Book where young people around the world are chosen to go to the moon

Question: I have a vague memory of reading a book where young people around the world are chosen to go to the moon (I guess like Charlie and the Chocolate Factory or a competition or something along those line). While I don't really remember much else, all I know is they got to the moon where there was a sort of moon base, where the people invited with their accompanying relatives live. I know it's vague but I'm not getting much else from my memory.
Comment: In Heinlein's 'Have Spacesuit, will travel' there's a competition to win a trip to the Moon.
Answer: It could be Arthur C. Clarke's Islands in the Sky.
young people around the world are chosen to go to the moon (I guess like Charlie and the Chocolate Factory or a competition or something along those line).
From Wikipedia:

At age sixteen, Roy Malcolm has made himself an expert in the history of aviation, so much so that he wins the Aviation Quiz Program, presented on television by World Airways, Inc. Because the prize was described as an all expenses paid trip to "any part of the earth" (rather than on Earth), Roy is able to request a trip to the Inner Station, which is considered part of Earth because its orbit lies under the one-thousand-kilometer limit of earth's legal territory.

While on the station, he goes to glasses and activities with several cadets his age, and takes part on several missions with them. Later in the book, they go on a rocket trip which makes a pit stop on the moon.
Comment: Good thought. They only orbit the moon though, and that by accident.
Comment: The story involves multiple young people from around the world, as well as some of their accompanying relatives. If I remember correctly, they weren't actually aware that what they won was a trip to space, I believe it was actually winning a trip to a new amusement park that no one has ever gone to, where one of the rides was going to this moon base, the main ride rather.


Martingale representation theorem for symmetric random walk

Question: Let $X(t)$ be a martingale w.r.t. filtration generated by Brownian motion $B(t)$. There is a well-known theorem that states that there is a unique adapted process $H(t)$ such that
$$ X(t) = \int_0^t H(s)dB(s).$$
Let's define step process $$M(t)=2 \sum_{j=1}^{\lfloor t \rfloor} 1_{\{ B(j)-B(j-1) > 0 \}} - \lfloor t \rfloor.$$ It's just a simple martingale, generated by fair coin tosses, just written in another way. One can see that $B(t)$ is the only source of uncertainty. What is $H(t)$ for $M(t)$ then? Or am I missing something and M.R.T. is not applicable here?
Comment: $M(t)$ is not continuous. But for distrete time, we can use the stoping time $T=\inf{t:B_t =\pm 1}$ and $H(s)=1_{s\leq T}$
Answer: I've caught myself into a trap here: the reason why I couldn't find a martingal representation is that $M(t)$ is not a martingale. For example:
$$ \mathbb{E}[M(1) | M(1/2)] = 2 \mathbb{P}(B(1) > 0 |B(1/2)) - 1. $$
And if $B(1/2) \neq 0$, which is true a.s., $\mathbb{E}[M(1) | M(1/2)] \neq M(1/2).$


Magnetic Flux Pinning Whipple Shields?

Question: Would it be possible to have a collapsible whipple shields that can extend far out of a spaceship and be held in places with super conducting magnets? My rough idea is to be able to absorb more damage without the bulk with standard whipple shields, while adding more layers/distance between each metal plate, which would enable the ability to rearrange amount of protection when in combat with other spaceships.
Comment: Please don't modify your question to invalidate existing answers. You may also want to keep in mind that we have a strict one question per post policy.
Comment: Whipple shields are for particle collisions. There's no reason to suspect they are any good against weapons.
Comment: How do super conducting magnets hold something in place?
Comment: @Daron - not quite. Modern systems are designed for objects of up to 3-5 cm in diameter, so not really a particle. But then again - you're correct that it will not work against weapons. Properly designed projectile would have no trouble penetrating multiple layers and still cause huge damage.
Answer: You definitely can use magnets to keep the layers of the shield in place but what benefit does an active system like this over a passive system?
In a passive system you get a significant reduction in weight compared to a solid protection system at the cost of taking up more space inside your launch fairing.
In an active system you cut into your weight savings by adding the components necessary for the active system, while needing constant application of power to keep the shield operational. This extra power requirement needs additional hardware, to provide the power and thermal infrastructure needed to keep the system constantly powered, cutting further into the weight savings. In exchange for this you can somewhat reduce the amount of space needed inside the launch fairing.
Engineering is about tradeoffs. Passive whipple shields are designed to balance the needs of protection, mass budget, and volume. An active system may provide more protection, but if the existing system already provides good enough protection, there isn't a reason to add weight and power requirements for even more protection.
Comment: My thought was that maybe the layers could be rearranged to change amount of protection in each area, which might help cause you know where the enemy spaceship. Also I thought the prospect of being able to extend the space between layers could be useful, to help reduce the rotational inertial to make the spacecraft more maneuverable, maybe for missile dodging?
Comment: The mass of a whipple shield designed to be as low as possible making it's effect on rotational inertia limited. However the further away from the center of gravity it is, the greater rotational inertia it will have. So limited effect but extending it would make rotating more difficult not easier. You could move the shield around as needed but since whipple shields are designed to protect against projectiles smaller than 1 cm and traveling up to 18 kilometers per second. You don't have the ability to track incoming debris or time to react.
Comment: I guess what I was going for was to have the shields extend out when engaging, then pull back in when running away or maneuvering towards away. Would that be viable at all?
Comment: I don't think it makes much sense. The particles a whipple shield is designed to protect against are too small to track, and moving to fast to react to. Given how light weight they are, and that there is no drag in space there is no reason to add complexity and weight to be able to extend and retract the shield.

Magnetic Flux Pinning Whipple Shields?


How do people in the matrix use their powers?

Question: How do people in the matrix use their powers? Like for instance when Neo stops all the bullets, or Agent Smith moves super fast, or when Neo flies, or when Morpheus jumps really high. How do these people use their powers? Do they have to actively think about what they want to do (i.e. if they imagine it, it happens), do they have to move their body a certain way, etc. Basically, I am asking how to people in the Matrix use their powers. I am not asking why they have powers, or how they work. I am asking what it is like for them to use their powers; how they trigger their powers. Is there any official source for this?
Comment: The same way you get to Carnegie Hall
Comment: How do they use their powers? _Very carefully._ Interesting to ponder though. How do you use your powers? How do you walk?
Comment: @PaulD.Waite - very good question; if you ever have a pressing need to appear to be very drunk when you are in fact completely sober, try walking while thinking _very hard_ about exactly how you're doing it and what the mechanics are :-)
Comment: So, basically, how do they gauge their jump so they can go from one building to another without falling short or flying right over it?
Comment: @DavidW sort of. I was asking how they learn to actually apply their powers. Most of them never had any powers before they were red pilled, so they wouldn't know what it is like to do things like stop bullets with their mind. It is really hard for anyone to imagine how how they would use things like telekinesis.
Comment: Since they know they can jump from one building to another, how do they jump from one building to another? Like if you were in the Matrix, and you knew you could fly, how would you fly wen you want to?
Comment: @Valorum i don't understand how they can practice something they dont know how to do.
Comment: As no-one in the real world can do those things we cannot answer the question. I don't believe there has been any acknowledgement by the The Wachowskis on how it is actually done
Comment: There is not "powers". People just do what they can do. Why they can? Because they believe they do. Neo (after his breakthrough) and Agents are kinda different then "people", They simply see what can be actually done using software inherent to Matrix.
Answer: "Powers" seem to be invoked by believing that one can do something, and then just deciding to do it.

Morpheus: Do you believe that my being stronger or faster has anything to do with my muscles in this place? You think that's air you're breathing now?… Again.

....

Morpheus: What are you waiting for? You're faster than this. Don't think you are, know you are…. Come on. Stop trying to hit me and HIT me.

....

Morpheus: I'm trying to free your mind, Neo, but I can only show you the door, you're the one that has to walk through it. Tank, load the jump program…. You have to let it all go, Neo, fear, doubt, and disbelief. Free your mind.

And later, when Neo starts coming into his power, and is moving faster than was believed possible:

Morpheus: He's beginning to believe.

One might question why one can't simply believe six impossible things before breakfast and hit the win button, but the world does still have rules.

Morpheus: I've seen an agent punch through a concrete wall. Men have emptied entire clips at them and hit nothing but air. Yet their strength and their speed are still based in a world that is built on rules. Because of that, they will never be as strong or as fast as you can be.

Basically, to do something, you have to actually be able to do it within the rules of the code. Neo, due to his connection with the Source, gets to essentially reprogram the rules as he goes (although he does still seem to have some limitations preventing him from being able to push an "I win" button), has a slight edge, but it seems he is still limited by his mortal preconceptions as to what can happen.
Comment: Similarly, the humans in the Matrix gain understanding of programming (which underlies how this world works), refine their own bodies, and eliminate their mental blocks, and they can leap between buildings or even fly, same as we do in dreams, but just doing it.
Comment: Honestly, a lot of it is a *wuxia*/cultivation kind of deal where self-refinement and attaining arcane knowledge of how the world *really* works gives you powers that essentially work intuitively. Within regular athletics, you see examplars who employ regular exercise to improve their body, knowledge of body mechanics to work most efficiently, and bypassing mental blocks that normally keep us from quite hitting our true potential. And then, they just run and jump, and they break a world record in the long jump.
Answer: As The Architect explained: Neo is The Anomaly - the personification of the imbalance in the equation, which is the option of the choice for all enslaved humans. This imbalance comes from the need to create perfect Matrix, which by definition cannot be perfect as it's used to host fallible human consciousnesses.
In other words: Neo is Humanity's champion (The Choice Personified) and his power is the ability to rewrite code in a way he sees fit (but as he's still human being, he is limited by his own imagination still) and that power comes - indirectly - from all humans in the Matrix. He uses that power by simply changing locally the rules of the Matrix to suit his needs.
It may be also called "a godmode" or "cheat codes" on the fly.
In  yet another words - Neo is something of a system process with access to the core of the system and all it's resources, which cannot be terminated - as in termination will cause crash, and consequently (yay! Channelling the Architect ;-) the death of all humans in the Matrix - and system's credentials to control that Neo process are no longer valid...
Comment: @lucasbachmann - And if we take Matrix: Resurrections as canon, then it stands to reason Machines use humans for creative work, as well. It's not just distributed computing, it's also solving complex problems that needs some OOB... As for The Spoon Kid - yep, he said as much.
Comment: Largely agree. I'd speculate- The human "batteries" are a distributed computer that are collectively imaging the Matrix as roughly an operating system and doing whatever computations the machines want/need as the primary program.  Anyone - like no spoon kid- strong enough should be able to alter the Matrix locally because they are the ones generating it.
Answer: I believe that it is based off of code that already exists. The code of the one (the prime program) existed 5 times before Neo came along. The code also included a way for people to free their minds (Zion is part of the control scheme). So, they do not break any rules, since the breaking of the rules is enabled by the Matrix itself. As such, everyone who can potentially use the powers have the code/knowledge in their mind of how to use them. Also, one could simply download a program into their head in order to use their powers.
Comment: the helicopter program for trinity I believe was such thing
Comment: They _could_ download something like that, but _do_ they?


