Bronze badge for sharing?

Question: Should we create a bronze badge on the public beta sites for helping to promote good questions from the sites with the new facebook/twitter links?  I'm thinking a little encouragement would go a long way, but I wouldn't want it to get abused (i.e. someone just shares a random question that isn't good to get the badge).
Comment: *"Promoting good question from the public beta sites..."* I would think that is *outside* the purview of Area 51; a badge for the *site*, I would say.
Comment: A badge on the beta site or on 51?
Comment: @waiwai933: The beta site...
Comment: -1, I don't think it's a good idea for the reasons you yourself and Robert pointed out :S
Answer: This has been completed with the introduction of the Announcer, Booster, and Publicist Badges.
Answer: You already get badge and reputation for referring people to commit for a website on Area51. Why would we need other badge for nearly the same exact thing ?
The reputation you get on Area51
Referred user with a confirmed email address commits to the proposal +5
Referred user participates in the beta  +25
The badge you get on Area51
(Silver Badge) Campaigner - Referred 25 committers to a proposal
(Bronze Badge) Lobbyist - First commitment referral
(Gold Badge) Activist - Referred 100 committers to a proposal
Answer: In order for it not to be abused, we'd have to tie it to something else happening, beyond simply sharing with the buttons. Not sure what that is or could be.
For example, Student and Teacher both require 1 upvote on either a question or an answer respectively. We don't give the badges for simply posting a question or answer.
(yes, we do give a badge for filling out your profile page, and other one-off events like first upvote and first downvote.)
Comment: and yet you implemented it anyway.  GOOD JOB!
Comment: @power well the "something else" was "count the number of unique incoming IPs"..

Bronze badge for sharing?


How do I get the Angler?

Question: I can't get Angler NPC in Terraria on iOS. I think this is because I made my whole world flat and put lizhared bricks on top for armour. When the Angler was finally released on iOS I made a beach of my own and hoped the Angler NPC would come but unfortunately he didn't.
Comment: I'm fairly certain you will need to generate a new world to get the new stuff
Comment: What @Aequitas said, but also because the Angler will only spawn in the ocean biome, and if the mobile version is like the desktop version that background indicates that you're not in the ocean biome, for whatever reason
Answer: According to the terraria gamepedia, the Angler will spawn at random and he is not guaranteed to spawn on either ocean immediately. You may have to simply wait for him to spawn.
Also, since he is a Town NPC a suitable 'house' is required in order for the Angler to appear. If you do not have an empty house built, the Angler will not appear. Ever.
A proper ocean biome must also be present for the Angler to arrive. If you don't have an ocean or have obstructed yours you can build an artificial ocean out of sand and water. If the 'Ocean' music is playing and the background is a vast sea of water with a green hill or two than the Ocean biome is present. If not, than you may have made a mistake.
Answer: I noticed that the Angler wasn't spawning on my old world too. So when I started up a new one and went straight to the beach to either the left or right I found him, laying there sleeping on the beach. So my hypothesis was you can only get him when you create a new world. Try it out and see what happens. After all, Terraria can get boring if you have nothing to do

How do I get the Angler?


Permissioned network with Substrate

Question: I trying to build permissioned network with Substrate. I don't understand diagram following. Let help me explain it. Thanks.
Comment: this is docs: https://docs.substrate.io/tutorials/build-a-blockchain/authorize-specific-nodes/
Answer: A permissioned network is one where only authorized nodes are allowed to perform specific network activities.
The diagram shows you a network with 4 nodes: Alice and Bob are the nodes predefined in the genesis specification for genesis state which basically means the 2 nodes that will start the network.
In a permissionless blockchain anyone can join the network by running the node software, but in the case you want to build you have to add a new node as a well-known node. In the diagram Charlie and Dave has been added lately into the network.
To add these nodes in Substrate is recommended to use the node-authorization pallet.

The node-authorization pallet is a prebuilt FRAME pallet that enables
you to manage a configurable set of nodes for a network.

For more practical knowledge I suggest you to go and build that network following these tutorials: Add trusted nodes and Authorize specific nodes.
After this you will understand better the diagram, how to run a network and how to use the node-authorization pallet to add nodes into your network.

Permissioned network with Substrate


R33 GTR Importation

Question:
The r33 gtr as I know of now has to be 25 years old to be imported to the USA. The r32 is already legal and is an example of this rule in effect. I looked up this NHTSA sheet and it says here that the r33 is legal to import, and they even assigned it an eligibility number! Did I read it wrong, am I right, or is there some catch I'm missing?
Comment: For a definitive answer - phone them and ask...
Comment: Web link to the pdf: https://www.nhtsa.gov/sites/nhtsa.dot.gov/files/documents/elig120115.pdf
Answer: I found a Facebook post from an import tuner that helps to explain the situation.  Basically, you can import one, but it must meet certain requirements that can be difficult to acheive.

When VCP17 was revoked and VCP32 was put in place of it that suddenly changed the requirements and eligibility of all R33 Skylines to later being specifically the 1996 to 1998 R33 GTS and GTR with dual-side air bags. This also further complicated the process because traditionally MotoRex Inc was faking the funk back in the day by avoiding OBDII through mis-declaring 1996 and newer Skylines as 1995 so they didn't have to do OBDII on them (taking the easy way out). Today there is no faking the funk under VCP32 and any R33 that is imported under VCP32 must be brought into conformance with the FMVSS and also EPA which means OBDII is required. While this may not sound like a complicated process it actually is and also very costly.

https://www.facebook.com/rivsuimports/posts/1811338242430133:0
I'm not an expert at importing JDM cars, but I believe what the post is getting at is that there was a VCP17 that was allowing their import, but that was nullified and replaced by VCP32, but thats only for models with side airbags and you must install a working OBDII diagnostics.  That sounds impossible to me, but the post at least makes it sound possible if not extremely difficult.

R33 GTR Importation


What are the positions of planets at the beginning of every Yuga (age)?

Question: How to find all planetary positions on starting of every Yuga?
Comment: Well asked question... Might be I am very interesting to see the answers.
Comment: @ssr1012 I hv edited out the "planet" tag u hv added as the "navagraha" tag is already present :)
Comment: I saw only "yuga" tag then only I added "planet"... Might be some confusion.
Comment: @user17854 can you highlight what research you have done in this regard ?
Answer: Executive answer to planetary positions at Krita, Treta, Dwarpara and Kali yugas = We don't yet have the capability to know beyond our current Kali yuga start of 3102 B.C.
The following site, although a blog, compiles excellent citations for this debate http://bharatkalyan97.blogspot.com/2021/11/kaliyuga-date-18-feb-3102-bce-friday-is.html
To get that figure, you need an ancient genius like Aryabhata who accurately, astronomically measured jyotish astrology with long lost books, and a succession of genius mathematicians. They would need for example;
ephemeris with longitude, preferably hour of birth for horoscopes, calculating naksatras, rasi and navamsa.
Due to complex fluctuations of Earth tilt, obliquity, precession and spin speed over millennia, we need computer software to pinpoint accuracy  with what we feed it and calibrate it with. Free programme Jaganath Horas is accurate but buggy, Sri Jotish Star software in experienced hands will not function with Greek-based sidereal sky.
So despite a plethora of planetary comments in the vedas, we'd need a.i.nstein (nailed it, coined it) to even begin to see older configurations. There are at least two NASA scientists working on it https://www.vice.com/en/article/wjwmgb/nasa-engineer-who-secretly-practices-astrology
So how do we know we are on the right track today? The calculation must conform to the right traditional calendar, for longer cyclic periods then Puranic time, cross correlating the Vedas and Puranas etc. If we instead try to conform to Christian time (6000 years since adam) it of course falls apart. It May weakly confirm with a geological idea, but it must also correlated with Vedic scripture and chronology.


Identity map between metric spaces continuous or not.

Question: How to counter this problem? Is it enough to show pre image of unit ball in some metric is open under another to show the continuity? I am not at all getting the path to proceed. And how to contradict when it's not continuous?
Comment: Tip: Questions are not well-received on here whenever the question asker doesn't mention any attempted efforts towards a solution. So always try to mention any attempted work even if it seems embarrassingly incorrect, otherwise other users are less motivated to help.
Comment: You can use epsilon-delta methods to prove continuity. Proceed as you would with the standard metric but substitute the appropriate metric to make arguments like $d_2(f,g)\lt\epsilon$.
Answer: We know that a continuous map between metric spaces is one which preserves limits: $f$ is continuous if and only if $\lim\limits_{n\to\infty} f(x_n)=f(\lim\limits_{n\to\infty} x_n)$ for each convergent sequence $x_n$. To prove that a map is not continuous, we can try to find a sequence that is convergent in one metric but not the other.
For example, I don't think that b) is true. Consider the sequence
$$f_n(x)=x^n.$$
Then $\lim\limits_{n\to\infty} f_n=0$ in the metric $d_2$, but not in $d_1$: it even fails to be Cauchy in $d_1$.
Of course, in general it is fine to show that the preimage of any open ball is open. There is some general theory that tells you that if the identity map is bounded - that is, if for every $x\in X_i$, $d_j(x,0)\le M\cdot d_i(x,0)$ -- then the identity map from $X_i$ to $X_j$ is continuous. This uses the fact that all these metrics come from norms, though, and takes some work to establish.
Comment: nice counterexample
Answer: a. Let $\varepsilon>0$ be given. Since $\sup_{x\in [0,1]} | \, f(x)-g(x)| \geq \int_0^1 |\, f(x)-g(x)|dx$, we may choose $\delta=\varepsilon$ so that $\int_0^1 |\, f(x)-g(x)|dx<\varepsilon$ whenever $\sup_{x\in [0,1]} |\, f(x)-g(x)|<\delta$. Therefore $id:X_1 \longrightarrow X_2$ is continuous.
b. Notice $\int_0^1 |x^n|dx \to 0$ as $n \to \infty$ BUT $\sup_{x\in [0,1]} |x^n|=1$ for every $n \in \mathbb N$. Therefore $id:X_2 \longrightarrow X_1$ is NOT continuous at the constant function $g \equiv 0$ because it is not sequentially continuous there. There are other counterexamples but this is the preferred (by at least 2 people in this thread).
c. I leave to you. The notion of sequential continuity seems good for demonstrating a counterexample to such statements, while the standard notion of continuity seems good for proving such statements true which is not exactly strict advice.

