Channel bonding modes

Question: I wanted to know where is the documentation of channel bonding implementation in Ubuntu because the bonding modes are described differently on the internet
I have Ubuntu Server 10.0.4 LTS and it's using 2.6 kernel
Thank you
Answer: The Ubuntu documentation related to bonding modes is available here.
Descriptions of Ethernet bonding modes:
Mode 0
balance-rr

Round-robin policy: Transmit packets in sequential order from the first available slave through the last. This mode provides load balancing and fault tolerance.
Mode 1
active-backup

Active-backup policy: Only one slave in the bond is active. A different slave becomes active if, and only if, the active slave fails. The bond's MAC address is externally visible on only one port (network adapter) to avoid confusing the switch. This mode provides fault tolerance. The primary option affects the behavior of this mode.
Mode 2
balance-xor

XOR policy: Transmit based on [(source MAC address XOR'd with destination MAC address) modulo slave count]. This selects the same slave for each destination MAC address. This mode provides load balancing and fault tolerance.
Mode 3
broadcast

Broadcast policy: transmits everything on all slave interfaces. This mode provides fault tolerance.
Mode 4
802.3ad

IEEE 802.3ad Dynamic link aggregation. Creates aggregation groups that share the same speed and duplex settings. Utilizes all slaves in the active aggregator according to the 802.3ad specification.
Prerequisites:

Ethtool support in the base drivers for retrieving the speed and duplex of each slave.
A switch that supports IEEE 802.3ad Dynamic link aggregation. Most switches will require some type of configuration to enable 802.3ad mode.

Mode 5
balance-tlb

Adaptive transmit load balancing: channel bonding that does not require any special switch support. The outgoing traffic is distributed according to the current load (computed relative to the speed) on each slave. Incoming traffic is received by the current slave. If the receiving slave fails, another slave takes over the MAC address of the failed receiving slave.
Prerequisites:

Ethtool support in the base drivers for retrieving the speed of each slave.

Mode 6
balance-alb

Adaptive load balancing: includes balance-tlb plus receive load balancing (rlb) for IPV4 traffic, and does not require any special switch support. The receive load balancing is achieved by ARP negotiation. The bonding driver intercepts the ARP Replies sent by the local system on their way out and overwrites the source hardware address with the unique hardware address of one of the slaves in the bond such that different peers use different hardware addresses for the server.
Comment: Thank you I already found this but it's not enough. I need something saying how each mode work in a much more specific way.
Answer: I think that the best source is this Kernel.org document, it also provides explanations about when is each type optimal, and so on.
The sections are long, I only paste short excerpts:

11.2.1 HA Bonding Mode Selection for Multiple Switch Topology
In a topology such as the example above, the active-backup and
  broadcast modes are the only useful bonding modes when optimizing for
  availability; the other modes require all links to terminate on the
  same peer for them to behave rationally.
active-backup: This is generally the preferred mode, particularly if
    the switches have an ISL and play together well.  If the    network
  configuration is such that one switch is specifically     a backup switch
  (e.g., has lower capacity, higher cost, etc),     then the primary option
  can be used to insure that the    preferred link is always used when it
  is available.
broadcast: This mode is really a special purpose mode, and is suitable
    only for very specific needs.  For example, if the two  switches are
  not connected (no ISL), and the networks beyond   them are totally
  independent.  In this case, if it is  necessary for some specific
  one-way traffic to reach both     independent networks, then the
  broadcast mode may be suitable.
...
12.1.1 MT Bonding Mode Selection for Single Switch Topology
This configuration is the easiest to set up and to understand,
  although you will have to decide which bonding mode best suits your
  needs.  The trade offs for each mode are detailed below:
balance-rr: This mode is the only mode that will permit a single
    TCP/IP connection to stripe traffic across multiple     interfaces. It
  is therefore the only mode that will allow a  single TCP/IP stream to
  utilize more than one interface's     worth of throughput.  This comes at
  a cost, however: the  striping generally results in peer systems
  receiving packets out     of order, causing TCP/IP's congestion control
  system to kick    in, often by retransmitting segments.
...
active-backup: There is not much advantage in this network topology to
    the active-backup mode, as the inactive backup devices are all
    connected to the same peer as the primary.  In this case, a     load
  balancing mode (with link monitoring) will provide the    same level of
  network availability, but with increased  available bandwidth.  On the
  plus side, active-backup mode     does not require any configuration of
  the switch, so it may     have value if the hardware available does not
  support any of    the load balance modes.
balance-xor: This mode will limit traffic such that packets destined
    for specific peers will always be sent over the same    interface.
  Since the destination is determined by the MAC    addresses involved,
  this mode works best in a "local" network     configuration (as described
  above), with destinations all on  the same local network.  This mode
  is likely to be suboptimal    if all your traffic is passed through a
  single router (i.e., a    "gatewayed" network configuration, as
  described above).
As with balance-rr, the switch ports need to be configured for
    "etherchannel" or "trunking."
broadcast: Like active-backup, there is not much advantage to this
    mode in this type of network topology.
802.3ad: This mode can be a good choice for this type of network  topology.  The 802.3ad mode is an IEEE standard, so all peers   that
  implement 802.3ad should interoperate well.  The 802.3ad  protocol
  includes automatic configuration of the aggregates,   so minimal manual
  configuration of the switch is needed     (typically only to designate
  that some set of devices is   available for 802.3ad).  The 802.3ad
  standard also mandates    that frames be delivered in order (within
  certain limits), so   in general single connections will not see
  misordering of    packets.  The 802.3ad mode does have some drawbacks:
  the   standard mandates that all devices in the aggregate operate at
    the same speed and duplex.  Also, as with all bonding load  balance
  modes other than balance-rr, no single connection will    be able to
  utilize more than a single interface's worth of   bandwidth.
...
balance-tlb: The balance-tlb mode balances outgoing traffic by peer.
    Since the balancing is done according to MAC address, in a
    "gatewayed" configuration (as described above), this mode will  send
  all traffic across a single device.  However, in a    "local" network
  configuration, this mode balances multiple    local network peers across
  devices in a vaguely intelligent  manner (not a simple XOR as in
  balance-xor or 802.3ad mode),     so that mathematically unlucky MAC
  addresses (i.e., ones that    XOR to the same value) will not all "bunch
  up" on a single   interface.
...
balance-alb: This mode is everything that balance-tlb is, and more.
    It has all of the features (and restrictions) of balance-tlb,   and
  will also balance incoming traffic from local network     peers (as
  described in the Bonding Module Options section,  above).
The only additional down side to this mode is that the network
    device driver must support changing the hardware address while  the
  device is open.

Channel bonding modes


Abaqus *ORIENTATION

Question: What do you actually type into the *ORIENTATION entry in an input file?
Is it a rotation matrix w.r.t. the Global axes?
Answer: The meaning of the *ORIENTATION keyword in Abaqus can be checked in the "Abaqus Keyword Reference Manual", which unfortunately is not made publicly available by Simulia.
As most Abaqus keywords, *ORIENTATION has a lot of optional parameters, which change its meaning and syntax. In most cases however the lines following this keyword are the $X, Y, Z$ coordinates of three points $a,b,c$: $c$ (optional if coincident with the global axes origin,i.e. $(0,0,0)$) is the origin, $a$ lies on the $x$ axis, $b$ is a point on the $xy$ plane.

Abaqus *ORIENTATION


How to determine XML type in Powershell?

Question:
I'm writing a script that is supposed to look at the content of a file and determine if it is a (well formed) XML or not. I found a page on [ss64.com][1] that this is quite easy to do:

>32 -is [int]
True

The thing is however that I can only test this by casting the left-side for XML files:
>[xml](Get-Content c:\Path\To\xml_file.xml) -is [xml]
False

...which in this case would be rather pointless: if the file is XML, the casting will already prove this, else throw an exception. I therefore wonder: is there any way to determine XML files in Powershell in a True-False way?
Answer: Try the -as operator:
[bool]((Get-Content c:\Path\To\xml_file.xml) -as [xml])

Answer: function Is-Valid-XML
{
    param ([string] $path)

    $xml = New-Object System.Xml.XmlDocument
    try
    {
        $xml.Load($path)
        $valid = $true
    }

    catch
    {
        $valid = $false
    }

    return $valid
}


Is it necessary/advisable to uninstall QGIS 1.8 first before installing 2.0?

Question: Who has installed QGIS 2.0 on Mac OSX 10.6.8 (Snow Leopard)? Is it necessary/advisable to uninstall QGIS 1.8 first? Have you seen any problems importing QGIS 1.8 projects to 2.0?
Answer: If you install the 2. version It will overwrite the 1.8 version (QGIS.app).
But,I use everyday versions 1.8, 2. and master on Mac OSX 10.6.8 without problem (often at the same time).
So, the only thing you need to do is to rename the version 1.8 in QGIS1.8.app , otherwise, it will be deleted (QGIS.app)
Comment: Great - it works. There are a lot of problems with 2.01; I'm sticking with 1.8 excepting one map where I need to increase marker line thickness.


Remounting MySQL under a new mount point

Question: /var/lib/mysql is mounted on partition / which is now full. Is there a way to mount this under /home/mysql for more disk space?
Answer:
Stop MySQL
Move MySQL data from /var/lib/mysql to /home/mysql
Point MySQL to the new data directory
Start MySQL

It could be implemented like this (WARNING: not tested)
/etc/init.d/mysql stop
mv /var/lib/mysql /home/.
ln -s /home/mysql /var/lib/mysql
/etc/init.d/mysql start

Always have a backup before you start moving the data directory.
Comment: Running this now! :D
Answer: Ideally, you'd give it it's own mountpoint on it's own partition, which can be mounted to /var/lib/mysql.
Failing that, you do have the option of using a loopback filesystem, though it definitely does introduce some I/O delays that MySQL may notice. Only you can know if increasing I/O latencies a bit will harm your expected performance.
The rough procedure....

Use dd to create a loopback file somewhere.

dd if=/dev/zero of=/home/mysql/datavol bs=1M count=4096

Attach the file to a loopback device

losetup /dev/loop0 /home/mysql/datavol

Format the device as you will.

mkfs.ext3 /dev/loop0

Mount it somewhere

mount /dev/loop0 /mnt

Copy your data

cp -a /var/lib/mysql/ /mnt/
Other methods may be used, this is just one.

Clear /var/lib/mysql
Remount to /var/lib/mysql

umount /mnt ; mount /dev/loop0 /var/lib/mysql

Ensure that directory is owned by the right user

chown mysql:mysql /var/lib/mysql

At that point, you should have a larger /var/lib/mysql actually hosted from your larger partition.


Can plantlets on Kalanchoe be removed?