Number of antenna to perform MUSIC algorithm

Question: When the MUSIC algorithm is used for direction of arrival (DOA) estimation it can identify up M-1 signals, where M is the number of array elements. MUSIC can also be used for spectral frequency estimation. Does MUSIC require more than one receiver (or more than one receive antenna) for frequency estimation of a single signal?
Comment: No, but for spectral estimation, MUSIC needs multiple observation samples in the discrete time domain, just like the similar requirement in the spatial domain for DOA estimation.
Comment: When you say multiple observation samples, do you mean contiguous samples? Or multiple contiguous observations? For instance, if you have a continuous wave signal that is always transmitting, how would you go about constructing the correlation matrix (Rxx) with one antenna?
Answer: In the field of array signal processing, the signal model is
$$
\mathbf{x}(n) = [x_0(n), x_1(n), \ldots, x_{M-1}(n)]^T
$$
where $M$ denotes the number of sensor and $n$ is the discrete time.
While in the field of spectral estimation, the signal model is
$$
\mathbf{x}(n) = [x(n), x(n-1), \ldots, x(n-M+1)]^T
$$
where $M$ is the number of samples of the observed signal, and $x(n-m)$ represents the previous signal before $m$ samples.
The autocorrelation matrix is
$$
\mathbf{R}_{xx} = E\{ \mathbf{x}(n) \mathbf{x}^H(n) \}
$$
and it is usually replaced by its estimation $\mathbf{\hat{R}}_{xx}$ in practice.
Comment: How well does this work if a signal has a bandwidth (e.g., .25 normalized frequency) ? All the examples I have seen are for ideal sinusoids, which create sharp peaks in the MUSIC spectrum.
Comment: @BigBrownBear00 I think it doesn't work well for a wideband signal. Similar with DOA estimation, what if a sound source has a certain width? Theoretically, signals with a bandwidth (line sound source or surface sound source in DOA) can be decomposed into infinite single frequency components (point sources in DOA).


Prove or disprove that a series is convergent

Question: I was given the following task which I struggle with.
Prove the following statement, or disprove it by giving a counter example:
if $\sum_{n=1}^\infty a_n$ is convergent then $\sum_{n=1}^\infty (a_n)^3$ is also convergent.  I think the statement is true. My intuition is that $|a_n|$<1 starting some index $N$ (since the series converges). Therefore, starting $N$, $|(a_n)^3|$ is smaller than $a_n$, therefore $(a_n)^3$ is "closer" to zero, and since the original series converges, the new one should also converge, yet I can't prove it (nor can I think of a counter example). Any help would be appreciated!
Answer: Consider the sequence:
$$1-\frac{1}{2}-\frac{1}{2}+\frac{1}{\sqrt[3]2}-\frac{1}{2\sqrt[3]2}-\frac{1}{2\sqrt[3]2}+\frac{1}{\sqrt[3]3}-\frac{1}{2\sqrt[3]3}-\frac{1}{2\sqrt[3]3}+\frac{1}{\sqrt[3]4}-\frac{1}{2\sqrt[3]4}-\frac{1}{2\sqrt[3]4}+\ldots$$
It clearly converges to $0$, but the sequence of its cubes does not (by combining terms you can see that the sequence of its cubes becomes $3/4$ times the harmonic series).
Comment: Since there are two identical answers to this question. I wonder if your solution is from yourself or from a solution book you happened to have?
Comment: It's my own. The motivation is that when you try to prove the statement using your method, you realize that you run into a problem with positive and negative numbers, so you try to make all the negative numbers "closer" to 0 than the positive numbers, so that cubing them makes their absolute value even smaller.
Comment: Good job. +1 for innovative answer.


League based on trophies

Question: How am I still in Crystal III League if the minimum trophies for crystal are 2000 while I just have 1981  trophies?
Answer: I guess it works like  this:
Crystal III
Trophy range: 2000 - 2199
Demoted to a lower league if trophy count drops to 1900
Promoted to a higher league if trophy count rises to 2200
https://forum.supercell.com/showthread.php/37226-How-Leagues-work
Comment: I believe this is the correct answer based on my experience. I always drop a league when I'm 100 trophies or more below the threshold for joining the league I was originally in.
Comment: Actually yeah I just experienced the same yesterday.. thanks @Ellesedil
Answer: When you are placed in a league, you are in that league until you graduate or flunk. You graduate by gaining the required number of trophies for the next league, and you flunk by dropping below the current #100 player in your league OR dropping below 100 trophies less than the required number for your current league.
Source: Personal experience, @Ellesedil
Comment: Do you really drop leagues if you drop out of the top 100? I've always noticed my league decreases when I drop 100 trophies below the minimum threshold for joining the league in the first place. For example, if I need 2400 trophies for Crystal 1, and I drop down to 2300 trophies or less, I'll drop down to Crystal 2.
Comment: You might be right, but then both conditions are in place. Editing. Just never happened to me before!

League based on trophies


Integral with Branch Cut: $\int_{0}^{\infty}\frac{\sqrt{z}}{1+z^2}$

Question: While evaluating the integral $\int_{0}^{\infty}\frac{\sqrt{z}}{1+z^2}$ we use branch cut.Such a shame that I couldn't draw the complex plane with diagram, but I hope you would understand my explanation to the problem. The two horizontal parallel lines above branch cut would result the same integral which we can add, the bigger circle (contour) would vanish because of jordan's lemma if we take the limit as radius tends to infinity, but why the smaller circle with centre at branch point 0 would vanish as its radius tends to 0.
Comment: $\left| \int_\Gamma f(z)\; dz\right| \le M L$ where $|f(z)| \le M$ on $\Gamma$ and $\Gamma$ has length $L$.
Comment: By the estimation lemma.
Comment: But, how would I prove it for this integral.
Answer: Estimation Lemma:
Let $M=\sup_{z\in\gamma}|f(z)|$ and let $L = \int|\gamma'(t)|dt$ be the length of the curve $\gamma$. Then
$$
\biggl\lvert\int_{\gamma} f(z)dz\biggr\rvert \leq ML
$$
By definition
$$
\biggl\lvert\int_{\gamma} f(z)dz\biggr\rvert = \biggl\lvert\int_{\gamma} f(\gamma(t))\gamma'(t)dt\biggr\rvert\leq\int\lvert f(\gamma(t))\rvert\lvert\gamma'(t)\rvert dt\leq\int M\lvert\gamma'(t)\rvert dt = ML
$$
As $\epsilon\to 0$, what does the length go to?
Comment: @RoshanShrestha if you aren't a math student, then professor should be okay with you saying integral goes to zero when epsilon goes to zero by the estimation lemma since he is a professor of physics, engineering etc.
Comment: Thanks @dustin and everyone others, since I am a student of physics
Comment: Sorry, but I am not a maths student, so I don't think I will be able to understand the topology statements, so it would be great if there are any other simple explanation for it with regard to this problem.
Comment: @RoshanShrestha the sup is the supremum which is the least upper bound. Let's call it M. Next let's let the length of the small circle be L. We have that the integral is less than or equal to ML where L is the length. When $\epsilon\to 0$, the length goes to zero. So the integral goes to what?
Answer: For the smaller circle about the origin, set $z=\epsilon \, e^{i \phi}$.  Then the integral about the smaller circle is
$$i \epsilon \int_{2 \pi}^0 d\phi \, e^{i \phi} \frac{\sqrt{\epsilon} \, e^{i \phi/2}}{1+\epsilon^2 e^{i 2 \phi}}$$
As $\epsilon \to 0$, this integral goes to zero as $\frac{4}{3} \epsilon^{3/2}$.
Comment: Now, that's the explanation I needed, Thanks.......
Comment: Shouldn't the integral be through an incomplete circle?
Comment: @PedroTamaroff: Technically, yes, but the correction gained by introducing such a complication would result in a smaller order error.


Help please with this SMD identification - SOT23-6 top marking F1TL

Question:
I need some help identifying this SMD part. It has F1TL marking and is on a 3-axis gimbal daughter board. There's 3 on the board and 2 are burnt. Thanks in advance. [6-pin SMD, 3x2mm in size approx.]
2 more image's, hope they help.I believe the gimbal ended up in the sea, the battery pack suffered, (replaced), I need to replace these so as to aid further testing. Strip down of the unit could be destructive.

