Question about the Inverse Fourier Transform of a particular function.

Question: I'm trying to find the IFT of the following function: $\hat{f}(\omega) = \frac{i\omega}{1+\omega^2}$. I know about the Laplace Transform, so right away I thought about looking for an inverse of $\hat{f}$ involving $\cosh$ and then do a change of variables. I managed to arrive at $f(x) = \sqrt{2\pi} \cosh(x) \theta(x)$, where $\theta$ is the Heaviside Function. Wolfram Alpha agrees that the Fourier Transform of that $f$ is $\frac{i\omega}{1+\omega^2}$. However, when I use Wolfram Alpha to calculate $\hat{f}^{-1}$, I have in return $g(x) = -\sqrt{\frac{\pi}{2}}e^{-x}[e^{2x}\theta(-x) - \theta(x)]$. Both functions, $f$ and $g$, are obviously not the same. Did I do something wrong? Is it normal for a function to have more than one inverse? In that case, if I was asked on an exam to find the IFT of a function, would both $f$ and $g$ be correct?
Comment: Are you sure $f \neq g$? Keep in mind that $\cosh (x) = \frac{e^x + e^{-x}}{2}$. Try doing some algebraic manipulations with this in mind.
Comment: maybe there was a data-entry error somewhere? considering $x>0$ and $x<0$, your g seems to be more or less $e^{-|x|}$; the Fourier transform of _that_ is not your $\hat f$.
Comment: Fair enough. When you say you did a change of variable, do you mean you did a change of variable from the laplace transform to the fourier? If yes, did you use the classical Laplace transform or the bilateral Laplace transform? That could account for the difference
Comment: @jgd1729 I know, they look a bit similar, but $f$ is zero for negative values and $g$ is not.
Comment: @jgd1729 I knew that $\mathcal{L}\{\cosh(at)\} = \frac{s}{s^2 - a^2}$. So, by multiplying $\cosh(x)$ by $\theta(x)$, I only needed to integrate on one side of the line, which allowed me to go from the Fourier case to the Laplace case. At the end, I changed $s$ for $i\omega$. Maybe I did something I shouldn't do, but still wouldn't explain why Wolfram Alpha agrees with the answer.
Comment: @Aeschylus I just went through the integrals involved and have found some insight, though not a full answer. According to Mathematica, if you calculate the Fourier transform of $f$ by taking the integral directly (rather than using the built in Fourier transform function) you do get $\hat{f}(\omega)$ where $Im(\omega)>1$. Meanwhile $g$ transforms to $\hat{f}(\omega)$ where $-1 < Im(\omega) < 1$. My background in transform methods is limited, so I may be missing something, but it seems that the difference comes from not being careful about convergence.
Comment: @Aeschylus why did you multiply $\cosh$ with a $\theta(x)$ though? The expression you call $g$ in OP is correct
Answer: Traditionally, the Fourier transform operator is considered over the space of finite energy signals, i.e. signals that have a well defined square integral,
$$\int_{-\infty}^{\infty} f(\tau)^2\,d\tau < \infty,$$
quotienting out those signals that differ on a set of measure zero. This ensures the convergence of the Fourier integral to produce the corresponding transform $\hat{f}.$ The (continuous) Fourier transform is one-to-one and onto on this space of signals. As far as I recall, one can relax these conditions a bit, but not to the extent of your $f.$ Your signal $f$ grows unbounded on the imaginary axis in exponential order. The Fourier integral,
$$\hat{f}(\omega) = \int_{\mathbb{R}} f(x)\,e^{i\omega x}\,dx$$
simply does not converge for real values of $\omega.$ Verify that,
$$\begin{aligned}
\int_{-\infty}^{\infty} f(x)\,e^{i\omega x}\,dx
&=\int_{-\infty}^{\infty}  \sqrt{2\pi} \cosh(x)\, \theta(x)\, e^{i\omega x}\,dx\\
&=\sqrt{2\pi} \int_{0}^{\infty} \cosh(x)\, e^{i\omega x}\,dx\\
&=\frac{\sqrt{\pi}}{\sqrt{2}} \int_{0}^{\infty} \left[ e^{x} + e^{-x}\right] e^{i\omega x}\,dx
\end{aligned}$$
The second term will converge to finite value, but the first term grows unbounded inside the integral; the integral over $\mathbb{R}$ is not going to be finite for any $\omega.$
Compare this with the Laplace transform of your signal,
$$\hat{f}(s) = \int_{\mathbb{R}} f(x)\,e^{-s x}\,dx$$
which does converge but for $\mathfrak{Re}\{s\} > 1.$ That is, the domain of convergence for the Laplace transform does not contain the imaginary axis. It follows that the Fourier transform of $f(x)$ is not well-defined and does not exist!
What Wolfram Alpha computes is the square-integrable signal that does have that corresponding transform. The function $g(x)$ decays in both the negative $x$ direction and in the positive $x$ direction exponentially. First rewrite $g$ as,
$$g(x) =
\left\{
\begin{array}{ll}
-\sqrt{\frac{\pi}{2}}\,e^{x} & \mathrm{if } x < 0,\\
\sqrt{\frac{\pi}{2}}\,e^{-x} & \mathrm{if } x > 0,
\end{array}\right.
$$
where I'm leaving out the point at $x=0$ because it simply won't matter for this discussion. Observe that $g(-x) = -g(x).$ Therefore we can reduce the Fourier integral of $g$ to,
$$\begin{aligned}
\int_{-\infty}^{\infty} g(x)\,e^{i\omega x}\,dx
&=
\int_{-\infty}^{0} g(x)\,e^{i\omega x}\,dx + \int_{0}^{\infty} g(x)\,e^{i\omega x}\,dx\\
&=
\int_{0}^{\infty} -g(x)\,e^{-i\omega x}\,dx + \int_{0}^{\infty} g(x)\,e^{i\omega x}\,dx\\
&=
\int_{0}^{\infty} g(x)\left[e^{i\omega x} - e^{-i\omega x}\right]\,dx\\
&=
2 i \,\int_{0}^{\infty} g(x)\,\sin(\omega\,x)\,dx\\
\end{aligned}$$
Substituting $g,$
$$\hat{g}(\omega)
=
2 i \sqrt{\frac{\pi}{2}} \,\int_{0}^{\infty} e^{-x}\,\sin(\omega\,x)\,dx
$$
If you squint hard enough, you'll recognize this integral as just the Laplace transform of $\sin(\omega x)$ evaluated at a point which will result in the correct transform up to a constant factor (the constant factor is because I didn't use the standard multiplier in the Fourier Transform). This Fourier integral converges for all $\omega \in \mathbb{R}.$
In summary, if you want to use transforms for your signal $f,$ you will need to use the Laplace transform to ensure you can invert it. Of course, again, you'll have to consider the class of signals you concern yourself with. Presumably, the use of the heaviside step function suggests you are concerned with one-sided signals where the one-sided Laplace transform does have that required property that you can go forward and invert and get a unique result.
Answer: The inverse Fourier transform of $i\omega/(1+\omega^2)$ is
$$
        \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{i\omega x}\frac{i\omega}{1+\omega^2}d\omega.
$$
This is well-defined because $i\omega/(1+\omega^2)$ is square integrable on $\mathbb{R}$, which guarantees that the Fourier transform is square integrable as well. This may be further decomposed as
$$
     \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{i\omega x}\frac{1}{2}\left[\frac{i}{\omega-i}+\frac{i}{\omega+i}\right]d\omega
$$
If $x > 0$, then $e^{i\omega x}$ decays in the upper half-plane. If $x < 0$, then $e^{i\omega x}$ decays in the lower half-plane. So, for $x > 0$, the above may be written as the limit of a contour integral that is closed in the upper half-plane, which allows us to evaluate the above as a residue at $\omega=i$, and that is given by
$$
         \frac{\pi}{\sqrt{2\pi}}e^{-x}.
$$
Similarly, if $x < 0$, then the integral may be closed in the lower half-plane, leading to the negative of a positively oriented contour integral evaluation at $\omega=-i$:
$$
     - \frac{\pi}{\sqrt{2\pi}}e^{x}.
$$
So it appears that the original integral evaluates to
$$
         \mbox{sgn}(x)\sqrt{\frac{\pi}{2}}e^{-|x|},\;\;\; -\infty < x < \infty,\; x\ne 0,
$$
where $\mbox{sgn}(x)$ is $1$ if $x > 0$, is $-1$ if $x < 0$, and is $0$ if $x=0$.
The original integral does not converge absolutely for $x=0$, but it does converge to $0$ as a Cauchy principle value integral. Regardless of the details at $x=0$, the inverse transform is in $L^2(\mathbb{R})$, and everything is just fine in that regard.
Verification: The Fourier transform of our function is
$$
  \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-i\omega x}\mbox{sgn}(x)\sqrt{\frac{\pi}{2}}e^{-|x|}dx \\
  = -\frac{1}{2}\int_{-\infty}^{0}e^{-i\omega x}e^{x}dx+\frac{1}{2}\int_{0}^{\infty}e^{-i\omega x}e^{-x}dx \\
  = -\frac{1}{2}\frac{1}{1-i\omega}+\frac{1}{2}\frac{1}{1+i\omega} \\
  = \frac{1}{2}\left[\frac{1}{1+i\omega}-\frac{1}{1-i\omega}\right]=\frac{-i\omega}{1+\omega^2}
$$
Oops! I'm off by a sign. I'll let you sort it out.


