsa-sha2-nistp256  Host github     HostKeyAlgorithms ssh-rsa     Hostname        github.com     Port            22     User            git     PubkeyAuthentication yes     IdentityFile    ~/.ssh/some-filename-here   Having that, I receive the following error (HostKeyAlgorithms is not overriden at all):  debug1: /home/username/.ssh/config line 14: Applying options for github &lt;...&gt; debug2: kex_parse_kexinit: ssh-ed25519-cert-v01@openssh.com,ssh-ed25519,ecdsa-sha2-nistp521-cert-v01@openssh.com,ecdsa-sha2-nistp521,ecdsa-sha2-nistp256 &lt;...&gt; Unable to negotiate with 192.30.252.130: no matching host key type found. Their offer: ssh-dss,ssh-rsa   It is similarly not working for the global PubkeyAuthentication no options with an override in a host configuration.  Also, the match doesn't help either:  match host github     HostKeyAlgorithms ssh-rsa   So, is there a way to selectively redefine those options?  NOTE: I'm using the openssh-7.1_p2-r1 on gentoo. What's is the origin of Orbital Angular Momentum of electrons in atoms? Consider the Hydrogen 1s electron. We know that, in the quantum picture, the electron isn't orbiting or rotating at all, rather we simply state that the electron is spread over the entire space with the probability of finding it being maximum a radial distance $R$ (=Bohr radius) away from the nucleus. This helps explain why the electron does not radiate EM radiation while in the atom.  But, with this understanding, I do not understand the source of the orbital angular momentum. Is it intrinsic like the spin? A clarification in a limit While reading this  I do not understand that how can we claim that there exist a polynomial with the properties, even if it exist then how is it same for the function and the inverse. Prove that $u\cdot v = \frac{1}{4}||u+v||^2 - \frac{1}{4}||u-v||^2 \forall u,v \in \mathbb{R^n}$ I am trying to prove the above statement but I'm not sure if my proof is correct. My proof is as follows,  Given $u\cdot v$, we know by the C-E Inequality that $|u \cdot v| \leq ||u|| \ ||v||$  Consider $\frac{1}{4}||u+v||^2 - \frac{1}{4}||u-v||^2$  We can manipulate this as follows $$\frac{1}{4}(||u+v||^2 - ||u-v||^2) = \frac{1}{4}(||u+v||- ||u-v||)(||u+v||+||u-v||)$$ Using the triangle inequality the above can be written as follows, $$...\leq \frac{1}{4}(||u|| +||v|| -||u||+||v||) (||u|| +||v|| +||u||-||v||) = \frac{1}{4} (4||u||||v||) = ||u||||v||$$  As such both $u\cdot v$ and $u\cdot v = \frac{1}{4}||u+v||^2 - \frac{1}{4}||u-v||^2 \forall u,v \in \mathbb{R^n}$ are $\leq ||u|| ||v||$  I'm not sure whether I've proven this correctly because I ended up proving that both sides of the equation are less than or equal to something entirely different (the magnitude of v multiplied by the magnitude of u). Is this acceptable? I don't think it is, since if both can be less than or equal to the magnitudes multiplied then they do not necessarily have to be equal to each other. If this is the case, can someone give me a hint on how to move forward with this proof? Why are some muggles allowed to know of the magical world? Why does my object not show up? Underscores in custom fields While trying to perform a WP_Query using meta_query(), I had some trouble because the custom fields created by a plugin had _ in the beginning of its name.   Although I noticed it was stored this way in the database, I had no idea why it was there and thought I had to reference it without the underscore (for the record, that is wrong).  Why do some custom fields start with underscores and others don't? What are the underscores used for? Are they mandatory in some circumstances? How can I recover deleted photos from an SD Card? My technophobe mother returned from a European trip to discover she'd erased all but 15 images from a nearly full 2GB card. As I am the computer guy in the family, she posted the card to me and asked that I make something magical happen.  The photos were taken with and deleted by a Nikon Coolpix. (I don't know the model.)  I am a Linux guy, but there is a Windows Vista box at my disposal.  What is my best strategy to try to recover something? Batch renaming files Identity for convolution of central binomial coefficients: $\sum\limits_{k=0}^n \binom{2k}{k}\binom{2(n-k)}{n-k}=2^{2n}$ It's not difficult to show that  $$(1-z^2)^{-1/2}=\sum_{n=0}^\infty \binom{2n}{n}2^{-2n}z^{2n}$$  On the other hand, we have $(1-z^2)^{-1}=\sum z^{2n}$.  Squaring the first power series and comparing terms gives us  $$\sum_{k=0}^n \binom{2k}{k}\binom{2(n-k)}{n-k}2^{-2n}=1$$  that is,  $$\sum_{k=0}^n \binom{2k}{k}\binom{2(n-k)}{n-k}=2^{2n}$$  My question: is there a more direct, combinatorial proof of this identity?  I've been racking my brains trying to come up with one but I'm not having much success. I pick a random subset $S$ of $\{1,\ldots,N\}$, and you have to guess what it is. After each guess $G$, I tell you the number of elements in $G \cap S$. How many guesses do you need? Using echo "20+5" literally produces the text "20+5".  What command can I use to get the numeric sum, 25 in this case?  Also, what's the easiest way to do it just using bash for floating point? For example, echo $((3224/3807.0)) prints 0 :(.  I am looking for answers using either the basic command shell ('command line') itself or through using languages that are available from the command line. Object not visible in render I have a model which you can see here (camera perspective).    In the second image, you can see how the render turned out.    I'm at my wit's end trying to figure our why it's not showing up in the render. If anyone needs more information, just ask and I'll get it to you. I've already checked the normals and they're not the cause. Strange artifacts along window borders after waking computer from sleep mode NB: Nvidia has released an official fix; please see ,  and/or .  Upon waking up Ubuntu 16.04 from Suspend, the following graphical glitches appear around all windows, context menus, and similar:      This is currently known to affect the following NVIDIA cards running the proprietary drivers:  NVIDIA GeForce 610M, driver version 375.39, 340.102 NVIDIA GT 630M, driver version 375.39 NVIDIA GTX 650, driver version 375.39 NVIDIA GT 740M, driver version 378.13 NVIDIA GTX 745, unknown driver version NVIDIA GTX 850M, driver version 375.39, 378.13 NVIDIA GT 940M, driver version 375.39 NVIDIA GTX 950M, unknown driver version NVIDIA GTX 960M, driver version 378.13 NVIDIA GTX 970, driver version 375.39 NVIDIA GTX 970M, driver version 375.39 NVIDIA GTX 980 TI, unknown driver version NVIDIA GTX 1060, driver version 375.26, 375.39, 378.09 NVIDIA GTX 1070, driver version 378.09  The above above list has been sourced from reports from other users here on AU,  thread, and  thread. NB: This is be no means an exclusive list. It probably effects nearly all Nvidia cards.  Furthermore, this issue has been replicated on a clean install where the only change was a software update and the installation of the latest NVIDIA drivers.  What is going on, and how can it be resolved? $C \otimes A \cong C \otimes B$ does not imply $A \cong B$ Let $R$ be a commutative unital ring and let $M$ be an $R$-module and let $S$ be a multiplicative subset of $R$.  Today I proved both of the following: $$ S^{-1} R\otimes_R S^{-1}M \cong S^{-1} M$$ and $$ S^{-1} R \otimes M \cong S^{-1} M$$  Now I'm slightly confused.      Either my proofs are wrong or $C \otimes A \cong C \otimes B$ does not imply $A \cong B$. But I can't come up with an example. Can someone give me an example? (Or tell me that my proofs are wrong.)