Technically for b. I used the fact that the functions $f_n(x):=x^n$ from the sequence of functions $\{f_n\}_{n=1}^\infty$ converge to the constant function $g \equiv 0$ in $(X_2, d_2)$ but the same is not true in $(X_1, d_1)$. And so $id: X_2 \longrightarrow X_1$ is not sequentially continuous at $g$ as opposed to identifying the pointwise limit $f(x):=\lim_{n\to \infty} f_n(x)$ which is not in the set $\mathcal C[0,1]$ as $f$ is not continuous at $x=1$.
If this seems like an issue, then showing that the chosen sequence of functions is not uniformly Cauchy on $[0,1]$ (Cauchy in $(X_1, d_1)$) avoids this. For this argument we make the following two observations, $|x^n-x^m|=|x^n||1-x^{m-n}|$ and the function $F(x):=\left(1-\frac1x\right)^x$ is nondecreasing on $[1, \infty)$ with $\lim_{x \to \infty} F(x)=\frac1e$. So with $\varepsilon=\frac1{4}(1-e^{-1})$ and for any $n\geq 2$, we may select $m=2n$ and have
$$\sup_{x\in[0,1]} |x^n-x^m| \geq \left(1-\frac1n\right)^n\left(1-\left(1-\frac1n\right)^n\right)\geq \varepsilon$$
which shows that the sequence of functions $\{f_n\}_{n=1}^\infty$ is not Cauchy in $(X_1, d_1)$ by definition.
Comment: For the part c, Cauchy Schwartz says, integral 0 to 1 of h < (int 0 to 1 of h²)^1/2. Hence like part a, choosing delta same as given epsilon we get d3 < delta implies d2< epsilon. Therefore id: X3 to X2 is continuous.
Comment: Regarding $a.$ If $d_1,d_2$ are metrics on a set $X$ and if there exists $K>0$ such that $d_2(f,g)\leq K\cdot d_1(f,g)$ for all $f,g\in X$ then the topology generated by $d_2$ is a subset of the topology generated by $d_1,$ so $ id_X:(X,d_1)\to (X,d_2)$ is continuous because the inverse of a  $d_2$-open set, which is itself, is also $d_1$-open.... In $a.$ we have $K=1.$
Comment: You got it, btw I didn't downvote your question. Someone else did and so I shared my tip. My upvote was preemptive but I would say you proved worthy. Most people seem eager to rush towards negative judgement these days (not to say some don't deserve it...rhymes with rump).
Comment: Oh but learn to write in latex too. Tip: right click>"show math as" AND for symbols there are pdf documents littering the web with thorough latex code catalogs.
Comment: Comparing topologies on the same set certainly can come in handy sometimes: https://imgur.com/cfOxqfs (an old assignment I did that plays off the same notion). I guess considering a and b together, we might say that an open bijection is not always continuous -_-


Parabolic, Hyperbolic, Elliptic

Question: The terms "parabolic," "hyperbolic" and "elliptic" are used to classify certain differential equations. The terms "hyperbolic" and "elliptic" are also used to describe certain geometries. Is there a connection between these usages, and, if so, what is it?
Comment: [Related](https://math.stackexchange.com/q/21525/8157)
Comment: In my mind the elliptic/parabolic/hyperbolic language all stems from conic sections -  the conic $a^{ij}x_i x_j + b^i x_i = 0$ and the PDE $a^{ij} \partial_i \partial_j u + b^i \partial_i u =0$ are classified near identically. Elliptic/hyperbolic geometries have models built from circles/hyperbolae, and volume growths given by circular/hyperbolic trigonometric functions.
Comment: It has to do with the characteristic curves of the equation, as an intuitive example, the wave equation has two families of nonparallel lines (degenerated hyperbola) and the heat equation has one family of lines (degenerated parabola). But I would like to read a more developed answer as well.
Comment: For "elliptic" in particular I had asked [a similar question](https://math.stackexchange.com/q/1548001/35416) in the past, albeit a bit broader by taking other uses into account as well.
Comment: Related: https://math.stackexchange.com/q/4294133/169085
Answer: I have just added an entry on my blog to explain this.
Let's consider the space $\mathbb{M}^4$, that is, the manifold $\mathbb{R}^4$ with coordinates $(x,y,z,t)$ but with the pseudo-Riemannian metric given in each tangent space by
$$
ds^2=dx^2+dy^2+dz^2-dt^2.
$$
This metric is left invariant by the group $O(3,1)$ when acting on $\mathbb{M}^4$. It also leaves invariant the cone
$$
t^2-x^2-y^2-z^2=0
$$
so is logical that we pay attention to it.
Now, following TRTR from Penrose page 423, we can consider the family of hyperplanes inside $\mathbb{M}^4$ given by
$$
z+t+\lambda(t-z)=2
$$
intersecting our cone in different 2-dimensional submanifolds $P_{\lambda}$ of $\mathbb{M}^4$.
Here we have a picture from that book, where we consider only $(x,z,t)$ for purposes of drawability.

If we analyse the case $\lambda=0$, we observe that we obtain the submanifold $P_0$, $E$ in the picture with a shape of a parabola, given by the embedding
\begin{align*}
  \phi \colon \mathbb{R}^2 &\longrightarrow \mathbb{M}^4\\
  (u_1,u_2) &\mapsto (u_1, u_2, 1-\frac{u_1^2+u_2^2}{4},1+\frac{u_1^2+u_2^2}{4})
\end{align*}
We can look for the (possibly pseudo) metric inherited by $P_0$. Since
$$
d\phi=\begin{pmatrix}
1 & 0 \\
0 & 1\\
-\frac{u_1}{2} &-\frac{u_2}{2}\\
\frac{u_1}{2} &\frac{u_2}{2}
\end{pmatrix}
$$
we conclude that $P_0$ consists of $\mathbb{R}^2$ with the euclidean metric:
$$
g=\begin{pmatrix}
1 & 0 \\
0 & 1\\
\end{pmatrix}
$$
That is, although we see it like a parabola, intrinsically is only the Euclidean plane, and that is the reason the latter is sometimes called parabolic geometry.
But, what happens for others $\lambda$? Let's see. We have a far more complicated parametrization of our embedded manifold:
$$
\phi(u_1,u_2)=(u_1,u_2, \frac{-1+{ \lambda }+\sqrt{-(1+{ \lambda })^{2}\left(-1+{\lambda}\left(u_1^{2}+u_2^{2}\right)\right)}}{2{ \lambda }},
$$
$$
\frac{2+{ \lambda }+\sqrt{-(1+{\lambda })^{2}\left(-1+{\lambda}\left(u_1^{2}+u_2^{2}\right)\right)+\frac{1-\sqrt{-(1+{\lambda})^{2}\left(-1+{\lambda}\left(u_1^{2}+u_2^{2}\right)\right)}}{{ \lambda }}}}{2(1+{ \lambda })})
$$
(computations made with Mathematica).
If we compute the (possibly pseudo) metric in this chart we obtain the inherited metric in $\mathbb{R}^2$
$$
g_{\lambda}=\begin{pmatrix}
\frac{1-{ \lambda } u_2^{2}}{1-{ \lambda }\left(u_1^{2}+u_2^{2}\right)} & \frac{\lambda u_1 u_2}{1-{ \lambda }\left(u_1^{2}+u_2^{2}\right)} \\
\frac{\lambda u_1 u_2}{1-{ \lambda }\left(u_1^{2}+u_2^{2}\right)} & \frac{1-{ \lambda } u_1^{2}}{1-{ \lambda }\left(u_1^{2}+u_2^{2}\right)}\\
\end{pmatrix}
$$
Next step is to show that for $\lambda>0$ we have an isometry into the usual sphere, and that for $\lambda<0$ we have an isometry into the hyperbolic space. This would justify the terms elliptic and hyperbolic for those geometries, as you can see at the picture above ($S$ and $H$ respectively). In order to keep it simple we can work with $\lambda=1$ and $\lambda=-1$.
For $\lambda=1$, we have
$$
g=\begin{pmatrix}
\frac{1- u_2^{2}}{1-\left(u_1^{2}+u_2^{2}\right)} & \frac{ u_1 u_2}{1-\left(u_1^{2}+u_2^{2}\right)} \\
\frac{u_1 u_2}{1-\left(u_1^{2}+u_2^{2}\right)} & \frac{1- u_1^{2}}{1-\left(u_1^{2}+u_2^{2}\right)}\\
\end{pmatrix}
$$
But if we look at a typical chart of the usual sphere in $\mathbb{E}^3$
$$
\psi: (x,y)\longmapsto (x,y,\sqrt{1-x^2-y^2})
$$
and compute the metric we obtain just
$$
g=\begin{pmatrix}
\frac{1- y^{2}}{1-\left(x^{2}+y^{2}\right)} & \frac{ xy}{1-\left(x^{2}+y^{2}\right)} \\
\frac{xy}{1-\left(x^{2}+y^{2}\right)} & \frac{1- x^{2}}{1-\left(x^{2}+y^{2}\right)}\\
\end{pmatrix}
$$
so they are the same.
And for $\lambda=-1$ we have
$$
g=\begin{pmatrix}
\frac{1+ u_2^{2}}{1+u_1^{2}+u_2^{2}} & \frac{ -u_1 u_2}{1+u_1^{2}+u_2^{2}} \\
\frac{-u_1 u_2}{1+u_1^{2}+u_2^{2}} & \frac{1+ u_1^{2}}{1+u_1^{2}+u_2^{2}}\\
\end{pmatrix}
$$
But if we look at the typical model for hyperbolic plane that consists of the pseudosphere embedded in $\mathbb{M}^3$:
$$
x^2+y^2-z^2=-1
$$
with a chart given by
$$
\psi: (x,y)\longmapsto (x,y,\sqrt{1+x^2+y^2})
$$
we obtain the inherited metric in $\mathbb{R}^2$ (keep an eye: inherited from the Minkowski metric, not the Euclidean one):
$$
g=\begin{pmatrix}
\frac{1+ y^{2}}{1+x^{2}+y^{2}} & \frac{ -xy}{1+x^{2}+y^{2}} \\
\frac{-xy}{1+x^{2}+y^{2}} & \frac{1+ x^{2}}{1+x^{2}+y^{2}}\\
\end{pmatrix}
$$
so we are done.

Parabolic, Hyperbolic, Elliptic


how do I install a network manager on Ubuntu?