Thank you for the help, I'll do some more testing and see if the units repairable or scrap. Thanks again.
Comment: What does it connect to? Voltages?
Comment: Paul Rimmer - Hi, That looks like a SOT23-6 package so I added that & the main part of the top marking into the title, to help future searches. Can you also add a photo showing more of the PCB, to give some more context? Also, have you tried doing any reverse-engineering of the schematic, to help show how the unknown components are part of the overall design? If so ,please [edit] the question & add that image. The [current photo](https://i.stack.imgur.com/2XXUr.jpg) shows the unknown device close to a connector. Are all 3 devices close to connectors? Do you know what event damaged the devices?
Comment: Paul Rimmer - Thanks for the update. (a) The photos show all 3 devices are near the same connector, so if the accident led to excessive / wrong polarity voltage on that connector, this could explain why 2 of the 3 devices were clearly damaged (and the 3rd one might be damaged internally too). (b) Now seeing the full PCB, it's not possible to reverse-engineer the relevant parts remotely using just the photos. However where we can see the PCB tracks going to those devices, we see that the tracks to pins 1 & 3 are thinner than others. That fits with the identification in [the answer](/a/609156).
Answer: Most likely it is a AO6601 MOSFET, a complementary n and p channel pair in a single package.
Here is a package marking description.
And here is a datasheet.
Comment: Nice and welcome.


circuit for quantum simulation?

Question: What would be the circuit for operation exp(iθZ⊗Z⊗Z⊗X) by only using CNOTs and single-qubit gates. And How we can improve the circuit to implement the operation exp(iφZ⊗Z⊗X⊗Z ).exp(iθZ⊗Z⊗Z⊗X).Are there are multiple ways to do it?
Answer: Try this circuit:

How did I come up with this? The shortcut is that I recognised the $Z\otimes Z\otimes Z\otimes X$ as the stabilizer of a graph state, and that has certain standard rules for building it: apply a controlled-phase between the place of the $X$ and each of the $Z$ terms. In other words, if I group those three controlled-phase gates as "$U$", then
$$
U(Z\otimes Z\otimes Z\otimes X)U^\dagger=I\otimes I\otimes I\otimes X,
$$
so
$$
Ue^{i\theta Z\otimes Z\otimes Z\otimes X}U^\dagger=e^{i\theta X_4}=I\otimes I\otimes I\otimes R_X(2\theta)
$$
(depending on your convention for what the rotation matrix means.) Also, note that $U=U^\dagger$, so inversion is easy.
Now, you actually asked for controlled-nots instead of controlled-phases. We can convert between the two using Hadamards, and you have a fairly arbitrary choice of which qubit will be the target for each of the gates. A particularly efficient version is to make the bottom qubit the target every time.

If you didn't have that shortcut, how might you have gone about constructing this? My first reaction was to look at the $Z\otimes Z\otimes Z\otimes X$, and apply Hadamard to the last qubit, so that you get $Z\otimes Z\otimes Z\otimes Z$. What does this observable detect? The parity of the number of 1s in a state. So, if you can find a circuit that computes the parity of the number of bits, you're effectively calculating the value of that observable onto a single qubit, so single qubit gates can add the phase that you need, and then you can uncompute the value. Controlled-nots are great at computing the parity of bit values.
As for your last question, for any Hamiltonian $H$,
$$
e^{i\theta H}e^{i\phi H}=e^{i(\theta+\phi)H}
$$
so instead of running your circuit twice, you only run it once, with an altered value of the phase rotation.
Answer: You can use a Suzuki-trotter approximation to your Hamiltonian $e^{i\theta ZZZX}$ using two-qubit gates as mentioned in Qiskit's operator-flow docs. Below is a sample code to perform the same (Ref).
<code>def compute_U_trot(H, time, trotter_steps, order=1):
    """
    Apply the Suzuki-Trotter approximation to the Hamiltonian H to
    compute a trotterized unitary time evolution operator.

    """

    U_trot = Suzuki(trotter_steps, order=order).convert(time * H)
    return U_trot

def convert_U_trot_to_qc(U_trot):
    """
    Convert the U_trot ComposedOp object returned by compute_U_trot()
    to a bare quantum circuit.
    """

    qc_trot = PauliTrotterEvolution().convert(U_trot)
    return (
        qc_trot.to_circuit().decompose().decompose()
    )  # add decompose methods so gates are not abstracted unitary blocks

H = PauliSumOp(SparsePauliOp(["ZZZX",],[1]))

time_param = Parameter("θ")
trotter_steps = 1

# Compute unitary time operator
U_trot_params = compute_U_trot(H, time_param, int(trotter_steps))

# Convert to trotterized unitary to circuit
qc_trot = convert_U_trot_to_qc(U_trot_params)
</code>
which gives the following circuit:

The other hamiltonians can be implemented in the same way.

circuit for quantum simulation?


SuiteCRM Permissions Issue

Question: On an Ubuntu 14.04.05 (Trusty) server running in AWS, I am trying to install SuiteCRM. The webserver is Plesk.
The directory is /var/www/vhosts/domain.com/httpdocs/crm
I cannot get the installer to go past the system check: it says '/cache' is not writeable. I have set chmod -R 775 on /cache and it still fails.
Browsing around, all instructions I can find say to chmod 755 or 775, yet neither work.
Owner is root (as is all the other files in the folder)
I don't know what to do next - any suggestions?
Oh, and the 'help' and 'installation guide' and 'support forums' links? USELESS - not a single reply at all after several attempts.
Answer: Ownership was the issue. No web directory should be owned by root! should have seen that myself (duh)
I did chown -R webadmin:psacln crm and the rest worked like clockwork.

SuiteCRM Permissions Issue


jssor thumbnavigator not completely hidden when $ChanceToShow=0

Question: In JSSOR 26.5, using $ThumbnailNavigatorOptions: {$ChanceToShow: 0} is supposed to completely hide the thumbnails.  However, the display:none is applied only to the thumbnavigator element, NOT the container that is injected around thumbnavigator.
As a result, the absolutely-positioned container renders atop the slider.  It's invisible because its only child (the thumbnavigator) is not displayed.  But it intercepts click events, so it is having an effect on my UI.
In one case, the navigator (bullets) are rendered "below" the hidden thumbnails, which makes the bullets not clickable.
In version 19.0.1. there was no wrapper injected around "thumbnavigator", so the problem was not there.  We moved from 19 to 26, and that is where I see the problem.
Solution is to put the display:none on the injected thumbnavigator wrapper, rather than on thumbnavigator itself.
Answer: Please resort html order to move bullet navigator above thumbnail navigator.

    ...

    ...

Comment: I can do that for now, but going forward, wouldn't it be better to put the display:none on the injected thumbnavigator wrapper, rather than on thumbnavigator itself?
Comment: I see. I think bullet navigator should be always above thumbnail navigator. Also, in some cases, there is no wrapper for thumbnail navigator.


Modified or Macauley Duration in python

Question: are there any existing python modules that can calculate Modified and/or Macauley Duration of a bond.
Answer: I calculate duration in Python using numpy, it's nice and simple:
def durations(cfs, rates, price, ytm, no_coupons, payments_per_year=2):
    import numpy as np
    mac_dur = np.sum([cfs[i]*((i+1)/payments_per_year)/np.power(1+rates[i],i+1) for i in range(len(cfs))])/price
    mod_dur = mac_dur/(1+ytm/no_coupons)
    return mac_dur, mod_dur

Comment: You have an error in your calculation your range function should start in 1 but because you are iterating in an array this will cause index out of range, when you say cfs[i]*i should be * (i+1) because the first i is 0 and you are loosing then the first number the same in your power method
Answer: Go talk to Fincad. Here is their page on integrating with scripting languages:
http://www.fincad.com/news-events/assets/pdfs/mar07/using-fincad-developer-scripting-languages.pdf
Their analytics libraries include bond analytics, and they have a spreadsheet product so you can test methods and results before implementing them.
Disclaimer: I work for a company who is a customer of Fincad's analytics.
Answer: You can use my script:def Duration (timetomaturity,nominalvalue,yieldrate,couponrate):    import math as m    yld=yieldrate/100    cpnr=couponrate/100    t=list(range(1,timetomaturity+1))    cfi=nominalvalue*cpnr    cfN=nominalvalue*cpnr+nominalvalue    cfl=[cfi]*(len(t)-1)+[cfN]    B=0 # B is the bond's present value    for k in range(0,timetomaturity):        B=B+cfl[k]*(m.exp(-yld*t[k]))    D=0 # D is the duration    for i in range(0,timetomaturity):        D+=(t[i]*cfl[i]*m.exp(-yld*t[i]))/B    return round(D,2),round(B,2)#Duration(5,100,1,1)#By Tural Valiyev
Answer: Example: Consider a 7% bond with 3 years to maturity. Assume bond is selling at 8% yield.
Most concise function for explicit formulas:
def durations_explicit(c, y, m, n):
    macaulay_duration = ((1+y) / (m*y)) - ( (1 + y + n*(c-y)) / ((m*c* ((1+y)**n - 1)) + m*y) )
    modified_duration = macaulay_duration / (1 + y)
return macaulay_duration, modified_duration

durations_explicit(c=.035, y=.04, m=2, n=6)
(2.75371702450442, 2.6478048312542497)

For a more complete answer with docstrings and accounting for the case when coupon rate per period equals yield per period
def durations_explicit(c, y, m, n):
    """Parameters:
        c = coupon rate per period
        y = yield per period
        m = periods per year
        n = periods remaining"""
    if c==y: # Shorter explicit formula if coupon rate per period = yield per period
        macaulay_duration = ((1+y)/(m*y))*(1 - (1 / (1+y)**n))
        modified_duration = macaulay_duration / (1 + y)
        print(f"Macaulay Duration: {macaulay_duration}")
        print(f"Modified Duration: {modified_duration}")
        return macaulay_duration, modified_duration
    else:
        macaulay_duration = ((1+y) / (m*y)) - ( (1 + y + n*(c-y)) / ((m*c* ((1+y)**n - 1)) + m*y) )
        modified_duration = macaulay_duration / (1 + y)
        print(f"Macaulay Duration: {macaulay_duration}")
        print(f"Modified Duration: {modified_duration}")
        return macaulay_duration, modified_duration

durations_explicit(c=.035, y=.04, m=2, n=6)
Macaulay Duration: 2.75371702450442
Modified Duration: 2.6478048312542497
(2.75371702450442, 2.6478048312542497)

Comment: I agree, but I wanted to stay consistent with "Investment Science" by Luenberger. He flips the usage of "yield to maturity" using "lambda", and using "yield per period" using "y" as in the formula.
You have to adjust mbudda's formula by dividing by semi-annual ((i+1)/2) or dividing his final result by 2 to get the same duration calculations as the explicit formulas. In this manner I thought it would be best to have the individual supply with their payment structure upfront.
Comment: Very nice. But your input definitions seem somewhat non-standard, for example coupon rate is usually expressed "per year", not "per period".
Answer: You can look at the answer I gave here: https://quant.stackexchange.com/a/61025/40827 which is based on potentially irregular cashflows. Ie you pass the exact dates or periods and amounts, you don't specify, say, a bond paying 5% every semester for the next 3 years.