Image problem with the sheet not grounded

Question: This is the question of the classic image problem in which a point charge $q$ is placed at height $h$ above a grounded plate.
Only this time the plate is not grounded.
When the plate is grounded the force on a charge in the region between the point charge and the plate is proportional to $2q$ because there is equal $-q$ on the sheet.
But due to induction an equal $+q$ charge should also be induced on the other side of the plate thereby causing the force to be proportional to only $q$.
We can say that this is wrong since the plate is grounded so the induced positive charge will flow down to the earth causing the total force to be proportional to $2q$.
But if this plate is not grounded then what shall be the case ? Will the force be proportional to $q$ or $2q$?
Answer: It doesn't matter if the plate is grounded or not, the field outside the plate is the same either way.  You know this because the only constraint on that field is that its parallel component at the plate must be zero.  That suffices to specify the image-charge field, as long as there is no net charge density on the infinite plate.  If there was a net charge density on an ungrounded plate, that just adds a constant field perpendicular to the plate, but if there's no net charge density (i.e., no infinite charge on the plate), whether the plate is grounded or not will simply result in a negligible charge distributed over the whole infinite plate.  In short, pulling charge from the ground simply distributes that charge over the whole infinite plate, with negligible effect.
Comment: But then the force should be proportional to q & not 2q since the effect of induced -q will cancel out the effect of  +q.
Comment: The answer is that the field is the same whether the plate is grounded or not, so that seems to be the case you are calling "proportional to 2q."  Note that q is also proportional to 2q, so the language is not precise, but I believe I understand what you mean.  Note that if you add a net but non-infinite charge to an infinite plate, it has no consequence whatsoever.
Comment: What I mean is that if the plate is grounded i can say that the induced +q flows to the ground and the negative induced q applies force in the region above the sheet along with the original point charge q ( hence force is proportional to 2q)
Comment: What I mean is that if the plate is grounded i can say that the induced +q flows to the ground and the  induced -q applies force in the region above the sheet along with the original point charge q( hence force is proportional to 2q). But if the plate is not grounded then an equal +q is induced on the sheet ( like any normal plate kept in front of a charge ) .  The induced charges will give opposite fields in the region on the side of the plate and the point charge and hence their result will be to cancel and  force will be only because of the original  point charge q(force proportional to q)
Comment: Sure, and if the plate is not grounded, all that happens is the charge on the plate moves around.  The net result is all the same fields outside the plate, because either way, the charges in the plate have to cancel the field parallel to the plate.  They don't have to cancel the field perpendicular to the plate outside the plate, and inside the plate that is cancelled by the surface charge.  The net charge on the plate never matters, it is distributed so as to produce no field anywhere.
Comment: @KenG What I think the op is referring to  is that the force on an arbitrary charge in the region above the sheet is due to the original point charge and the induced negative charge on the top of the plate in the case when the plate is grounded (as he says the positive charge flows to earth) . But in the case when it is not grounded the potential in the region above the sheet due to sheet is 0. The force then is only due to the point charge
Comment: No, that cannot be correct.  The conducting plate must move charges until there is no field parallel to the plate.  A nongrounded plate has to do that as well, and the grounded plate does the exact same thing, it merely also pulls in a cancelling charge that does nothing to the fields.


Choosing a country's color for its slice of a pie chart?

Question: I'm writing some software that involves a breakdown of data by country. For it, I plan on using a pie chart where each slice corresponds to the activity in a certain country. I'm wondering - is there any recommended way to choose which color goes with which country? The USA seems like red white or blue, but that's the same as the UK and Australia. Or would it best to just choose distinct colors, not worrying about them being obvious which country it is?
Comment: I hope you are not dealing with more than three – five tops – countries at once. Pie charts cannot handle more than that.
Answer: USA, UK, Chile, Czech republic, Russia, Serbia, Philippines all have Red, White and Blue as primary colors in their flags. It might be an overkill to identify the best color to represent each country for a pie chart.
However, if you would like to build a color pallet, one source you can check out is the Passport colors of various countries > https://www.passportindex.org/index.php
You can also look at the Sports uniforms that countries wear for global events like Olympics.
Answer: I would choose just random colors, depending on your target audience, it may discourage, or could create an offensive feeling.
Remember, color affects emotion, therefore, depending on your data you are going to show, choose randomly but in similar hue, or saturation.
I would also consider to choose what type of other UI it will be interacting, like background colors, and overall theme of your UI.
Answer: I would personally chose the colours that you think suit the country,or colours related in their country flag. Don't use asteriotypes they can sometimes offend people.
Answer: If you wanted to get really detailed, you could write a little program that would determine the dominant color in the flag. If two countries flags have the same dominant color you could move down to the second or third etc.
Answer: As you can see here, publishers of world maps don't agree on any standard colours, but if you chose to reuse a palette from one of these map publishers, at least your palette would conform to some standard, rather than an entirely new scheme.
Another simple theme you can consider is political: left wing countries can be red, right wing countries blue, and middle (liberal) countries yellow.  How these will be interpreted might depend on your audience.  In the USA the phrase "red states and blue states" embodies this idea.
In the past it was also common for British maps to show Commonwealth countries in pink.
To ensure that your pie chart is accessible to color blind people, it should make sufficient use of brightness, or texture, or have clear borders, so that the slices can be distinguished.
Taking that to its extreme, you could consider using miniaturized and tiled country flags as the texture for each country!
Answer: I suggest to steer clear from the country colors, because so many countries are sharing the dominating colors of the flag — US, UK, France and Russia all have the red-blue-white flag, so if you use the connection flag color — country, it can be extremely confusing.
Just use the palette that rhymes with your UI coloring. Colors in the palette should all equally be different from each other, so they don't blend in the chart.
Palette tools: colorschemedesigner.com, color.adobe.com.


Type with nested table in oracle

Question: I want to create a type with a nested table. I have a table called assignment_tbl which I want to put a type assignment_status. This assignment_status contains primitive types but also contains a table called corrections. All my basic objects are compiling fine, just trying to build the assignment_tbl is a problem as im not sure how to use the nested table statement.
CREATE TYPE correction_info_objtyp AS OBJECT
(
   correction_date   DATE,
   raised_by         VARCHAR(50),
   problem           VARCHAR(255)
);

CREATE TYPE correction_info_ntabtyp AS TABLE OF correction_info_objtyp;

CREATE TYPE assignment_status_info_objtyp UNDER status_info_obj
(
   corrections     correction_info_ntabtyp
);

CREATE table assignement_tbl
(
  assignment_id       NUMBER(6),
  assignment_title    VARCHAR(40),
  elements            elements_v,
  assignment_status   assignment_status_info_objtyp
)NESTED TYPE assignment_status STORE AS assign;

-- I know this nested type isn't going to work because it's not a table,
-- But inside the assignment_status_info_objtyp is correction_info_ntabtyp which is
-- a table and needs the nested type statement

I've tried to follow the example here with the planets and satelites, where they have a nested table in a nested table, but for my situation i have a nested table in a type because I only ever want one instance of assignment_status in the assignment.
Any help, hints or even a link to an example or a page in a book would be of fantastic use.
Exhausted first 20 results of Google already.
Comment: I'm not sure if this is doable. It's time for me to sleep now, but i'll revisit in the morning.
Comment: I've found an [example](http://docs.oracle.com/cd/B19306_01/appdev.102/b14260/adobjcol.htm) where they have created a nested table within a VARRAY, maybe I could have a VARRAY of size (1), which would be close to what I want, I just want to follow the best practice though.
Comment: From looking at the [oracle docs](http://docs.oracle.com/cd/B28359_01/appdev.111/b28371/adobjcol.htm#CHDGJHEE) which specifies the combination of possible collection types, the one I proposed doesn't exist. I think I will go with a varray(1) I cant think of less painfull alternative.
Answer: I managed to create an array of size one instead of using a object type. Through thorough reading I found no mention of whether a nested table within a type was possible, although  varrays of nested tables are; Nested tables of varrays are also allowed.
CREATE TYPE correction_info_objtyp AS OBJECT
(
   correction_date   DATE,
   raised_by         VARCHAR(50),
   problem           VARCHAR(255)
);

CREATE TYPE correction_info_ntabtyp AS TABLE OF correction_info_objtyp;

CREATE TYPE assignment_status_info_objtyp UNDER status_info_obj
(
   corrections     correction_info_ntabtyp
);

CREATE TYPE assignement_status_info_v AS VARRAY(1) OF assignment_status_info_objtyp

CREATE table assignement_tbl
(
    assignment_id       NUMBER(6),
    assignment_title    VARCHAR(40),
    elements            elements_v,
    assignment_status   assignement_status_info_v
)VARRAY assignment_status STORE AS LOB assignement_status_store; -- contains the nested table

Comment: I have implemented this, also managed to insert a row, it was monstrous. Ouch my head.

Type with nested table in oracle


can't publish a question "This looks like spam."

Question: I'm trying to publish this question but I can't as the body is marked with "This looks like spam.". Any idea how I can fix my post? Thanks
Comment: This has to be a bug. The question is clearly fine: it's clear, detailed, well formatted and everything!
Comment: As a temporary workaround, can you post a stub, and then try and edit the full question in? Alternatively, if you can post this as a gist on github or something and link to it, one of the mods or any high rep user can just copy it into your question. If you do this, please also raise a moderator flag, choose the "needs moderator attention" option and just link us to where you have put the full text of the question so we can edit it for you .
Comment: This reminds me of a similar experience I had way back in 2016.  https://meta.stackexchange.com/q/284812/307622 I bet combination of bad keywords (camera, stream) plus new account result in the block.
Comment: Might be worth looking at https://meta.stackexchange.com/questions/375108/cant-post-question-on-stack-overflow-this-looks-like-spam and those links. If I had to guess, your block may be due to the number of "stream" words. I happen to have seen a large influx of streaming-related spam recently (on another SE site), so I wonder if there's an SE mechanism that is blocking your post for that reason.
Answer: My blind guess for a google service would be:
The url fragments and keywords collect too many negative point. Basically what Wildcard commented.
Stackexchange seems to use a similar system.
So...
It physically hurts me to say this but:
Try posting the stuff in code tags as an image.
Comment: Maybe even combine [Luuklag's advice](https://meta.stackexchange.com/questions/375108/cant-post-question-on-stack-overflow-this-looks-like-spam#comment1249585_375108)  of *"cutting part by part from your post until it works"* and then pasting the remainder as an image?
Comment: ... or changing it so that it does not contain the offending terms, once you know what they are.
Comment: No, please don't post as an image. That really isn't helpful. Just post a stub, and then edit the code in, or flag for mod attention and ask us to add it, but posting an image is really not helpful.


Why Cant I Sculpt Together These Joined Pieces?

Question: I'm in the middle of making my own game and right now and I'm trying to make a model for my main character(A Cat). But it seems I can't get this ear to work properly? Whenever I try to use the sculpting tool to join the two meshes together, I can't. No matter the brush they just don't want to touch each other. That makes no sense to me, they are joined, aren't they? I grabbed them all and used Ctrl + J, so why aren't they acting like one single object? Did I do something wrong?
Comment: can u upload ur blend file?
Comment: You'll need to remesh them first, so the seperate meshes will be merged. (Look at the top right of the 3D view, where it says "Remesh".
Answer: They have been joined so they exist under the same mesh data block, but it looks like they are still separate islands.  Sculpting works on one island of verts that is a closed shape.
You could perhaps separate them back out, move them such that they intersect just a bit and run a union boolean on them.