Question: I uninstalled my network manager trying to fix a problem.  Now I have no network manager and therefore no internet.  So I need one and I need to install it?
Comment: I edited my answer taking into consideration you do not have an internet connection.
Comment: Side-note for those looking at this only to create a VPN connection: In my case, I did not need the network-manager. I had an ovpn file ready. I could import it as a new vpn in the networks settings.
Answer: I did not realize the OP could not connect to the internet. You don't need a network manager to connect to the internet. Don't waste your time looking for a different computer and transferring the files. If you have an Ethernet cable connect the computer directly to the modem and then:
sudo nano /etc/network/interfaces

Append the following lines to the end of the file:
allow-hotplug eth0
iface eth0 inet dhcp

Ctrl+X to exit and Y to save. Then:
sudo ifdown eth0
sudo ifup eth0

You should get an internet connection. If you do not have a DHCP server use the following instead (if you don't know if you have a DHCP server, then you probably have one so nevermind the following):
iface eth0 inet static
address 192.168.1.5
netmask 255.255.255.0
gateway 192.168.1.254

Replacing the address, netmask and gateway values with the ones you are used to.
If you cannot possibly connect the computer directly to the modem, you can also manually connect to a wireless AP, though it's much more complicated.
After connecting to the internet:
sudo apt-get install network-manager

or wicd instead of network-manager - it's the one I use.
If you are not on the sudoers list issue su to become root and then do all the commands without sudo.
Comment: @Alex you didn't try what @Alaa said: `sudo -i` gives (if really needed) a true root shell.
Comment: `su` needs a root password to be set, which is not set by default, and most new users wouldn't know how to set it, and shouldn't really need to. Why become true root when you have your friendly neighborhood `sudo`? I never advise anyone to become true root unless absolutely needed, and I would do it by `sudo -i`, not `su`.
Comment: @Alaa I advise it because of two reasons: First, one might not be on the sudoers list, which is the default for non-root users in a lot of distributions. Second, you don't have to type `sudo` all the time. I'm used with Debian, and Debian-based distributions usually have a root password set on installation by default, if I remember correctly that was the case on Ubuntu too, at least on earlier versions.
Comment: my /etc/network/interfaces file informs me that ifupdown has been replaced by /etc/netplan ...
Answer: To install NetworkManager that is installed by default when you installed Ubuntu,  just press Ctrl+Alt+T on your keyboard to open Terminal. When it opens, run the command(s) below:
sudo apt-get install network-manager

Thanks to vasa1, I failed to see the no internet part.  So what you can do is visit Ubuntu Packages, from a machine that has Internet connectivity (friend's, work, etc.) go down to the bottom of the page, and download Network Manager that corresponds to your Ubuntu version and architecture (32, or 64 bit), (in my case its for 13.04 Raring) and once downloaded, take it over to your system, and install it from the downloaded file.
Comment: But OP doesn't have an internet connection now. So how would `sudo apt-get install network-manager`work (unless the relevant debs are in `/var/cache/apt/archives`)?
Answer: Happened to face same problem. That's how i solved it. I happened to remove the network-manager and yep now I have no internet. So I came to know after that best shot is http://packages.ubuntu.com/. So but which package I asked myself.
So what i did i executed command again
sudo apt-get install network-manager.
Its obvious it fails, but I came to know what package it was trying to download.
e.g in my case if failed while downloading http://in.archive.ubuntu.com/ubuntu/pool/main/n/network-manager-applet/network-manager-gnome_0.9.8.8-0ubuntu4.3_amd6. So that's it, go to your friend system. Hit this url .It will download deb package.
Now copy this package and paste it in /var/cache/apt/archives. And now run again sudo apt-get install network-manager and that's it.
I am hoping all dependencies are already there. So by this way you can install network manager without using any boot device
Answer: The easiest way is boot from an installation media and then use chroot.

Boot from an ubuntu installation media.
Mount your system drives: sudo mount /dev/sdX /mnt
chroot into your system: chroot /mnt /bin/bash
Install networkmanager with sudo apt-get install network-manager
Reboot your system


Effective masses for different direction

Question: Assume we have an indirect semiconductor where the effective mass becomes anisotropic in different directions. Usually, one talks about a mass in parallel and perpendicular direction referring to momentum (or k-)space. So if I have a semiconductor sample for which I know the growth direction is lets say the parallel, I know which mass I have to consider when I look along this axes in real space. But what mass acts along the perpendicular direction in real space? Since real and k-space are connected via a Fourier transform, the mass perpendicular to the parallel mass in real space is not the one perpendicular to the parallel mass in k-space. So how can I find the mass acting perpendicular to the growth direction in real space when I know the mass along the growth direction and the mass perpendicular to that but I know it in k-space?
Answer: If you refer to a layered structure along growth direction (e.g. quantum well, superlattice), you can calculate your mass for different states and positions from the 2nd derivative of the wavefunctions.
Perpendicular to it you would have the dispersion (and therefore mass) of a free particle, since you would usually assume that your layer planes are infinite in two dimensions.
If not, you have to solve the Schrödinger equation for more dimensions and again evaluate the mass based on the bending of the wavefunctions.


Does modprobe.d blacklist always need the prefix "blacklist-"?

Question: In the german Ubuntu-Wiki it says, that the name of a blacklist in modprobe.d can be chosen as prefered but the filename has to begin with "blacklist-". Is that right? Because in a forum I read a blacklist was named "broadcom.conf". Will this work?
Answer: From the Red Hat documentation it would appear the name of the file containing "blacklist" is optional.
36.3.2. Blacklisting a driver, specifically step 6:

6)  Create a new file under /etc/modprobe.d/ that contains the command blacklist name_of_driver. Give the file a descriptive name that will help you find it in future, and use the filename extension .conf. For example, to continue to blacklist the driver foobar when the root device is mounted, run:
$ echo "blacklist foobar" >> /mnt/sysimage/etc/modprobe.d/blacklist-foobar.conf

Looking through several of my systems I could find no example where the command blacklist was contained in a file other than ones named blacklist-.....
I'd try it but I'd expect it to work in any file name. The command blacklist ... would seem to be what's important.
Answer: The file name in /etc/modprobe.d/ has to end with the .conf extension.
The content of the file, that is the directive to not load a certain module, has to use the keyword blacklist, like so:
less /etc/modprobe.d/bluetooth.conf
blacklist btusb
blacklist bluetooth
See the Archwiki entry for details.


What academic level would one need to be at to fully understand papers published on the twin prim conjecture?

Question: Specifically, what academic level would one need to be at to fully understand Goldston-Pintz-Yildirim's work on twin primes? Undergraduate, Graduate, or PhD?
Comment: I would say undergraduate is enough - this article is very well written and is self-contained. A plus is that there isn't much specialized vocabulary in this article. (when compared to algebraic geometry, for example)
Comment: Why the votes to close?
Answer: This is a rather subjective question, and it depends on specialization: I suspect there are "pre-docs" (even pre-"grads") studying number theory who would better understand the paper than some PhD-level mathematicians, in say, group theory. Mathematical maturity, and area (and level) of specialization is of greater importance than the degree one holds. It also depends how deeply you want to understand any published paper, including the paper you refer to (and what you mean by "fully" understand.)
A good look at the references will help indicate what prior knowledge and/or familiarity with such work is assumed, as will the abstract. References will often include work that may be a bit more general, and you can backtrack to that work, to check and see what level of familiarity is assumed to understand that work.
Answer: Having an undergraduate degree in math from MIT, I'd have to sit down with this paper for a long time before I could understand it... if ever...
At this level of math I think it's less about what general academic level someone has and more about his or her specific area of expertise.
Answer: Based on the link you have attached It doesn't seem like you need too much. The first 10 or so pages all made perfect sense to me (then again I've read like every prime related article on Wikipedia)
For this overall:

Read Up on Primes on Wikipedia and click each link, one by one studying each of the pages, each time you come across a word or a term that you don't understand google it, or copy and paste it into Wolfram Alpha and learn it.
Buy yourself a basic number theory book (ex: Art of Problem Solving Introduction to Number Theory) to get yourself started on understanding how a lot of this is done.
Read up on some standard High School AP Statistics, and read up some mathematical notation ex: Infinite Product Notation, Infinite Sums, Inclusion - Exclusion...

Thats all the math you'll need to understand the paper. Now you'll just have to move through it slowly, reading each line one by one, and if ever you find a line that makes no sense look it up.
GOOD LUCK!
Comment: timipueo basically sums it up very well... this type of document is geared towards those who are experts in the subject of primes... That has nothing to do with a degree etc... just a matter of how much time you have spent studying primes


Tensor product is to flat as Hom is to ?

Question: Sorry if I'm missing something here, but what do we call $M$ if the functor $H_M:N\mapsto Hom(M,N)$ is exact? Is this in fact equivalent to being flat through some adjointness properties?
Comment: Why the close votes? -- This is a 6-year-old question which has a highly-voted accepted answer. Of course it's just a question about definitions, which would likely be closed if asked nowadays, but anyway.
Comment: Projective module?
Answer: We call such modules projective.  If you take $N\mapsto Hom(N,M)$ then you get injective modules.  This is fairly basic, and covered in any homological algebra book, and mentioned on wikipedia.
Comment: Thanks for the reference. I'm still learning about tensor products from Atiyah-Macdonald, and I was just wondering what the analogous concept was :)
Comment: It's all there in Lang, I suppose. An advantage with a bigger algebra book is that you have everything in one place.
Answer: It might also be helpful to know that projective is equivalent to being a summand of a free module (apply $Hom(M,\text{--})$ to a presentation of $M$), and hence projectives are flat.
The converse is not true in general (e.g. $\mathbb Q$ is flat as a $\mathbb Z$-module, but not projective), but for finitely presented modules over commutative rings, flat and projective are equivalent.
Answer: I'm pretty sure that $M$ is called projective in this case, and if $N \rightarrow Hom(N,M)$ is exact then $M$ is called injective. I might have it backwards, though.


Why is my white wire hot?

Question: I have (1) 12/3 wire coming in to the box for the dimmer switch, lights started blinking, 3 lights on this circuit and then they finally went out.  When I checked the wires red, black and white, white to ground read 120 volts, red and black to ground read nothing, white is the only one that's hot what would cause this could squirrels in the ceiling bit through the black or red and caused white wire to read hot now? Ty for any information
Comment: Is it a simple single pole dimmer switch or a three way switch(two switches control the lights?  Is the white wire connected to switch?
Comment: Odds are good that if you flip *the other switch* the wire that's hot will change. Since it seems likely that it's a 3-way switch and the white is one of the travelers. But sure, could be squirrels.
Comment: Can you post photos of the insides of the boxes involved please?
Comment: There is only one switch and how do I post a photo from my phone,ty.
Answer: You have a switch loop
"One cable into the box" is the dead giveaway.
And an old-style one, from the sound of it. This is using white for always-hot, as Code requires.  The other two are switched-hot, to allow separate control of two things, e.g. two banks of lights, or light+fan.
It's possible the last installation fed both switched-hot wires from the same dimmer, or the fan is missing.  In such a case, it may be possible to rewire the switch loop to use white as actual neutral, black as always-hot and red as switched-hot, in the post-2011 wiring style.
Comment: And, any attempt at a rewiring should be _proceeded_ by a new question here including pics of the wiring in this box as well as the wiring at each of the light fixtures involved. (At a minimum, it will ensure the OP has "before" pics for when things go pear shaped afterwards.)

Why is my white wire hot?


Outlet to Outlet/Switch Combo

Question: So I want to add a switch/outlet combo to an existing outlet line, I want the new switch to control the outlet in the fixture with it while keeping the existing outlet I had already always active. There are two conductors/one ground to the existing outlet. Do I take the neutral from the existing outlet and tie that to the hot screw of the new switch then run from the neutral of the new outlet to the neutral of the feed line?
Answer: Existing outlet should have black (hot), neutral (white) and ground. All grounds always go together and are never switched, so we can ignore them for this answer (but do connect them all together).
Change the hot & neutral going into the outlet to pigtails. Each single wire goes into a wire nut along with a short wire (same color) going to the outlet as well as a wire (same color) going to the new switch box.
The new switch box will have two sets of wires coming in:

Black/white pigtailed as noted above to the old outlet.
Black/white going to the new, switched, outlet. Black normally goes to the brass screw, white normally goes to the silver screw. Don't use the backstab connectors as they are less reliable than screw terminals.

