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# October 17

## Plane Projection of a Hyperbola

Two planes intersect at an angle strictly between 0° and 90°. One of them contains a hyperbola, which we project onto the other intersecting plane. Is the result still a hyperbola ? And if the answer is affirmative, does this mean that the answer to this question is also a yes, inasmuch as any triangle can be seen as the projection of an equilateral one onto a plane at a given angle ? — 79.113.235.46 (talk) 06:13, 17 October 2016 (UTC)

If one of the axes of the hyperbola is parallel (or normal) to the line of intersection of the two planes: yes (by a change of variables in the conic equation). I suspect it's true in only that case (an example of an obviously non-hyperbolic projection of a hyperbola), and so is not useful for the triangle question. --Tardis (talk) 00:54, 18 October 2016 (UTC)

## Unitary Transformation

Let ${\displaystyle I=\{0,1,\dots ,2^{n}-1\}}$, ${\displaystyle A=\{|y_{i}\rangle =\sum _{x_{i}\in I}{(-1)^{a_{i}}|x_{i}\rangle }\mid a_{i}\in I\}}$, ${\displaystyle f(|y_{i}\rangle )=\oplus a_{i}}$. (the XOR of the phases' exponents). We say that ${\displaystyle |x\rangle =|y\rangle }$ iff ${\displaystyle \forall i:|x_{i}|^{2}=|y_{i}|^{2}}$ (two vectors are equivalent iff all of their amplitudes are equal).

Is there any unitary transformation ${\displaystyle U}$ that satisfies: ${\displaystyle \forall |a\rangle ,|b\rangle \in A:f(|a\rangle )\neq f(|b\rangle )\implies U|a\rangle \not \equiv U|b\rangle }$?

Thanks in advance! עברית (talk) —Preceding undated comment added 06:27, 17 October 2016 (UTC)

• The definition of A sounds fishy, but in any case, a prominent notice at the top of the page states: We don't do your homework for you, though we’ll help you past the stuck point. TigraanClick here to contact me 11:09, 17 October 2016 (UTC)

## Measure of angle formed by absolute value graph

Let a be a positive real number. What is the measure of the angle formed by the graph of ${\displaystyle f(x)=a|x|}$? GeoffreyT2000 (talk, contribs) 16:03, 17 October 2016 (UTC)

• Just compute the angle between ${\displaystyle f(x)=ax}$ and the vertical line, and double that. Are you familiar with tangent (trigonometry)? TigraanClick here to contact me 17:04, 17 October 2016 (UTC)
• The angle formed between ${\displaystyle f(x)=ax}$ and the vertical line has the same value as the angle formed between ${\displaystyle f(x)=ax}$ and the horizontal line. Logic dictates that it must be half the angle formed between the horizontal line and the vertical line. 175.45.116.99 (talk) 04:48, 19 October 2016 (UTC)
• The IP's answer is only true for a = 1. However, for a > 0, the angle between y = ax and the x-axis is tan−1a, the requested angle is 2(π / 2 – tan−1a) = π – 2tan−1a. As expected, as a → 0+, the angle → π. As a → +∞, angle → 0, and at a = 1, angle = π / 2. If you wanted an angle of π / 3, solve for a to get a = √3. EdChem (talk) 05:25, 19 October 2016 (UTC)
... and, of course, all of the above assumes identical scales on the axes. Dbfirs 15:54, 19 October 2016 (UTC)

# October 18

## Triple bar: real or hooey?

Our article on, say, triple product contains lots of instances of ${\displaystyle \equiv }$, and I don't get the point. The cross product of two vectors is straightforward, if annoying, to work out as pure algebra, and if I worked it out as an equation, I'd use a plain vanilla equals sign in that equation. I looked up the triple bar article and Identity (mathematics) and I don't really see anything, outside of specialized contexts, where it has any general significance I can understand. To me the sign is confusing because there have been times when I've sat in a classroom and seen the sign used in the same way as "<-" in a computer program (or "=" in languages that use "==" for equals). So I have to look and see, are they saying these things work out to be equal, or are they defining them to be equal?