How can I get the json object which represents a Yahoo! pipe

Question: It seems that Yahoo pipes are represented using JSON. I want to download these JSON objects for some research purpose. Usually a Yahoo pipe is rendered in a browser editor thru a url like this: http://pipes.yahoo.com/pipes/pipe.edit?_id=XgRo96h13BGtJWvS8SvLAg, but you can't get the corresponding JSON object to this Yahoo pipe. Does anyone know how to get JSON objects representing Yahoo pipes and store them in any persistent form?
Answer: It is possible to get hold of a JSON description of a Yahoo Pipe using a URL of the form:
<code>http://pipes.yahoo.com/pipes/pipe.info?_out=json&_id=PIPE_ID
</code>
The pipe2py python library demonstrates how to grab the JSON description of a pipe and "compile" it to a Python equivalent that can be run on your own server.
The post Exporting Yahoo Pipe Definitions, Compiling Them to Python, and Running Them in Scraperwiki describes how you can use pipe2py in the Scraperwiki environment to compile and execute pipes on Scraperwiki using pipe definitions imported directly from Yahoo Pipes, or exported from Yahoo Pipes and then stored locally in a Scraperwiki database table.
Answer: When I load that page in a browser I can see that it makes an ajax request for:
http://pipes.yahoo.com/pipes/ajax.pipe.load?id=XgRo96h13BGtJWvS8SvLAg&_out=json&modinfo=true&rnd=7560&.crumb=MjvGjpzhPLl
That's your object but I'm not sure if I'm answering your question of how to "get it". If you need to get it through a program you would need a script that loges into pipes and extracts that url.
Comment: I've recently found my way to download pipe jsons. Actually the most tricky part is to notice that the /ajax.pipe.load action behind the front-end url as you mentioned. Once you found out the action, you can use any download utilities to download pipe jsons. I was using wget. Besides, downloading pipe jsons requires cookie for authentication.
Answer: A quick way, while not automated, is to use an HTTP analyzer. Here's a process for getting the object using HttpFox (I use v0.8.9) for Firefox. With the analyzer running, load the edit page for a pipe, like the one you linked:
http://pipes.yahoo.com/pipes/pipe.edit?_id=XgRo96h13BGtJWvS8SvLAg
Look at the request with a URL that starts with:
http://pipes.yahoo.com/pipes/ajax.pipe.load?id=....
Next, explore the content of the request (there's a 'Content' tab in HttpFox). That's the JSON object representing the pipe structure.
Comment: I've recently found my way to download pipe jsons. Actually the most tricky part is to notice that the /ajax.pipe.load action behind the front-end url as you mentioned. Once you found out the action, you can use any download utilities to download pipe jsons. I was using wget. Besides, downloading pipe jsons requires cookie for authentication.
Answer: Use pipe.run?[your pipe id here]&_render=json as opposed to pipe.edit
So in your case to get the json it would be - http://pipes.yahoo.com/pipes/pipe.run?_id=XgRo96h13BGtJWvS8SvLAg&_render=json
I guess how you implement the client is dependent on what you like writing in/what other functionality you need.
You could also do it the other way around and use the web service service module to post the data to a script that can extract the json and persist it to a database. You could check out json.org.


What ukulele chords would go well in a song with a C-F-G7-C progression?

Question: I am a beginner ukulele/guitar player and although I'm not a great musician I enjoy making up my own songs. I've started improvising a bouncy, twee little tune with a C-F-G7-C progression but not sure what other chords I can add in to spice it up. Hopefully easy ones... Any suggestions? Thanks!
Comment: Pretty much any chord in the key of C will work.
Comment: You can add the minor chords Dm, Em, Am, and play a C7 instead of C.
Answer: The world is your oyster. You could add any chords you like, but some choices are much more likely to give an effect you will like the sound of!
As @Dom says in his comment, other chords from C Major will work well. This is because C, F and G are the primary triads of C Major. The other triads are: Dm Em and Am. (There is also Bdim, a diminished triad, but you might want to leave this for now. Besides, it is contained within your G7 chord anyway.)
Beyond these triads there are a number of ways to find other chords that might work for you:

you could extend these C Major triads by adding, for instance, sevenths. This would give: Cmaj7 Dm7 Em7 Fmaj7 G7 Am7
you could "decorate" your basic chords by adding 9ths or 6ths to them, or by alternating them with suspended chords (e.g. sus2 or sus4 chords). An example would be playing C Csus4 C Csus2 C C6 instead of six C chords
you could "borrow" some chords from keys closely related to C Major, or even use these chords to modulate to a new key. For instance, for a darker turn to your chord sequence, you could head towards A Minor, the relative minor of C Major, by using chord V of A Minor, which is E. Alternatively, borrowing a D Major chord from G Major, the dominant of C Major, might work well. This is called a secondary dominant. (Chord V is known as the dominant, so D is the dominant of the dominant, G.) Bm could be "borrowed" from G Major, too. You could borrow chords from F Major, the sub-dominant key of C Major. This would usefully give you Bb and Gm chords.

These are just a few obvious ideas, but hopefully should suggest some avenues for experimentation. Don't rule out less obvious choices: there are plenty of great chord sequences that don't follow the kind of "rules" outlined above. A personal favourite of mine is a sequence of major chords, whose root notes are related by the notes of a minor scale. Smells Like Teen Spirit by Nirvana is an obvious example. Its introduction uses these chords: F Ab Bb Db (although not in that order...) Taking C as your home key, this would suggest using: C Eb F Ab, and maybe Bb, D or G too...
The possibilities are endless!
Comment: Ah yes! Fm, minor iv chord, nice! Very "music hall". Ummm, Bm would work well in a descending progression, for example: C - Bm - Am - G. But, to my ears this does establish G as the home key quite strongly... Good approach to Em, too...
Comment: Kind of. Using chord V of G Major, which is D, would emphasise the modulation more strongly.
Comment: in which context would you suggest the OP to use Bm (thus vii minor)? Would be interested to know as i am experimenting with non conventional chords as well. I would also use Fm as part of a rather fancy ending/plagal cadence.
Comment: Interesting. .Thanks @Bob-Broadley. So that sounds like Bm can be used for modulation from C to G?
Answer: Using C, F and G7 gives you the main chords that will be found in almost all tunes played in C major. There are three minors which also belong to that key, Am, Dm, and Em.(Actually the three relatives of the original chords). They will all be good to use. The remaining chord straight from C is B diminished, which isn't quite as usable as the other 6, but is made up from B, D and F.You could also use D7, especially if G followed, or E7, especially if Am followed.
Comment: perhaps the OP can also add A major to this chord list, right before moving to Dm when appropriate. It is not as commonplace as the D-G pair or E-Am pair, but it gives some flavour i think ☺


Is there a way to reset iCloud drive?

Question: My iCloud sync is totally broken. I have a set of files on web icloud drive app which are not downloading to my devices (iphone, macbook pro) and when I add files to iCloud folder on my devices they are not uploaded. Also my Apple Books are not syncing too. Probably other things are broken. Only thing that syncs is Photos. Can I somehow reset the iCloud drive entirely and start from scratch? Btw, I tried every possible proposed solution on the net for the sync problem to no avail.
Answer: There is not a specific feature of iCloud drive that says "Reset all settings." however paying attention to the warnings when you sign out or turn off iCloud drive will reveal the trick.
So what you need to do is (FIRST!) make sure anything on iCloud drive you have copies of somewhere. Then you will need to turn off iCloud drive on all devices you are signed into iCloud in.
On every device you turn off iCloud drive on you should get a warning about removing files (it's been a while so I don't recall the exact wording). Tell iCloud drive to delete the files.
Once you do that on all your devices iCloud drive should be empty. Personally I'd wait a day before turning it back on but, realistically, that is not likely to matter. Just turn it back on, add your files and watch them sync.
Now if it STILL fails, I would call Apple support and tell them what you have done, why and ask them to fix it for you.

Is there a way to reset iCloud drive?


JabRef 3.2 and automatic formatting on save

Question: I just upgraded JabRef from 2.10 to 3.2 on OS X 10.11. JabRef 2.10 automatically formatted the BibTeX file on save, which was the number one reason I used JabRef. JabRef 3.2 does not do it, and I could not find any preference to enable this feature. Is this feature still available? If so, please could you explain me how to enable it?
Comment: I think my question is not so different from several other questions on TeX.SX. Besides, there is no (easy to find) answer to this question on TeX.SX or elsewhere, so I think a public answer could be beneficial to others. Nevertheless, if this question does not comply with the policies of TeX.SX, I would be happy to delete it.
Comment: Welcome to TeX.SX! This question is probably better directed towards the authors of JabRef!?
Comment: Related discussion: [BibTeX format differs between machines](https://github.com/JabRef/jabref/issues/823).
Answer: The reason for this was a long discussion of the developers with users who have upgraded from 2.9.2 to 2.10. JabRef changed the serialization format and users had issues with their version control: There were users working with 2.9.2 and users working with 2.10. The bibliography was completely changed, even if there was only a minor change in one entry. After a long discussion (see https://github.com/JabRef/jabref/issues/116), the developers decided to keep the entries as they are. If an entry is modified, the entry is written as a whole.
Regarding your concrete issue, you have to do a bit cheating to JabRef to let it think that you modified all entries. Do following steps:

Click "Search" in the menu
Click "Set/clear/rename fields"
Enter "cheat" as field name
Select "All entries"
Ensure that "Set fields" is selected
Enter "cheat" as value in "Set fields"
Click "OK"
Click "Search" in the menu
Click "Set/clear/rename fields"
Ensure that "cheat" as field is set
Select "Clear fields"
Click "OK"
Hit "CTRL+s" to save

All entries should have been rewritten now. Tested with the latest build available at http://builds.jabref.org/master/.
If you want to have this feature with a single click, please file an issue at https://github.com/JabRef/jabref/issues.
More information on sharing via JabRef is collected at http://www.jabref.org/faq/#sharing.
Comment: I just want to add that the most recent JabRef development version (http://builds.jabref.org/master/) contains an option to reformat all entries on save (under Preferences, File). So the above work-around is no longer necessary.
Comment: @AlessandroRossini It looks like JabRef 3.3 will be released within the next two weeks. If unexpected problems occur, the release should at least be made in march. No absolute guarantees, though :)
Comment: In the database properties you can specify the sort order of the entries in the .bib file.
Comment: That is good news. I tested today's build (i.e., JabRef-3.3dev--snapshot--2016-02-18--master--99f8905.jar) with Java 8u73 on OS X 10.11, but I get a NullPointerException in thread "AWT-EventQueue-0" when I try to open Preferences. I will try again with future builds. When do you expect to release JabRef 3.3?
Comment: Would it be possible to add an option to sort all entries on save (as in JabRef 2.10) in addition to the one to reformat them?