Wire nut the two whites together. You don't need them now, but you will in the future if you ever put in a smart switch.
The black from the old outlet goes to the switch ("line" if you have "line" & "load"). This is the hot.
The black going to the new, switched, outlet goes to the switch ("load" if you have "line" and "load"). This is the switched-hot and you may want to mark it differently (e.g., red or some other color besides black, white or green) on both ends so that anyone looking at it later will know that it is a switched-hot and not a regular hot.

Outlet to Outlet/Switch Combo


If the lecture room is usually full, is it appropriate for a TA to compete with the seat of students?

Question: There is one topic in the course that I am TAing that I felt interested to re-learn (since I forgot most things about it). I also had some questions I want to ask about the topic (that could be beneficial to students). As in the question, knowing that the room is frequently overflowing, is it not appropriate to "steal" the seat of students to sit in the lecture?
Comment: This seems to me to be two completely separate topics...1) taking a seat as a TA in a room where all the seats are filled (I assume you mean literally? as in, students would have to stand), and 2) asking questions when you are the TA rather than a student.
Comment: @Bryan this is just to say that If I occupy the seat, I won't just "sit" there, I also plan to do something potentially beneficial to the class as well. The primary question is still regarding taking a seat from students as a TA.
Comment: Do you mean literally taking a seat so someone has to stand? or taking a seat as in someone has to sit further to the back?
Comment: I will not be able to know in advance if the lecture room will be actually full. Perhaps, I have the option to observe the situation a bit, if it is indeed full-seated, then I could sit on the stair or sth, tho not ideal.
Comment: More information, in the last lecture even some students appeared to sit on the stair/stand. It was not even just full, it was overflowing. But I dont know if it is just because it is still add/drop period.
Comment: @ychung If the lecture room is exceeding room capacity, that is a fire code violation and building regulation concern. You should 1) report this repeating incident and 2) request a larger lecture room (or request the lecturer do it).
Answer: Ask the professor
The professor should be the one to make this decision, weighing the potential positive and negative impacts on the students.
As a TA, you should in any case ask the professor before planning to attend lectures, whether the course is full or not, and you should particularly ask whether it is okay for you to ask questions.  In some classes, the professor might think having a TA present would be distracting, and may or may not share your view that your questions would be "beneficial to the students".
If you do end up asking questions, it would be good to check in with the professor periodically to see if they are still finding it helpful.
If the professor asks you not to attend the class, then it is your responsibility to find some other way to get up to speed on the material (i.e. reading on your own).  If you do not think that will be possible, then you should ask to be assigned to a different course.
Comment: Thank you for the answer. I agree the best is to ask the professor directly. Over the years, I have observed that it is pretty common in my university where the TA sit in the course and even raise question (for the courses I have taken so far).
Answer: If I was in the situation with a room this full, I'd probably find a stool or chair and sit to the side or behind the professor rather than taking a "normal" seat. I'm sorry you are in a position with such limited seating for the class size.
I think it's completely normal and desirable for TAs to attend lectures regularly, but I'd also agree with @NateEldridge in that you should probably run your questioning by your professor in some way. There are a lot of questions a TA might have that aren't really relevant to the course, and you run the risk of undermining or appearing to undermine the professor in front of their students, which, even if it is in some way deserved, is unlikely to work out positively for anyone involved.
Answer:
is it not appropriate to "steal" the seat of students to sit in the lecture?

No, it is not appropriate.  In most cases it is appropriate and potentially slightly helpful for a TA to be present in a class when they are not teaching it.  But the TA should allow the students to take priority.  The class is for them.


Can we define a data partitioning in K clusters, by cutting the branches of the tree at some levels in the tree below the root node?

Question: Assume we have a dendogram (hierarchical clusterisation tree), can we define a data partitioning  in K clusters, by cutting the branches of the tree at some levels in the tree below the root node?
Answer: Yes absolutely, in fact that's exactly how one obtains actual clusters with hierarchical clustering.
With hierarchical clustering it's possible to decide the number of clusters (i.e. the level of "precision") after obtaining the tree (dendogram).


generating function for $F_{5n}/(5F_n)$

Question: Let $F_n$ be the $n$-th Fibonacci number with $F_0=0$, $F_1=1$, and $F_k=F_{k-1}+F_{k-2}$ for $k\ge2$. Prove
$$\sum^{\infty}_{n=0}{a_nx^n}=\frac{1-4x-9x^2+6x^3+x^4}{1-5x-15x^2+15x^3+5x^4-x^5}$$ where $a_n=F_{5n}/(5F_n).$
Comment: You can do it fairly mechanically using Binet's formula $F_n = \frac{\phi^n - \varphi^n}{\phi - \varphi}$, although you might consider that cheating. More interesting might be to find a combinatorial proof.
Comment: This sequence satisfies the recurrence relation $a_{n+4} - 4a_{n+3} - 19a_{n+2} - 4a_{n+1} + a_{n} + 5 = 0$.
Comment: @ChipHurst how have you derived that?
Answer: This seems quite tough so I'm thinking you must have some results from a course text book that are relevant. Anyway, tackling it from the back end, the given generating function's denominator can be factorised;
$$\frac{1-4x-9x^2+6x^3+x^4}{1-5x-15x^2+15x^3+5x^4-x^5}$$
$$=\frac{1-4x-9x^2+6x^3+x^4}{(x^2-7x+1)(x^2+3x+1)(1-x)}$$
and then partial fractions yield,
$$\frac{2-7x}{5(x^2-7x+1)}+\frac{3x+2}{5(x^2+3x+1)}+\frac{1}{5(1-x)}$$
These quadratic denominators in a generating function scream Fibonacci;
$$\frac{1}{x^2+3x+1}$$
$$=1-3x+8x^2-21x^3+55x^4-144x^5+\dots$$
$$=F_2-F_4x+F_6x^2-F_8x^3+F_{10}x^4-F_{12}x^5+\dots$$
and
$$\frac{1}{x^2-7x+1}$$
$$=1+7x+48x^2+329x^3+2255x^4+15456x^5+105937x^6+726103x^7+\dots$$
$$=\frac{F_4}{3}+\frac{F_8}{3}x+\frac{F_{12}}{3}x^2+\frac{F_{16}}{3}x^3+\frac{F_{20}}{3}x^3+\dots$$
and these and their ilk in themselves would be interesting to explore.
Moving to the front end, $F_{5n}$ is always a multiple of 5, so I can see why the 'divide by 5' is in there, and it's in the partial fraction decomposition too. This can be proved by induction, and is a well known result;  e.g. http://www.math.utep.edu/Faculty/duval/class/2325/104/fib.pdf
The $F_{5n}$ can be broken down into an expression involving $F_n$ and $F_{n+1}$ with the neat algorithm on page four on the following document;
http://www.maths.qmul.ac.uk/~pjc/comb/ch4s.pdf
Wikipedia has this decomposition for $F_{4n}$
$$F_{4n}=4F_nF_{n+1}(F_{n+1}^2+2F_n^2)-3F_n^2(F_n^2+2F_{n+1}^2)$$
I found an $F_{5n}$ decomposition (Falcon, Plaza 2007) as;
$$F_{5n}=F_{n+1}^5+4F_n^5-F_{n-1}^5+10F_{n+1}F_n^3F_{n-1}$$
I don't know if that's going to help.
There is an interesting result;
$$\frac{F_{kt}}{F_t}=\sum_{i=0}^{(k-3)/2}(-1)^{it}L_{(k-2i-1)}t+(-1)^{\frac{(k-1)t}{2}}$$
k is odd and > 2 and L is the Lucas number. (Vajda 85)
This and 300 more Fibonacci relationships are at http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibFormulae.html
It may not be relevant but I note that the convolution of
$$\frac{1}{x^2+3x+1} \times \frac{1}{x^2-7x+1}$$
is
$$(1.1)+(1.7+(-3).1)x+(1.48+(-3).7+8.1)x^2+(1.329+(-3).48+8.7+(-21).1)x^2$$
$$+(1.2255+(-3).329+8.48+(-21).7+55.1)x^3$$
That is
$$1+4x+35x^2+220x^3+10556x^4+\dots$$
and the $\frac{1}{(1-x)}$ will be taking the partial sums of this.
Update : $F_n^4$ has the generating function
$$\frac{1-4x-4x^2+x^3}{1-5x-15x^2+15x^3+5x^4-x^5}$$

generating function for $F_{5n}/(5F_n)$


What is Ethernet diff. pair referenced to between connector and magnetics

Question: This is a little bit confusing part. I clearly understand that in ethernet cable fields are coupled between wires of differential pair. On PCB on IC/PHY side it is also known that differential pair fields are coupled to ground (reference) plane - coming from that, that on PCB diff. pair is actually just two single ended transmission lines close by.
And then there is this confusion land where cable connects to PCB and goes to transformer...
To what are the diff. pair lanes coupled to ? To ground, but why would they, as magnetics on both ends of cable give then isolation. Or do they still couple to each other ? If so, then every Zdiff. calculation tool is unusable as those assume some reference plane.
Comment: Electrical isolation doesn't magically stop coupling of AC signals (sent differentially or otherwise) from coupling with a localized ground/earth plane or any other neutral plane that has capacitance to earth.
Comment: In practice, or at least good practice, the termination resistors / transformers are very close to the connectors, so the PCB lines are very short, and any deviation in impedance from the cable to the PCB lines is swamped out by the impedance of the termination resistors / transformers. However, I am not entirely clear what you are asking about, so what I just said may not address your concern.
Answer: The whole point of using a differential pair for signalling is that there is no need for a reference plane to compute the odd mode (the signalling) impedance. When you do the calculation, there is an implicit reference plane mid-way between the two conductors. As the conductors are driven with opposing voltages, the voltage mid-way between them is always 0.
You can also define an even mode impedance, which is the same signal being carried on both conductors, and this is with respect to some remote reference plane. In the case of ethernet, this even mode signal is poorly driven and received, as it has to couple capacitively across the isolating magnetics at either end. This is the signal responsible for EMI via conducted emissions, so on many cables it's attenuated by lumps of ferrite on the cable that raise the even mode impedance further.
Answer: The PHY-side Ethernet wiring has both intra-pair differential and common-mode impedance coupling to ground (generally, 100 and 50 ohms, respectively.)
The field-wire side of Ethernet only has intra-pair differential coupling (100 ohms) and no common-mode impedance to any reference (more on that below.) The field-wire cable in fact has no DC path to ground at all, and can be any voltage with respect to ground, even a potentially damaging one.
(Power-over-Ethernet exploits Ethernet's field-wire isolation by inserting DC offsets on the field wires to transmit power. That's another discussion, but illustrates the point.)
In practice, the Ethernet magnetics has the field-side pairs terminated using what's called a Bob Smith termination: each field wire ties to a 75 ohm resistor, with a common tie to a high-voltage cap to ground. This terminates the field wires into a 75 ohm single-ended impedance to a 'virtual ground', with the added benefit of reducing the possibility of DC buildup.
The magnetics also includes common-mode filtering which further improves rejection of any transients that couple onto the field wires. A robust design will also include ESD protection for the PHY-side pairs to further reduce the effects of any transients that make their way through the magnetics.