Question: I'm wondering if the plantlets on kalanchoe can be removed occasionally without affecting the plant in some anyway. To my knowledge, these structures on the leaves are basically a means of natural propagation. So if they don't serve any other function, can the plantlets occasionally be removed so that resources for the plant are directed more to plant growth rather than plantlet formation?
Answer: That is the goal of those plantlets: With time, they detach easily and fall-off. Best time to harvest them is when they mature. As an indication, even the lightest touch will dislodge them. The way the leaves face upwards increases the chances the plantlet will land with its base facing the ground - sort of parachute - and that's where it starts to root.
Comment: I appreciate your contribution. What I meant to say, though, is is it okay if I remove the plantlets? The plant is quite young and keeps shedding them off. So many of the plantlets are taking root at the base of the mother plant and I don't need that at the moment.
Comment: If you want just to take them away, that's fine.
Comment: Okay thank you.

Can plantlets on Kalanchoe be removed?


Have Ghostscript 9.06 rasterizing EPS files with same colors as of 8.70

Question: For an online design application I'm using Ghostscript to convert some kind of files into PNG and/or PDF (depending on user request).
Consider this EPS file:
http://designer.realtimedesigner.com/images/1/cliparts/99010.eps
With GS 8.70 I used to convert it to PNG this way:
<code>gs -dTextAlphaBits=4 -dGraphicsAlphaBits=4 -sDEVICE=pngalpha -dEPSCrop -sOutputFile=./result.png -r458 ./99010.eps
</code>
I noticed 8.70 has some bugs when the resolution is higher, which may happen in my application. So I gave GS 9.06 (latest version) a try.
However, by using the very same command, the output is different.
With 8.70 the main rasterized color is a pure RGB black (0,0,0), while with 9.06 it's 35,31,32.
This is a problem for me, because for some specifics of my application I need the RGB values to be exactly the same for further processing.
I cannot understand while there's this difference between the versions, perhaps I'm missing some option that could be set in 9.06?
My goal, at the end, is to rasterize EPS files to PNG via GS 9.06 with colors converted to same RGB values as of 8.70.
Thanks in advance for any help/suggestion you may provide.
Answer: 9.06 uses a colour management engine (in fact the whole 9.x series does). To control this you need to supply ICC profiles which determine the mapping between the input and the output, there is more information in the documentation.
In the short term you could use -dUseFastColor, but if you care about colour reproduction, you should investigate the colour management more thoroughly.
Comment: This is interesting. Will explore this solution very soon and will post again when I'll have some result in case everyone else will be looking to solve the same problem


GPRS on iPhone Simulator

Question: I want to activate/enable GPRS on the iPhone programmatically. What are the APIs I can look into? There is no "Network" option in the settings application on the simulator so do I need to test out the application on the device itself?
Answer: Using the SDK, you can't "activate" any sort of network access. Simply try to connect to the remote server.
If the device can connect, it will. If it cannot, you need to detect this and display a warning to the user.
Apple's sample code has plenty of examples on how to detect if a network is active.
Answer: The Simulator always connects through the host Mac's internet connection. If you are asking how to test GPRS performance on the simulator then you should buy a USB GPRS radio and test rhrough that. You could also use the built in ipfw to throttle network bandwidth to the simulator and simulate GPRS bandwidth (note that latency would still be that of your hows connection). Throttled provides an easier UI than bare ipfw: http://www.intrarts.com/throttled.html
As August says, you can't switch the network on and off from code however, you can test the connection and suggest the user goes and switches on the connection.
There is good sample code from Apple for testing the connection availability and type: http://developer.apple.com/iphone/library/samplecode/Reachability/index.html

GPRS on iPhone Simulator


Bethe approximation for factor graphs

Question: I am confused at computing Bethe approximation for factor graphs in here. It generalizes Bethe approxmiation in a pairwise case. However, I am wondering why (75) goes to (78) with (76):

We can verify that
$$\hat b_i(x_i)\propto \exp\left[(d_i-1)^{-1}\sum_{a\in N_i}\lambda_{ai}(x_i)\right]$$
$$\begin{align}\hat b_i(x_i)&\propto \exp\left[(d_i-1)^{-1}\sum_{a\in N_i}\lambda_{ai}(x_i)\right]\\
&=\prod_{a\in N(i)}\exp\left[(d_i-1)^{-1}\lambda_{ai}(x_i)\right]
\end{align}$$
Since (76) involves $m_{\text{fact}\rightarrow\text{singleton}}$, it's tempting to use the second equation in (76), and thus we have
$$\begin{align}\hat b_i(x_i)&\propto \prod_{a\in N_i}\exp\left[\log \prod_{c\in N_i\backslash a}m_{c\rightarrow i}(x_i)^{\frac1{d_i-1}}\right]
\end{align}
$$
Then everything is messed up. I do not see why after canceling $\exp$ and $\log$ we obtain $m_{a\rightarrow i}(x_i)$.
There are several references where $m_{a\rightarrow i}$ all comes out mysteriously. I am looking for help in this derivation.
Other refs:

p. 28 here:

p. 24 here:
Answer: after some efforts with a friend of mine and this book chapter,
I think I kinda worked it out.

Bethe approximation for factor graphs


applying the mean value theorem to inequalities

Question: So I'm ask to prove using the mean value theorem that $\sin(x) \ge x-\frac{x^3}{6}$
I understand that the mean value theorem works because both sides of the equation equal zero when $x=0$.
To start I set $f(x)=\sin(x)$ and $g(x)=x-\frac{x^3}{6}$ and let $F(x)=f(x)-g(x)$ and try to show that $F'(x)>0$ for all $x>0$.
Therefore, we have $F(x)=\frac{F(x)-F(0)}{x-0}(x-0)$
By the mean value theorem, we know there exists a $y>0$ such that $F'(y)x=(\cos(x)-1+\frac{x^2}{2})x$
If we apply MVT again for some $z>0$, we get $F'(z)x=\frac{F'(y)-0}{x}x=(-\sin(x)+2x)x$
It seems pretty obvious that $F'(z)x>0$ but just for safe measure I'll apply it again: There exists a $c>0$ such that $F'(c)x=(-\cos(x)+2)x$
$\therefore$ since $x>0$ and $2-\cos(x)\le 1$ (by the bounds of cosine), $F'(c)>0$ for $c>0$
$ \therefore F(x)>0\equiv f(x)>g(x)$ if $x>0  $
I have my final tomorrow so any tips to make this more concrete would be appreciated
Answer: You have the right idea. Show in succession, using MVT, that for $x>0$,
$x>\sin x$, $\cos x>1-x^2/2$, $\sin x> x-x^3/6$.


Test of equality between coefficients of different IVs across two different models

Question: Say I have a categorical variable concerning the voting intention for ten parties, and I want to run two separate (logit) models for the voting intention of two parties in particular, Party A and Party B. I've thus created two different DVs: a dummy for A (Vote for A=1, All others=0) and a similar dummy for B. These are my two very simple models (dataset is the same for both models):
Model 1= logit VotePartyA EconomyIndex Sociallndex Ideology age i.gender i.education i.job i.area
Model 2= logit VotePartyB EconomyIndex Sociallndex Ideology age i.gender i.education i.job i.area
My hypothesis concerns the first two IVs, both continuous (and standardized). Party A is economically-oriented, whereas Party B is socially-oriented. Their preferred policy Index is positive and significant, but I also expect that the effect of the EconomyIndex on the vote for Party A will be greater than it will be the effect of the SocialIndex on the vote for Party B. To test this hypothesis  I've run a test of equality between the coefficient of EconomicIndex for Party A (in Model 1) and the coefficient of SocialIndex for Party B (in Model 2), i.e. on Stata, after using suest, I've run:
test [Model1_PartyA]EconomyIndex=[Model2_PartyB]SocialIndex
My question is very trivial: is it correct to validate my hypothesis using a test of equality of coefficients of two different variables across two models that have the same independent variables but - clearly - differ on their DVs?
I've found a lot of examples for comparing the coefficients of the same variable across different models, or comparing different variables in the same model, but I am struggling to find something similar... So I was wondering if I messed something up, which probably I did.
Many thanks in advance!
Answer: I think the index function coefficients from the logits themselves are not comparable, it would be much better to compare the average marginal effects instead. The scales are not the same across the two models, so you need to go back to probabilities to compare.


google talk android device sign is not shown by sony xperia ray while logged in google talk in it what can I do about it?

Question: When I log in to google talk in my sony xperia ray (android 4.04 ICS) and see by any other account how I am online it shows like I am online in a personal computer somewhere. It shows a circle green or red. No android sign is there. How can I fix that? I want people to know I have android mobile and I am logged in through my mobile not my personal computer so that it signifies I am travelling.
Answer:
Launch the Google Talk app
Open the overflow menu (three vertical dots) and choose Settings.
Tap the account you're using.
Check the Mobile indicator option.

Note: If you're signed in to your Google account on a PC (like if you left your browser open and logged in to Gmail) the mobile indicator may not be displayed.
(For reference: Android version 4.1.1, Google Talk v. 4.1.1 - 424425)
Comment: I do not know which is the overflow menu. By that you must mean the menu where all the apps shortcuts are present. Yup went there. Menu - > Settings -> Accounts & Sync -> Account I am using -> It has only options to check sync chrome, contacts,drive, gmail (no mobile indicator) :(
Comment: The overflow menu is the three vertical dots.
Comment: You need to do this in the Google Talk app, not in the general device settings. I'll update my answer.
Comment: Yup it was already checked. It shows VideoCamera sign when I log into the Google Talk app and none of my laptops or PCs are on. Is it because I have a front camera for video chat? But your answer seems correct so. (but no android sign)


low disk space on "PBR_image" in ubuntu 14.04

Question: i have recently installed ubuntu 14.04 LTS
when i boot my laptop then a dialog box appears which shows the following output
low disk space on "PBR_image"
the volume "PBR_image" has only 313.4 MB disk space remainig
what should i do now
Answer: You should always avoid PBR at all costs. Canonical only provides official support for Rolling Rock, available in the yourlocal-liquor-store-ppa repository.
As far as your hard drive (I assume) is concerned, it's low on disk space. You need to get a new one (external one is okay if you want to dump some files to it), delete some stuff to make some room on your drive, or migrate to a new, larger hard drive.


What is Epsilon Convergence in Probability?

Question: I understand that the formula for probability of convergence is $P[|X_n − X_\infty| \gt \epsilon ]\to  0$ and I can solve problems using the formula. Can anyone explain it intuitively (like I am a five year old), particularly in regards to what $\epsilon$ is?
Comment: We studied $\varepsilon$ in 9th grade in Analysis class in high school. It was one of the toughest thing to understand for me. I don't think you can explain it normal 5 y.o.
Comment: Not literally to a five you old.... Just as clearly as possible
Comment: I, for one, do not understand this in the context of limits. $P[|X_n − X_\infty| < \epsilon ]\to  0$ would allow one to choose an $\epsilon$ no matter how small such that the difference  $P[|X_n − X_\infty|$ is less than that would converge, but $P[|X_n − X_\infty| > \epsilon ]$ would not. So, somebody please explain.
Answer: Sequences of random variables.
Intuition comes from metaphors.  The following metaphor, which models random quantities by pulling slips of paper out of a container, captures all the essential mathematical elements while glossing over a technical condition ("measurability") needed to make sense of situations with uncountably many tickets.