Does "mind" has any physical reality in biology? Or is it just an assumptive concept?

Question: Biologically,  Brain controls our thinking,  ideas,  decisions everything along with controlling each body parts. My question is,  is there anything real as "mind "?  If it's controlled by brain then how does brain controls the mind?
Comment: This might be a better question for the Psychology SE.  Neuropsychology studies the relationship between the brain and the mind, to an extent.  Neurology can touch on this, however, because self-report of experiences (as unreliable as it may be) is often the easiest way to get information about the mind.  A patient describing depersonalization isn't describing anything that you can measure easily without talking to the person, for one.  I'll try to answer below.
Comment: @TrixieWolf - Psychology SE does not exist as separate entity anymore. Please refer users to Cognitive Sciences instead.
Comment: @AliceD  Thanks.  I apologise for not checking first (I belong there more than here anyway).
Comment: @TrixieWolf - Please don't apologize :) I have been pointed out to it myself by others :) Still going through your answer - it looks good, but have to dig it through before upvoting :)
Answer: The mind is an abstraction which arises from physical and chemical processes within the brain.  For a crude analogy, think of the brain as an incredibly complex, self-modifying, multithreaded program.  The mind would then be the abstraction arising from the behavior of the program as it runs.  Consciousness would be the abstraction arising from the behavior of one particular thread in the program: a thread which has access to a limited buffer of its own (that thread's) behavior, and control over limited input and output from the body.  All of the other (unconscious) threads may affect the conscious thread, but the conscious thread has very limited access to information about the behavior of the other threads.
Since the conscious part of the mind has access to information about its own behavior, there is actually some recursive looping going on when it comes to thinking and feeling.  The feeling that this goes in circles somehow (which you seem to be expressing) is a common intuition, and it's almost certainly a correct one.  But the manner in which this occurs is not well-understood, and it's definitely not well-understood within biology at the macro level because the brain is unbelievably complex and plastic.  We really don't have a clue how to begin modeling the brain's processes.  This is one reason artificial intelligence is still extremely crude today (despite what Elon Musk and Bill Gates may try to say).
If you want to read a scientifically-inspired theory of from whence consciousness comes, and of how it arises from the brain, I would recommend the book Godel, Escher, Bach by Douglas Hofstadter.  Douglas does a good job of combining computational theory with biology, music, and art to illustrate a theory of consciousness the reader can grasp even if you don't have a strong background in any of those subjects.
Comment: +1 - You should definitely check out Cognitive Sciences too! But there's definitely overlap, so don't just run away - expaND :)
Comment: *The Emperor's New Mind* by Roger Penrose is another book on the subject that's worthwhile.


How far does Marlin travel to find Nemo?

Question: Given that the movie "Finding Nemo" takes place in a world resembling our own (with locations like Sydney Australia). Where is the reef (Marlin and Nemo's home) located? And, given the landmarks Marlin travels through (ie: the "EAC"), how far does Marlin travel from that reef to get to Sydney Australia?
Comment: Interesting question. And a follow-up might be: Could a clownfish actually make that distance?
Comment: Anywhere is within walking distance, if you have the time.  ;o)
Answer: What a fun question!
Ok, so Marlin and Nemo live in the Great Barrier Reef which is located off the coast of Queensland in north east Australia.

From there he wanders around and hops aboard the East Australian Current (EAC), which luckily enough passes right through the Great Barrier Reef and passes by Sydney in the south east of Australia.

Using Google Maps' measuring tool to plot a (very) rough course between the Great Barrier Reef and Sydney we find that Marlin traveled the equivalent of 1275 miles (2040 KM).

For simplicity sake we'll just take he maximum speed reached by the EAC is 90 cm/s (about 3.24 km/h) to calculate the trip taking about 630 hours or 26 days. That's a long time to stay on a surfer Turtle's shell :)
Comment: You have pretty pictures and I don't.  :o(
Comment: @JohnnyBones - Amusingly enough we got roughly same the answers too :D
Comment: Congratulation for wining [additional bounty as consolation](http://meta.movies.stackexchange.com/questions/1848/rewards-for-the-best-answer-from-the-2nd-quarter-of-2015)  prize for being 3rd highest score in event.
Comment: @AnkitSharma - Yay!
Comment: There is a flaw in this as far as the movie not the answer. If it took Marlin 26 days to find nemo this wouldn't work cause the dentist's niece Darla was coming in a week to an appointment with her uncle. Nemo was her present. She came approximately 5-7 days after Nemo was captured/taken. That would leave 19 days unaccounted for the time taken for Marlin to get there. Marlin got there the same day that Darla came to see her uncle.
Answer: Nemo and Marlin live on the Great Barrier Reef.  According to MapQuest, there is approximately 1,500 miles of coastline between the shores off the Great Barrier Reef and the capital city of Sydney, which is located on the southeast corner of the continent.
Marlin, on his journey, hops into the EAC (Eastern Australian Current) which runs down the coast from the Great Barrier Reef and ends in Sydney.  So far, this all checks out.
As for Walt's follow-up, Clownfish are small animals that could maybe reach a top speed of 3MPH.  The EAC also moves at approximately the same speed, so that would double Marlin's speed to reach Sydney.  At that rate, swimming non-stop, it would take about 250 hours, or about 10½ days to make the trip.  However, on the way back, it would take at least 21 days, assuming they stayed out of the EAC.  And that's non-stop.
Comment: Sydney is not the capital of Australia
Comment: @OrangeDog Sydney is the capital of the state of New South Wales


"Administer Taxonomy" is needed to make a user select a term

Question: I need a user to select a term from a given vocabulary but the field is only available to a normal user if I enable "Administer Taxonomy" for them. Is there a way around this? Shouldn't I be using taxonomy for this?
Answer: You don't need to give an user the permission to Administer Taxonomy in order to use a vocabulary. Assuming you are using Drupal 7, as the name implies Administer vocabularies and terms it's a permission that allows one to edit and/or *delete a term from any vocabularies. It can also be specified per vocabulary.
Answer: You can install "Vocabulary Permissions Per Role":

You have set up some vocabularies (with some terms) like Color (Red, Green, Blue) and Shape (Toroid, Cube, Sphere). Maybe you even have a vocabulary that controls who can view a specific node (with terms) like Access control (Editors, Colleagues, World). Here comes trouble: your precious editors want to add more Colors, or reorder the terms of the Shape vocabulary (probably they want to see Sphere before Cube, who knows,) and you do NOT want to grant your editors the administer taxonomy permission, as it grants too much power (and that implies too much sharing of responsibility). Without this permission they were NOT able to do the mentioned tricks: adding terms to a vocabulary but not to another, reordering/editing terms of a vocabulary but not another's, etc.

https://drupal.org/project/vppr


Children joining train later UK

Question: I need to travel from London to the north and my children aged 5 and 8 will join me on the train at Milton Keynes, delivered there by their grandparents. If I book the ticket for all three of us from London it works out cheaper than a single for me from London to MK and then all three from MK to destination. But is it possible for the kids to join at MK if their ticket is from London, or will it cause problems going through the barriers and boarding at MK?
Comment: What about two singles from MK to destination and one from London? That being the actual travel that is planned.
Comment: As @Muzer alluded to in their answer, I would also be concerned about how you're going to meet up with your children on the train. Is Milton Keynes manned, and will the staff allow the kids through the barriers and escort them to the train, or permit their grandparents through too? Be aware that the train would stop for only a couple of minutes. Also, I've had several occasions where you cannot access all carriages of a train from inside it.
Comment: I agree with @Jontia. Most tickets, if they're not Advance fare, will allow what's called "break of journey", allowing you to exit at any intermediary station along your valid route, and continue your journey later. This can also be done as many times as you want.
Comment: *will it cause problems going through the barriers* Important thought
Comment: To clarify a bit of my comment further, the platform where the children are boarding from may be crowded. Even knowing beforehand which carriage you're on, they may find it difficult to be able to board that carriage. Furthermore, the carriage interior may be crowded, making it difficult to get to that carriage within the train, or the carriages may not be linked internally (if the train is becoming a split service, for instance). There's a lot of variables.
Comment: @user25730 at the moment trains in the UK are generally pretty empty.  Milton Keynes is a major staffed station and staff are generally helpful in letting passengers be escorted to the train by a non-traveller if they need assistance (e.g. children, but I've seen other examples)
Comment: Normally family+friends railcards are good; mine paid for itself in one trip.  However they're only valid while travelling together, so don't be tempted to get one if your tickets are different.  You could get one if you end up booking all together from MK with a separate journey (no railcard) London-MK.  Also station ticket office staff are usually very helpful at the planning/buying stage. At my station they're also bored from lack of customers so would have plenty of time to help (especially outside rush hour)
Answer: It depends on the fare type.
If this is an "Advance" ticket - an inflexible ticket which requires you to travel on a specific train - then this is not, in theory, valid. There is, of course, the possibility of station staff taking pity on you, but it's not something you should rely on.
If, on the other hand, you're buying flexible tickets - (Super) Off-Peak or Anytime Singles or Returns - this would almost certainly be valid, but I'd have to know your destination to be sure (a few destinations have "break of journey" restrictions on the fare, specifically ones in North Wales).
Other things to consider:

I assume you're travelling Avanti West Coast and want to ensure you sit together? If so could you book through their own website and specifically choose your seats? Not sure how COVID affects this ability though. That way you could buy a single London-(Destination) for yourself and two child singles Milton Keynes-(Destination) for your kids. If all else fails, due to COVID social distancing you should be able to book a table seat for yourself (if one is available) and then have your kids sit with you, even if they have reservations elsewhere on the train.
If the worry is because you're actually using a Family and Friends Railcard, bear in mind that this is not valid if you're not travelling with kids, so your London-Milton Keynes portion would be invalid, even if you have also bought child tickets (if you don't physically have children with you, it's not valid). In this case if you want the discount your only option is to do as you suggested - a ticket London to Milton Keynes, then three tickets Milton Keynes-(Destination). If you're not considering a Family and Friends Railcard, this is something you should be considering - they cost £30 for one year or £70 for three, and give you 1/3 off adult tickets and 60% off child tickets for up to 4 adults and 4 children, as long as one of the (up to two) named cardholders is travelling along with at least one child. If you make regular rail journeys with your kids this is almost certainly worth it; and depending on how much you intend to spend sometimes it can be worth getting it even for just one journey.
If you're travelling London Northwestern Railway they don't offer seat reservations so this should not be a concern.
Make sure their grandparents leave plenty of time at the station, as there are ticket barriers at Milton Keynes Central. I'm sure that once the situation is explained they will be let through, or at the very least the children can be handed off to a member of staff who will see them safely onto the train, but I'd want to avoid any last-minute dashes for this reason.
Comment: People have been forced to buy new tickets by on-train revenue protection staff when 'joining late' using Advance tickets (which are only valid between the named start & end stations), and saying 'the barrier let me through' won't cut any ice.
Comment: I'm pretty sure that if the fare type does not permit joining late (e.g. Advance fares) the ticket won't operate the barriers at MKC.
Comment: @MichaelHarvey eh, I'd be surprised if that's the case. Ticket barrier programming rarely matches actual ticket validity in my experience... usually they have a vastly simplified set of rules.
Comment: Furthermore, I'd ensure the grandparents stay with the children to the train. My experience with ticket barriers has been normally good, but I've had a few instances where a ticket barrier has just "eaten" my ticket and not let me through, or where I've put my ticket in to open it and someone has barged in front of me to go through, with the barrier closing behind them (not good at a destination, as then you're left ticketless behind a barrier).
Comment: @user25730 yep, definitely not suggesting the kids be sent to use the barriers on their own! My point was either the grandparents should be allowed through the barriers to see them onto the train, or if the barrier staff don't want to let them through, a member of station staff should be able to escort them through the barriers and to the train.