What is the shape of the initial universal wavefunction?

Question: In the many-worlds theory and Bohmian mechanics, the universe has an "initial" "universal" wavefunction which then evolves according to the Schrodinger's equation and determines the future of the universe. My question is that what the shape of this initial wavefunction is more likely to be? Can it be any random integrable/differentiable function over the configuration space with equal probabilities, or it is supposed to have anisotropies?
Answer: The universal wave function is a term introduced by Hugh Everettin in his Phd. thesis The theory of the universal wave function.
The wave function of the universe is given by The Hartle–Hawking state is the wave function of the Universe—a notion meant to figure out how the Universe started—that is calculated from Feynman's path integral. More precisely, it is a hypothetical vector in the Hilbert space of a theory of quantum gravity that describes this wave function.
There have been criticism associated with its mathematical validity and what application does it serve. So, in my opinion we cannot give a shape to the universal wave function.
For more info read this dissertation where the word Universal wave function was used -dissertation


InnoDB cluster error: ReferenceError: cluster is not defined

Question: I was setting up the MySQL InnoDB cluster, when I got this error. I created the cluster from the primary node, the cluster.status() was giving output as 'ONLINE' at this point with only one node part of the cluster (all good till here). But when I tried adding a node in the cluster I got the error:

"Cluster.addInstance: Cannot add an instance with the same server
  UUID (db3fdca5-3e67-11e9-98d3-42010a940021) of an active member of the
  cluster '10.148.0.38:3306'. Please change the se rver UUID of the
  instance to add, all members must have a unique server UUID.
  (RuntimeError)"

To resolve this error, I edited the auto.conf and restarted the second instance (did not touch the primary instance at all) Now if I try to check the cluster status (or any commands like cluster.addInstance()) I get the same error as given in the title.
Answer: The auto.cnf file is automatically generated, you cannot modify this file.
In order to change the UUID, you must delete auto.cnf and restart the MySQL Server. MySQL will generate a new UUID and store it in a new auto.cnf file.
Please check the following documentation section: https://dev.mysql.com/doc/refman/8.0/en/replication-options.html#sysvar_server_uuid
Cheers,
Miguel


Best approach to 2D racing game AI?

Question: I have no experience in AI at all, but I would like to know how I could do this?
Even if the AI just followed a path how could I do this?
Comment: How could you do what, precisely?
Comment: It depends on a lot of other factors - e.g. if there's no car-car collision you can just hard-code car movements or use 'ghosts' from replays of previous races.
Answer: My first thought naive approach would be to create a list of points  corresponding to the coordinates of the track curves (probably more points would be needed to prevent the AI keep going against walls after an unplanned collision with another car). When the AI car is close enough to the next point, you would put it (the point)  in the end of the list. All the AI car would need to do is to always go in a straight line in the direction of the next point in the list.
Comment: Yeah, basically waypoints. You have the car use basic [steering behaviors](http://www.red3d.com/cwr/steer/) to go from one waypoint to the next.
Answer: Yes, make a type of path that the car follows. First, you need to make the car just drive. (Make the car drive forward forever). Create a rectangle at each turn on the track then make if statements for when the AI car intersects them. When the AI intersects them(the rectangles) make it rotate according to where it needs to turn.

Best approach to 2D racing game AI?


Facilities for asynchronous texture loading

Question: How would one go about loading textures asynchronously while a 3D application is running? What I do understand is that OpenGL contexts are not thread safe and that I should separate them on different threads.
But my primary problem is picking a proper multithreading facility/framework to actually implement this with Windows and C++, I've heard a lot about C++11 including threading support in its standard lbrary, but could someone lay out just the basic steps?
What would be the safest way to do it? And how does one update the state of the other context that it registers the changes done on the other thread? I suspect glFlush and glBind*?
Comment: Actually OpenGL and OpenGL contexts **are** thread safe. You can just have a single context be active in more than one thread at a time.
Answer: The most time-consuming part of texture loading is typically the disk access and any format conversion, both of which are independent of OpenGL and thus can safely take place on another thread. Once the texture is read into memory and in the desired format, the actual copy into an OpenGL buffer is fairly quick.
Going into the details of threaded programming is entirely too complex for this answer, but there's a good bit of documentation around and once it clicks in your head, it's pretty easy (like pointers and memory).
The general concept here is to create a list of texture holder objects (containing, say, the file/name, an initially-null buffer, and a loading-complete flag) and pass that to your loading thread when it's created. The loading thread then goes through the list, opens each file, loads it into memory and attaches the buffer to the list entry, then sets the loaded flag and maybe increments a counter. The main thread takes the newly-loaded texture, copies it into an OpenGL texture, and increases the progress bar or whatever your loading indicator is. Once all the textures in the list have buffers and are flagged as loaded, the other thread's work is complete and it can be stopped (or left alive to load future textures).
The main advantage to that model is due to not having to share the actual graphics context. In APIs that can be thread-safe (DirectX), there's a performance penalty to doing so, and OpenGL requires a decent bit of work on your part to have multiple contexts or make sure you're sharing them properly. The heavy lifting when loading textures is typically the file read and parameter checking, unless you do format conversion or rotation, which may dwarf even disk access. The actual copy to video memory is highly optimized and not likely to be a bottleneck (if you're worried about that, try profiling in a tool capable of recognizing GPU call cost, and see). None of this work depends directly on OpenGL, so you can push it off to another thread with very little worry.
If you're on Windows specifically, there are built-in threading functions that can be used, working off of a simple callback model (provide a function and initial params, in this case your texture list, and make the API call). I'm not personally familiar with C++11's threading support or how it works in Visual Studio, if at all, so you'd have to check on that.
Comment: Nice answer.  Yes, C++11's `std::thread` works on MSVC++.  It's usage is *very similar* to `boost::thread`.
Answer: The answer by @ssube is correct about the complexity breakdown of the task, but assumes that "OpenGL requires a decent bit of work on your part to have multiple contexts or make sure you're sharing them properly", and I disagree.
An easy solution is to create a main context (for drawing) and a secondary context (for texture loading), at the start of the program, for example with:
m_hRCDrawing = wglCreateContext(m_hDC);
m_hRCSecondary = wglCreateContext(m_hDC);

And then sharing the data between the m_hRCSecondary and the m_hRCDrawing contexts can be done with:
wglShareLists(m_hRCSecondary, m_hRCDrawing);

Finally, when you are about to read a texture from a "texture loading" thread that has no GL context, you can simply call:
wglMakeCurrent(m_hDC, m_hRCSecondary);

after which any assets loaded within this thread are shared by the drawing thread's context.
A slightly more thorough explanation is available here.


How Stack Overflow questions appear on Google as soon as you post

Question: As soon as you post a question on Stack Overflow, in 5-10 seconds it starts to appear on Google Search Results. How that works? Is that the result of a very high page rank or there is some kind of alliance between Google and Stack Overflow? If you don't believe, you can try it to make sure.
Comment: i just tried Yahoo and Bing. the same story... they index StackOverflow pages very fast... I'm so jealous =)
Comment: Optimised content does not explain the speed of indexing. A high PageRank and known high post rate will speed up crawling but even that can't get you a guaranteed almost instant index. I would suspect some sort of push notification that Google subscribes to. E.g. an RSS feed and PUBSUBHUBUB service.
Answer: I think the question is how they rank not just appear.
It says here that:

"The Golden Rule of SEO: Content is King" is the first thing to
  consider. It's quite simple: have good content. That's the primary
  reason Stack Overflow has a chance to rank at Google, simply because
  of how page ranking works.
Websites that have no optimization or guidelines concerns whatsoever
  but got nice stuff have a great chance of ranking high, because there
  will be people linking to good content. And most sites trying to do
  SEO without content won't rank well because nobody will link to them.
With that been said, Stack Overflow does a lot of work from a long
  time ago to keep clean HTML code.
One good example is how rel='canonical' is applied in
  here to keep search engines away from linking to duplicate content. For
  instance, let's get this page source:
<code><link rel="canonical" href="https://meta.stackexchange.com/questions/14056/seo-in-stackoverflow">
</code>
That is an optimization that will not help to directly rank in the
  search engine, but it will make the results the engine can find much
  more relevant and through that it might generate more links (from
  people who found what they were looking for) to the good content it
  already have, thus actually indirectly improving its rank.
Comment: No they aren't. Actually, the explanation is clear. SO posts are SEO Optimize. Or shall I say more optimize than CodeProject. That I think answers the question.
Comment: ok, i know that stackoverflow uses a clean html and it's content is great. the page rank is also very high which is 7. the same thing with codeproject.com. a very high page rank, and a great content. but new articles on codeproject will not appear on google without few hours. i'm just surprised how fast google indexes the stackoverflow pages...
Comment: Wait, are you searching the same or exact keywords? same with code project?
Comment: exact keyword. it doesn't matter if i'm searching the exact keyword or some, what really matters is Google indexes too fast, like stackoverflow and google uses one single database... =)
Comment: SO developers understand how google indexing works giving the advantage over their search engine.
Comment: got it. thanks. so that is all about SEO Optimization, if the page is highly optimized for SE, and the page rank is high, SEs index the page so fast. is that it?
Comment: Yes.. Exactly..
Answer: See: Why does Google index pages of my blog in seconds, while pages on the rest of the website take longer to get indexed
When something appears in Google in seconds, it is generally because there is something notifying Google about this new content.  http://pingomatic.com/ is a service that can be used for this purpose.  WordPress integrates with it by default which allows blogs powered by WordPress to get listed in Google very quickly.
Answer: I think it has something to do with how unique most of the questions on S/O are. They're not typically questions being written about elsewhere.
Comment: i don't think that is the only reason. there must be some trick to make google index the page in 5-10 seconds. that is an incredible thing.


Where does Ulysses store its local content?