Modified or Macauley Duration in python


Living room breaker will not turn on after pressure wash

Question: Front of the house pressure washed. Water got in the 2 outlets outside that caused the living room breaker to trip. We waited 24hrs for it to dry out. Nothing. Replaced with a new breaker and still no power in the living room or porches. Its been over 48hrs and its still not working.
Comment: Did the breaker have a "test" button on it?  If so, it's GFCI.  You have a simple cause/effect issue here. The house got pressure washed and now some electrical doesn't work. The obvious place to start is with all outdoor fixtures and outlets. Are the outdoor outlets GFCI? Somebody who didn't know what they were doing might have wired some inside outlets/fixtures from the load side of the GFCI.   Based on the symptoms, you have water in your outdoor fixtures and/or outlets.   Lastly **breakers seldom fail**,  you are throwing parts at this without a proper diagnosis.
Comment: Did you open the outside outlets so they could actually dry, or just wait?
Comment: We opened the outlets and took the bulbs out of the porch lights.
Comment: You should replace the outside outlets (with some with RCD if you're in the US). Also check the light fixtures outside, those may well be flooded.
Comment: What you mean by nothing?  The breaker won't turn on or no power at the outlets?  Is there any GFCI outlets on the circuit?  Outside outlets are suppose to be protected by GFCI,  Outdoor GFCI outlets are known to fail sooner than indoor GFCIs.
Comment: This just happened to me, so I went and turned on and off all the surge protectors I had, I also turned the breaker for the microwave on and off as well as the back bedroom (where it had partial power) and that did the trick. When you said 'there are weird pathways for wiring, you weren't kidding.
Answer: Hire an Electrician
It's time to get a licensed pro to fix this.
Yes, this is a DIY forum, but for someone who only knows to wait 24 hours and then replace a circuit breaker, it's time to get real help.  At the very least, the exterior portions of the circuit need to be disassembled and inspected.  If the cables were submerged or otherwise water damaged, they might need to be replaced.
Comment: +1 if your thought is ever "replace a circuit breaker" (basically always wrong and probably **dangerously wrong** if that's your thought) it's time to stop thinking about doing anything yourself and call a professional.
Comment: Yep, throwing parts at problems without a proper diagnosis is problematic.  It happens A LOT repairing cars, even socalled professional mechanics will do that.  I've helped others fix cars after the "pros" worked on them and charged hundreds of $$$ and have too many   horror stories to get into here.
Comment: "Hire" is probably the easiest answer, however, if the OP gives us more detail, we can certainly help trouble shoot, and that will most likely be cheaper _and_ the OP will learn some new skillz!
Comment: You might have simply disconnected the wire from the old breaker, to see if it stays on by itself.  If it does, and pops when you connect the wire, it is almost certainly working perfectly.   There are some logical steps you can follow to divide and conquer this problem.  I'm afraid I agree with this answer because you don't seem to have the knack for basic fault finding in a fairly simple system.  You'll waste a lot of money and time and you could get hurt or hurt someone else if you don't learn the basics before trying random things.
Answer: There is still one thing that isn't clear to me.
You said you replaced the breaker but still no power on that circuit.  Does that mean the new breaker immediately re-trips when turned on, or does it remain on?

If the breaker immediately re-trips, you have a short somewhere in that branch circuit.  This is unlikely to be caused by pressure-washing, especially since you already looked inside the exterior outlet boxes.  Could be a result of something that was put back incorrectly, however.

If the breaker remains on, then something on that branch circuit is getting power, just not all of it.  If that is the case, I would check in other rooms, basements, garages, etc. for another GFCI outlet that is tripped.  Sometimes branch circuits travel an odd route and serve a lot of devices on the way.  In one of the houses I used to own, all the exterior lights, exterior outlets, and the entire unfinished basement shared one 15A circuit.  The GFCI for that circuit was..... drumroll........ in the crawl space.  Took us forever to find that.
Comment: Just wanna thank you a year later for this simple explanation/ debugging guide. We lost power to our fridge and dishwasher with all the breakers working fine. Lo and behold, they share a circuit with some outdoor outlets by the deck , which I was washing with the hose earlier today.
Answer: If you can replace a circuit breaker, you can open up a receptacle box and inspect it / clean it out.
That's what you have to do on all the receptacles, because obviously pressure washing either placed or activated materials that are not curing by being given 24 hours to dry. You may have a mud-wasp condominium now soaked with power-wash solvent.  So-called "weatherproof" boxes aren't quite, and they're certainly not powerwash-proof.
Replacing breakers is tricky. You need to make sure you are getting a breaker type correct for your panel - they are not interchangeable across brands, though they seem to be since they "seem to fit".  (they don't clamp on right, which the critical problem; this can create an arcing fire at the bus stab).
Further, if a breaker was GFCI or AFCI, it is so for a reason and must not be downgraded.
Comment: Per comment response above, sounds like the receptacle boxes were opened, but the light fixtures were not `We opened the outlets and took the bulbs out of the porch lights` so opening the porch light fixtures seems reasonable as a next step - or finding where the outside connects to inside and disconnecting it in "isolate the trouble" fashion.


NUMBERS: Lookup hourly rental rate based on date of transaction

Question: I have a resource that has an hourly rental rate. That hourly rate has changed (increased) over time. To capture that rate history, I have created a table with the date of the rate change and the new rate that looks like this:
Date of rate change     | Rate
2015-01-01              | $60
2017-01-01              | $65
2017-07-13              | $68
I am trying to create a formula that will look up the hourly rate based on the date of the transaction/rental and then determine what the rental rate would have been on that date. I figure there must be some way to do this with a vlookup but from the documentation, it looks like that function requires an exact match and I want to match based on the input date being after the date in the lookup table not an exact date.
Any ideas?
Answer: Ok, so now had chance to test, use this formula :
<code>=IF(C2<=A4,B4,IF(C2<=A3,B3,IF(C2<=A2,B2)))
</code>
The date tested is in cell C2 and the three dates are in A2, A3 and A4, while the three rates are in B2, B3 and B4.
The logic has been reversed ie test for the oldest date first.
You could also use names for the 3 rates, but it will work the same.


Is there a way to realize two finger dragging and three finger rolling?

Question: Three finger is more dull than the two finger, which is more suitable for simple up and down scroll sliding, while the two finger occupy less area of the touch panel, and the resistance during sliding is less than the three finger, which is more suitable for more complex drag actions. Is there any way to exchange the default three finger drag and two finger scroll?
Comment: Have you tried Better Touch Tool?
Comment: @Gilby I've tried BTT. But as Brandon Lewis answered, it doesn't seem to work.
Answer: Unfortunately, no. The only thing that might theoretically make that possible is BetterTouchTool, but at the moment, it cannot remap scrolling and screen-swapping.
Comment: Thanks! I tried BTT as you suggested. Actually it does able to remap scolling and I made a three finger scrolling profile. Unfortunately, it scrolls pixel by pixel


Why rogued certificate from hash collision is harmful

Question: I cannot understand why rogued certificate is thought harmful though It can be used to impersonate some website. The attacker can craft a rogued certificate with different identities and public key but the same signature through hash collision, such as md5 collision. But the attacker cannot aquire the private key corresponding to the public key in the rouged certificate. So in the handshake stage of SSL, the attacker can't decrypt the PreMasterKey encrypted by the public key, then the negotiation will be fail.
Can anyone tell me where I was wrong. Truly appreciated.
Comment: What makes you think that the attacker cannot acquire the private key?  Depending on how to generates the collision, the public key can be either a value he selected (in which case he can know the private key), or a known random value (and a nontrivial fraction of random 2048 bit values can be factored, being a smooth number times a large prime).
Comment: [Read this.](http://www.win.tue.nl/hashclash/rogue-ca/) It talks about exactly how security researchers actually created a faked CA certificate and legit certificate colliding pair, got the legit certificate signed by a real CA (who was not in on this), and then put that signature on the CA cert to get a real signed CA cert for which they had the private key and which, had they not intentionally made the CA cert long-expired, they could have used in actual attacks.
Answer: Very basically, the hash collision it can be used to fool CA's to sign a certificate that stores one set of information in the certificate's attributes, while there the requester uses it to create one or more certificates with different attributes.
The attributes in the certificate are as important as the public key that is being signed. They contain, for instance, the information of the holder. For web certificates the most important holder attribute is the domain name of the server. It is of course of vital importance that the name cannot be changed. Another part of the attributes is the key usage, which defines how the key should be used. One very important bit is the bit that defines if the certificate can be used as CA certificate. If it can be set then the user may simply start acting as an intermediate CA, issuing certificates himself, for any domain.
The public key stored in the certificate is only one part of information that is signed. That you can also sign another public key is not so much of a problem as the private key still needs to be controlled by the party sending the certificate request. This would only be a problem if the location of the private key is somehow part of the security properties covered by the certificate.
All in all, you seem to be focusing too much on the Public Key part and too little on the Infrastructure part of PKI. The attack isn't really a problem during the handshake of TLS, the attack is on the trusting of the certificate. The PKI method of establishing trust is used by TLS, it is not defined by it.
Comment: Follow the link @cpast gave me, I found the paper **"MD5 considered harmful today"**. It's really useful. Now I know the attack scenario where rogued certificate is used.
Comment: Side node: it may be that the maximum path length is controlled by the CA certificates. In that case a well implemented client should not accept a leaf certificate for which the path length is out of range.