Children joining train later UK


Wireshark TLS communication shows green HTTP packets when using CCProxy proxy server?

Question: Client <--> CCProxy <--> Server
I have IE configured to use a proxy server for HTTP and Secure connections. I also have a proxy server is set up with CCProxy. It is my understanding that CCProxy cannot be configured with its own certificate and private key. I used IE to connect to an HTTPS website and captured it with Wireshark. The captured packets for the Client<-->CCProxy communication show the TLS handshake and Application Data as green packets. The Packet Detail window shows both Hypertext Transfer Protocol and Secure Socket Layer rows for these packets. Following the stream shows that the data is still encrypted.
For comparison, I have an Apache proxy server that is configured with its own certificate and private key. That Wireshark capture shows grey packets for Client<-->Apache TLS communication and the Packet Detail window only shows Secure Socket Layer.
Can someone please explain why the TLS communication with CCProxy shows the packets as green HTTP?
Answer: It turns out the extra HTTP layer in the frame was because it was connecting over port 8080, so Wireshark assumed there was HTTP in the connection.


EMR Spark History Server UI accessed through ssh tunnel displays no logs

Question: The /var/log/spark/apps/ folder was deleted on our EMR cluster.  I created a new hdfs folder with the same name and changed the permissions to 777. Now each spark application is successfully writing logs to this hdfs folder.
However, something else was in that folder that allowed the Spark History Server that you can connect to through ssh tunneling to display the list of application logs.  It worked just fine prior to the folder getting deleted, but now it does not display any spark application logs (complete or incomplete), even though hdfs dfs -ls /var/log/spark/apps/ shows that the folder is full of logs.
The Spark History Server accessed through the EMR AWS Console still works, but this is less ideal as it significantly lags behind the Spark History Server accessed through an ssh tunnel.
What other item do I need to restore to this folder so that the Spark History Server opened through ssh tunneling shows these logs?

On a Windows computer, the following PowerShell code still opens the Spark History Server UI correctly, but the UI does not show any logs:
Start-Process powershell "-noexit", `
"`$host.ui.RawUI.WindowTitle` = 'Spark HistoryServer'; `
Start-Process chrome.exe http://localhost:8158 ; `
ssh -N -L 8158:ip-10-226-66-190.us-east-2.compute.internal:18080 firstname.lastname@example.com"

Note:
I have also stopped and restarted the Spark History Server.
sudo stop spark-history-server
sudo start spark-history-server
Also:
sudo -s ./$SPARK_HOME/sbin/start-history-server.sh
Answer: Changing the permissions fixed it.
hdfs dfs -chmod -R 777 /var/log/spark/apps/


Is battery capacity proportional to Ampere-hour in every case?

Question: Suppose I have two batteries one inside the laptop with 2.5 AH giving a backup for around 2 hours. The battery is external inside a UPS system with 7.5 AH. Since it has thrice the capacity of the previous one, can I conclude it will give backup for 6 hours?
Answer: You have to compare the energy content which is equal to Ampere hours times voltage. Also with the UPS system you have losses in the voltage converter.
Comment: True, but you should mention that you'll be doing an educated guess based on this. Battery chemistry, age, the current you drain, temperature all affect the system. To compensate for that without the hassle of calculation take **at least** 20-30% off the time you get.
Answer: No, you cannot, since you have not specified the voltages of the two batteries.
Ignoring converter losses, the relevant quantity is not amp-hours but watt-hours. So let's say the laptop battery is 18 volts and 2.5 amp-hours, while the UPS is 12 volts and 7.5 amp-hours. The energy available from the laptop battery is 18 x 2.5, or 45 watt-hours. The energy available from the UPS battery is 12 x 7.5, or 90 watt-hours. All else being equal, the UPS will provide twice the duration of the laptop battery.
All else, of course, is not equal. The laptop battery power goes through a set of DC-DC converters to provide the actual voltages used by the circuits, which have an efficiency less than one. The UPS battery power goes through an inverter with its own inefficiency, and the resulting AC goes to the laptop where it is converted to the required internal voltages. Consequently, you'd expect the overall efficiency of the UPS battery to be less than the efficiency of the laptop battery. Exactly how much less this is, and its effect on the ratio of the two durations, is not something which can be figured out from first principles.


What are the applications of matrices in real world?

Question: Matrices are considered very important in mathematics. What are some examples of applications of matrices to real world problems that would be understandable by a layman?
Comment: Have you looked at the [Wikipedia article](http://en.wikipedia.org/wiki/Matrix_(mathematics)#Applications)?
Comment: The link given by T.Bongers coyly talks about text mining. I would say that every time you use Google, a huge matrix is involved. And that's one of IMHO least interesting applications of matrices.
Answer: One would be hard pressed to find a tool in mathematics that is more widely used for real world applications than matrices and linear algebra.

Physicists and engineers to model physical systems and perform precision calculations needed for complex machinery to work.  Electronics networks, airplane and spacecraft, and chemical manufacturing all require fine tuned computations arising from matrix transformations.
Routing problems and other problems in operations research make extensive use of very large matrices.  There are entire subfields of this discipline related to finding the fastest, most accurate solutions to various matrix problems, for purposes of managing very large supply chains.
Many internet and computer programming companies also use matrices as data structures to track user information, perform search queries, and manage databases.  In the world of information security, many public key cryptosystems are designed to work with matrices over finite fields, in particular those which are designed with speed of decryption as a goal.

These are only a few examples, but let me assure you that nearly any quantitative line of work will make extensive use of matrix equations.
Comment: yeah. i can find the usage in terms of equations, but i want the generic use of the matrix if used any where, be it even of small usage.
Comment: @Saurabh What exactly do you mean by a "generic use"?
Comment: generic use in the sense we use additions for calculating our cash, counting the people in a queue. The same way use of matrix for a common email@example.com
Comment: I've worked as a chemical engineer for 20-years  and, 10-years ago, earned a degree in electrical engineering.  I must say that I have managed to have a successful career in two disciplines and the only times I used matrices was in classes.  I would like to hear the details of your claims that chemical engineers and other engineers use matrices.  Do you have a specific example?
Answer: If you have system of nonlinear equations and linearize them, you get a linear system, which can be represented by a matrix.
So, matrices are relevant whenever you can model the real world with equations, which is very often.
Answer: A simple answer is matrix notation can be used as a way to solve a set of linear equations.  Linear equations are those that consist only of unknowns to the first power multiplied by constants, added together, and set equal to another constant.
In the sciences, this may be used in chemistry to solve material and energy balance problems, electrical problems using Kirchoff's loop current and node voltage laws, and mechanics problems using translational and rotational equilibrium equations.
They may also be used in finance models to solve for supply and demand curve intersections and asset allocation problems.
These kinds of problems occur so frequently that most math software libraries and spreadsheets have built in routines to solve them if the number of unknowns is equal to the number of equations.  Look for "solving simultaneous equations', 'Gaussian Elimination', and 'Matrix Inversion' in the documentation.
This is such a handy approach that ways to extend the usefulness have been found.  For example, the constants in the first paragraph can be replaced by parameters of any mathematical form you have, so long as they do not contain any of the unknowns.  By a technique called Laplace transforms, systems of linear differential equations may be converted into systems of linear algebraic equations and again solved using matrix methods.  This allows the solution of some changing systems such as oscillating mechanical systems (pendulums, manometers), AC electrical problems, and process and mechanical control.  A matrix (transformation matricies) may be used to change the scale of an object, to model reflections across lines and points, find new or old locations after rotation of coordinate axes, and even do some simple derivatives in calculus.  A matrix and its inverse can be used together to move a vector during rotation of axes.  More applications can perhaps be found in linear algebra texts.
But none of these answers may fit your everday life.  Consider how you felt about addition and subtraction when you were first learning them and now after you have become comfortable with them.  You now think about which one to apply anytime you have a problem in life involving the quantity of things.  It is the same with multiplication and division, fractions and decimals, roots and powers.  You may have actually come up with new ways of using them.  For matricies, can you make it fit the form in the first paragraph?  Make it a practice to consider how you can use any of your mathematical tools to situations you encounter and you find the best answer to your own question.


Dryer timer switch motor

Question: Can I put a toggle switch in place of a dryer timer motor to turn the dryer on ?
I have tested for voltage coming into the dryer, it has voltage. I did all the standard tests on all the dryer components, Is it possible to put a switch temporarily in place of the timer switch motor to run the dryer ?
Comment: Thats a bit hard to answer. Model details and pictures would help
Answer: I've never had one completely in bits to really check carefully, but the most basic ones have two* switches in the timer - one for the motor & one for the heater; so you can have that 10 minute cooldown phase at the end. We're going to need pictures to even guess at which circuit is which. You'll need a multi-meter to test open/close positions [with the power off].
Once you overcome that, then you'll have to tackle how the wires are attached. Usually they're on spade connectors, which you'll have to wreck to remove, unless they're also soldered, in which case you'd have to cut them short - meaning you might struggle to re-connect.
All in all, I'd wait until your new timer arrives.
*…and possibly a third, for the timer's own motor, though that may be shared with the dryer motor.

Dryer timer switch motor


Share Home Directory between Arch and Windows

Question: Would it be possible to share the home directory between Arch and Windows with both installed on the same device? With encryption?
Answer: Linux and Windows home directories are quite different, so sharing them may not be in your best interest. But you can share a directory dedicated for this purpose by using a filesystem readable by both operating systems, such as FAT32 or NTFS. For encryption you can use VeraCrypt.