Question: I'd like to back up my local content with my own scripts, but I'm not familiar with how macOS apps work. For example, I can't find where Ulysses stores its local content.
Or more generally: where do macOS apps usually store their files?
Answer: It's really up to each program to decide if it stores in the home folder / documents and lets users manage things or if it makes a database in plain view (like Photos or iTunes) with aspects of the database elsewhere or entirely hide the filesystem and manage all files / syncing.
In the case of Ulysses - the developer has decided to manage things in a bundle/database and either sync it to iCloud via a local cache or store the content locally in your user library folder - out of view of normal expectations in ~/Documents or ~/Desktop
Content stored On My Mac
If you do not sync your local content using iCloud (i.e. it's in the On My Mac section in your Ulysses sidebar), you can access each group in your Ulysses library at the following file path:
~/Library/Group Containers/X5AZV975AG.com.soulmen.shared/Ulysses/Documents/Library/Groups-ulgroup/
Each group belongs it its own -ulgroup directory and will have an (unhelpful) hash for a name—for example, cba24c5b16a949c388b0204c9aa28867-ulgroup.
Within each -ulgroup directory, your documents will each have their own .ulysses file. And you guessed it, they also have a hash for a filename. Note that .ulysses files actually contain both a Content.xml and a Text.txt with your content in them.
Content synced via iCloud
You will have a hard time accessing your Ulysses iCloud-stored library from Finder, but I seem to be able to access mine fine using the Terminal if you're comfortable doing that. It's stored similarly as I described above, but here is the file path you'll be looking for instead:
~/Library/Mobile Documents/X5AZV975AG~com~soulmen~ulysses3/Documents/Library/Groups-ulgroup
A note on iCloud. The com~soulmen~ulysses3 is the application bundle/group name provided to Apple by the developer when they created the iCloud entitlements for this application. The X5AZV975AG is Apple's provided identifier that combined with the reverse dns type name space, makes each iCloud bucket unique but common across everyone that uses that app.
Comment: ***Easily Find Your Files:***
**Go here** `cd ~/Library/Group\ Containers/X5AZV975AG.com.soulmen.shared/Ulysses/Documents/Library/Groups-ulgroup` **Search all the files:** `grep -Ri "PHRASE"` - "oven to 350" is better than "oven" **It prints the location of the file, and the whole line too!** You will find what you're looking for very quickly. ***Other grep options:*** `-v` **everything except PHRASE** `-n` **line #'s -m 2 first 2 matches**
Answer: There's a way to do this where even if you lose access to Ulysses you will still have your files in a way that is useful for you and won't make you cry.
Move your files in Ulysses' iCloud and "On My Mac" sections to two external folders. The exact directory structure gets copied & the sheet names are retained; it took me about a minute.

Once you've set it up to only use external folders you can do normal things with the files.
For comparison:

Everything Ulysses keeps in the ~/Library area is all over the place. Hopeless. I thought I was going to have to make every folder (group) by hand and export each "sheet" one by one.
No one should have to go through this to move a directory structure full of text files from one place to another my God. Comma key is broken apologies.


Guidelines to setup sound buttons

Question: Are there any guidelines for positioning of sound buttons (play, pause, mute)?
Comment: What platform is this for? Web, moble, desktop? What are you making?
Comment: @jonshariat Web generally. I am making a social game.
Comment: Oh then I would add that you dont want to play audio right off the bat. Many people play at work, school, library, coffee shop.
Comment: Is this for video files or just sound files?
Comment: @Ben D For background music.
Comment: @jonshariat That is useful too, add it to your answer :-)
Answer: General Guidelines:

Play and pause should toggle
Volume control should be separate from mute (dragging the volume down is not clear on how to stop the sound)
Play button should be in close proximity to the timeline.
UI design should blend into the background of what is being viewed.
Small, but big enough click area for users to miss with their selecting device.

For many different design patterns of audio controls, check out this wonderful collection by Toxboe on http://www.UI-Patterns.com
Comment: We had a good question a while back about whether play and pause should toggle. http://ux.stackexchange.com/questions/1215/separation-of-play-and-pause-buttons-for-playing-media
Comment: Thanks for your answer, but I need information on positioning. I have seen the buttons in corners, on top, in the botton etc.
Comment: Check out my edit above, there is a great collection of ideas here -> http://ui-patterns.com/users/1/collections/347
Answer: I'm not sure what the standard is for this, but when I'm annoyed by sound I always look in the upper left, then right-hand corners for a mute button.
Failing that, I look for a settings area.
You should also consider where you have put other user controls, and whether or not it makes contextual sense to group this with them.

Guidelines to setup sound buttons


Diffeomorphism variation of the Christoffel symbol

Question: Under an infinitesimal diffeomorphism the Riemann metric changes by the Lie derivative
$$
\delta g_{\mu\nu} = ({\mathcal L}_\xi G)_{\mu\nu}=\nabla_\mu \xi_\nu+\nabla_\nu \xi_\mu
$$
and under a change of metric the Levi-Civita Christoffel symbol changes by
$$
\delta {\Gamma^{\alpha}}_{\beta \mu}=  \frac 12 g^{\alpha \lambda} (\nabla_{\beta} \delta g_{\lambda \mu}+ \nabla_{\mu} \delta g_{\beta \lambda}-
\nabla_{\lambda} \delta g_{\beta \mu})  $$
If one plugs the variation of the metric into the variation of the Christoffel sysmbol there are many terms, but in Bardeen and Zumino's famous paper on anomalies ((Nuc Phys B 244 (1984) 421)
they assert (eq 4.10, but in my notation) that
$$
\delta {\Gamma^{\alpha}}_{\beta  \mu}=\xi^\tau \partial_\tau {\Gamma^{\alpha}}_{\beta  \mu}+ \partial_\beta \xi^\sigma {\Gamma^{\alpha}}_{\sigma   \mu}+ \partial_\mu \xi^\sigma {\Gamma^{\alpha}}_{\beta    \sigma} - \partial_\sigma \xi^\alpha  {\Gamma^{\sigma}}_{\beta   \mu}-\partial^2_{\beta\mu}\xi^\alpha
$$
This last equation makes sense as an infintesimal co-ordinate transformation or  Lie derivative and greatly simplifies their later algebra. Is there a straightforward way to  see the equivalence of the two computations? It's not plug and chug, but seems to lead into wasteland of  Bianchi identities.
Comment: Ive posted an answer to to my  question on physics stack exhange under the heading ``Variation of Christoffel symbol and Lie derivative''
Answer: If you consult Sec 3.1 of Wald's book on GR, you will see how Christoffel symbols can be interpreted as tensors parametrizing the difference between the coordinate connection $\partial$ and the Levi-Civita connection $\nabla$, symbolically $\Gamma=\nabla-\partial$ (cf. Wald's (3.1.7)). You are interested in $\Gamma'=\nabla-\partial'$, where $\partial'$ is the new coordinate connection. Writing this in another way, $\Gamma' = (\nabla - \partial) + (\partial - \partial') = \Gamma + (\partial - \partial')$. Both terms in the last equality are tensors, so once you know them in one coordinate system, you know them in all coordinate systems, following the usual tensor transformation properties. The last formula in your question is essentially $\Gamma'$ written in the primed coordinates, to first order in the difference between the original and primed coordinates. The $\Gamma$-dependent terms are just the tensor transformation rule for $\Gamma$ interpreted as a tensor, while the $-\partial^2\xi$ term is precisely the $(\partial - \partial')$ difference. I'll leave checking that last claim as an exercise.
Comment: Thanks a ton! I did eventually figure it out on my own. The problem is that it is easy to get signs wrong when using the general argument --- and I think that B and Z did exactly that as there should be  no minus sign before the $\partial^2_{\mu\nu}\xi^\alpha)$
Comment: If you are happy with the answer, you may want to click the check mark next to it, to mark it as correct. Otherwise, the question will periodically appear on the front page as unanswered.


Salesforce certified Administrator or Salesforce certified developer?

Question: There is a couple of certificates available to take to further your knowledge and develop your career with salesforce and I was wondering which certificate is more prosperous to get . What are the main differences between being a developer and an administrator. I though both of them would have to deal with apex and basic functions. Any information on this subject will be useful. Thanks
Answer: Salesforce's Certifications have changed a lot recently. For instance, when I got Certified they were Structured a little like this:

Dev 201: Salesforce Administrator
Dev 301: Salesforce Advanced Administrator
Dev 401: Salesforce Developer
Dev 501: Salesforce Advanced Developer

I only tell you this because you may come across material for these exams which are now out of date.
These are still Administrator and Advanced Administrators for Admins, but Platform Developer I and Platform Developer II for developers. Others have been introduced as well such as an App Builder Qualification.

What are the main differences between being a developer and an administrator

In a nutshell, an Administrator will:

Manage users, data, and security
Maintain and customize Sales Cloud and Service Cloud applications
Build reports, dashboards, and workflow

A developer will:

Design the data model, user interface, business logic, and security for custom applications
Develop custom applications using Apex and Visualforce
Be familiar with the development lifecycle from development to testing, and have knowledge of the available environments

I was wondering which certificate is more prosperous to get

I'm assuming by prosperous you mean lucrative financially, in which case It'd almost certainly be the Developer qualification (out of the 2, of course). But, there'd be no harm in doing a job hunt for how much someone with a Developer Qualification gets paid compared to an Administrator.

I though both of them would have to deal with apex and basic functions

Apex, no. You're only expected to know Apex for your Dev501/Platform II qualification. Doing Dev401/Platform I for example you'll only look at the creation of objects, fields, security settings etc... Effectively, the "Clicks not Code" philosophy. Basic functions is a little vague to give you clarity on.
I did a bit of research following Juris Upenieks' comment and indeed found that:

Platform Developer 1 focus on coding skills. It also focus on Introduction to Lightning.
There is no prerequisite for Platform Developer 1 certification now unlike DEV 501 for which we had to be certified developer was prerequisite.

I strongly recommend you take a look at the Certification Site which has a lot of material, study guides, recommended courses which will tell you exactly what each qualification entails.
Hopefully, this gives you some idea as to what to expect.
Comment: @JurisUpenieks thanks for letting me know about this. Wasn't aware Platform Developer I was more coding based!
Comment: Please edit the part about Apex and Platform Developer 1, I am currently preparing for this exam and there is pretty much a lot to know in terms of Apex (see Study guide)
Comment: That's useful information I've just started with Apex and Java and it takes quiet a bit of time to get it just right, so everything works, especially in the dev console. I found Admin work more "easy" if i can say so. But developer certification seems better as a career option.
Answer: The 201 Administrator will cover profiles, permissions, page layouts, user setup, etc.  The most advanced thing you'll learn is the formula field which references other objects.  Here's a link to the study guide so see all topics covered.
The 401 Developer will be less setup and more build custom applications using the point-and-click capabilities of the platform, design the data model, user interface, business logic, and security for custom applications and design reports, dashboards, and portals just to name a few.  Here's a link to this exams study guide.
To dabble in each, I would recommend your free training or view the Admin Beginner and Developer Beginner at Trailhead.  I set up my own developer org and gone through some of these trails and they're really cool.


How energy is transferred between charges?