Consider a tickets-in-a-box model of a sample space $\Omega$: the name of each element $\omega\in\Omega$ is written on a slip of paper (a "ticket") which is put into the box.  Elements with greater probability are named on more tickets.
A random variable $X$ is a consistent way of writing a number on each ticket.  "Consistent" means that all the tickets for any particular $\omega$ all get the same value of $X$, written $X(\omega)$.
A sequence of random variables $X_1, X_2, \ldots, X_n, \ldots$ therefore can be conceived of as a sequence $X_1(\omega), X_2(\omega),\ldots$ written on each ticket (again in a consistent way).
$X_\infty$ is another random variable, which is one more number written on each ticket.
Events and probability.
Let $\epsilon$ be any real number.  We will say more about it below.
The event $|X_n - X_\infty| \ge \epsilon$ describes all the tickets $\omega\in\Omega$ for which the values $X_n(\omega)$ and $X_\infty(\omega)$ differ by $\epsilon$ or more. It's a subset of the tickets in the box.  These tickets form some proportion of the box: that proportion models their probability, $\Pr\left(|X_n - X_\infty| \ge \epsilon\right)$.
Limits.
Every assertion about a limit is a form of mathematical game.  When we write that some sequence has a limit $L$, what we mean is we can play a game against a hypothetical opponent (who is doing their best to make us lose) and we will always win.  In the limit game, your opponent names some positive number--usually a tiny one--which we will call $\delta$.  You win if you can remove a finite number of elements from that sequence and show that all the remaining elements are within a distance $\delta$ of $L$.  As in any game, you may calibrate your response to your opponent's move: the elements you remove are allowed to depend on $\delta$.
Limits in probability.
Let's apply the limit game to the assertion $\Pr\left(|X_n - X_\infty| \ge \epsilon\right)\to 0$.  Because this assertion involves an unspecified quantity $\epsilon$, your opponent may also specify its value.  That makes the game as difficult as possible for you to win.
So, no matter what values of $\epsilon$ and $\delta \gt 0$ your opponent specifies, your response will be to cross out some finite number of the random variables $X_i$ on the tickets.  For every remaining random variable $X_n$, let the tickets where $X_n(\omega)$ differs from $X_\infty(\omega)$ by $\epsilon$ or more be the "bad" ones for $n$.  You win the game provided the proportions of bad tickets are always less than $\delta$ (for all the $X_n$ that remain).
A little thought reveals the subtlety of this game: the bad tickets for $n$ do not have to have any relationship to the bad tickets for $m$ (where $n$ and $m$ designate any of the remaining random variables you didn't cross out).  In other words, on any given ticket the values $X_n(\omega)$ can bounce all over the place.  The limit in probability is a statement about what's written on all the tickets in the box but it is not a statement about what might be written on any individual ticket.
Comment: @mickeyf You're welcome.  I paid attention to the OP's comment at http://stats.stackexchange.com/questions/242793/what-is-epsilon-convergence-in-probability/242812?noredirect=1#comment461789_242793.
Comment: I am five years old. I have no idea what you are talking about. You lost me at "a sample space Ω". You would have lost my less precocious classmates at the first sentence. Thanks for trying.
Answer: Since we're talking about convergence - specifically, in this case, $X_n$ converging to $X_\infty$ - we want to show that $X_n$ gets really, really, really close to $X_\infty$ as $n$ gets larger and larger.
Think of $\varepsilon$ as any really small positive number; say you think $\varepsilon = 0.01$ is good enough. Then in order to show that $X_n$ is really, really, really close to $X_\infty$, we want to show that $X_n$ falls inside $(X_\infty-0.01,X_\infty+0.01)$ for sufficiently large $n$. (Sufficiently large $n$ just means that there is some $n'$ such that for every $n > n'$, $X_n$ is within plus or minus $0.01$ of $X_\infty$ with probability 1.)
But say that I'm not convinced that $X_n$ converges to $X_\infty$ because $\varepsilon=0.01$ just seems too big for me. So instead, let $\varepsilon = 0.0001$. Then I'm convinced that $X_n$ converges to $X_\infty$ (or that $X_n$ is really, really, really close to $X_\infty$) if we can show that, for sufficiently large $n$, $X_n$ falls inside $(X_\infty-0.0001,X_\infty+0.0001)$.
Suppose you have a lot of friends who pick $\varepsilon$ to be smaller and smaller. The idea behind convergence is that for any $\varepsilon > 0$, no matter how small $\varepsilon$ gets, showing that $X_n$ falls inside $X_\infty\pm\varepsilon$ for sufficiently large $n$ demonstrates that $X_n$ converges to $X_\infty$.
In the most basic terms, $\varepsilon$ is just a small positive number. As it relates to convergence, you want to be able to show that for any $\varepsilon > 0$ (so that all of your infinite friends with different $\varepsilon$ values are convinced), the sequence that converges will, at some point, get within plus or minus $\varepsilon$ of the limit to which you believe the sequence converges. If you cannot show that your sequence falls within $\varepsilon$ of the believed limit for some $\varepsilon$, then the sequence cannot converge to that limit.
Comment: That's generally the case, Matt, but not always true. As a trivial example, let's say your sequence is $\{2,2,2,...\}$ and you want to show that this converges to 2. No matter how small your $\varepsilon$ is, $n=1$ will suffice to prove that it converges.

However, it's important to point out that, in order to prove something converges within this context, you must be able to show this for _*all*_ $\varepsilon>0$, no matter how small. It is not sufficient to pick one $\varepsilon$ and say it converges. For example, consider the sequence given by $X_n=sin(n)$ and let $\varepsilon = 10$.
Comment: Could you say the smaller $\epsilon$ , would require a bigger $n$ to prove it converges?


2005 suzuki gsxr 600 won't start

Question: My bike simply won't start after I replaced the side stand relay, the economy, check all the fuses. It turns over but doesn't start. All my relays work. My fuel pump works. But when I turn the key, the fuel pump does not kick in.
My tachometer does not reset, and the display says CHECK. I tried the dealership thing, still NO CODE displayed.
Answer: Check you have fitted the new relay correctly. Check that you did not disturb anything else doing the work. Check the supply and fuses. Check the stand switch for function and adjustment - maybe the relay was not the real or only problem...

2005 suzuki gsxr 600 won't start


What are the depicted measurement devices called in English?

Question: Please check the image. What is it called in English?
Is it as 'a measurement indicator'?
How about this one?
Comment: By the way, if you are trying to find out what something in a picture is called in English, you can use Google.  Go to http://images.google.com/ and drag the image into the search box, and Google will give you its best guess what it is.
Comment: They're both *tape measures* (or, as @shin says, *measuring tapes*); one's a carpenter's tape measure, the other's a tailor's tape measure.
Comment: It is a measuring tape.
Answer: Measuring tape, or tape measures.
The first image is for carpentry and other construction and surveying, the second one is for tailors or others to measure in order to fit clothes for you.
https://en.wikipedia.org/wiki/Tape_measure
Answer: In practice, the terms "tape measure" and "measuring tape" are not interchangeable (in American English).
The first image shows a "tape measure".  As Nihilist_Frost points out, construction workers use tape measures to measure more-or-less straight distances.
The second image shows a "measuring tape".  As Nihilist_Frost points out, seamstresses use measuring tapes to measure curved or flexible things, like distances along people or cloth.
Where possible, American surveyors tend to use laser levels and laser distance measuring equipment instead of tape measures.  Laser measurements are less sensitive to temperature, and levels can prevent errors due to slopes.


Can I omit 'to' after 'and'?

Question: For example

I want to go and leave you.
I want to go and to leave you.

Which one would be accurate?
Comment: A better way would be *"I don't want to be with you anymore"*, if that is what you mean.
Comment: I want to leave you, and go.
Answer: Neither is wrong, certainly not for common use of English. The second sentence is a much less common construction, in my experience as a native speaker in the United States.
The first sentence suggests that both actions are combined in some way:

I want to go, and by going I will leave you.

Or, rephrased:

I want to leave you, which I will do by going.

The second suggests that the speaker wants to do both things, specifically:

I want to go, and I also want to leave you.

There is also a common use of the phrase go and which usually does not convey much information other than that someone chose to do something:

Why did you go and do that?

In those cases go and could be completely omitted and the meaning of the sentence would not really change at all. I don't think that that pattern is in place in the first sentence in the question, but I wanted to distinctly mention it because it's likely that you will come across this specific usage of go and at some point.

Can I omit ‘to’ after ‘and’?


Rails Formtastic Polymorphic Relationships

Question: I'm having issues with Formtastic not displaying information from a related table that has a has_many polymorphic relationship.
There are four models at play: NPC, Player, Ability, and Xref_Ability. But I'm only concerned with two at the moment: Player and Xref_Ability
I've gone against the naming convention for cross-reference tables because xref_ability is tied to NPCs as well as Players. See the model relations for Xref_Ability and Player below:
Player
class Player < ActiveRecord::Base
  attr_accessor :characterable
  ...
  has_many :xref_abilities, :as => :characterable
  #has_many :xref_abilities, :through => :characterable
  accepts_nested_attributes_for :xref_abilities
end

Xref_Ability (cross reference table for abilities for NPCs and Players alike)
class XrefAbility < ActiveRecord::Base
  attr_accessible :ability_id, :characterable_id, :characterable_type

  #Validation
  validates :ability_id, :uniqueness => {:scope => [:characterable_id, :characterable_type]}
  #Relations
  belongs_to :ability
  belongs_to :characterable, :polymorphic => true
end

I've added the accepts_nested_attributes_for as many people seemed to have experienced issues without this included. However, once I include that, I couldn't use :xref_abilities in my input for loop; receiving the error: uninitialized constant Player::Xrefability. So I attempted to use :characterable and it returns null:

  <%= semantic_form_for @Player do |f| %>

<%= @Player .name %>

    <%= f.inputs %>
    <%#= f.input :characterable, :collection => XrefAbility.all %>
    <%= f.inputs :for => :characterables do |abil| %>
      <%= abil.input :ability_id %>
    <%  end %>
  <% end %>

You can see I tried collection and that returned a drop down with all records... Not what I was looking for. I also tried doing .find_each instead of .all with conditions, which gave me a no block (yield) error.
I know it has something to do with the polymorphic relationship and how Xref_ability is not exclusive to Player. Nor does it have a direct relationship to Player. If I knew how to do this, then maybe I could solve the issue. I couldn't find anything on the gitHub so suggestions appreciated.
Answer: Try with this modified stuff it should work.

  <%= semantic_form_for @Player do |f| %>

<%= @Player .name %>

    <%#= f.inputs %>
    <%#= f.inputs :characterable, :collection => XrefAbility.all %>
    <%= f.semantic_fields_for :characterables do |abil| %>
      <%= abil.inputs :ability_id %>
    <% end %>
  <% end %>

Comment: semantic_fields_for and :for can be used interchangeably. It had the same result: null.