Is there a way to defend this usage, to say that yeah, anyone looking at this knows this triple-bar thingy has to be there instead of an = sign, and everyone knows what that signifies? Wnt (talk) 18:01, 18 October 2016 (UTC)

Huh, that looks like simply a mistake in the article. Equal signs all around seem to be warranted. But now, I feel like I may be missing something, as well... TigraanClick here to contact me 18:09, 18 October 2016 (UTC)
I don't think you all are missing anything. The article used equal signs until April 2016, when someone went about tweaking the equations with no explanation or edit summaries. Per MOS:MATH#NOWE, a simple equal sign is preferred over ${\displaystyle \equiv }$ or ":=". That an equation serves as a definition should, in my opinion, be explicitly indicated in the prose rather than rely on a possibly obscure symbol. If no one, objects, let's revert to equal signs. --Mark viking (talk) 19:30, 18 October 2016 (UTC)
I'm pretty sure triple product is intending to use ${\displaystyle \equiv }$ to mean roughly "equal by virtue of explicit prior definition or assumption.", as opposed to e.g. "equal because we did the same thing to both sides". And not necessarily doing anything with any consistency. But I agree with Mark that WP MOS is pretty clear on this, and we should probably revert to all '='. SemanticMantis (talk) 20:16, 18 October 2016 (UTC)
My interpretation was the triple bar was meant as identity as opposed to equality. The distinction is subtle and it's much better to spell out in words what is meant rather than using an obscure symbol. For example if "${\displaystyle \sin ^{2}\theta +\cos ^{2}\theta \equiv 1}$" is meant to mean "${\displaystyle \sin ^{2}\theta +\cos ^{2}\theta =1}$ for all ${\displaystyle \theta }$" then just write it out that way instead of confusing half of your readers. In the triple product article the "for all a, b, c" is implied by context so the distinction isn't needed. A bigger question is whether ${\displaystyle \equiv }$ as identity should be in any WP article other than to mention it as being used by "some authors". --RDBury (talk) 21:57, 18 October 2016 (UTC)
As someone who used to use the triple bar back when I published research, I disagree with the above discussion and the preference stated in the MoS. A double bar means equals, maybe for some value of a variable or maybe for all values of parameters and variables. A triple bar means more specifically the latter, and so should be preferred on grounds of clarity in my opinion. And ${\displaystyle f(a,b)\equiv g(a,b)}$ is more succinct than, and hence preferable to, ${\displaystyle f(a,b)=g(a,b)}$ for all a and b. Loraof (talk) 23:39, 18 October 2016 (UTC)
This distinction continues to evade me. Do we write a = a + 0 or do we write a ${\displaystyle \equiv }$ a + 0 or a = a + 0 "for all a"? I mean, the cross product is just a shorthand for some algebra, ${\displaystyle \mathbf {u} \times \mathbf {v} =(u_{2}v_{3}-u_{3}v_{2})\mathbf {i} +(u_{3}v_{1}-u_{1}v_{3})\mathbf {j} +(u_{1}v_{2}-u_{2}v_{1})\mathbf {k} }$. At least it's a double bar in cross product where I stole this math code from. Or is that a triple bar when we say it is equal ... or identical... to something else? Wnt (talk) 23:52, 18 October 2016 (UTC)
Essentially, the triple bar is used for equations that are always true, whereas the double bar is used for those that are not always true (and you can solve for the values that make it true). So you would write sin θ = 1 with the double bar (it's not always true, but you can easily find solutions, like θ = π/2); but sin2 θ + cos2 θ ≡ 1 needs the triple bar because it doesn't matter what θ is, the equality always holds. Double sharp (talk) 02:58, 19 October 2016 (UTC)
So E=mc^2 needs to be "E${\displaystyle \equiv }$mc2? I feel like when working out math problems we use quite a bit of "identity", and I'm loath to give up my ascii equals sign for all of it. As much as it is used in triple product, by this standard there are many other sections and instances (like under "Proof") where the triple bar still needs to be put in to replace the double. Wnt (talk) 11:45, 19 October 2016 (UTC)
There probably are situations where it is helpful for the reader if the text tries to distinguish identities (equations that are true for all values) from definitions and from constraints (equations we want to be true and where we want to find the specific values that make them true). However, I don't think the triple bar is widespread enough to be used without explaination, and I fear it might confuse the reader rather than help him. (For what it's worth, in situations where it is important to distinguish identities from constraints I have seen an equals sign with an exclamation point above used to denote constraints, like ${\displaystyle \sin \theta {\stackrel {!}{=}}0}$. Similarly, sometimes a question mark above an equals sign is used to denote identities we want to show to be true. I have only seen this usage on blackboards, for what it's worth.) Tea2min (talk) 12:16, 19 October 2016 (UTC)
The double-bar/triple-bar distinction is only useful when it is important to distinguish constraints from identities. When that is not the issue at hand, it is often not used, because it is then not serving any useful purpose. Oh, and I would also confirm what Tea2min has said about ? and ! modifying equals signs on blackboards. Double sharp (talk) 13:14, 19 October 2016 (UTC)
If the idea is to use the triple bar for every expression that evaluates to true no matter what values the variables around it have (i.e. as a shorthand for "forall..."), it looks pretty ridiculous to me. I could be convinced it is helpful as a shorthand for "analytic equality", i.e. an equality that is deduced purely from logical axioms and the form of the variables of the LHS and RHS (type but not value in a programming language). But it would take a lot of convincing, a better definition, and anyways, it is the MOS that needs convincing (good luck with that). TigraanClick here to contact me 13:22, 19 October 2016 (UTC)
The article references that I can access use just an ordinary equal sign. Does anyone object if we put the article back to that format (with a mention in the text that the identity is true for all values if anyone would possibly read it differently)? Dbfirs 15:52, 19 October 2016 (UTC)
I think article should match MOS. While I understand the complaints against the content of MOS, I think it is still best to follow something consistently. As it stands, we have three (or more) notions what that symbol means. Any given (good) math text that uses the symbol can simply define it at first usage, but that is not how we are supposed to approach math articles on WP. SemanticMantis (talk) 18:14, 19 October 2016 (UTC)