Question: When we move an electron(e1) towards another electron(e2, not fixed), the work we do is stored as potential energy in e1 but from where does e2 get energy to be repelled?  What is source of potential energy in e2?
Answer: Let us clear up that there exist two theoretical models that fit the data and observations we have and are also predictive of new data. The classical electrodynamics, represented by Maxwell's equations and is used for dimensions larger that the microscopic ones , and quantum electrodynamics, which is the quantum theory for elementary particles and atoms and molecules, and electrons are elementary particles.
At present physics theories assume that the underlying level of nature has to be described by quantum mechanical equations, from which level classical theories emerge.  The classical  models are mathematically consistent with the quantum mechanical models in the overlap region of variables.
So if you are really asking about electrons and not point charges of the classical physics, the electrons interact and are modeled with QED , a perturbative expansion of the crossection of electron electron scattering /interaction, it is modeled with Feynman diagrams. The first order ones are

So an electron interacts with an electron with an exchange of a virtual photon, and that is what generates the repulsion , not action at a distance. Important to note that all quantum calculation predictions for interactions are probabilistic, the probability of the interaction happening is determined by the calculations, and the calculations fit the data very well.
For everyday macroscopic usage classical electrodynamics is adequate to describe the behavior of charges as described in the other answer.
Comment: Really nice answer.
Answer: Since e1 and e2 are continously interacting with each other through the electrostatic force, there is no sense in speaking about the potential energy of only e1 or only e2. One can only speak about the potential energy of the system of e1 and e2. You can think about this by looking at the equation for the electrostatic potential $$E_\text{pot} = -\frac{1}{4\pi\epsilon_0}\frac{e_1 e_2}{|\vec{r}_1-\vec{r}_2|}$$ and you see that it depends on quantities of both particles that interact with each other.
Therefore, the work that we do by moving e1 is stored in the system of e1 and e2 as a mixture of potential energy between the charges and kinetic energy of e2(the kinetic energy of e1 is fixed since we move it). Usually, one talks about the easy case when e2 is fixed and only potential energy plays a role. However, the calculation in our case is still possible, but may need another approach than just energy conservation. It turns out that this is just another form of the Kepler problem that we know from astronomy.
Comment: Why we talk only about system's potential energy, not individual ones?
Comment: As I pointed out by giving its formula, potential energy is a property that needs at least two charges. In school, you usually talk about the potential energy of one particle in the field of the other one, but this is the same thing.


Ho To Disable ssl3 in OSX Server (for mail and Web hosting)

Question: how do i Disable the ssl3 cipher in OSX Server so as to not be vulnerable for POODLE ?
Comment: Which version of OS X server app? Since the "poodle" vulnerability is a problem where data is leaked on the client side, what specific client are you looking to secure? Web clients running on a machine with server.app configured, etc...
Comment: i actually have very old stuff running: Lion Server app and Snow Leopard Server. and it is about securing the mail functionality, which means it includes postfix, Apache and Dovecot as far as i reckon
Answer: Depends on the Server...
Apache:
SSLProtocol All -SSLv2 -SSLv3
NGINX:
ssl_protocols TLSv1 TLSv1.1 TLSv1.2;
Postfix:
smtpd_tls_mandatory_protocols=!SSLv2,!SSLv3
Sendmail (sendmail.mc):
LOCAL_CONFIG
O CipherList=HIGH
O ServerSSLOptions=+SSL_OP_NO_SSLv2 +SSL_OP_NO_SSLv3 +SSL_OP_CIPHER_SERVER_PREFERENCE
O ClientSSLOptions=+SSL_OP_NO_SSLv2 +SSL_OP_NO_SSLv3
Dovecot (In /etc/dovecot/local.conf or /etc/dovecot/conf.d):
ssl_protocols = !SSLv2 !SSLv3
If you use virtual Servers with SSL, please note that you have to do this on all instances!
Answer: If you're configuring VirtualHosts using Server.app, it will always set SSLProxyProtocol -ALL +SSLv3 +TLSv1 in your site configs. You can edit this line in /Library/Server/Web/Config/apache2/sites/*.conf, but watch out in case Server.app undoes your changes after an edit.
To fix in future VirtualHosts, you could try editing the default template in /Library/Server/Web/Config/apache2/sites_disabled/0000_default_default.conf, but watch out - upgrading Server.app through the app store may restore the original default config. Previously default_default.conf.default was the template, but this doesn't seem to be the case in Mavericks.
If you are ignoring Server.app and doing manual configuration, you may be able to get away with configuring once by installing a config fragment in /Library/Server/Web/Config/apache2/other/ (or /etc/apache2/other/ if Server.app is not installed at all). Here's mine:
/Library/Server/Web/Config/apache2/other/die_poodle_die.conf:
<IfModule mod_ssl.c>
    SSLProtocol All -SSLv2 -SSLv3
</IfModule>


What is the maximum amount of items you can have in your inventory in Minecraft

Question: So I just had a thought.
If I can make one Shulker Box and put a Shulker box in it with a Shulker box in it with a Shulker box in it repeat as many times as you want to, how much could I fit in my inventory? I think I can do infinite, just as long as it doesn't take up a lot of memory.
Answer: If you could put shulker boxes inside each other, you could actually have an infinite amount of items inside one shulker box, as long as you have loaded shulker boxes inside each other.
Now here we are going to solve how many items you can actually fit into one inventory.
There are 36 inventory slots with one available slot for shields, arrows, maps, totems of undying, etc., and 4 slots for armour. First, lets add up the armour and shield slots. That bring us up to 5 items. Now, if we fill a shulker box up with items, we can fit up to 27 slots, with 1728 items. Because there are 36 spaces in your inventory, we have to multiply the amount of items in the shulker boxes by 36. By doing that, we can get 62,208 items. If we include the shulker boxes themselves, we can fit up to 62,244 items and then finally we can add the remaining 5 slots from shields and armour. This totals up to a grand total of 62,249 items that you can fit into your inventory at one time.
Note: I'm not adding a shulker box into the off-hand because you are playing minecraft bedrock edition, but if you do want to add a shulker box into your off-hand, you can get up to another 1729 including the extra shulker box. This leads to another grand total of 63,978 items.
Comment: I edited a typo in one of the numbers and a few of the sums which were off by 2, hope you don't mind. You could mention that 1728 = 27 slots times 64 items (1 stack), it might not be obvious for everybody
Comment: Would you count holding an ender chest in your inventory, full of shulker boxes full of iron blocks, that can then be crafted into 81 times more nuggets than blocks? You can hold 139968 iron nuggets in a single shulker box if you use the crafting trick. multiply that by 27, and you get 3779136 items in an ender chest.
Comment: An interesting addendum to your amazing answer: You can put shulker boxes in the 2x2 crafting grid, giving another 6912 items, but I doubt that counts as it is not in the inventory.
Answer: I'm coming from Java edition, but I'm pretty sure the game locks you out from doing just that, but assuming you could, I'm also curious how much you could fit before running out of memory.
Afaik Minecraft stores NBT data as a dictionary, and since you're using Bedrock, that's using C++ which is a good bit more memory efficient.
Let's say the average length of an item name is roughly 10 characters, and every item name is preceded by "minecraft:". This means that the name field is, as a very rough estimate, 21 bytes plus 8 bytes for the pointer on the stack. Then I'm pretty sure the Count is stored as a Short, which is 2 bytes, meaning 1 stack of items with no tags or other NBT is about 31 ish bytes. C++ probably also has to store a pointer for each item so that totals as 39 bytes. If you're looking for a quick answer then this multiplied out would be about 25.6 million items able to be stored in a Gigabyte.
I'm not saying this is anywhere close to the answer but that's probably within an order of magnitude
Comment: There's also damage values, NBT data, lots of other random entries, etc. for each item


SharePoint development with visual studio

Question: i have installed SharePoint 2007 and VS2008 extension on my computer.i have created a site but i don't find source code for solution file.how i development for my SharePoint site on visual studio?
Answer: Marcus' article is a good one. However I cannot recommend using Visual Studio Extensions for WSS (VSeWSS) due to the bugs and problems involved.
A good alternative is WSPBuilder. This community project integrates into Visual Studio makes it simple and easy to see exactly what you are doing and gives you control over how your code is deployed into SharePoint.
Also see the Getting started developing for SharePoint question for links to more resources to help get you started. I strongly recommend getting a good grounding in how SharePoint works first. It takes more effort initially but will save you a lot of time and confusion later. A good book to start with is Inside Windows SharePoint Services 3.0.
Answer: Try the following link, it's a beginners guide of how to develop for SharePoint using VS2008:
Beginning SharePoint Web Part Development
This article should cover all you need to get along. Good luck! :)


Why are the leaves on my annuals turning purple?

Question: I've got a bunch of annuals and they aren't doing very well. The leaves and flowers are turning brown and purple. Any ideas?
Comment: Can you give some information on how much light they are getting (under a tree, half day, by an east wall, etc.), How much you're watering (or it's been raining), What texture the soil is, or any other relevant info you think of? This makes it much easier to make a correct diagnosis. Thanks!
Answer: This looks to me like nutrient deficiency; Phosphorus in particular plus a few of the micronutrients. If you've never used compost (decomposed organic matter) now would be a good time. In addition, try fish emulsion for fertilizer.  It is a slower release, organic product 5-1-1. Makes plants hardier, more vigorous, more healthier.  Then in a week or two give them Osmocote 14-14-14.  They won't ever need more (the annuals in your picture, I mean).
Also, when growing annuals, the best thing you can do is cut the flowers off.  Annuals have but one job in life and that is to make babies, seed.  Once they set seed, it is time to die.  The more brutal you are with cutting flowers off, the more vigorous, bud-producing, large, healthy the plant becomes.
I used to do entrance color install and maintenance for homeowner associations.  I would cut off every! flower.  I had lots of horrified homeowners perplexed.  I told them to come back next week, same time.  They did and were blown away by the profusion of flowers.
I see seeds on your plants, so part of the problem could be your plants have done their job and they are dying quietly.  The fish fertilizer might help, definitely the compost will help.  (Evil grin) Cut flowers off!  Cut all those seed heads off! All that energy goes back into the plant to use to make itself bigger, stronger and more likely to produce lots of potential seeds. I hope this helps extend your plants' lives.
Comment: Phosphorus deficiency symptoms can also be a sign of possibly too much water or that the soil is too cold. Both of those, especially after the plant has had a good warm spell can lead it to think it's getting time to drop seeds. Think of how the plant tells that fall is coming on. When that happens, quite a few plants show similar signs to this.
Comment: A lot of good points except for "Science doesn't know why but this stuff does more than add nutrient".  If you can't find a reference for it then it is just an opinion. Can you rephrase this or provide a reference?
Comment: Nope. Sometimes that is the truth.  Science does studies, finds results but can't say why.  There will always be mysteries.  Sigh.  I'll try to find something more on this.
Comment: I agree with kevinsky. Please rephrase that statement, or add references.
Comment: So I cut most of the flowers off, added compost, and applied fish fertilizer. They are looking a bit better. I'm guessing they are getting too much water.


Check the exact location of each other any moment of time

Question: we have two android items one is nexus 5 phone with 3G connection with my younger brother and other one is nexus7 tablet with wi-fi only (with me) and both are in different cities.We both are connected to internet through our device .can I check each others exact location any moment of time? if we can then which App we need to use in india?
Answer: From Google+ help

Google+ allows you to share your current location with the people and
  circles you choose across Google products such as Google+ and Google
  Now. Once you've enabled Location Sharing, Google shares your best
  available location.
You can share your pinpoint or city-level location with others, and
  you can do so on a circle-by-circle basis. For example: you can share
  your pinpoint location with family, and your city-level location with
  your co-workers. You can also change who can see your location or
  completely turn off Location Sharing at any time.
To use Location Sharing, you need to:
Turn Location Reporting on from a compatible device [Android] [iOS].
Enable Location Sharing and choose who can see your location.