OpenAI Gym equivalent for supervised and/or unsupervised learning

Question: OpenAI Gym has really normalized the way reinforcement learning is performed. It makes it possible for data scientists to separate model development and environment setup/building and to focus on what they really should be focusing on.
Quoting from Gym website:

Background: Why Gym? (2016)
Reinforcement learning (RL) is the subfield of machine learning
concerned with decision making and motor control. It studies how an
agent can learn how to achieve goals in a complex, uncertain
environment. It's exciting for two reasons:

RL is very general, encompassing all problems that involve making a sequence of decisions: for example, controlling a robot's motors so
that it's able to run and jump, making business decisions like pricing
and inventory management, or playing video games and board games. RL
can even be applied to supervised learning problems with sequential or
structured outputs.
RL algorithms have started to achieve good results in many difficult environments. RL has a long history, but until recent
advances in deep learning, it required lots of problem-specific
engineering. DeepMind's Atari results, BRETT from Pieter Abbeel's
group, and AlphaGo all used deep RL algorithms which did not make too
many assumptions about their environment, and thus can be applied in
other settings.

However, RL research is also slowed down by two factors:

The need for better benchmarks. In supervised learning, progress has been driven by large labeled datasets like ImageNet. In RL, the
closest equivalent would be a large and diverse collection of
environments. However, the existing open-source collections of RL
environments don't have enough variety, and they are often difficult
to even set up and use.
Lack of standardization of environments used in publications. Subtle differences in the problem definition, such as the reward
function or the set of actions, can drastically alter a task's
difficulty. This issue makes it difficult to reproduce published
research and compare results from different papers.

Gym is an attempt to fix both problems.

I was wondering if there is something equivalent for supervised or unsupervised learning. I believe such a tool will be useful for the same reasons. Am I missing something? Or is there there something already out there?
Answer: One standard way to represent data for both supervised and unsupervised learning is the tidy dataframe. Tidy dataframe is putting data into tabular form where each row is an observation and each column is a feature. Once data is in that form, machine learning modeling can take place.


Finding potential at a point in a circuit

Question:
$I_1=20/7 A$ , $I_2=2/7 A$ and $I_3=18/7 A$
In this circuit can we calculate the value of absolute electric potential (or simply potential) at specific points like B or E or A etc ??
My textbook assumed potential of E as $0$ and then calculated potential at B = $-\frac{30}{7}$ volts. How can we arbitrarily assume potential at different points, is it allowed (when?) ?
Answer: It is actually a fundamental principle in physics, that there is no absolute scale for voltage, i.e. you can set your zero point anywhere and define all other voltages in relation to that. It is the same with energy.  The only thing that matters physically is voltage difference.
It is usually a good idea to set V=0 at the lowest voltage point in the system, probably at the lower voltage end of one of the batteries
Answer:
How can we arbitrarily assume potential at different points, is it
allowed (when?) ?

Yes we can arbitrarily assign any point as being zero potential because, no matter what point we assign, the potential differences will be the same between points in the circuit and that, in general, is all we care about.
But while it may be arbitrary, there are in fact logical points to assign zero potential to make circuit analysis easier, in particular, node analysis where the objective is to determine node voltages.  Note that E (which is electrically the same as F and D) is a common node for all three branch currents I1, I2, I3. You will understand how this makes it easier to solve for the node voltages at A, B and C when you learn node analysis, if you haven't already.
You may ask, why would we want to know the node voltages with reference to the common point E (F or D). A practical application is trouble shooting electric circuits. If you know what the voltages should be at the points, you can connect the black (common) lead of your voltmeter to E (or F or D) and connect the red lead to those points to verify the correct voltages exist.
Hope this helps.


Equivalence class plain English

Question: Can someone please explain in plain English what is meant by an equivalence class? I've been reading it over and over again and it just doesn't make any sense to me. I've tried looking up definitions online as well but they're too difficult to understand. If it's of any help, I know what an equivalence relation is and I know what congruence modulo are as well.
Comment: You will likely find [this post](https://math.stackexchange.com/q/2525064/191378) to be quite useful.
Comment: Possible duplicate of [Why do we care about equivalence relations?](https://math.stackexchange.com/questions/2525064/why-do-we-care-about-equivalence-relations)
Answer: here's an analogy that'll hopefully shed some light :
consider the equivalence relation 'lives in the same country' defined over the set of all the inhabitants of planet earth.
it is indeed an equivalence relation because :

every person lives in the country that they live in
if a person 'A' lives in the same country as a person 'B' then 'B' lives in the same country as 'A'
if a person 'A' lives in the same country as a person 'B' and the person 'B' lives in the same country as a person 'C' then 'A' lives in the same country as 'C'

now think of the equivalence class of a person 'X' as their country
example : I'm Algerian my equivalence class with respect to that equivalence relation is simply Algeria
Answer: When objects are "equivalent" it's a way of saying they are "the same". An equivalence class is a collections of objects that are all "equivalent" to each other, that is, we treat them all as being somehow the same object (using our equivalence relation to determine this).
Comment: The above description may be used to argue that a subset of an equivalence class is also an equivalence class. I'd make it more precise to avoid this.
Comment: Thank you very much! I think it's when it was first introduced to me in congruence modulo that it became very confusing as there were multiple elements to the set. Idk, I'm just a slow person, but once I get it, I get it
Answer: Given an equivalence relation $R$ on some set $S$, $R$ partitions $S$ in a very natural way. Each part of the partition is a subset of $S$, all elements of which are considered equivalent (e.i. related by $R$). Those are the equivalence classes.
For instance, let $S$ be some set of people, and let the relation be "has same colour shirt as". Then the equivalence classes will be "everyone with red shirt", "everyone with black shirt", "everyone with blue shirt" and so on.
You can even go the other way, first deciding what you want your equivalence classes should be, and then define the relation as "belongs to the same class as".
Answer: An equivalence relation on a set $X$ is a partition of $X$, that is a way to divide $X$ into subsets which are non-empty and pairwise disjoint. Each of these subsets is an equivalence class.
Answer: As the name suggests, an equivalence class is a set of objects that are all in some sense 'equivalent' to each other.
For example we can take a bunch of objects and put all objects with the same color in an equivalence class.  Or: having the same weight. Or same shape. Or ....
Answer: A classic equivalence relation and corresponding equivalence classes you've surely seen before is the set $S=\mathbb{Z}\times(\mathbb{Z}\backslash\{0\})$ with equivalence relation $(a,b)\sim(c,d)$ if and only if $ad=bc$, although you've most likely thought of $S$ as the rational numbers and $\sim$ as equality, i.e.:
$$\frac{a}{b}=\frac{c}{d}$$ if and only if $$ad=bc$$
An equivalence class is just a way of grouping objects that, though different, are the "same" in some way. $\frac{1}{3}$ and $\frac{3}{9}$ are, in fact, different under the above definition, but we treat them the same under this equivalence relation, as they belong to the same class.
Answer: The analogy I always use is buckets.
The items in each bucket may be different, but up to some description are the same. Let's take people that live in certain states.
The bucket for Texas, for example, will include people from San Antonio, Austin, Houston, Bryan, McAllen... all over! But for the sake of classification, they're all Texans. They're all in the same equivalence class. The same can be said for a bucket (equivalence class) corresponding to Ohio: Canton, Dayton, Akron, Columbus... all people from Ohio!
Let's put it mathematically. The first example of equivalence classes people see is in modular arithmetic. Let's consider modulo 12 (the usual clock). Numbers with the same remainder upon division by 12 are placed in the same equivalence class - bucket - since they're essentially the same. In the class corresponding to 3, written $[3]_{12}$, includes $3, 15, 27,$ and so on. these numbers all have the same remainder when divided by 12.
In short - equivalence classes just classify objects that are, pretty much up to some descriptor, the same.
Answer: Let $S$ be the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10 \}$. For any two elements $a, b$ from $S$ let's say that $a\sim b$ if $a$ and $b$ give the same remainder when divided by $4$. It's not difficult to check that this relation is reflexive, symmetric, and transitive, so it is an equivalence relation. If $a\sim b$, we say $a$ and $b$ are equivalent.
Now let's split up $S$ into smaller subsets. In each subset, we will put all of the elements of $S$ which are equivalent to each other (under our definition of equivalent).
So the elements of $A=\{1, 5, 9 \}$ are equivalent to each other. Also the elements of $B=\{2, 6, 10\}$ are equivalent to each other, and the same can be said about $C=\{3, 7\}$ and $D=\{4, 8\}$. The sets $A, B, C, D$ are the equivalence classes induced on $S$ by the relation $\sim$. These sets have the proprety that they are all disjoint from each other, and if you unite all of them yo get back $S$. Therefore these sets are called a partition of $S$. Always, equivalence classes of any equivalence relation $\sim$ on a set $S$ partition $S$.
Now we could keep referring to these sets as $A, B, C, D$, which is convenient for this example (because these sets are easy to list explicitly). However, when the sets are not so small, we need a different method to name the equivalence classes. We might refer to $A$ as "the equivalence class which contains $5$", or, in short, "the equivalence class of $5$". Note that $A$ has $3$ possible names using this naming method: $A$ could be called the equivalence class of $1$, of $5$, or of $9$.
We name $B, C,$ and $D$ in a similar way.

Equivalence class plain English


What is the chance to get a parking ticket in half an hour if the chance to get a ticket is 80% in 1 hour?