Optimal full-suspension design for mountain bikes?

Question: What is the best up to now high end all-mountain full suspension design? By "best" I mean a design that balances pedaling efficiency, strength/durability, and (lateral) stiffness and weight as best as possible. What should I look for when trying to get such a bike?
Comment: The most efficient, strongest, most durable, stiffest and lightest suspension is no suspension at all :)
Comment: Suspension linkage is often designed around a target shock (and front fork height, BB height, etc.), which has it's own characteristics as well.  So choosing a bike by looking at the linkage arrangement in isolation is a bad idea.
Comment: This Question is (still) too subjective for QA site such as this. Giveaway is the need to use "best". See if you can word you question without polling for subjective answers.
Comment: The "balance" is only going to be achieved under a specific, narrow set of operating conditions.
Comment: I think the reason your question is closed, and the reason you go not serious (even if subjective answers), is that there is far more to a bikes performance than simply looking at the geometry. Some geometries are cheap to make, meaning better components for the same dollars, e.g.  It may be better to have a simple swing arm and an anti-bob shock than a full FSR 4 link system and a cheap shock. Some are heavy, but might get you lighter frame materials for the same $$$. Some more suited to carbon than Alloy etc...
Comment: hah :) Thank you... it makes sense :) but what I meant was a good compromise between all of these characteristics for AM suspension (eg 130-150 mils of travel)
Comment: Have taken the liberty of clarifying your question a bit; I think this eliminates any confusion over the word "best"; but please revert my edits (or further change them) if I've mistaken your meaning. However, if you want to further add factors such as (say) maintaining control, that could be helpful.
Comment: OK..:) thank you cmannett85, mattnz. +1 These all were good points..could be answers too
Comment: Optimal is just a synonym for "best". Maybe instead ask what the advantages and disadvantages of the suspension systems on the bikes you're looking at are? (Better to ask about 2–4 choices than every choice ever on the market) "Best" is very subjective, heavier, stiffer, bouncier, angles, etc are much less subjective
Answer: Not only is the answer subjective - the "Best" will be different for each rider based on things such as budget, rider weight and fitness and skills, riders tolerance for bumps, riding style (speed, need for control etc), riding terrain, location, brand whoring, day of week, and possibly religion.
However based on you questions criteria, If you must have an answer, I have to go with @GordonM
Edit : ........
Note: This question addresses the original post, which has subsequently been edited.
Comment: While I agree in principle and personally loathe riding full suspension bikes, the question clearly lays out the criteria for what would be the "best" suspension. I believe that this answer does not answer the question on its own terms. And if you feel the question is unanswerable and that a balance such as is described is impossible, you could explain why you feel this is the case.
Comment: For instance, me at 65kg and having a 'controlled' riding style can use a less robust and lighter suspension than a 95kg unskilled rider who likes to play a little loose.  This is just one of so many variations that it makes the question unanswerable with the information provided in the OP. "balance" - What does he mean by that  - does the OP prefer stiffness to weight, travel over stiffness, strength over durably (usually, but not always the same) - we all understand that its a balance, but as the OP does not state any preferences, how can we guide him?
Comment: This answer isn't really an answer. If you think the question is unanswerable, then the place to say that is the comment section under the question. If you think GordonM's comment had the best answer, all you need to do is give his comment an upvote.
Answer: Legs & Hands
These are by far the BEST suspension you can have in any ride.

Legs: Fix yourself on the bikes, make a firm grip or the pedals (or use SPD :P), slightly bend it
Hands: Good grip with elbows pointing out and good balance

Thats it!!! Best suspension is now with you, the one which you can fully customize and control.
I am not saying suspension is not needed.
Having ridden no-suspension, front-suspension and full-suspension bikes I can definitely say that suspension is definitely a very effective addition to your bike, but there cannot ever be a best suspension on any ride I think.
A good suspension with good technology will perform better, but if you know how to use it properly.

Imagine yourself dodging a pit sideways, if you use your suspension correctly it will give you good height, but slightly over and under use and your front wheel will skid on the lift or on the land.

My suggestion would be don't buy a full suspension at first (it will give far too much comfort with no skill). A good bike with lockable front suspension should be good enough to start. Learn to use you legs on bumps, and pits, then try locking your front suspension and learn to use your hands as suspension.

Remember: A smooth sea never made a skillful sailor
Comment: I dont understand the downvote. Would you please care to explain?
Comment: I downvoted you because your answer didn't address the question at all.  If possible I would have downvoted you again for: "don't buy a full suspension (it will give far too much comfort with no skill)" - is that why some of the world's best XC riders and ALL of the best DH riders are on full floaters?
Comment: I have been riding hardtrail bike with rigid fork for many years.And I know perfectly well how to get over obstacles and absorb hits using my legs and arms. Maybe my question isnt defined well and I apologise for that but I am thinking of full susp bike and I dont know what frame should i choose. **I wanted just simple answer like** : dw-link is stiff and efficient design but it is not as strong as single pivot (dont know if its true or not but just for example); four bar can be heavier than others but it is more efficient; single pivot can be stornger and more durable but less effective...vpp
Comment: i cant downvote since i have less than 125rep but if I could i would. My qestion clearly wasnt good but although most of your statements are generally true, your answer is too uncostructive, too general and too far away. To sum up it doesnt have much in common with suspension geometry and desgns. P.S Sorry if my English is bad.


Batching together similar length sequences to avoid padding and packing

Question: I am training an RNN in PyTorch to produce captions for images. It's a pretty standard architecture – the image is processed by a pre-trained InceptionV3 to extract features, the recurrent module processes the words seen so far and then its result is merged with image features and fed through a linear layer to produce a probability distribution over the vocab to predict the next word. A network trained this way can auto-regressively predict a whole caption for an image (it's seeded with an artificial  token at the beginning to have a preceding sequence to start off with).
To train the network every caption is decomposed like this in the training dataset:
X, y
<START>, a
<START> a, man
<START> a man, is
<START> a man is, sitting
<START> a man is sitting, on
<START> a man is sitting on, a
<START> a man is sitting on a, chair
<START> a man is sitting on a chair, <END>

The result is that there is a lot of input sequence length variation in the training data which is a problem since we would need a lot of padding in batches. I am aware that there are techniques of counteracting this (pack_padded_sequence in PyTorch) but I came up with another idea.
I am modifying the Sampler object that's used to provide indices for the DataLoader to construct batches in such a way that the Sampler orders all of the example indices by input sequence length, groups it into batch-size chunks, and shuffles the chunks before feeding them to the DataLoader. As a result the sequences in individual batches are almost always the same length – not because of any padding but because of the fact that we ordered the examples by length before grouping them. There will be some batches that will encompass examples on a border of length groups so some padding will be necessary but it will be minimal (just one zero for a group of lengths [5, 5, 5, 6, 6, 6] for example) and the number of such groups would be negligible (~0.002% of the batches in my case).
My question is – does this method introduce some kind of drawback that I am not aware of? I find it simpler than having completely random batches and padding and packing them. I can't tell however, if it would introduce some kind of bias to the model. My intuition tells me that this should not be problem since the sequence lengths for batches would vary greatly across training (since I am shuffling the index chunks), but I wanted to get a second opinion. I guess the question boils down to -> is sequence length variation within the batch important for proper training? Or, for this specific case -> is it important that the net process examples decomposed from a single caption within the same batch or close to each other time-wise?
I'd be happy to train the network with both solutions and check the results myself but it's quite time-expensive and I'd prefer to avoid some stupid stuff if I can.
Answer: You can read that this was done in "Attention is All You Need", for Transformers: "Sentence pairs were batched together by approximate sequence length.".
But, with RNN you don't need any padding at all when using packedSequences, so I don't think your idea will be efficient, if I understand correctly.


Magnetic field lines are not perfectly straight between magnetic poles

Question: Why is fringing between magnetic poles inevitable?
The diagram given below is the incorrect representation of magnetic field lines between two poles:
Comment: Because magnetic field lines repel each other. They try to stay away from each other.
Answer: There can't be such straight field lines between magnetic poles as the magnetic field can always form closed loops. However the tangential vectors  through each loops point in the same direction which is illustrated in the figure. Remember that the magnetic field is a pseudo-vector. Consider a solenoid for example.

We know that the magnetic field forms closed loops around a current carrying wire. (Can you possibly say the direction of a closed loop?) More the current more will be the field line density. But that's not the reason we draw straight lines between two opposite poles of a magnet. You can only specify the direction of field at a point. That we do by drawing a tangent through the point on the loop. So if you take a region inside a solenoid, or in between the north pole and south pole, you can draw the direction of field at that point by drawing tangent on the loops at that point which will be parallel pointing in the same direction
What that has been illustrated in your picture is just the direction in which the magnetic field acts, not the field lines.


Nasal /i/ in French exists?