What's the limit for the size of iPhone apps?

Question: I have an app which is quite huge. It has tons of images and sounds. In total it's 30 MB in size.
1) Is there an official limit for the app size?
2) Are there other "practical" limits I should consider?
Answer: Reading around, there are reports that the absolute maximum size for an application is 2GB (http://discussions.apple.com/thread.jspa?threadID=1605342).
Other than that, if your app is over 10MB, users will only be able to download it over a wifi connection, having the impact that users won't be able to download your application easily when they are out and about.
Answer: This might sound obvious, but there are two size limits to keep in mind here - neither are absolutely defined.
One is the limit for the application bundle. I don't know if there is a real limit (other than the device's flash storage limit), but this should rarely be an issue. Keep in mind the 10Mb threshold that the other Phil mentions though. There may be something in the 2GB limit too, I don't know.
The more important limit is on the size of the loaded binary image and it's runtime memory usage. The iPhone is obviously a constrained device in this area. The original iPhones and iPod touches had 128Mb of total RAM, with no paging. Current models have 256Mb, but it's a bit early to limit yourself to that subset of the installed base if you can help it.
Assuming 128Mb, of that a fair chunk is used by the system itself and its background tasks (and people want user background tasks too!?!).
In a conversation with an Apple Engineer he said that apps should be written to use no more than about 25-30Mb! Up to 60Mb may be available but cannot be counted on (but you could use this extra for caching stuff that you can purge as necessary).
Answer: 1) I believe that you shouldn't ask for maximum size, as you have to aim for minimum possible size for your apps. As iPhone has limited memory (both storage & RAM), you should not create a huge app. (I doubt an app > 200MB can pass through Apple's app check.)
Yeah, Phil is right, when you app is over 10MB, it can only be downloaded via wifi or iTunes. Quite a lot users may want to download anywhere they like using 3G networks, so >10MB is not recommended.
Also, take care those users with iPhone 2G. If you produce a >10MB app, it takes them a day to download your app (if they are lucky enough to have no disconnection at the middle of download...)
2) practical limit ... K.I.S.S = Keep It Simple and Small.
That's it. Hope it helps.
Comment: There are several apps in the App Store well over 200 MB. The TomTom app is 1.2 GB, and I've seen several offline Wikipedia readers that are 800+ MB.
Comment: K.I.S.S = Keep it Simple Stupid :)


Finding mixed Nash equilibria in continuous games

Question: I'm taking my first (graduate-level) game theory class. I understand how to find Nash equilibria in simple games, such as those given in finite tables, and can see (usually) how to find the mixed equilibria in those cases.
However, most of our problems have been discussing continuous games such as auctions, firm pricings with continuous prices, etc. Is there a good method of knowing when there exist and finding the mixed Nash equilibria in such cases?
The pure equilibria are generally easy to find, but the only thing I can think of in most cases is to check uniform distributions over all possibilities.
Comment: There is no really general method. What is usual done is that you guess the support of the strategies, plug in the condition that everything in the support should give the same payoff and hope that you can solve the resulting functional equations.
Answer: In general, when you face with the problem of finding mixed Nash equilibrium in a 2-player game, you must use the best response functions (BRF). With the BRFs you can solve both finite table games and continuous game.
Let $S_1$ and $S_2$ be the sets of strategies for player 1 and 2 respectively, and let $x_1 \in S_1$ and $x_2 \in S_2$ the strategies played by each player. The payoff functions are $f_1(x_1, x_2)$ for player 1 and $f_2(x_1,x_2)$ for player 2.
The BRF of the player 1 $\beta_1(x_2)$ is a "function" which return the best strategy(ies) that player 1 must choose when player 2 plays a given strategy $x_2$. It is a "function" since for every $x_2$ there can be more than one best strategy for player 1. It is more correct to say that $\beta_1(x_2)$ is a set.  The same construction is used for $\beta_2(x_1)$. Namely we have the following:
$$ \beta_1(x_2) = \{ x_1 \in S_1 : f_1(x_1,x_2) \geq f_1(y, x_2) ~ \forall y \in S_1\}$$
$$ \beta_2(x_1) = \{ x_2 \in S_2 : f_2(x_1,x_2) \geq f_2(x_1, z) ~ \forall z \in S_2\}$$
Once you have build the BRFs, you have to solve the following system:
$$ \left\{ \begin{array}{l} x_1^* \in \beta_1(x_2^*) \\ x_2^* \in \beta_2(x_1^*) \end{array}\right.$$
All the solutions $(x_1^*, x_2^*)$ are Nash equilibria. They can be mixed or pure.

Example with continuous strategies
A classical example is the Cournout duopoly. Two firms produce the same good and they act on the same market. They must decide the quantities $x_1$ and $x_2$ of good that they have to produce. Produced quantities must not be greater than the demand $D$. We have that $S_1 = S_2 = [0, D]$.
The firms are different in the sense that they have different cost of production (say $c_1$ and $c_2$ are the unitary cost for firm 1 and firm 2 respectively). Payoff functions are:
$$f_1(x_1,x_2) = k(D - x_1 - x_2)x_1 - c_1x_1$$
$$f_2(x_1,x_2) = k(D - x_1 - x_2)x_2 - c_2x_1$$
where $k$ is a positive constant.
The way to evaluate the BRF is to maximize the payoff functions w.r.t. own strategy when the opponent strategy is fixed. We use the derivative to maximize (note that, since $k>0$, then each payoff function has second derivative negative and this guarantees that the stationary point is a local maximum):
$$\frac{\partial f_1}{\partial x_1} = k(D - x_2) - 2kx_1 - c_1$$
$$\frac{\partial f_2}{\partial x_2} = k(D - x_1) - 2kx_2 - c_2$$
and we equate them to 0 in order to find the maximum:
$$ \left\{ \begin{array}{l} \frac{\partial f_1}{\partial x_1} = 0 \Rightarrow x_1 = \frac{k(D-x_2)-c_1}{2k} = \beta_1(x_2) \\ \frac{\partial f_2}{\partial x_2} = 0 \Rightarrow x_2 = \frac{k(D-x_1)-c_2}{2k} = \beta_2(x_1) \end{array}\right.$$
The last equations on the right holds since, for every $x_2$ ($x_1$) fixed we can find the best $x_1$ ($x_2$) that firm 1 (2) can adopt. It is worthwhile to note that in this case the BRFs are real function, since there is a 1-on-1 correspondence between an opponent strategy and the best response to it.
At this point, we can solve the system:
$$ \left\{ \begin{array}{l} x_1* = \beta_1(x_2^*) \Rightarrow x_1^* = \frac{kD-2c_1+c_2}{3k} \\ x_2* = \beta_2(x_1^*) \Rightarrow x_2^* = \frac{kD-2c_2+c_1}{3k} \end{array}\right.$$
and you obtain the Nash equilibrium!

About the usage of BRF with finite table games
When you are in this case, you have $2$ payoff matrix, say $A, B \in \mathbb{R}^{2 \times 2}$. From these you can build you payoff functions:
$$f_1(x_1,x_2) = [x_1 ~~~(1-x_1)]~A~[x_2 ~~~(1-x_2)]^T$$
$$f_2(x_1,x_2) = [x_1 ~~~(1-x_1)]~B~[x_2 ~~~(1-x_2)]^T$$
At this point, you act as in the previous example. Note that $S_1 = S_2 = [0, 1]$ instead of $\{0, 1\}$ because you have to extend to a continuous situation if you want to find the mixed equilibria.
In this case, when you use the BRF, you will find (if there exists at least one) mixed Nash equilibria. Some times it will find also pure equilibria, but in general you have to restrict the maximization on the border of the set $\Delta = \{ (x_1, x_2) : x_1, x_2 \in [0, 1] \wedge x_1+x_2 = 1\}$.
Comment: I don't see the point of your Cournot example. There you have a single-valued best response and an equilibrium in pure strategies. At a nondegenerate mixed strategy equilibrium, best replies can never be single-valued. As for the rest, you just restated that NE are fxed points of the BR correspondence. Since  this does not tell you how to find fixed points, this is of little help.
Comment: You restated the definition of NE, but this does not give rise to any way of finding NE. And your example does not adres OPs question.
Comment: Sorry, I got it late... I think I understand your point of view. You are saying that finding fixed point is not in general a trivial task, that's right?
Comment: The set of mixed best replies to a profile is always the convex hull of the pure best replies in finite games. For games with a continuum of strategies, essentially the same holds true (almost all pure strategies in your support have to be best responses).
Comment: @MichaelGreinecker ok, but I still don't get the point. I proposed the use of BRF as a method for finding Nash Equilibria. It's clear that there is no standard way to determinate a BRF and it is also clear that there are no standard techniques to find out the way to achieve a larger knowledge of the game you are facing with and hence you can find NE more easily I think.
Answer: Computing Nash equilibria "in general" is a hard problem. You can model it as a complementarity problem (look it up). There are good solvers out there for such problems, a good one I saw many years ago was the PATH solver (search PATH solver for complementarity problems). There may be better solutions these days.
However, in many cases, you can prove that the equilibria has certain structure (see, for example, supermodular games). That structure can simplify things a lot (but it depends on the structure, for supermodular games, things become much easier). Look up papers on computing Nash equilibrium.
The question is also if you need to find just one Nash equilibrium, or all. You can try, like someone mentioned, guessing the support (you can eliminate strictly dominated strategies) and using the fact that in equilibrium each strategy "component/action" yields the same payoff to find the equilibria.
Depending on what you want to do, you may want to find the set of correlated equilibria first (that's easier) and go from there.
As for when Nash equilibria exists, well, check Nash's theorem (there are some stronger existence theorems, especially in some special circumstances, again, check supermodular games for some cool examples). But from Nash's theorem, the sufficient conditions for existence are a compact, convex set of strategies and continuous utility functions.
I'm posting this as a brain dump, so please forgive small inaccuracies.