Question: This sounds more like a brain teaser, but I had some kink to think it through :( Suppose you're parking at a non-parking zone, the probability to get a parking ticket is 80% in 1 hour, what is the probability to get a ticket in half an hour? Please show how you deduce the answer. Thanks!
Comment: As a Frenchman, 80% chance in 1 hour seems really high to me, would be rather 20% chance in 1 day.
Comment: Where I'm from, between the hours of 3PM - 6PM, M-F, the probability of getting a ticket and towed is 100% within 10 minutes.
Answer: It really depends on what model is assumed. However, if the idea is that no matter how long you leave your car there, you have a $20$% chance of getting through any given hour unscathed, you can treat it as an exponential decay problem. Let $p(t)$ be the probability that you do not get a ticket in the first $t$ hours. Then $p(1)=0.2$, $p(2)=0.2^2$ (a $20$% chance of making it through the first hour times a $20$% chance of making it through the second), and in general $p(t)=0.2^t$. The probability of not getting a ticket in the first half hour is then $p(1/2)=0.2^{1/2}=\sqrt{0.2}\approx 0.4472$, and the probability that you do get a ticket in the first half hour is about $1-0.4472=0.5528$.
Comment: +1 for "It really depends on what model is assumed."  Without making assumptions, all we can really say is that the probability of getting a ticket in 20 minutes is less than or equal to 80%.  An alternative model is there is an approximately 80% chance that the ticket-writer is working today.  If the ticket-writer is working there is an approximately 100% chance that you will get a ticket within 5 minutes.
Comment: Parking attendants are not memoryless.
Comment: @smdrager, I don't understand.  There are uncountable infinite number of models to choose from.  Anyway you are not provided with any data to calculate likelihood.
Comment: @Dave: You're assuming something like full-time checkers with regular routes. Try assuming instead random checks.
Comment: @Rock: How would you do it? You can't say that your chance of getting a ticket in two hours is $0.8^2$, since it should obviously be **more** than $0.8$, and you certainly can't say that it's $1.6$, since that's greater than $1$. It's much easier to see how your chance of **not** getting one keeps shrinking.
Comment: I'm not seeing the logic behind this model. If there is an 80% chance of getting a ticket in 1 hour, wouldn't this suggest that the parking checker covers 80% of the route in an hour? If so, then the checker covers 40% of the route in half an hour.
Comment: @Zarrax, then the challenge is calculating the likelihood of choosing the correct model for determining the probability of getting a ticket. Assume there are N models for determining probability, what is the likelihood of choosing the correct model? You could probably then use it to calculate a margin of error for your model.
Comment: @Robert Mastragostino If there's an 80 percent chance you get ticketed in an hour, then some possible reasons would be that this is the time of day they usually ticket around there, or they enforce only 4 days a week, or the ticketer is on break 20 percent of the time and the route takes an hour, or they do have 1 1/4 hour rounds and this is 80 percent of it. Without any extra info you can't say what accounts for the other 20 percent, but it won't be exponential in any of these scenarios. Since there are so many options I don't think there's one "right" model.
Comment: @emory: It has to be possible to write down a model, otherwise it isn't a model.  So for this sense of the word model, the set of all models must be at most countably infinite.
Comment: Thanks, Brian. I kinda get the idea but what confuses me is why can't we consider the probability for having a ticket like 0.8^t but have to do the other way around? Thanks!
Comment: @BrianM.Scott could you say that the probability of getting a ticket is 0.8 to the power of 1/nbhours ?
Comment: I personally think neither of the models in the answers here are particularly realistic. If you watch "Parking Wars" for example, you see they target certain high-yield areas at certain times of the day.. So I don't think the parking enforcer circles the same area uniformly over and over again, nor is it a random occurrence. But if you need a numerical answer you have to choose *some* model.
Comment: @Zarrax, I definitely see your point as far as realism goes, but isn't this solution implied by the question? 80% in one hour, without extra qualifiers, seems to imply 80% in any hour. In those other situations, knowing there are 80% odds in an hour wouldn't be enough to solve the problem, though I agree that they're more realistic. What other model would be 'most reasonable' (ugh fuzzy term) to assume given no extra information?
Comment: It doesn't matter if the answer is 40% or 55%.  Either way, the probability is way to high to risk not putting money in the meter.  The ratio of the parking ticket to the meter fee is usually much, much greater than 2.  So if you're asking this question, the answer is, put money in the danged meter.
Comment: Just expect that there's 80% chance of bird dropping falling onto the car in the next hour. Find the probability that it will happen in the next 30 minutes. An exponential distribution is memoryless, so I would use it.
Comment: @emory, I just took the question to the meta level; it wasn't really intended to be a serious answer. What is the likelihood people will think that? According to upvotes 1 in 4.
Answer: If my probability of getting a parking ticket in a half hour is T, then
$$T+(1-T)T$$
are the odds I got one in 1 hour. The first term says that I got it in the 1st half hour (so it doesn't matter what happens after that), and the second term says that I got it in the 2nd half hour (so not in the first). Solving and taking the sensible solution:
$$2T-T^2=.8\implies T^2-2T+.8$$
$$T=1-\sqrt{.2}$$
Which are roughly $55.3\%$ odds in a half hour.
Comment: @Rock you could instead choose this as your answer.
Comment: You've made a Markov assumption, which may or may not be valid.
Comment: @BenVoigt fair enough, but it's either this, assume extra conditions that weren't specified, or be unable to do anything at all. You always have to assume some model anyway, and in the absence of other information a Markov assumption seems to be the natural choice. Of course not a statistician though, so I wouldn't be surprised if I'm begging to be corrected here.
Comment: @Joshua Now I have it. The accepted answer gives a general solution which I guess is what the question potentially asked for.
Comment: Hi @RobertMastragostino, just wanted to let you know that the ["Italian Language & Usage"](http://area51.stackexchange.com/proposals/42949/italian-language-usage) proposal has restarted. Since you committed to the previous proposal, maybe you are still interested. See you! (I hope this comment is not considered as spam: if so, I beg your pardon)
Comment: @Luicius it probably is, but it's appreciated. Thank you, though you might want to delete the comment just to be safe.
Answer: Another way of looking at the exponential model: If $P$ is the probability you get lucky and don't get ticketed in the first 30 minutes, then $P^2$ is the chance you luck out twice in a row and don't get ticketed in the first hour. You know $P^2 = 0.2$. So $P$ is the square root of that or $0.447$. So the chance you are unlucky in the first 30 minutes and get a ticket is $1 - 0.447 = 0.553$ or $55.3$ percent.
Comment: For me the smoothest solution !
Comment: A very clear explanation!
Comment: Thanks.. but to be clear I'm just explaining the exponential model in layman's terms, and I am not suggesting I agree this  model is necessarily the "correct" one for the problem.
Answer: A simple but not unreasonable model is that the ticket person cycles through her/his round in roughly constant time.  Suppose that you park illegally for exactly an hour at a randomly chosen spot in the round, and at a start time uniformly distributed with respect to the cycle. Since your probability of getting a ticket in an hour is $80\%$, the cycle length is $1.25$ hours. Thus your probability of getting a ticket if you park illegally for half an hour is $40\%$.
Comment: For the people who wonder how your model can give something different: The other answers assume that getting no ticket in the first half of an hour does not increase the chances of getting one in the second half of the hour. In a place where 80% of cars get a ticket within the hour, this is not a very realistic assumption. So, we can rejoice and expect that the probability is closer to 40% than to 55%.
Comment: Another explanation to this: Assume there are 100 illegal cars, so the parking warden can handle 80 tickets per hour (we assume linear distribution of tickets per time) and thus 40 tickets half an hour. If you are one of that cars during that half of an hour ... your risk getting caught is 40/100.
Comment: IMO from a modeling perspective this is by far the most reasonable answer given. The exponential model seems unreasonable since clearly your hazard (aka force of mortality or failure rate) ought to be increasing over time.
Comment: Agreed, @guy.  With an exponential decay model of tickets, no matter how long you leave your car on the street, there is some nonzero chance that it will not get a ticket.  That's not the behavior of any conceivable real-world method for writing parking tickets (which generally involve somebody walking down the street and ticketing every illegally parked car).
Answer: There is another way to look at this problem.
Most of the answers of ~55.3% (including the currently accepted answer) assume the "meter reader" is checking cars COMPLETELY at random.  In this model, there is an extremely small but non-zero chance you could leave a car in the same spot for a month and not get a ticket (although it's up there in "lottery" winner and struck-by-lightning chances).  In fact, the chance of getting a ticket using this model is always less than 100% for a finite amount of time.
However, meter readers typically work in a linear pattern starting at one point and then looking at ALL the cars in a line down one side of a block.  If we consider the realistic pattern that cars are checked in a linear fashion instead of cars being checked randomly, the answer becomes much simpler.  Therefore, when they completely cover an area (in a finite amount of time), your car has a 100% chance of getting a ticket.  If you are 100% likely to get a ticket within a finite amount of time, then the p(t)=1-0.2^t answer for getting a ticket is invalid.
So lets consider linear coverage.
If you have an 80% chance of getting a ticket in one hour, the meter readers in your city walk 4 out of 5 blocks per hour (in 60 minutes).  Assuming they follow a pattern which has complete coverage and fairly evenly sweeps the whole parking area, they could have 100% coverage of all parking spots in 75 minutes.  That means no matter where you park, within 75 minutes you will get a ticket.  Under the linear coverage model, p(t)=0.8t.  So in half an hour, you have a 40% chance of getting a ticket.
This is a case of not trying to get too mathematically clever because the simplest answer could actually be the correct one.
The real answer depends whether your meter people ticket cars randomly or sweep areas linearly with even coverage (and gets more complicated with real life where they sweep some areas much more often than others but still get 100% coverage over a finite amount of time).
Answer: Not enough data to come up with an answer. You would have to know the pattern that the parking warden takes to get the correct answer (first half hour, second half hour), also this depends on which part of the time unit the parking warden would take up, for example 1pm-2pm, 1.05pm-2.05pm etc
80% can reasonable be assumed for an hour, but for half an hour less so.
I would therefore conclude that there is an 80% chance that you will get a ticket as there is no way to distinguish the half hour from the full hour because of lack of data for the movements of the parking warden.
Comment: Why be pessimistic? Maybe it's your lucky day and there's a zero percent chance. :)
Comment: If your a pessimist, there are no unexpected nasty surprises, you will never feel the pain of disappointment, and reality will hold no fear. Abide by the law, there will be a 100% chance of not getting a ticket. :)
Comment: You know that there is an 80% chance of getting a ticket in a hour. Since the full hour encompasses the half hour in question, then it is definably an 80% chance through set theory. Without further data about the parking wardens movements, it is impossible to suggest anything better.
Answer: This sounds like a Poisson Process to me... so, assuming that:
Since we don't have additional data, I will asume that the expected probability of being spotted once per hour is $p=0.8$. This is a Bernoulli experiment (either you get spotted or not). I will model the probability of being spotted by a machine, person, or some other authority that can issue a ticket ;-)
If $\lambda$ is the expected times that you get spotted every hour, then $\theta=\frac{1}{\lambda}\ $ is the expected time between events.
The number of events given in $t$ can be modeled by a random variable $N \sim Poisson(\lambda·t)$ where $t$ is the amount of time (in hours) elapsed, and the time between events can be modeled by a random variable $X\sim Exponential(\theta)$.
So, the probability of getting $n$ tickets in $t$ hours, with an expected probability of $\lambda$ that you get a ticket in one hour is:
$$Pr\{N=n\ |\ \lambda, t\}=\frac{e^{-\lambda·t}·(\lambda·t)^{n}}{n!} $$
Since we need to calculate the probability of being spotted at least once, we can calculate $1-Pr\{N=0\}$. Then, for $n=0$, $\lambda=0.8$ and $t=0.5$, then:
$$1-Pr\{N=0\ |\ \lambda=0.8, t=0.5\}=1-\frac{e^{-0.4}·(0.4)^{0}}{0!}=1-e^{-0.4}\approx0.4918 $$
You should notice that, even if you "get spotted" more than once per day, this model still holds, since this process is memoryless.
Further reference: Ross, Sheldon, "Introduction to probability models", Ninth edition, Chapter  5.
(I really hope this is not your homework)
Comment: A Markov chain is a much better assumption, since I've never heard of multiple tickets being issued to a car (that remains parked in one place) in the same day.  Therefore "got 1 ticket" is a terminal state, there is no chance of moving to two tickets.  The problem stated that **there is an 80% chance of getting a ticket, not that the mean number of tickets is 0.8.**  Your answer confuses these two quantities.
Comment: You did it wrong.  $\lambda$ is not $0.8$.  To get $\lambda$, you need to reverse the calculation that you did get right.  If the _probability_ of getting a ticket in an hour is $0.8$, then $1-e^{-\lambda t}=0.8$ where $t$ is one hour, so the _rate_ $\lambda$ is $1.61$ tickets per hour.  Then we use $1-e^{-\lambda t}$ with $t$ equal to half an hour.  We get $1-e^{\lambda{1\over 2}hr}$ is $0.553$.  The same as the exponential examples above, since it is the same memoryless assumption.
Comment: @BenVoigt: Unfortunately, in some cities, you can be ticketed multiple times, in the same spot, if your car is still illegally parked.
Comment: With no further information, the expected value of the tickets per hour is 0.8 (is a Bernoulli experiment). I think the Poisson Process hypothesis holds because all you know is that you may "get caught" with a given probability in a unit of time. Of course you can't be caught more than once a day, but that doesn't implies that you can't be "spotted" _n_ times a day.
Comment: That said, I'm correcting my answer, because I calculated the probability of "being spotted" once per interval, which is not what we are looking for. The probability we're looking for is "what is the probability of 'being spotted' at least one time", which is 1 - Pr{"Not 'being spotted'"}.
Comment: @Barranka: apparently it's not a hw. but thanks for giving your distinct Poisson solution, which I thought about too.
Comment: @MarkAdler: another proof of the beauty of Math :)
Comment: @MarkAdler Thank you Mark... I will soon correct my answer to include this calculation for the value of $\lambda$. And yes, Math is beautiful! :-)
Answer: I agree that it depends on the model. Otherwise some fair of independence is presume in suggesting either 'half hours are equi-likely'
Although if you specify that is its an arbitrary hr, and any arbitrary half hour segment (rather than the first half of every hr) that might be different. Otherwise in every hr one inspector might give a zero percent chance for each first half hour, and 80 percent for the second 'half hour'.
But once its specified that
' to be arbitrary hr, or any arbitrary half hr segment'or the average or expected probability value".
Or that
" that the model or probability, value does  not pertain a one shot case (ie one is not considering a singular case scenario)- where it only applies to a singular occasion for some specific hr ."
Then I presume that this, would not be  be a counter-example to the claim (the second half - hour, is also the first half  of the full next hour segment from t/2 to 1.5 t where t = 1 hrs), and that the conventional answer sounds reasonable.