Question: So, hear me out. I am going nuts over this. I hope I can explain myself appropriately: Whenever I hear a female French native speaker, I notice some sort of nasalization of /i/ after certain phonemes such as /d/ as in the word "dire" or even present in the combination /ui/ when saying "aujourd'hui". This may also be true for men--and I may be missing more examples of its occurrence--but I'm positive something is going on there and I can't quite put my finger on it. I've looked up info on French phonology (including some posts on this site) and haven't had any luck. This became even more evident after coming across a Mexican tiktoker who tries to teach people about languages and she herself speaks French but overdoes that "nasalization" and thus, instead of accurately pronouncing it--as her female native counterparts would--she just sounds as if she's got a cold and all sounds nasal AF, so maybe I'm not that wrong, it's a thing. Hope someone can shed some light on it and prove I'm not delusional. Merci d'avance !
Comment: Can you share the tiktok link, so we can have an idea ? Also, maybe what you hear is a "e prepausal" : a sort of quirk of some native speakers who will add a vowel sound at the end of some phrases. It can definitely sound like a nasal "in" or "en" sound. I would not dare to say female speakers use it more...but...
Comment: Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.
Comment: On the _e prépausal_: [From Castello-Lopes on Canal+](https://www.youtube.com/watch?v=_W3jHGgJW5A).
Comment: The only nasalized vowels in French are /ɛ̃/ ; /œ̃/ ; /ɔ̃/ and /ɑ̃/.  /i/ is only nasalised when combined to a following consonant. Non natives can mispronounce of course, it can depend on the existing sounds in their native tongues, on what they can hear & reproduce, etc. Don't trust a non native whose credentials you are not sure about. [Here's a lesson](https://www.verbotonale-phonetique.com/voyelles-nasales-fle/) on nasalised vowels by a trustworthy lecturer of FLE. His whole site on the sounds of French and how to teach them is very good, it is aimed at people who teach FLE.
Comment: FLE = Français Langue Étrangère/ *French as a foreign language*
Comment: Ok, I get it. It's about perception. When you say aujourd'hui, the phonemes in d'hui can have a "lighter" d sound or heavier d sound. It can sound like the dw in dweeb (more d) or it can be like the sound of the word yes, oui, separated out. If you are talking delicately, like some women, the ui can be out there with s lighter d sound. Or, as with most people, that dwee sound can be like what you are calling "nasalisation". But it is not about nasal constriction. It is about reverberation in the nose. Similar to the n the word não in Portuguese.
L
Comment: Portuguese does have a nasal /ĩ/ - it is worth making sure that the speakers in question are not Portuguese expatriates or their children/grandchildren. This is quite a numerous group.
Answer: So, after some time spent delving into it and analyzing different samples of spoken French, I came up with a hypothesis that still needs to be tested--and I hope you can give me a hand with that.
I'm quite positive that the phoneme I'm talking about occurs in Swedish as well (its incidence may as well be higher). Here's what I found on such phoneme (for Swedish, that is):
One of the varieties of /iː/ is made with a constriction that is more forward than is usual. Peter Ladefoged and Ian Maddieson describe this vowel as being pronounced "by slightly lowering the body of the tongue while simultaneously raising the blade of the tongue (...) Acoustically this pronunciation is characterized by having a very high F3, and an F2 which is lower than that in /eː/." They suggest that this may be the usual Stockholm pronunciation of /iː/. (Ladefoged & Maddieson, 1996, p. 292)
And this as well:
Patterns of diphthongs of long vowels occur in three major dialect groups. In Central Standard Swedish, the high vowels /iː/, /yː/, /ʉː/ and /uː/ are realized as narrow closing diphthongs with fully close ending points: [ɪ̝i ʏ̝y ɵ̝˖ʉ̟ ʊ̝u] (McAllister, Lubker & Carlson, 1974; cited in Ladefoged & Maddieson, 1996, p. 295). According to Engstrand (1999, p. 141), the second element is so close as to become a palatal or bilabial fricative: [ɪ̝ʝ ʏ̝ʝʷ ɵ̝˖βʲ ʊ̝β].
Here's a link to a video of a female French native speaker: https://www.youtube.com/watch?v=5-_EZDKFv7U
I think she realizes that same phoneme when uttering certain words such as "aujourd'hui" (0:37 in the video)--which could be some sort of a diphthong pattern where /i:/ is somehow affected by the preceding phoneme /ɥ/--and then again at 1:02 when she says "traduire" (same mechanics involved as the example above).
I definitely hear more constriction in the realization of that /i/. It may only occur in the phoentic diphthong pattern /ɥi/ that is phonemically realized as /ɥɪ̝ʝ/ (if that even exists, of course); I don't have any other examples at the moment, but I'll share them with you in case I come across any in the future.
Comment: Welcome! I'm a native speaker of a regional variation, I don't know anything about phonetics or these other languages and I'm not sure what OP is talking about but I did listen at 0:37 and 1:02. To me it sounds very typical and metropolitan, high pitch voice & very high, the highlight of these words, almost "watery", "flickering", which is all the opposite of something nasal I would think. Is it a diphtong? Maybe it's just a bit long or because she's trying to pronounce maybe a bit slower than her typical speed, and in the case of _traduire_ the e is very soft, it sounds quite powerful? It is.

Nasal /i/ in French exists?


Proof of well ordering principle for the set of positive integers with directly using the principle of induction and not strong induction

Question: Can we prove well ordering principle for the set of natural numbers (positive integers ) with directly using the principle of induction  i.e. $( S \subseteq \mathbb N ,1 \in S \space \&\ n \in S \implies n+1 \in S) \implies S=\mathbb N $ and not using principle of strong induction ? The proof I know first prove principle of induction implies principle of strong induction and then  uses strong induction to prove well ordering principle , but I want a direct proof using  only induction . Thanks in advance
Comment: Why do you want that? You can just fold the two sub-proofs together -- that is, is you have a proof of $\varphi(n)$ by long induction, you can trivially word the same argument as a proof of $\forall k(k\le n\to\varphi(k))$ by mathematical induction instead.
Answer: You might consider this to be a bit cheating, but here is a proof using only induction and one additional inference.
Lemma. If $A\subseteq\Bbb N$ has a maximal element, then $A$ has a minimal element.
Proof. First note that $A$ has a maximal element, so it is non-empty. Now we prove by induction on $n$: If $\max A=n$ then $A$ has a minimal element.

If $\max A=0$ then $A=\{0\}$ and $0$ is also $\min A$.
Assume that for $n$ the claim holds. If $\max A=n+1$ then either $A'=A\setminus\{n+1\}$ is empty, in which case $A=\{n+1\}$, so it is also the minimal element; else $\max A'=n$ and by the induction hypothesis it has a minimal element, $k$ and $k\leq n<n+1$, so $k=\min A$.

Therefore if $A$ has a maximal element, it has a minimal element. $\square$
Theorem. The principle of induction implies that every non-empty set of natural numbers has a minimal element.
Proof. If $A$ is a non-empty set, pick some $n\in A$, then $A'=\{a\in A\mid a\leq n\}$ has a maximal element $n$. We proved by induction that $A'$ has a minimum element $k$. If $m\in A$ then either $m\leq n$ in which case $m\in A'$ so $k\leq m$ or $m>n$ and in which case $k<m$. In either case $k=\min A$. $\square$
Answer: I claim that for all $n$, for all $A \subseteq \mathbb{N}$, if $n \in A$ then $A$ has a least element.
The base case is $n = 0$. Clearly, if $0 \in A$, then $0$ is the least element of $A$.
Now suppose the claim holds for $k$, and suppose we have $k + 1 \in A$. We consider two cases. The first is that $k + 1$ is the smallest element of $A$. In that case, we're done.
The other case is that there is some $j < k + 1$ such that $j \in A$. Note that $j \leq k$. Let $A' = A \cup \{k\}$. Then $k \in A'$, so by the inductive hypothesis, $A'$ has a least element $w$. Since $w \in A \cup \{k\}$, we see that either $w \in A$ or $w = k$. If $w = k$, then we have $k \leq j \leq k$ and therefore $w = k = j \in A$, so either way, we know that $w \in A$. Since $w \in A \subseteq A'$ and $w$ is the smallest element of $A'$, it is also the smallest element of $A$.
Answer: Let $S \subseteq N$ be non-empty, and define:
$R = \{x \in N : x \le  y, \forall y \in S\}.$
Then $0 \in R$ since  $0 \le y$ $\forall y \in N$, in particular $\forall y \in S$.
Since S is non-empty, there is a $y \in S$; this implies $y + 1 \notin R$: otherwise
we would have $y + 1 \le y$, [Perhaps you can disprove this with PEANO'S AXIOMS]
Thus $R$ contains $0$ but $R \neq N$; the induction axiom then implies that there
must exist an $x \in R$ such that $x + 1 = s(x) \notin R$ [$S : N \rightarrow N$ is a successor function]. We claim that $x$ is a smallest
element of $S$.
First, $x \in R$ implies $x \le y$, $\forall y \in S$> So, we only need to show that $x \in S$.
Assume $x \notin S$; then $x \le y$ ,$\forall y \in S$ implies $x < y$ (because we can't have
equality), hence $x + 1 = s(x) \le y$,  $\forall y \in S$, which by definition of R shows
that $x + 1 \in R$ in contradiction to the construction of $x$.


Subgroups as isotropy subgroups and regular orbits on tuples

Question:
Is there some natural or character-theoretic description of the minimum value of d such that G has a regular orbit on Ωd, where G is a finite group acting faithfully on a set Ω?

Motivation:
In some cases, there are so many kinds of orbits of G on Ω that every subgroup of G appears as a point-stabilizer.  However, these actions are quite rare, even in the case of a point group of a group of Euclidean isometries. (The only such actions in 2D are C1, D1, and Cp for primes p.  This follows because rotation subgroups only stabilize the origin.)
  (example for D4 by Bill Cook).
If we allow the group to act on subsets too, then we may have better luck.  See for instance, the lovely example of dihedral groups and cyclic groups acting naturally on the Euclidean plane as Rosette groups.  Here one can take various natural subsets of the polygon to get all subgroups.
In general a finite point group of a group of Euclidean isometries has a regular orbit, and the blocks of that orbit suffice to specify all subgroups (there is a 1-1 correspondence between subgroups containing a point stabilizer and blocks of the orbit given, and this correspondence is given by the set-wise stabilizer).  This is true more generally for small groups acting on large vector spaces (if V is n-dimensional over K and |G|−1 < 1+|K|+…+|K|n−1, then G has a regular orbit on V in any faithful, linear action of G on V; in particular finite groups acting on infinite vector spaces are fine).
However, determining which medium-ish linear groups over medium-ish vector spaces have regular orbits is really hard, as far as I know.  I wondered if it was much easier if we allowed one to replace V by V ⊕ V.  Certainly we can replace V by VV (taking d=|V|), but that is a bit extreme.

Is there some reasonable value of d that can be read from the permutation character of G on Ω, from the (Brauer) character of G on V, or some other natural invariant of the action of G on Ω?

An expression for the minimum would be nice, but even some decent bound on d in terms of invariants would be fine.
To identify Galois groups, one also looks for certain invariants for polynomials, and I believe what I am asking is relevant to how complicated such invariants need to be.  GAP for instance takes d ≤ 5, I think (and used to take d ≤ 3, I think), but it has lots of "backup" routines to identify various other subgroups, possibly for efficiency, but possibly because one cannot choose d that small in general.
Comment: Jack, this seems to be an excellent research question, maybe you should post it on http://mathoverflow.net/ too. Have you contacted Marty Isaacs on this? He is on this StackExchange.
Comment: Very nice question, Did you reach any conclusion ?
Answer: This has been open for nearly nine years and is the top-rated unanswered question on Group Theory. So it deserves an answer, even though the questioner appears to no longer come round these parts.
First, something easy. Let $V$ be a $kG$-module, which we will assume is faithful.
Lemma: if $G$ possesses at most $m$ subgroups of prime order, and $|k|\geq m$ then $G$ has a regular orbit on $V$.
The proof is clear: each element of prime order centralizes a proper subspace of $V$. Since $V$ is not the union of $m$ proper subspaces, there is a point $x$ not centralized by any element of order $p$. Thus $x$ lies in a regular orbit. (You can obviously do much better than $m$ here, but this is the easiest to state.)
This deals with $k$ of characteristic $0$, of course. The best results are when $\gcd(|G|,|k|)=1$. Here we use a result of Halasi--Podoski from Every coprime linear group admits a base of size two. This states that any faithful module has base size $2$, i.e., there exist two points, $x,y\in V$, such that $C_G(x)\cap C_G(y)=1$. Thus if $W$ is a faithful $kG$-module and $W\oplus W$ is a submodule of $V$ then $G$ has a regular orbit on $W\oplus W\leq V$.
Thus if $V$ is a faithful $kG$-module then $V\otimes V$ contains two non-isomorphic faithful modules, namely $\Lambda^2(V)$ and $S^2(V)$. The tensor product of the two contains two copies of $\Lambda^2(V)\otimes S^2(V)$, so there is always a regular orbit in $V^{\otimes 4}$.
If $V$ is a permutation module then $V=1\oplus W$, and $W\oplus W\leq V\otimes V$. Hence there is a regular orbit in $V^{\otimes 2}$. This of course generalizes to $V$ an arbitrary sum of modules, but we have to be a little careful to make sure that $V\otimes V$ is faithful.
I would guess that even when $V$ is simple that $G$ has a regular orbit on $V\otimes V$, but this has not been proved, as far as I know.