Do I have to capitalize ESP in sci fi fiction sentences?

Question: Some of my fiction sci fi characters are ESP sensitive and the term is used a lot. Capitalized ESP stands out in my sentences like yelling so I am using esp,lower case. Is that acceptable? Will a potential publisher object filing my novel in the trash?
Comment: kinda depends on the pronunciation. Saying /EE-ES-PEE/ is written . If she is saying /esp/ it's written .
Comment: I feel like abbreviations really have a negative impact on the story unless they are used for (company) names and such. If you must use the abbreviation at least make sure you explain this to the reader the first time you use it. That helps the reader understand and accept it.
Comment: ESP is an acronym of Extra-Sensory perception. Acronyms are usually capitalized. Not capitalized a well-known sends a signal to publishers. It says you don't know how to use the language in case how to present acronyms. Potential publishers won't like it. It may look like shouting, but it's not. Although If you call a character an esper, that's a word not an acronym.
Comment: @Boondoggle Sorry pronunciation has nothing to do with it. It's an acronym and they're usually capitalized. Although some acronyms have become words in their own right. ESP is not included among them.
Answer: ESP stands for "Extrasensory Perception". It is an abbreviation. The correct way to write it is therefore in All Caps: ESP, similar to how one writes NASA, USA and DNA.
A potential publisher wouldn't trash your novel for a capitalisation error, but it is quite possible they wouldn't understand what you mean by 'esp', similar to how one might look at a misspelt word, and be unable to figure out what it was supposed to be.
Comment: "esp" might also be mistaken for "esp.", the abbreviation of "especially".
Comment: How about RADAR (RAdio Detection/Direction And Ranging), LASER (Light Amplification by Stimulated Emission of Radiation), and so on? Just because it's an abbreviation or acronym doesn't mean it should *necessarily* be written in all caps; usage evolves over time. Note: I'm not arguing that you're wrong for the specific case, but I'm not sure you can generalize quite as far as you do.
Comment: @MichaelKjörling Radar, Laser etc. have evolved to become words on their own right - the common user would know what they are without necessarily knowing the word started out as an abbreviation. There's also the distinction between acronyms, which are read as a single word, like NASA and AIDS (with the last in the process of undergoing the same evolution as 'laser'), and initialisms, which are read by letter, like USA, DVD, and ESP. With an initialism, one can hardly "forget" it's not a word on its own right.
Comment: @celtschk Yeh, that was my first thought too.
Answer: As stated in the other answer, ESP should be in all caps. One way around this problem is to come up with a term that people would use in every day usage to refer to someone with this ability. This happens all the time in science fiction and fantasy (and, it should be said, in real life). Alfred Bester, Isaac Asimov, and Robert A. Heinlein used to refer to Espers for people with ESP powers, and I think Phillip K. Dick used the term "teeps" for people with telepathic powers in "The Hood Maker" ... there are many examples to draw from.
Answer: Another possibility is to "reclaim" the abbreviation and turn it into a noun or verb. This would allow you to use lowercase in particular sentences.
For instance, the group of espers was adept at esping other people. (I know I've read science fiction stories where the lowercase espers was used.)
Answer: If it is referred to as "ESP" specifically, it should be capitalized. Reading "esp" lowercase feels like reading a shortened "especially" in a text message. But you might avoid an overabundance of ESPs by allowing the narrator and/or characters to call it something else, especially if there are many of them that have it. They might have "powers," a "sixth sense," "radar," or use another story-specific noun/verb like "esper" (pronounced ess-purr) as mentioned in other answers, to refer to ESP or those with ESP in the story. You could even use the context of your sci-fi world to avoid "ESP"... if there's an invented language, name it in that, translate it once, and use your name from then on. Or, the history of a discovery or technology that explained or enabled the individuals with ESP might be discussed, and from its name derived another for the phenomenon itself.


Does storing a Bag of Holding in a Glove of Storing activate the tear in the astral plane?

Question: Does storing a Bag of Holding in a Glove of Storing activate the tear in the astral plane? And is there a difference between a Glove of Storing and a Pathfinders Pouch?
Answer: You can put one inside the other safely.
Only Bags of Holding and Portable Holes interact in a way that causes a rift into the Astral Plane. All other items that create extradimensional spaces simply cease to function and become inaccessible when placed inside another such item or space (such as the space created by the spell Rope Trick)

A number of spells and magic items utilize extradimensional spaces, such as rope trick, a bag of holding, a handy haversack, and a portable hole. [...] Such items do not function, however, inside another extradimensional space. If placed inside such a space, they cease to function until removed from the extradimensional space. For example, if a bag of holding is brought into a rope trick, the contents of the bag of holding become inaccessible until the bag of holding is taken outside the rope trick. The only exception to this is when a bag of holding and a portable hole interact [...]

There is a difference between a Glove of Storing and a Pathfinder Pouch
The Pathfinder Pouch makes a rather small extradimensional space that can be used to store a handful of items, and does not detect as magical, nor can any items contained within be detected. A Glove of Storing, on the other hand [HA], does not actually store an item in an extradimensional space, it merely shrinks a single item to become to small to be seen, and attaches it to the palm of the glove. It also does not have any protection against being detected as magical, nor does it provide any protection of that sort to an item it is storing. Theoretically, you could put on a Glove of Storing store some sort of item in it, like a greatsword or a crossbow, and then remove the glove and place it inside a Pathfinder Pouch in order to smuggle something larger than the pouch would normally allow somewhere.
Answer: From the Bag of Holding description:

If a bag of holding is placed within a portable hole, a rift to the Astral Plane is torn in the space: bag and hole alike are sucked into the void and forever lost.

Looking at the description of the Glove of Storing it says it shrinks the item down, rather than shifting it into another dimension.

The item is shrunk down so small within the palm of the glove that it cannot be seen.

As for the Glove of Storing vs Pathfinders Pouch, the Glove stores a single object, up to 20lb and requires a free hand, the Pouch stores any number of items, up to 10lb and just takes up space on a belt.
Comment: Is that a "no", then?
Comment: Reading the linked description of the Glove of Storing, it looks like indeed you can safely drop a Bag of Holding into the Glove of Storing. I wonder if this means you could then put the Glove of Storing into a second Bag of Holding without creating a rift to the Astral Plane, since the Glove of Storing has established a "barrier" of sorts between the two Bags of Holding?
Comment: @RobertF It wouldn't do anything, as a *Bag of Holding* only tears a rift if it is placed within a *Portable Hole*. That being said, the *Glove of Storing* specifically states that it does not suppress the effects of any item stored within, so if you stored a *Bag of Holding* in a *Glove of Storing* then tossed it into a *Portable Hole* it would tear a rift open.
Comment: Welcome to RPG.SE! Take the [tour] if you haven't already, and check out the [help] for more guidance.


default protection seller long or short credit risk?

Question: A default protection seller is long/short credit risk?my guess it is short the credit risk, anyone can help clarify?
Answer: A default protection seller pays in case the credit event is triggered. Therefore, the risk he faces is the show up of the credit event. Therefore, he is short credit risk.
Comment: This is not correct. Selling protection is the same as being long credit, as it is the same direction as owning a bond from the issuer outright given you are betting the company will not default.
Answer: It may help to consider the more familiar analogies with someone who is long a risky bond (or, in contrast, is short a risky bond, which can also be done as an alternative to buying CDS protection).
Someone who is long a risky bond loses money if the bond defaults; and also loses money (unrealized loss) if the bond's credit spread widens. They make money (unrealized) if the bond's credit spread tightens.  They earn the accrued coupons if there is no credit event, which is probably more than they pay for financing the bond, so the carry is positive.
Conversely someone who is short a risky bond makes money if the bond defaults; and also makes money if the bond's credit spread widens.
The view of someone selling CDS protection is similar to the view of someone long a bond, but the carry is somewhat different. After the upfront fee is paid, the protection seller is paid the running spread, and that's pretty much the entire (positive) carry. If the upfront fee is paid to the seller (i.e. market standard quote is wider than the running spread), then the seller also invests the upfront fee and earns some interest. If the market standard quote is tighter than the running spread, then the protection seller actually pays the upfront fee to the protection buyer.
The view of someone buying CDS protection is likewise similar to the view of someone short a bond. After the upfront fee is paid, the protection buyer pays the running spread making their carry negative. If the CDS spread in the market widens, the buyer makes unrealized gain - they can unwind their CDS for more money. And if there is a credit event, then the buyer generally makes lots of money, and the seller generally loses a lot.
(Further, a fixed-coupon bond has interest rates risk. Someone who does not want that (wants only to take a view on credit) would hedge the interest rates exposure. But a CDS has very little interest rates risk, and is more convenient for expressing view on credit risk alone.)


jcarousel not working in Google Chrome

Question: I am having a problem with the jcarousel in my Magento webshop. I bought a templete for my webshop so that I did not have to do any programming myself since I don't have any knowledge about programming.
On the bottom of the product pages of my webshop a jcarousel is shown in which related products are shown. This function works well in every browser except Google Chrome. Here none of the products in the carousel can be seen.
Someone on an other forum told me it had to do with the width of the carousel. That this particular line was causing the trouble:

Since I have no knowledge whatsoever on programming, I don't know how to adjust the width. Can any of you tell me if this is indeed the problem and how I can solve it?
To give you an idea what it is all about, here's an url of a page on which it doesn't work:
http://supplementendeal.nl/index.php/creatine/scitec-crea-bomb.html
Answer: Well the problem is that a width of 179769313486231600000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000px
is being inserted into the li tag of the unordered list. Without the actual javascript its hard to say how to fix this.

jcarousel not working in Google Chrome


How to analyze the effects of two treatments on two groups as measured by pre, post, and delayed post tests?