Methinks we should move that whole conversation to the article TP, and give page watchers a few days to react (or do the change now, but be prepared for a possible on-sight revert; we cannot really claim a meaningful consensus from here). TigraanClick here to contact me 16:41, 19 October 2016 (UTC)

I suspect that it would be only Dangvugiang who would disagree, but we don't want to start an edit war. I've copied this content to the talk page of the article in the hope that we get further opinion. If Dangvugiang doesn't respond, would it be permissible to revert his change to "equiv" back to plain "equals"? Dbfirs 07:29, 21 October 2016 (UTC)
If you scan that user's contributions, 100% of them are to article space, and there is not a single edit summary (not counting the auto-generated ones for section headings and undo's). So I wouldn't hold my breath waiting for him/her to engage on the question. Leave a note on the talk page (and on the user's talk page too; won't help probably but it's worth a try), and if there's no response, after a decent interval (which in my opinion can be pretty short; I don't have a lot of patience for this sort of editor), change it back. --Trovatore (talk) 07:45, 21 October 2016 (UTC)
Thanks. Yes, I see what you mean. I've left a courtesy note on his talk page so that all opinions can be taken into account. Dbfirs 08:27, 21 October 2016 (UTC)
I've seen this often enough on Wikipedia that I don't think we should make this about one article or one editor. There's a broader issue here and I've seen people here weigh in with points toward both sides. My preference remains for simple equals without philosophy but I'm not declaring a jihad over it. Wnt (talk) 11:09, 21 October 2016 (UTC)
I agree that the issue is much wider, but the question was raised here about a relatively recent change to the article mentioned above. Our article at Identity (mathematics) uses a plain "equals", not "equiv". Dbfirs 12:05, 21 October 2016 (UTC)

## equation of circle

how to find equation of circle which touches two parallel lines 3x-4y=7and 3x-4y+43 and center lie on the line 2x-3y+13=0.SonishaShrestha (talk) 23:31, 18 October 2016 (UTC)