Having dealt with the coprime case, we look at the general case. Here the base size can be greater than $2$. Some recent papers of Melissa Lee give bounds on base sizes for quasisimple groups, but as far as I know this looks quite hard. Note that a result of Auslander--Carlson states that if $V$ is absolutely indecomposable and of dimension divisible by $p=\mathrm{char}(k)$, then $V\otimes V^*\otimes V$ contains $V\oplus V$. This means that in that situation we do obtain two summands isomorphic to $V$ in $V^{\otimes 3}$, whenever $V$ is self-dual. Thus we may exploit base sizes again. In general, of course, knowing composition factors of $V^{\otimes n}$ doesn't help you too much, because you need composition factors of the socle to guarantee that you achieve a regular orbit.
My guess would be that if $V=W_1\oplus W_2$ and each $W_i$ has base size $2$, then with few exceptions $V$ possesses a regular orbit.


Correcting information in original text

Question: I am publishing historical letters. My research has uncovered some mistakes (e.g. name Jackson instead of Johnson). I'd like to retain the original text, but supply the correct information. I'd like to avoid footnotes, as I have endnotes, and using both is confusing to the reader. But endnotes for corrections are cumbersome. How can I preserve the original and include the correction in a way that's helpful and not intrusive to the reader?
Comment: I don't think footnotes and endnotes are confusing.  Information needed quickly like Jackson or Johnson (or possibly an identifier of a person) are useful at the bottom of the page. (I've edited a book of letters and thus been through this.)  Endnotes (sources) would be numbered; footnotes are asteriks etc. and "restart" with each page.
Comment: Hello and welcome to EL&U. I haven't voted on your question, but the down-votes might be because footnotes and endnotes aren't really aspects of the English language ([Academia.SE](https://academia.stackexchange.com/help/on-topic) might be more relevant). However, the answers below do provide English conventions that address your question. You might want to try rewording your question so that it's relevant to English.SE, though please take care not to invalidate the answers.
Answer: This can be done using sic and recte, or just sic as described by Wikipedia:

Alternatively, when both the original and the suggested correction are desired to be shown (as they often are in palaeography), one may give the actual form, followed by sic in brackets, followed by the corrected form, preceded by recte, in brackets. The word recte is a Latin adverb meaning "rightly".

An Iraqi battalion has consumed [sic] [recte assumed] control of the former American military base, and our forces are now about 40 minutes outside the city.

According to the Journal of Seventeenth-Century Music Style Sheet, there should be no punctuation, for example no colon, before the corrected word when using recte. Sometimes only sic and the correction are in brackets, becoming as in the last example "[sic assumed]" (i.e. recte is omitted).

Here is an example in the wild:

Cerithium pbeliscus [sic] [recte obeliscus] Brug
Rizal's Conchology

And here's another example with a different format:

Court of William [sic, recte John (Stafford)] archbishop of Canterbury, Humphrey [Stafford] duke of Buckingham and others, feoffees of John [de Mowbray] duke of Norfolk. Tues. after Michaelmas, 28 Hen. VI
The National Archives
Answer: Pop a [sic] in the quote! After that, then you can clarify right after within or without parentheses.
Example:

Bill Nye has opinions about many presidents like Andrew Jackson. This
  was evidenced like with when he said "Andrew Johnson [sic] was a
  terrible president."

Here's some more info if you'd like:
http://data.grammarbook.com/blog/definitions/sic/
Answer: I don't think using both footnotes and endnotes is confusing to the reader.  In a collection of historical letters, you will probably have an endnote for each letter that provides its archival source and possibly other information, or possibly an endnote for the source of a group of letters.  You would probably use numbers (1, 2, 3) for them, renumbering with each chapter or possibly not at all.
To avoid distractions in the flow of the writing, footnotes using symbols restarting on each page (*, etc.) are convenient for the reader.  A reference (to use a modern example) to "Ted Kennedy" would have a footnote that says "Senator Edward M. Kennedy, D-Mass." for example.  A reference to the 1816 eruption of Mount Tambora that caused "A Year Without a Summer" would have a footnote "Tambora actually erupted the year before, in 1815."
If there's literary value to the writing, and especially if you are trying to reach a general audience, you don't want to put these details in the text.
Another alternative, used increasingly by publishers afraid of discouraging sales to the general public, is to avoid any numbers or symbols in the text, and instead to use a snippet of text (and the page number on which it appears) in endnotes.  This approach is extremely hard on the reader, who doesn't know whether there's additional information provided or not, and has to scan an entire page for the text to which a note refers. David McCullough's best-selling biography John Adams uses this approach; he quotes quite a few letters, but only in part, and he has his own text in which he can provide information incorrect or incomplete in the correspondence.

Correcting information in original text


How to clean settled grease from kitchen cabinets?

Question: I like to fry a lot (I am not fat though :)) so there is quite a lot of grease and oil fume buildup on my kitchen cabinets.  I tried spraying and wiping, to no avail.  Any idea how I can clean settled grease and oil fumes from my cabinets effectively?
Answer: A standard treatment for kitchen surfaces is to wipe it with a solution of TSP, an acronym for trisodium phosphate See http://www.ask.com/explore/what-tsp-cleaner.  It forms an strong alkaline (basic) solution that converts the grease to soap, making it easy to clean off.  The reaction is called saponification.  The same reaction takes place in drain cleaners and oven cleaners.
Be advised that the TSP will do the same thing to oils in your skin!  Use rubber gloves, eye protection, and whatever other precautions you find on the package.
A "Just Remembered" Edit:  TSP, and other alkaline solutions will attack aluminum surfaces!! Be warned...
Comment: TSP is great stuff for this situation. If you want something less caustic, aggressive, try some Dawn dish soap. I have had good success degreasing many items with it.
Answer: look for pinksolution - non-toxic and I was amazed at the results. No detergent removed old grease like this one. Good for all cleaning around the house. It's very concentrated and a tub will last a very long time. Bought it from Costco in Canada, but look on their website (pinksolution.ca)


When SIGTERM is received by a process, what happens to its children?

Question: I am wondering what happens when a process receives SIGTERM and it has children spawned (e.g. JVM that has /bin/sh commands running).
Is there some sort of automatism or does the parent simply exit and then the children get a SIGTERM too? Do they terminate?
Comment: Do you mean whatever Linux process (in which case this will depend on how the dev designed the handler) or specifically JVM and its *shutdown hooks* ?
Answer: What happens when a linux process receives SIGTERM (and, what can incidentally happen to child-processes) is left entirely to the programmer's will.
The process can simply be instructed to ignore (and pursue its business as usual), abruptly die leaving its child-processes orphaned, or execute some in-process built piece of code (the associated software interrupt handler) that can eventually propagate the signal to child-processes or take even greater care.

The Java Virtual Machine has got its own particularities which make so it just cannot stand as an… exempli gratia… ;-)
SIGTERM ? Just ignore ! thanks to the Xrs extra option, other code will handle more appropriately.

Applications embedding the JVM frequently need to trap signals such as
SIGINT or SIGTERM, which can lead to interference with the JVM signal
handlers. The -Xrs option is available to address this issue. When
-Xrs is used, the signal masks for SIGINT, SIGTERM, SIGHUP, and SIGQUIT aren't changed by the JVM, and signal handlers for these
signals aren't installed.

What should happen to whatever child ? Well… it'll all depend on the embedding application.
SIGTERM ? Just… shutdown… gracefully !
Any Java Program can attach a shutdown hook to the machine. i.e. code that the JVM should execute before going down. Something as simple as :
<code>public class ShutDownHook
{
public static void main(String[] args)
{
    Runtime.getRuntime().addShutdownHook(new Thread()
    {
    public void run()
    {
        System.out.println("Shutdown Hook is running !");
    }
    });
    System.out.println("Application Terminating ...");
}
}
</code>
This enabling the developer to handle ad libitum the termination of whatever child process and even let some of them abruptly die as a consequence of the machine going down.

The JVM (and VMs in general) differ greatly from standard linux processes in that the handling of some signals (and SIGTERM in particular) is likely to be coded (and customized) outside of the virtual's machine internal signal handler. This incidentally enabling this code to run even… if the VM died.


What does it mean for a CPU to have native USB ports?

Question: I'm planning on building my own computer soon, somethinng I've never done before. The CPU I want to get for it is an AMD Ryzen 5 5600. Something in its specifications page confuses me, though. It says that it has 4 native USB 3.2 gen 2 ports, and no other native USB ports. That said, the CPU itself doesn't have any ports; those are found on the motherboard. So, why does the specs page talk about native USB ports? The motherboard that interests me the most has six USB ports, but only two of them are gen 2, with the others being gen 1. Would I still be able to use them?
Comment: https://www.guru3d.com/articles-pages/amd-ryzen-5-5600-review,3.html  AMD has integrated USB 3.2 Gen2 support into the CPU, which means you get four 10 Gbps USB ports at SuperSpeed et cetera
Comment: A CPU is a chip (integrated circuit). It is the motherboard that has the USB controller and ports.
Comment: I would guess that it's saying it has 4 separate USB 3.2 Gen 2 host controllers, as opposed to a single host controller connected to a 5 port USB hub chip. So it's saying the 4 ports don't share bandwidth on a single bus; they each are their own bus and thus get full bandwidth.
Comment: It means that there are USB controller(s) integrated with the CPU (processor) as a single chip/package, a variation of System on a Chip, SoC. A motherboard could still add additional (external) USB controllers. In actual usage you probably would not be able to tell the difference whether a USB controller is "*native*" or external to the CPU. But the 3.2 Gen 2 capabilities are what may be salient. The faster transfer speeds of 3.2 benefit from a (physical & logical) closer connection to the CPU.
Answer: That the CPU has "native" USB ports is the same as the CPU hosting PCIe ports.
In other words there are USB ports connectors on your motherboard that rather than going through one of the motherboard chipset integrated circuits it connects directly to the CPU. The motherboard hosts the connector and wires, but not the actual controller.
What that means is that there is potentially one or more less "hops" between the USB connector and your CPU. Data comes to and from the CPU directly.
Its the same with PCIe, on modern systems one or more PCIe ports connects directly to the CPU, with other ports connecting to the motherboard chipset.
It's the left hand side of the following image. There are USB ports coming off the X570 chipset (the motherboard, shown in red) and there are ports that are controlled and hosted directly by the CPU on the right.
Comment: essentially the controller is part of the CPU package, rather than having a seperate controller chip somewhere on the motherboard.
Comment: Hmm, I see. Then I assume that, in order to use these ports, I'd have to hook them up differently than usual, right? Since they are hooked up directly to the CPU, rather than going through the motherboard.
Comment: You'd have to read the motherboard manual to be certain. They may well be connected directly to the backplate so all you do is fit the motherboard in a case and voila. Or they may be wired up to headers to connect to the case ports. The manuals should tell you what goes where.
Comment: It sounds to me like USB being moved to the north bridge (which is mostly if not all on-chip these days).
Comment: @Mokubai I see now. Thanks! And thanks to everyone else, too. Your answers were very helpful.