Turning on Location Reporting alone will not automatically share your
  location; both of these steps need to be taken for Location Sharing to
  work. Manage Location Sharing Turn Location Sharing on or off on
  Android
Open the Google+ application.
From Home, touch Menu > Settings.
Touch the account for which you'd like to turn Location Sharing on or off.
Select Location sharing.
Touch the switch next to Location Sharing to turn it On or Off. Turning off Location Sharing turns off sharing for both your pinpoint

location and city-level location.
      To share your precise location, touch Choose people to share with on the "Pinpoint Location" card.
      To share your current city location, touch Choose people to share with on the "City Location" card.
      Select OK.
If you don't edit your "Pinpoint Location" or "City Location" sharing
  settings, by default, your location is shared with "Only you."
  Where does my location appear?
People you choose to share location information with will be able to
  see your current location on Google products. For example, on your
  Google+ profile, it's displayed below your name and profile photo on
  Desktop, Android, and iOS devices. On Android and iOS, they'll be able
  to see your location on a map in the Locations destination in the
  Google+ app.
Answer: I think what you are after is Google's device manager app (search in play store). It lets you see the location of other Google devices registered to your account. It's primary aim is for device security. E.g. location if stolen or lost, remote wipe, make the other device ring.
If the two devices belong to different people, you can have more than one Google account set up on your android device (via Settings -> Accounts). Just get the other person to add their account to your device (but turn off the data syncing options unless you want all their email, contacts, etc) if you want to view their location in device manager.
My wife and I use it all the time to see where the other is, e.g. find out when to expect them home, or where I am while out cycling for safety. Also viewable in a web browser from a pc.
Comment: This seems like an awfully kludgy way to do things, especially considering you have to give someone else your account credentials (and, obviously, access to all the information in your account). What's the benefit of doing this when Google+ has it built-in, and there are [other apps](http://www.cnet.com/news/six-ways-to-share-your-exact-location-with-family-and-why/) for it as well?
Comment: I think g+ requires you to manually check in or share your location. Device manager allows another device to query your location without any input required on your end.
Comment: Once you enable location sharing it updates it automatically. It's essentially a replacement for Latitude. Other third-party apps work similarly.


Is there any way to receive email notifications when someone accepts/merges my pull requests?

Question: Is there any way to receive email notifications when someone accepts/merges my pull requests?
Answer: Github doesn't provide for such an option yet in their accounts/notifications page.

perhaps you can contact the repo admin to setup a service hook
Answer: This does work in the meantime:

We have just added notifications when a Pull Request is merged and when an Issue/Pull Request is closed or reopened.

Source: Github blog


Does it matter whether Catwoman chooses to save Batman or not?

Question: I'm at what I assume is the end of the 3rd Catwoman mission. I've beaten down the guards, I have two suitcases full of loot, and I'm apparently having a moral dilemma regarding whether I should leave Batman buried under a pile of rubble or flee Arkham City with my ill-gotten gains.
Though she apparently doesn't have any affect on the main story, I'm not sure which I should choose in order to get the most out of the DLC.
Are the two options merely an illusion of choice and end up with you taking the same path, regardless of which you choose? More importantly, are there any achievements associated with either choice?
Answer: Actually, since we are talking spoiler, check this video to see what happens:

Basically everything is ruined, Joker becomes an immortal and takes the city. The game ends, with Barbara Gordon crying out loud over a radio line.
But, if you watch it to the end, the game will rewind to the point of the decision, and let you choose again. :)
Comment: I ended up choosing it just to see what would happen! I can see why it would seem kind of out of the blue for her to appear if you don't have the DLC though. Also, if you just link bare youtube links, they'll auto-embed. I took the liberty of editing the link so it would show here. :)
Comment: what happens in the end of the good decision
Comment: @kenneth, the game continues normally, until you finish it.
Answer: Well, when I had to choose, I chose to leave Arkham City, but it re-winded back to when I had to chose. So, I had to chose to save Batman. Then, from there, you're Batman again and you have to stop Protocol 10


What is the keyboard shortcut to go to inbox in outlook mac 2011

Question: What is the keyboard shortcut to go to Inbox folder in Outlook 2011 on Mac?
Or if the current focus is on some email in the email list how to switch focus to side panel which lists folders? And vice-a-versa.
Answer: Tab key?
did you try the Tab key - maybe together with Shift, Alt or Command?
I don't have Outlook:Mac, but in MS Windows you can use Tab (or Shift-Tab backwards) to cycle through the window panes.

As this is not always very easy and quite limited, I'd suggest a more general and flexible approach to use MS Outlook mostly with keyboard shortcuts.
Maybe not the exact answer to your question, but it finally lets you do even more than you asked:
VBA macros and Special menu
Similar to you, I was missing shortcuts in Outlook 2007 on MS Windows which let me change the current folder/view without using the mouse and finally chose the following solution, which works very well for me:

write some short VBA macros which activate the views/folders, I'm using most often
create my own Menu in the Menu bar (I called it "Special")
Added those VBA macros to this menu and
defined a short key for every one of them (by typing & before the corresponding letter in the name of the menu entry)

With Alt+s I activate my "Special" Menu and then by typing the short key for an entry, I can easily activate the macro and let Outlook switch its perspective.
I'm currently not on the PC where I made this setup, so I can add more details next week, if you're interested.
EDIT: it seems that Outlook 2011 for Mac does not support VBA (see http://answers.microsoft.com/en-us/mac/forum/macoffice2011-macoutlook/does-mac-outlook-2011-support-vba-macro-programs/c6515223-1431-424f-851d-b7f5c6dc7df8). Alternatively, you could create Apple Scripts. I am not sure how to integrate them into the MS Outlook GUI - maybe with the Apple Script menu. However, Quicksilver (see below) could let you define keyboard shortcuts to launch those scripts from within MS Outlook (or even from within another application, if you wish).
Use Quicksilver or a similar tool
On MacOS X there is also the excellent (and free and open source!) software Quicksilver (http://qsapp.com/) which lets you define your own shortcuts, they are called Triggers (tutorial: http://vjarmy.com/archives/2006/01/quicksilver_gold_trigger.php )
Comment: I will give quicksilver a try. On microsoft forum it is mentioned that vba macros are not supported for outlook mac 2011.
Comment: thanks for your feedback! See my edit (maybe AppleScript+Quicksilver) is an alternative for you. If that works for you, you can "accept" my answer by clicking on the check mark below the up/downvote arrows. Learning and using `Quicksilver` is really helpful if you want to get rid of unnecessary mouse-usage.
Answer: As suggested by Martin I played some more with using TAB.
1) If focus is on list of emails then shift + TAB takes focus to side panel
2) Pressing option + up arrow selects the top most folder in the side panel which happens to be Inbox folder.

So in two steps I am able to select the inbox folder without using mouse.
Comment: I think this is as close as it gets.  I'd still like to be able to move to a different folder without relying on the side panel.  My goal is to completely hide the side panel to maximize the footprint of message list and reading pane.
Answer: There is no built-in keyboard shortcut, so you have to create an Applescript and then assign a shortcut using the Keyboard control panel.
Step 1
Create the Applescript. Here's the code I used. Feel free to copy and paste.
on run {}
tell application "Microsoft Outlook"
    tell the first main window
        set view to mail view
    end tell
    set selected folder to inbox
end tell
end run

Step 2
Create your shortcut. Rather than re-create the wheel and explain how to create the shortcut, I'll link you to the page where I learned how to. Go here.
Answer: Cmd-Sh-[ will move the focus to the sidebar pane with inbox and other folders, from there you can use up/down arrows to navigate.
My overall Outlook Mac 2011 shortcut buffet has:
cmd-1       mail view
cmd-2       calendar view
cmd-arrow   forward or back in calendar view
ctl-cmd-2   work week view in calendar
cmd-sh-[    previous pane
cmd-sh-]    next pane
cmd-opt-n   new message (any view)
cmd-sh-m    move mail item to folder
cmd-sh-j    mark as junk
ctl-1       flag follow up today
ctl-0       flag complete
cmd-opt-'   flag clear
cmd-f       find in mail

In the  mail sidebar I use several custom folders in the server inbox and a local archive folder, as well as a few folders added to Favorites which can be helpful.


Removing light switch from 3-way setup

Question: I'm definitely lacking in electrical skills, so please help me out on the right procedure here. I have two switches controlling one light. I'd like to REMOVE SWITCH A and completely control with SWITCH B. I've attached the switch wiring as pictures below, the red wire on switch B is one wire shared with another, unchanged switch in that box:
Answer: Turn off all power before starting any work. OK switch B can remain in place but you have to determine which of the two orange travelers is hot when switch B is in the top (ON) position, then disconnect and tape the other orange wire. Remove switch A and wire nut the yellow wire to the hot orange that you did not disconnect from the other switch. Tape the other orange. You'll need to get the correct cover plate for switch A to cover up the unused gang.
It's always a good idea to put a note in each of the two boxes you worked in describing what you did for future reference. It might just help you, or a new owner, in the future.
Answer: If all you want to do is disconnect the switch A and put a blank in it's place then remove the wires from the switch and connect the yellow wire to either orange, and cap off the extra orange. If that makes switch B upside down then swap the oranges at switch A location.
If you intend on eliminating the junction box for switch B or eliminating the conduit between the two then better pictures are needed of wires entering conduit are needed to see if that is possible.

Removing light switch from 3-way setup


On a strictly cash basis

Question: Can anyone explain why we say on a strictly cash basis (and not strict)?  Is cash here an adjective?  Is cash basis a noun phrase or compound noun?
Answer: Cash basis is a compound noun; it is frequently written as 'cash-basis'. The word preceding it ('strictly') is modifying a noun, so it is an adjective.  The adjectival form of of this word is 'strict'. So, we should say:

... on a strict cash basis.

Strictly is an adverb, which can modify a verb, an adjective, or another adverb; but an adverb cannot modify a noun. If you really want to use 'strictly' instead of 'strict', then you need to use a sentence such as:

We strictly deal on a cash basis.

In this case, 'strictly' modifies the verb 'deal'. Ultimately, both sentences mean essentially the same thing, but you should not merge them by allowing 'strictly' to modify the compound noun.
However, it is not uncommon, although grammatically incorrect, to hear sentences such as, "We deal on a strictly cash basis', spoken in English speaking countries. One day, given enough time, this incorrect phrase may garner enough acceptance that it will be regarded as grammatically correct, and yet another grammar 'rule' will have been modified.
Comment: If it is not uncommon to hear "on a strictly cash basis" spoken by native speakers of the language, then it can be argued that the rule has already *been* modified.
Comment: @James K  '... it can be argued that the rule has already been modified'  Yes, up to a point I believe that such an argument can be made. However, I do not believe that enough people have adopted this modification to make it grammatically acceptable. In saying this, I accept that official grammar rules often lag behind the way people usually speak.
Comment: cash basis is not a compound. Cash functions like an adjective.

On a strictly cash basis