Question: I taught two different classes using two different teaching approaches. Each class was given pre-test, immediate post-test, and delayed post-tests to assess their understanding of the topic. The pre-test was used to analyze their existing knowledge. I would like to compare the immediate post-test score between the inductive teaching approach group with the deductive teaching approach group(to see which one is more effective after immediately teaching). Then, I want to compare the immediate post-test score with the delayed post-test score of both groups, to determine which teaching approach is more effective one week after teaching. So which statistical method should I use? My sample is 30 students. 14 students in the control group, 16 in the experimental group. I'm a novice, so please help me.
Thank you.
Comment: If I got it right, you have class A (14 students) attending inductive teaching approach and class B (16 students) attending deductive teaching approach. Each class will be tested with pre, post and delayed post tests. Is that correct?
Comment: @Pitouille Yes, that's correct.
Answer: Below is an overview of the data that you have collected (or will be collecting):
Class Control: Inductive teaching approach (n=14)

Student
Score Pre
Score Post
Score Delayed

Student 1
x
y
z

...
...
...
...

Student 14
a
b
c

Class Experimentation: Deductive teaching approach (n=16)

Student
Score Pre
Score Post
Score Delayed

Student 1
x
y
z

...
...
...
...

Student 16
a
b
c

For your first assessment (post-test), one approach that I could suggest would be:  for each group and each student, calculate the difference of score between Pre and Post. You will get 2 samples $X_1$ and $X_2$ of "differences" (i.e. increase or decrease in score within each class and by student) and then perform a Welch's t-test on the obtained samples to determine whether the means of both approaches are statistically significant:

null hypothesis: means (variation of score in average) are equal
alternative hypothesis: means (variation of score in average) are not equal

in R, t.test(X1, X2) will perform a Welch t-test by default with unequal variance. The output of the test will also provide you with the confidence interval. You can follow the same approach for the second test.
For this type of test, you can also consider Mann-Whitney. There is study from Winter and Dodou that you can find in the stats literature that could be relevant to your problem. Here is a link on a site that covers this issue: https://blog.minitab.com/en/adventures-in-statistics-2/best-way-to-analyze-likert-item-data-two-sample-t-test-versus-mann-whitney
You can also have a look at @Glen_b's answer which discuss the approach: Mann Whitney or two tailed t-test


What is the meaning of "Ganpati Bappa Morya"?

Question: I heard this many times but don't know what it means.
Is Ganpati another name for Ganesha?
What is bappa and Morya?
Answer: Regard less of true story of Moraya Gosavi, that doesn't make anysense to say "Ganpati Bappa Moraya"... Actually splitting the word Moraya as "Mhora ya" means "come ahead and bless us" make sense here.
Thus "Ganapati Bappa Moraya" means "Lord Ganesha Bless us"
(Here Bappa is used for Lord...)
If in case you're interested in reading one of the famous prayers with english translation, follow the link. Ganpati Aarti (with Translation)
Answer: Morya Gosavi  was a prominent saint and the greatest devotee of Lord Ganesha. Morya is associated with the name of lord Ganapati as veneration to this great devotee.
Comment: Thank you. At least you explained who is Morya but I still don't know the meaning of "Ganpati Bappa Morya!".
Comment: Morya Gosavi story has no substance and is a internet driven copy-paste phenomenon.
Answer: Morya is changed verson of mor+ yaa
Which is again changed from  samor+yaa
Samor= in front of some one
And yaa=come
That means morya=samor yaa= be in front ( of us)
= guide us
Comment: you should cite sources!
Comment: downvoted due to lack of sources and impossible logic. why `Sa` would be deleted?
Answer: It is considered as a chant but have some Marathi meaning you need to consider.
Morya Mean
1 Go ahead (for Visarjan ceremony happen.)
2 It is Marathi ward mohre ya mean pudhe Vha (for Visarjan ceremony happen.)
3 It look like lolhapuri Language word mhore ya(go ahead)
ref: Page on wordpress.com
Quora Answer to Morya

jodiif.wordpress.com
http://www.karmayog.org/
yahoo QnA
Another Yahoo QnA
Answer:
Yes, Ganapathi is another name of Lord Ganesha which means the Lord/Owner of all Shiva Gana (The army of Lord Shiva).
Bappa means Father/Lord and Morya refers to the greatest devotee of Lord Ganesha. To understand the meaning completely, "Ganapathi Bappa Morya" alone is not sufficient. The entire stanza "Ganapathi Bappa Morya, Purchya Varshi Laukar ya" is needed, which means "O Lord/Father Ganesha, come again next year fast".


How to convert raster negetive pixel value to positive

Question: I was working on ground water Level. I used raster math minus function  and the Output raster  grid data contains negetive cell values. ma resulting raster contains values ranges from High : 0,187477 to Low : -1,14927. But I do not want any negetive values.I just need to alter the negetive sign to positive vales.How can I do that using arcgis10.2
Comment: Would you be able to edit your Question to provide more details of the steps that you have performed, leading up to where you are stuck, please?
Answer: Try using the abs function in raster calculator
Comment: I guess the pixel values are different because of the conversion from negative to positive so your previous low value of -1.14927 is now a high of 1.14927. Perhaps instead of removing the negatives you could add a number to the pixel values? This would allow you to change the negatives to positives and keep the difference between your high value and low value
Comment: Thanks a lot. it works but I have found a Little Problem with the Pixel values. ´The Output raster has Little different Pixel value than the original. The outraster Pixel values ranges from High:1.14927 and low:0.0379639. I tried in the Environment to use compression None. but it did not work: -) Do you have any idea how it works !!


Set active bone in pose mode from Python script

Question: I'm trying to use the function bpy.ops.pose.constraint_add() on bones through a script, but stupid thing is that I can't find how to set the bones as active ones which is needed to call this function.
Comment: My bad, I forgot to mention that this is under Blender 2.8
Answer: Set bpy.types.ArmatureBones.active
It's a matter of setting the active bone of the armature (the data part), and hence the active pose bone.
Python console code. An armature object is active and in pose mode. Long windedly sets the active bone.
>>> C.object
bpy.data.objects['Armature']

>>> C.object.data
bpy.data.armatures['Armature']

>>> C.object.data.bones["Bone"]
bpy.data.armatures['Armature'].bones["Bone"]

>>> C.object.pose.bones["Bone"]
bpy.data.objects['Armature'].pose.bones["Bone"]

>>> C.object.pose.bones["Bone"].bone
bpy.data.objects['Armature'].pose.bones["Bone"].bone

>>> C.object.data.bones["Bone"] == C.object.pose.bones['Bone'].bone
True

>>> bone = C.object.pose.bones["Bone"].bone
>>> C.active_pose_bone
bpy.data.objects['Armature'].pose.bones["Bone.001"]

>>> C.object.data.bones.active
bpy.data.armatures['Armature'].bones["Bone.001"]

>>> C.object.data.bones.active = bone
>>> C.object.data.bones.active
bpy.data.armatures['Armature'].bones["Bone"]

>>> C.active_pose_bone
bpy.data.objects['Armature'].pose.bones["Bone"]

Use bone.select = True to have bone both active and selected.
Note There is no need for the operator.
Example adds a copy rotation constraint to each selected pose bone.  The cube is used as the constraint target.  Run script in pose mode, else context.selected_pose_bones will be None
import bpy

context = bpy.context
scene = context.scene
cube = context.scene.objects.get("Cube")

if cube and context.object.mode == 'POSE':
    for pb in context.selected_pose_bones_from_active_object:
        # use if already existes, else create
        cr = (pb.constraints.get("CubeRot")
                or pb.constraints.new(type='COPY_ROTATION'))
        # set some properties
        cr.name = "CubeRot"
        cr.target_space = 'LOCAL'
        cr.target = cube

Comment: Thanks, that was quite simple after all. The API seems so different in 2.8 that i'm quite lost with it. Thanks again
Comment: Isn't "Use bone.select=True to have bone both active and selected." wrong? I believe "object.data.bones.active=bone" is correct.
Answer: i want to add a little snippet for clarity. Thanks to batFinger for the detail. It was quite tough to find out.

import bpy
#The desired Bone
boneToSelect = bpy.data.objects['Rigify'].pose.bones['spine'].bone
#Set as active
bpy.context.object.data.bones.active = boneToSelect
#Select in viewport
boneToSelect.select = True

Comment: I am wondering how you have the armature object be the armature currently in pose mode for and not an armature that has to be named 'Rigify'.
Comment: I figured it out.  boneToSelect = bpy.context.active_object.data.bones['spine']


Faster way to do multiple embeddings in PyTorch?

Question: I'm working on a torch-based library for building autoencoders with tabular datasets.
One big feature is learning embeddings for categorical features.
In practice, however, training many embedding layers simultaneously is creating some slowdowns. I am using for-loops to do this and running the for-loop on each iteration is (I think) what's causing the slowdowns.
When building the model, I associate embedding layers with each categorical feature in the user's dataset:
        for ft in self.categorical_fts:
            feature = self.categorical_fts[ft]
            n_cats = len(feature['cats']) + 1
            embed_dim = compute_embedding_size(n_cats)
            embed_layer = torch.nn.Embedding(n_cats, embed_dim)
            feature['embedding'] = embed_layer

Then, with a call to .forward():
        embeddings = []
        for i, ft in enumerate(self.categorical_fts):
            feature = self.categorical_fts[ft]
            emb = feature['embedding'](codes[i])
            embeddings.append(emb)

        #num and bin are numeric and binary features
        x = torch.cat(num + bin + embeddings, dim=1)

Then x goes into dense layers.
This gets the job done but running this for loop during each forward pass really slows down training, especially when a dataset has tens or hundreds of categorical columns.
Does anybody know of a way of vectorizing something like this? Thanks!
UPDATE:
For more clarity, I made this sketch of how I'm feeding categorical features into the network. You can see that each categorical column has its own embedding matrix, while numeric features are concatenated directly to their output before being passed into the feed-forward network.