How accurate does the alignment of molecules in DNA have to be?

Question: We have learned that the same information is saved in the DNA in the same place. So when DNA splits and then connects again (during reproduction); how does it maintain the integrity of the information? In other words, do they split and come back like a zipper, or does something else happen?
Comment: It is not clear what you want to ask
Comment: I think he's asking if whether the DNA post-division needs to be the the exact same spacial configuration as in the mother cell.
Comment: DNA and nucleotides don't separate and come and go on their own. Spefific enzymes mediate these processes. So it's not like the two strands of DNA suddenly separate and nucleotides wandering around just come and hybdridize. I suggest you to do some research on DNA replication. If you're interested in that, please edit your question, and we'll be happy to answer.
Comment: I still don't understand this question and am unsure why it was reopened. It's not even clear if the edits reflect the original intent. This doesn't seem answerable unless the OP clarifies what they are asking.
Answer: In the case of replication the parent strands do not pair up back into the original helix. Instead, as your picture and the one below show, the complementary bases are paired up on one of the parent strands, resulting in two copies, each with one parent strand and one new one.

Source: Kathy's webpage on DNA
Answer: When I was reading the question, I had a flash of insight into what I presume the question the OP is asking: how do we maintain "integrity" of DNA when we get a set of DNA from mother, and the other set from father (ignoring mitochondrial DNA and other neat things). I would assume this is the question due to the key word reproduction, but anyway,I am here to provide an answer.
As an answer:
I believe the OP has a misconception. DNA of the parents do not pair up in the progeny. Rather, we have 23 pairs of chromosomes, or 46 total chromosomes. What does this mean? The set from the mother and the father do not come together, like a zipper, and have certain bases that just happen to be different (due to single nucleotide polymorphisms and whatnot). Instead, there are two sets of DNA, each with a double helix.
I think this may be what AliceD is also conveying, but I wished to add some more points such as bringing to numbers to help clarify. If I misunderstood, please clarify!


Want to get back in touch with professor so he won't forget me but don't know what to say

Question: I want to get back in touch with professor so he won't forget me but don't know what to say! I took a class with him, but that's about it. I would like to keep in touch, so that he will remember me in the future but it's already been a few months, I just have no idea what to say, as I do not have anything to say really (Haven't been able to catch up with any research or so due to work). Any advice?
Comment: Can you clarify why is it important to you that he remembers you?
Comment: I would like to do a PhD in 3-4 years and will either need a reference or I would also like studying with him. However, it's all quite unclear as of now and I would like to build the relationship first. I just don't know how to get more on his radar, as it were.
Answer: I enjoy it when former students let me know what they are working on and how what they learned in my course is assisting them in their career.
I would suggest an email sharing this information and thanking the professor for teaching such a useful course.
Answer: Had a very similar situation recently, with an academic I worked with over 15 years ago.
The email I sent was related to his current work (research displayed on his professional website - this is an important 'hook' I found), and also linking it to the research that we collaborated with all that time ago and the classes I took from him earlier still.
After the usual greetings, I firstly asked about about some of his current research that was of interest.  Then in the next paragraph, I wrote a reminder of when we worked together in the context of how I have worked from there (once again linking it to his research). Then wished him well and signed off.
He remembered me - and answered my questions and we are now back in dialogue.


Why is my model showing different shading on a flat area around a hole?

Question:
I tried to make a Scythe, and I wanted to cut holes in the shape, but I'm getting these dark areas around the hole. This happens a lot and it forces me to change my design and I don't want to change my design this time.
Comment: Questions on Stack Exchange are supposed to be useful for people finding it in the future, not just for the person asking the question. "Why s this happening" isn't a very useful title. I would not search on google for "why is this happening?"
Comment: There isn't any reason to change model design in these cases. Just don't use Ngons + Subsurf. And if subsurfing no need to cut so many vertices circle, 8 vertices would be enough.
Comment: Please try to describe your issue as part of the title and on the text of the question. Read: [What is the problem with posting an image (or link) and asking "How do I do this?"](https://blender.meta.stackexchange.com/questions/2449/what-is-the-problem-with-posting-an-image-or-link-and-asking-how-do-i-do-this)
Comment: Look at this (possible duplicate) https://blender.stackexchange.com/questions/74922/how-can-i-make-holes-on-a-wrist-strap/74926#74926
Answer: It looks to me like you simply have smooth shading turned on. In edit mode, in the 3d view, the toolbar on the left (in your picture it says "Add Meshes" at the top), go to the transform tab, and in the "Edit" section, look for a label called "Shading:". Under that, there should be two buttons, one says "smooth" and the other says "flat". Click the one that says "flat" and see if those dark patches go away.


Does anyone know a safe and reliable grammar add-on for Scrivener?

Question: I was finally able to change from US English spelling to Canadian spelling, but my biggest problem is with grammar. I have tried to enable a grammar tool, but can't find one I trust. It would be wonderful if the darn program included a grammar tool of its own. Heck, even Word does that.
Comment: Also, I wouldn't go looking for grammar add-ons per-say.  If you look through some of the questions, people have been asking about grammar software and usually the software was given with the advice that you should still proof read/edit find someone who will.  Grammar software are reliable to a certain point but the other issue is they may change something that was purposely done with poor grammar.
Comment: Apparently none of us use Scrivener either... I don't know of any such add-ons. I recommend you check in at the Scrivener forums; they're more likely to have help for you: https://www.literatureandlatte.com/forum/viewforum.php?f=22
Comment: you guys are tough... anybody home? ;)
Comment: While Writers is a worldwide site, a lot of users are from the US where we've just had a four-day holiday weekend.  Activity's been down from that, sorry.
Comment: @ggiaquin: I find astounding how many viedos are out there showing how to use whichever grammar add-on and which clearly show people blindly following advise. These add-ons are incredibly useful to catch typos and to point out **potential** grammar problems. But the user must *always* go through each problem and...
Comment: (oops, pressed enter by mistake)... and analyse each potential problem. More often than not, it isn't.
Comment: Thanks for the responses. I returned to Word. I am used to it and it at least asks me about spelling. It does have some basic grammar. Now, I need to figure out moving my word doc to Scrivener.
Answer: ProWriting Aid works with Scrivener. I've used the grammar software mostly on MS Word and like it a lot. I have yet to try it on Scrivener because, at the moment, my Scrivener projects are not in English.


Can I run mongodb under a Managed Service Account

Question: I'm trying to run MongoDB under a Windows Managed Service Account. This works fine for memcached on the same server, but when I run the MongoDB account I get this:
Windows could not start the MongoDB ervice on Local Computer.

Error 1053: The service did not respond to the start or control request in a timely fashion.

Adjusting the file permissions so the account can write (or have fullcontrol) inside the MongoDb directory (where binaries, logs and data is stored gave me this:
Windows could not start the MongoDB on Local Computer. For more information, review the System Event Log. If this is a non-Microsoft service, contact the service vendor, and refer to service-specific error code 100.

The log file now says this:
exception in initAndListen: 98 Unable to create/open lock file: C:\data\db\mongod.lock

So, obviously I had a remaining mongod.exe running.
Answer: Permissions, file locks and all such things must match.
I figured the above out as I was asking the question and detailing every step.
Thanks for the rubber ducking! :)