ASP.NET Lifecycle and long process

Question: I know we need a better solution but we need to get this done this way for right now. We have a long import process that's fired when you click start import button on a aspx web page. It takes a long time..sometimes several hours. I changed the timeout and that's fine but I keep getting a connection server reset error after about an hour. I'm thinking it's the asp.net lifecycle and I'd like to know if there are settings in IIS I can change to make this lifecycle last longer.
Comment: http://stackoverflow.com/questions/118261/long-running-code-within-asp-net-process
Answer: You should almost certainly do the long-running work in a separate process (not just a separate thread).
Write a standalone program to do the import. Have it set a flag somewhere (a column in a database, for example) when it's done, and put lines into a logfile or database table to show progress.
That way your page just gets the job started. Afterwards, it can self-refresh once every few minutes, until the 'completed' flag is set. You can display the log table if you want to be sure it's still running and hasn't died.
This is pretty straightforward stuff, but if you need code examples they can be supplied.
Comment: I would go further and make it an NT Service.  Database as you suggest for communication with web server, is fine, or use WCF.
Comment: If it's a process that's run infrequently, and exits when done, I'm not sure what the advantage of a Service would be.
Answer: One other point to consider which might explain the behaviour is that the aspnet_wp.exe recycles if too much memory is being consumed (do not confuse this with the page life cycle)
If your long process is taking up too much memory ASP.NET will launch a new process and reassign all existing request.  I would suggest checking for this.  You can do this by looking in task manager at the aspnet_wp and checking the memory size being used - if the size suddnely goes back down it has recycled.
You can change the memory limit in machine.config:
<code>  <system.web>
    <processModel autoConfig="true"/>
</code>
Use memoryLimit to specify the maximum allowed memory size, as a percentage of total system memory that the worker process can consume before ASP.NET launches a new process and reassigns existing requests. (The default is 60)
<code><system.web>
    <processModel autoConfig="true" memoryLimit="10"/>
</code>
If this is what is causing a problem for you, the only solution might be to have a separate process for your long operation.  You will need to setup IIS accordingly to allow your other EXE the relevant permissions.
Answer: You can try running the process in a new thread. This means that the page will start the task and then finish the page's processing but the separate thread will still run in the background. You wont be able to have any visual feedback though so you may want to log progress to a database and display that in a separate page instead.
You can also try running this as an ajax call instead of a postback which has different limitations...
Since you recognize this is not the way to do this I wont list alternatives. Im sure you know what they are :)
Answer: Extending the timeout is definitely not the way to do it. Response times should be kept to an absolute minimum. If at all possible, I would try to shift this long-running task out of the ASP.NET application entirely and have it run as a separate process.
After that it's up to you how you want to proceed. You might want the process to dump its results into a file that the ASP application can poll for, either via AJAX or having the user hit F5.
Answer: If it's taking hours I would suggest a separate thread for this and perhaps email a notification when it is ready to download the result from the server (i.e. send a link to the finished result)
Or if it is important to have a UI in the client's browser (if they are going to be hanging around for n hours) then you could have a WebMethod which is called from the client (JavaScript) using SetInterval to periodically check if its done.
Comment: Is the separate thread also affected by the lifecycle because I've put it in a separate thread and it still stops about 45 minutes into the process.
Comment: A background thread shouldn't be affected by the page life cycle, but it might be affected by an asp.net recycle (see my new post)
Comment: After your initial page request completes, your separate thread is disconnected from any request. When the application pool recycles or cleans up, it sees a hanging thread and terminates it.

ASP.NET Lifecycle and long process


How is particle creation (or annihilation) possible in non-relativistic many body physics?

Question: How is that, in many-body physics, particle creation and annihilations are possible even though it is a non-relativistic theory?
Comment: The question is pretty vague, and asserts something that sounds false as a general statement. But anyway, trying to guess what might be relevant --eE.g., in low-energy nuclear physics we talk about creating quasiparticle excitations, which are basically particle-hole excitations. I believe this happens in condensed-matter physics as well. A lot of the formalism is similar to what you see with particle-antiparticle excitations, even though the calculations are nonrelativistic.
Comment: Particle creation and annhilation comes about by advancing from a Hilbert space to a Fock space.  This doesn't require relativity in principle, it's just that standard QM is inconsistent when used to describe a relativistic theory.
Comment: You mean by non-linear effects or inhomogeneities (like lattice defects or dopant atoms)?
Answer:
Can we have particle creation or annihilation in non-relativistic physics?

The answer is an emphatic yes. It is quite a routine in non-relativistic (NR) many-body physics.

How is particle creation or destruction possible without relativity?

Absence of relativity does not allow creation (and annihilation) of particle-antiparticle pairs (or mass) with disappearance (or appearance)
of energy. Non-relativistic physics, therefore, forbids energy turning into mass or vice-versa. Since relativity is not incorporated in non-relativistic many-body theory, mass creation or mass destruction is not possible.

What is happening then?

The energy supplied to a Fermi vacuum (which is simply the Fermi level) "creates" an electron-hole pair in the lattice. This electron-hole pair "creation" is not same as the electron-positron pair creation from the vacuum of a relativistic quantum field theory. Had it been so, one would have to supply at least $2m_ec^2$ amount of energy to the Fermi vacuum to create the electron-hole pair which is the minimum energy needed to create an electron-positron pair in relativistic physics. In metals, energy goes into the electron (already present at the Fermi level), which then absorbs it to get excited at a higher level. You can call this appearance of the electron above the Fermi level and the hole below it as the "creation" of the electron-hole pair. The typical incident energy $h\nu(\ll m_ec^2)$ is sufficient to really create an electron-hole.

Addendum

Moreover, particle creation may not require relativity if the particle is gapless excitation. This is the case for phonons and magnons which are essentially non-relativistic concepts. Any tiniest bit of energy will create them. The energy supplied goes to the lattice vibration and there is nothing like energy being converted to rest mass.
If we look at the "second-quantized" Schrodinger equation, the
Hamiltonian doesn't contain the rest energy. This is because, in non-relativistic physics, the mass cannot change to energy and therefore, mass is eternal and can be ignored in the Schrodinger equation. On the other hand, relativistic quantum field theories start with Lagrangians that include the rest mass from the very beginning.
Comment: maybe one  should stress that the particles created are not elementary  standard model  particles, but mathematical definitions of particle states.  the creation and annihilation operators on harmonic oscillator states for example. Assigning the name "particle" to the creation does not make it an elementary particle. It is just counting the number of states.
Comment: @annav I agree.
Answer: I don't know if that is the case, but you may have been misguided by the formalism.
For convenience, the language of second quantization is often utilized in non-relativistic many body physics. So you may have stumbled upon creation and annihilation operators in that context. Nevertheless, even if written in second quantization, the Hamiltonians of many non-relativistic particles preserve the number of particles.
A concrete example: the Hamiltonian of $N$ non-relativistic bosons with pair interaction (symmetric) potential $V$ is usually written on $L^2_s(\mathbb{R}^{3N})$ (the symmetric $L^2$, under permutation of the three dimensional variables $x_i$, $i=1,\dotsc,N$) as:
$$H=\sum_{i=1}^N-\frac{\Delta_i}{2M}+\frac{1}{N}\sum_{i<j}V(x_i-x_j)$$
On the $N$-particle sector, $H$ agrees with the second quantized operator on $\Gamma_s(L^2(\mathbb{R}^3))$ (symmetric Fock space over $L^2(\mathbb{R}^3)$):
$$\tilde{H}=\int_{\mathbb{R}^3}a^*(x)\Bigl(-\frac{\Delta_x}{2M}\Bigr)a(x)dx+\frac{1}{2}\int_{\mathbb{R}^6}V(x-y)a^*(x)a^*(y)a(x)a(y)$$
So, as you see, the operator $\tilde{H}$ uses the formalism of creation and annihilation operators $a^\#(x)$, but the operator leaves each fixed particle sector invariant, as expected in a non-relativistic theory.
Comment: Maybe I misunderstood but a non-relativistic hamiltonian doesn't always conserve particle numbers, for example, the BCS-type hamiltonian


Does two Higgs doublet model predicts two Higgs bosons?

Question: Does 2HDM predict two higgs boson? Why two doublets are needed? I need a simplified answer as i am new to BSM
Comment: It predicts 5 higgs bosons including the usual one. Usually called h (light scalar), H (heavy scalar), A (pseudo-scalar), H+ (charged positive Higgs boson) and H- (charged negative Higgs boson). I'd answer but don't have time to look up a citation. Not entirely obvious in 2HD models if the SM Higgs is h or H.
Answer: You may guess it by counting the number of degrees of freedom. One doublet corresponds to 4 real degrees of freedom. However the Higgs breaks $U(1)\times SU(2)$ gauge symmetry to $U(1)$. This requires the appearance of three Goldstone modes - one neutral and two charged ones conjugated to each other. Those 3 Goldstones are not physical and instead neutral $Z$ and charged $W^\pm$ bosons get longitudinal polarizations. This leaves us with only one neutral Higgs boson in the physical sector.
Two Higgs doublets give us 8 real degrees of freedom. But then 3 of them again correspond to unphysical Goldstone modes. Thus we are left with 5 physical modes (not two!) - 3 neutral bosons and charged boson and its antiparticle.
2HDM models are one of the simplest extensions of the Standard Model. You also have to introduce at least two Higgs doublets into the supersymmetric extensions like MSSM where you can't give masses to both up and down quarks with only one Higgs doublets. Another motivation comes from the axion models where you can't construct the nontrivial Peccei-Quinn symmetric potential for the one Higgs doublet.