Can we do this without iterating through each embedding matrix?
Comment: Would love to hear if there's a more efficient / cleaner way to do this too
Comment: Me too, but I haven't found anything so far :( in this case, tensorflow's compiled computation graph is more efficient but I'm disappointed because I like pytorch better.
Answer: just use simple indexing [, though i'm not sure whether it is fast enough
Here is a simplified version for all feature have same vocab_size and embedding dim, but it should apply to cases of heterogeneous category features
xdim = 240
embed_dim = 8
vocab_size = 64
embedding_table = torch.randn(size=(xdim, vocab_size, embed_dim))

batch_size = 32
x = torch.randint(vocab_size, size=(batch_size, xdim))

out = embedding_table[torch.arange(xdim), x]
out.shape  # (bz, xdim, embed_dim)

# unit test
i = np.random.randint(batch_size)
j = np.random.randint(xdim)

x_index = x[i][j]
w = embedding_table[j]

torch.allclose(w[x_index], out[i, j])


Why is it important for a stack to be a FILO structure for the services of interrupts in a microprocessor?

Question: I am trying to understand stacks and their implementations, but the most thing that has been bugging me is "Why is it important for a stack to be a FILO structure for the services of interrupts in a microprocessor?" What do I need to understand first before I can understand the Stack concept. Can anyone please give me a comprehensive explanation on this. Thank you
Answer: FILO is useful because it allows for data within the stack to be restructured before it is accessed. Specifically, FILO allows things to be reliably reversed. If you have a situation where inner elements are returned before outer elements (as might be common with recursion) then the FILO is perfect.
http://www.i-programmer.info/babbages-bag/263-stacks.html?start=2 Has an excellent explanation of the utility of the FILO paradigm.
As for interrupts, again the utility is that subroutines will call an interrupt before anything else.
There are actually a few reasons that FILO can be useful for interrupts. http://www.eee.metu.edu.tr/~cb/e447/Chapter%204%20-%20v2.0.pdf has an introduction to the concepts. One example would be when it is desirable for registers to be pulled from a program and restored to a program in reverse order.
This article has an animation that demonstrates the exact utility of FILO:
http://users.ece.utexas.edu/~valvano/assmbly/stack.htm
Answer: When an interrupt is being called it stores the current content + next instruction address on stack.
suppose you encounter one nested interrupt (i.e interrupt within interrupt)
then you will require to store things in sequence like data + next address of 1st interrupt on encountering 1st interrupt, executing few instruction there and again encountering next interrupt, here you will require to store data + address of 2ndinterrupt.
while returning you will go back to last address where interrupt was called from there again completion of that interrupt you will return to 1st interrupt (FILO).
Hence we use FILO.


What is the name of this vine?

Question: I had this plant growing in my greenhouse. I brought a couple of plants in over winter and now new leaves have emerged so that I am puzzled about what this is. Do you know its name?
Comment: Thanks for the info . My greenhouse is happy and very healthy @stormy. I've grown many plant to what i didn't know what they are and never had any issues. I wasn't asking to what i should and shouldn't be doing / allowing in my greenhouse so that bit of advisement was not needed. Thanks for the possible identification
Comment: Damien, please tell us more.  Definitely need more information.  You really should NOT  allow strange plants, smokers anything purchased from stores or given by friends into your green house.  What have you been growing?  This plant is a trilobed plant of some kind.  More pics of your green house and perhaps your suspicions...
Comment: I have to disagree @stormy, this is a *Passiflora* of some kind and I cannot think of a reason why you wouldn't want one (aside from obvious disease which is not evident in this image). They are incredible plants.
Comment: @Georgeofalltrades what the heck are you disagreeing about?  What I said about greenhouses is correct sweetheart. Basic warning...my goodness have you got a...grins.
Comment: I wasn't ID ing your plant...I always go to the precursors first.  Truly a super plant.  Your comment about bringing in plants that you did not know triggered my knee reflex action about green houses...just to amplify a tried and true rule...I am glad your greenhouse is happy and that you are a more than intermediate gardener even a master gardener.  I quit worrying about knowing other's qualifications and just BLURT.  Sorry, did not mean to offend you!  I was just asking questions that would help me help you.  Do be careful allowing strange plants and soil into your greenhouse esp. smokers!
Answer: I think this is Passiflora edulis, a vine that has whole basal leaves followed by somewhat lobed leaves and then trilobed leaves. It also has tendrils and exhibits gutattion.
See this question for different pictures of it.
Comment: Probably right about *Passiflora* not 100% sure about *P. edulis* though. The leaves look exceptionally glossy, almost to the extent of *Prunus laurocerasus* and *P. edulils* tends to more regularly trilobate.
Comment: That said, *P. edulis* is more common than other candidates (and of course could have been grown from the seeds of shop bought fruit), so applying Occam's razor, gives *P. edulis* as the answer.

What is the name of this vine?


NMOS and PMOS current sources

Question:
I'm really confused with the representation of PMOS device shown in the figure. Since conventional current direction in PMOS is from source to drain, shouldn't the current source be pointing upwards(from node Y to VDD)?
Comment: source is connected to VDD for PMOS
Answer: Simplest way to remember current direction is by the little arrow indicator on the transistor, for NMOS it is pointing out of the drain thus current flows from source to drain.
And for PMOS the arrow is into the source, so flows from source to drain.
It is important to remember that the source is indicated by the arrow location. So in the case drawn, NMOS source is on the bottom and PMOS source is on the top.
Comment: Your first sentence is wrong (both in location of arrow and direction of current flow), but you seem to have gotten the description right later. For those not familiar with the subtleties of the symbol differences, it is important to note that you can be deceived by looking at the direction of the arrows on the symbols. See [MOSFET symbol - what is the correct symbol](http://electronics.stackexchange.com/q/69311/25328).
Answer: Mosfets can be confusing at times. The main difference between the pmos and the nmos is whether you need to apply a positive or negative Vgs to form a channel. The current will always flow from the higher potential to the lower potential (so from vdd to gnd) and never the other way around. In fact when you look at the device physics, a mosfet is pretty symmetrical. Using names like drain and source are just conventions. The arrow is therefore pointing in the right direction.
Comment: As @Tut referred to in his link https://electronics.stackexchange.com/questions/69311/mosfet-symbol-what-is-the-correct-symbol there has been some confusion with various MOSFET symbols, but I think the more prevalent convention has the arrow replicating the body diode, and not the direction of forward-biased conduction.

NMOS and PMOS current sources


Does ground bar block my knockouts?

Question: Can I use the knockouts right next to my ground bar, or does it permanently block them?  I'm wondering if I am allowed to run conductors across/over the ground bar and if I can put EMT connectors in these knockout holes.
Comment: Will these be THHN wires or NM-B with the interior cable clamp?  Are there any other ground bar sites in the panel?  They typically give you 2-3 sites (site=screw holes pre-tapped to mount the bars, though you can always drill and tap your own).
Comment: This will be THHN.  Yes, there are other ground bar sites.  I already have a 21 slot ground bar at the top and I probably don't need another, but I went ahead and bought another.  Just trying to understand the consequences of bar placement.
Comment: Are these bars made for this panel? A little odd to see only 1 mounting screw on a bar that long.
Comment: Yes.  There are multiple locations for the bar.  Some locations allow two screws to be used (center hole then has a nub in it), others allow single center screw to be used (two end holes both have nubs in the to prevent the bar from turning).
Answer: I have used many panels just like this I usually don't move the buss but it is easy to do the 2 screws are usually 10-32 self forming screws that hold the buss in place move it if you would like, code has no requirement on the position.
If you can't or don't feel comfortable with that location move the buss.
I always have a set of drill/taps for box modifications they drill and tap in one quick operation if using a tapped hole it's a lot easier to put the screws in and code legal with a 32 pitch screw.
Never use self tappers inspectors look for those and will red tag a job if they see one. Self forming machine screws are different and have the 32 pitch most commonly 10-32 but I have found some 8-32's used by different mfg's.

Does ground bar block my knockouts?


Subpage numbering on only specific pages in InDesign

Question: We are creating spare part manuals in InDesign where some of the pages are for only a few different parts where we want to change their page numbers as for example from 7 to 6.1 (because page 7 is originally a slightly changed version of page 6) and then continue to count the main pages from where we left. This is not applied to all the pages but just to a subset of them. Is this possible and if yes how can we achieve it?
Comment: Are you asking about the logistics of how to accomplish this task in a specific program, or whether a reader will understand the numbering convention?
Comment: I'm asking about how it can be done in Indesign CC 2015. I guess I should also add Indesign CC 2015 as the tag (if it exists).
Answer: You would do this with section starts and prefixes.
Open the Pages palette and double-click on page 6 to make it active. Open up the flyout menu of the palette, and select Numbering and Section Options.
Leave the Start Section square checked. Click on the radio button which says Start Page Numbering at and put in 1. Yes, one.
Under Page Numbering, there is a Section Prefix line. Put in 6- or 6. or whatever you want it to be.
This makes all the subsequent pages 6-1, 6-2, etc. or 6.1, 6.2, etc. as you prefer.
When you come to the end of the 6- pages, you click on the next page, make that a section start, and under Start Page Numbering at, put in 7.
(As a separate tip, if you find difficulty in printing or exporting a PDF because of the section starts, count your physical pages and use a plus sign before the number to indicate an Absolute page number. In this example, +7 would give you the page named 6-2, not the page named 7, because that could be 10 pages farther along.)
Comment: Many thanks. It worked but instead of changing the start section on page 6 I've did it on page 7 because that page was my first subpage to be numbered as 6.1.
Comment: Btw double-clicking on a page in the Pages palette doesn't bring up the flyout menu on Windows version of Indesign CC 2015. You have to right-click on the page to bring out the flyout menu.
Comment: @Montag451 Double-clicking doesn't bring up the flyout menu on Mac either. I'll edit to clarify. :)