Is there a word to describe the positioning of words directly below other words, as in translations

Question: I was hoping to use a word to say that in a song lyric, the translation of the original language appears directly below the original language throughout the song, without having to say all that.  I thought "subscribed" made sense (sub=below; scribed=written), but can't find any dictionaries that give that definition to the word.
Comment: Because human language is a **combinatoric art rather than a crossword puzzle**, we normally do and should use multiple words to describe elaborate concepts.
Answer: According to dictionary.reference.com, interlinear is:

situated or inserted between lines, as of the lines of print in a book: a Latin text with interlinear translation.
Answer: What you are describing is subtitling, which according to dictionary.reference.com is:

3.
  Movies, Television.
a.
  the text of dialogue, speeches, operas, etc., translated into another language and projected on the lower part of the screen.
b.
  (in silent motion pictures) a title or caption.
Answer: Juxtaposition is not very specific to music but could work.
According to Google, the word means:

the fact of two things being seen or placed close together


Card game for children

Question: My nephew is 4 years old. I want to teach him a simple card game (my intention is that in some years he will be able to play bridge). For now, I just want to make him familiar with the cards. What simple game appropriate for really young people do you propose?
Answer: Here are my suggestions of games that teach various card game rules, ranked in order from least difficult to most difficult.
Go Fish has simple rules and can help you learn basics of card games, such as

Identifying matching sets of cards
Identifying card names and suits
Waiting your turn
Communicating with opponents when required (and being truthful about it)
Drawing and holding cards without revealing them

War kicks the
competitive level up a bit and has more rules. In addition, you learn card value progression, including the hierarchy of J-K (which can be explained fairly simply if the player is familiar with monarchy, fictional or otherwise).

Competition
Card values
Winning a round without winning the game

Bullshit is a matching game, like Go Fish, but it adds the concept of a bluff and shows that following the rules can sometimes feel like cheating, since duping your opponent is truly the objective of the game. This is useful when later learning games like poker. Also, this can introduce the concept of 'house rules', and hones your ability to agree on extra rules once the game is in progress and tensions are high. Requires 3 or more players.
Gin Rummy has quite a few rules in theory, but is quite simple to play most of the time. It provides practice for:

Identification of runs
Using the discard pile to draw from
Keeping score
Comment: I can add my hyperlinks once I have some rep points.
Comment: Are you not allowed to add more hyperlinks because of low rep?  I did not know that.
Comment: You have to have at least 10 rep. That's okay, because I have that now. :)
Answer: I started to play "Tschau Sepp" with my son who is also four years old. This is a swiss version of Mau-Mau. It is similar to Uno but uses a regular set of cards. He likes to play it but still needs some help as he often misses that cards in his hand have the same value (number) as the played card. Therefor we often play with our hands shown.

Card game for children


What is the distinction between "连笔字" and "行书"?

Question: I see that "连笔字" refers to "not lifting your pen from the paper while writing" (from ArchChinese) and 行书 refers to semi-cursive. Is there a distinction between them, or are they synonyms?
Answer: 连笔字,就是平时用的"行楷/行书"字体,因为笔画呼应、关联很强,写起来就要"连笔",通俗的称谓连笔字。
https://www.163.com/dy/article/FPQ64EJI0534F94V.html
Answer: 连笔字 is a kind of style, whereas 行书 is a font family.
some font families (行书、草书 etc.) all have the style of "连笔". If people's handwriting is just scribbled and does not conform to any font style, many strokes are connected, which can also be called 连笔字.
行书 is a font family. It has its own standards, not only "连笔"。
Answer: 連筆, accurately called continuous lines/strokes in english, is the act of writing without lifting the pen.  You could still be writing every single stroke, or writing a shortened version with less strokes like those commonly seen in cursive chinese.  While you can see it with calligraphy brush writing, you could also see it in someone's daily handwriting with a pencil etc.  It doesn't have to be beautiful or artistic characters, as long as the strokes you do write are connected together.
行書, accurately called walking/strolling script in English, is a type of chinese calligraphy cursive.  It is almost exclusively used with brush calligraphy, and most people find it hard to read due to the severe shorthand of stroke counts many characters undergo.  If you have not specially studied it to learn what each character should become you will have troubles with writing or reading it. There may be a few accepted versions of how to shorten a specific character, but they are all preestablished and very specific, changing the curve of a line or connecting two strokes that aren't usually could potentially make it a different cursive character.
A final comparison: continuous strokes is the literal act of your strokes connecting to each other, regardless of formal or ugly handwriting, writing medium, writing out every character in regular script or cursive... etc
walking script is a type of cursive that frequently combines what would be multiple strokes into single strokes, sometimes by leaving pen on the paper to connect them, but more frequently by creating a new single stroke to represent the multiple strokes. A 12 stroke character becoming a 3 stroke character as an example, you would still lift your pen fully between each of the three new strokes in a standard process.
It is possible to encounter both in the same writing, but definitely not guaranteed.  While 行書 with 連筆 does exist, I think you are more likely to encounter 連筆 in semicursive or regular script chinese overall: the larger amount of strokes gives a better reason to connect lines for speed, and the overall legibility is far less hindered by the stylistic choice. Compare full cursive with continuous strokes; the legibility for regular people gets very low, except maybe on characters written very large.


A concrete realization of the nontrivial 2-sphere bundle over the 5-sphere?

Question: Since $\pi_4 (PU(2)) = \pi_4 (SO(3)) = {\mathbb Z}_2$, the two-element group,
we know  that half of  the two-sphere bundles over the 5-sphere $S^5$ are trivial
and the other half are non-trivial and all isomorphic.  Can you  write  an explicit concrete realization  for this non-trivial bundle?   I have in mind something along the lines of  the Hirzebruch surface
(see eg. http://en.wikipedia.org/wiki/Hirzebruch_surface)
$\Sigma_1$ (or $\Sigma_k$, $k$ odd)  which   realizes  the unique topologically non-trivial $S^2$ bundle over $S^2$. (Since $\pi_1 (SO(3) = {\mathbb Z}_2$ again ``half'  the two-sphere bundles over the 2-sphere $S^2$ are trivial
and the other half are non-trivial and all isomorphic.)
Comment: I'm trying to pull one back from $S^5/S^1 = \mathbb{CP}^2$, but it seems like the nontrivial bundle there comes from $\pi_3(Diff(S^2))$, not from $\pi_4$, grr.
Answer: Is this concrete enough?  Recall that $\mathrm{SU}(3)$ fibers over $S^5$, with fibers equal to $\mathrm{SU}(2)$ and that this fibration is nontrivial.  Let $S^1\subset \mathrm{SU}(2)$ be (any) subgroup and let $B = \mathrm{SU}(3)/S^1$.  Then $B$ fibers over $S^5$ with fibers $S^2$.  If $B$ were trivial, then $\mathrm{SU}(3)\to S^5$ would be trivial as well, but it is not, since $S^5$ is not parallelizable.
In more detail:  Regard $\mathrm{SU}(3)$ as the set of triples $(e_1,e_2,e_3)$ of special unitary bases of $\mathbb{C}^3$.  Define a map $\pi:\mathrm{SU}(3)\to S^5\subset\mathbb{C}^3$ by
$$
\pi(e_1,e_2,e_3) = e_1\ .
$$
This is a smooth submersion with fibers isomorphic to $\mathrm{SU}(2)$.  Let $B$ be the set of pairs $(v,L)$, where $v\in S^5\subset\mathbb{C}^3$ and $L\in\mathbb{CP}^2$ is a line that is Hermitian orthogonal to the line spanned by $v$, i.e., $L$ is a line in $v^\perp\simeq\mathbb{C}^2$.  Then $B\to S^5$ given by $(v,L)\mapsto v$ is a smooth $S^2$ bundle over $S^5$.  If $B$ were trivial, there would be a section of $B$ over $S^5$ and hence a smooth mapping $\lambda:S^5\to\mathbb{CP}^2$ such that $\bigl(v,\lambda(v)\bigr)\in B$ for all $v\in S^5$.  This would define a smooth complex line bundle $\Lambda$ over $S^5$, and, since every complex line bundle over $S^5$ is trivial, there would be a nonvanishing section of this line bundle, i.e., a mapping $\sigma:S^5\to S^5$ such that $\lambda(v) = \mathbb{C}\cdot\sigma(v)$ for all $v\in S^5$.  However, then there would exist a unique mapping $\tau:S^5\to S^5$ such that $\zeta(v) = \bigl(v,\sigma(v),\tau(v)\bigr)$ is a special unitary frame for all $v\in S^5$, i.e., $\zeta:S^5\to \mathrm{SU}(3)$ would be a section of the nontrivial bundle $\mathrm{SU}(3)\to S^5$.
Comment: A more succinct description of $B$ is that it is the pullback to $S^5$ of the projectivized tangent bundle of $\mathbb{CP}^2$.
Comment: @Eric: Yes, but that doesn't make it obvious that $B$ is nontrivial as a bundle over $S^5$.
Comment: Thanks as always Robert.  You nailed it, and explained it very clear! Even better, the bundle you gave (in Eric W's incarnation) is exactly the one we wanted to understand (non)triviality of-R


How to punctuate abbreviation of a hyphenated middle name (Charles-Moore) on a plaque?

Question: How do you punctuate the abbreviation of a hyphenated middle name (Charles-Moore) on a plaque? Some thoughts include: C-M ; C - M ; C.-M. ; C. - M.
Comment: A hyphenated middle name? That's a first (for me that is). I would either do C. M. or just C. and assume the contraction includes Moore.
Comment: To TsSkTo -- This was a first for me, as well. I appreciate your suggestion.
Answer: I can't find a rule for this in the New Oxford Style Manual (2012), although the book gives the following rules for initials:

Initials before a surname are separated by full points, with a space after each:

J. S. Bach     -      E. H. Shepard    -      Hunter S. Thompson

although some modern designs, particualrly those of newspapers, omit
  the full points and spaces:

MR James  - PJ Harvey -   George W Bush

Normally, names given entirely in initials have points but no spaces
  (J.A.S., E.H.S., J.R.R.T.). When people are commonly known by their
  free-standing initials, these forms have neither points nor spaces
  (FRD, LBJ).

The general approach is supported by Rabinowitz and Vogel (The Manual of Scientific Style).
The Chicago Manual of Style (16th Ed, 2010) states that:

If two or more initials are used as an abbreviation for an entire name, no
  periods are needed.

Kennedy and Johnson soon became known as JFK and LBJ.

The compilers go on to state clearly:

Initials standing for a hyphenated given name should also be hyphenated.

Jean Paul Sartre; J.-P. Sartre; Sartre

However, this applies specifically to French names used in a US-English context, and they go on to say:

Since there is considerable variation in French usage, the guidelines and
  examples above merely represent the most common forms.

If cultural differences in approach are significant factors in making a choice, and linking orthography to common usage isn't sufficient to allow a sensible decision to be made, I would suggest looking at examples of usage in context.
For instance, the Wellcome Library abbreviates Kraft-Ebbing to KE. This would seem to be archiving shorthand;
the Journal of the History of Ideas Vol 62, No 1 abbreviates Kraft-Ebbing to K-E

which would seem to fit the academic style of this journal;
and the Austrian Biographical Encyclopedia (Österreichische Biographische Lexikon 1815-1950) abbreviates Kraft-Ebbing to K.-E. which is the most formal version of the three. Given the intended use, assuming maximum formality, I would personally go with C.-M..
Comment: Thanks very much for this information. After I posted this question here, I had found the Chicago Manual of Style guidance. It is nice to have you second this advice.