Heh, that will get ya thinking. The first thing I'd look at is the perpendiculars. You know which way to go from the line to the center of the circle -- exactly perpendicular to it. Per the first equation, y = 3/4 x + something. So the three lines you specify have slope 3/4, 3/4, and 2/3. Oh, they're parallel lines! Well, that makes it easy - you know the circle is on a line in the middle, i.e. average 7 and 43. Now you just have to see where that line intersects the other line, solve the two equations simultaneously, and you're done.
I'll stop there because the Refdesk has a sort of spoiler policy about homework, but it wouldn't take much to get someone to finish up if you really are confused, but by now you should be done. Wnt (talk) 23:57, 18 October 2016 (UTC)
If they weren't parallel lines, I think you'd take the negative inverse of the slope to get a perpendicular, i.e. -4/3. Then you can say that the circle is centered on the third line and the radius extends outward in the two different directions by equal amounts ... ? ... PROFIT! Maybe I have to think that over again. Meanwhile, the triangle article explains an inscribed circle is on each of the bisectors of the angles, but this seems to invoke more trigonometry (arctangent or something) than I'd prefer. Wnt (talk) 00:07, 19 October 2016 (UTC)
Actually, it is rather easy to prove that a circle tangent to two non-parallel lines has its center on the angle bisector of the angle formed by the lines (without trigonometry). This simplifies the algebraic brute-forcing of the problem a lot. (As a general rule, geometry problem can always be solved by calculus means, but it is much more dirty than geometric solutions.) TigraanClick here to contact me 10:50, 19 October 2016 (UTC)
Using a free tool like Geogebra might help your geometric intuition, find thesolution via an (interactive) construction, which you can use to verify solution you've found by algebraic computation.--Kmhkmh (talk) 00:09, 19 October 2016 (UTC)
Regarding the case with three non-parallel lines: I made a crappy graphic showing various equal angles and right angles and lines and such, then realized it was all irrelevant. The way to find the circle in the arbitrary case is to take the two bounding lines and solve simultaneously to get the intersection point. Then use the second formula in distance from a point to a line to get the x and y values for the nearest point on the third line (which the circle is centered on). Then use the first formula in distance from a point to a line, i.e. the distance from the point to the line, with either of the first two lines, to find the radius of the circle. Not a single sine or cosine or god forbid arctangent in sight. Wnt (talk) 12:02, 19 October 2016 (UTC)
• By "3x-4y+43" do you mean 3x-4y=43 or 3x-4y+43=0? It's fairly obvious that the center lies on the line 3x-4y=a, where a is 25 in one case and -18 in the other. So you have two simultaneous linear equations determining the center. —Tamfang (talk) 21:19, 19 October 2016 (UTC)
The + and = are on the same button, so I didn't think that was a hard one to guess. :) Wnt (talk) 21:23, 19 October 2016 (UTC)

# October 19

## Area of an ellipse

After a long search, I just found this website matematicasvisuales, and I learned from this a very intuitive way to prove the ${\displaystyle \pi ab}$ ― as intuitive as the method to prove the area of a circle.

I must ask, how come everyone associate this topic with high advanced mathematic, and the only way you can find everywhere for proving this includes Calculus, Integration etc.? The greeks had proved the circle's area formula the old "intuitive" fashion way, and that's how we mostly learn it.

In the Area its written like this:

The area formula ${\displaystyle \pi ab}$ is intuitive: start with a circle of radius ${\displaystyle b}$ (so its area is ${\displaystyle \pi b^{2}}$) and stretch it by a factor ${\displaystyle {\frac {a}{b}}}$ to make an ellipse. This scales the area by the same factor: ${\displaystyle \pi b^{2}\left({\frac {a}{b}}\right)=\pi ab}$ .

That is so not-intuitive; just like ${\displaystyle 2\pi R}$ doesn't claim a word about the area being ${\displaystyle \pi R^{2}}$ . Sorry. יהודה שמחה ולדמן (talk) 19:25, 19 October 2016 (UTC)

The fact is, this is not a rigorous proof, unless you go into advanced mathematics anyways: you have to be careful about what it means to "stretch" or "shrink" and why area is multiplied by the same factor. The circle's area formula has an "intuitive" proof but intuition is not enough; to make the Greeks' proof sound one must resort to mathematical analysis techniques anyway. There's no avoiding it.--Jasper Deng (talk) 19:33, 19 October 2016 (UTC)
That's not what I said; the wiki-page of Area said "stretch"! I didn't say the proof is rigorous either; and even if it isn't, why aren't the circle and ellipse treated the same way and not been taught together in school, or at least be taught together only through Calculus?
And by the way, I used a little Analytic geometry together with the circle and ellipse equations, and proved that the areas of the Trapezoid/Triangle cross sections relate to one other respectively as ${\displaystyle {\frac {a}{b}}}$ or ${\displaystyle {\frac {b}{a}}}$ . יהודה שמחה ולדמן (talk) 20:12, 19 October 2016 (UTC)
But "stretch" and "shrink" are equally intuitive then, and the two "proofs" by "stretching" (as in the article) and "shrinking" (as on the page you linked) are equivalent. Both formulae are conventionally derived rigorously in elementary calculus classes, and no way of computing the area of either (without knowing the area of at least one of them) avoids the use of calculus of some sort.(By the way, "calculus" is not a proper noun, don't treat it as such)--Jasper Deng (talk) 20:17, 19 October 2016 (UTC)
Calculus exists, no one argues about that. But the area formula of the circle was derived thousands of years ago, even if our 350 (or so) year-old calculus had "confirmed" it. Rigorous or not, People used it ever since.
But I realized we ran of topic ― my main question is why isn't this non-rigorous method been taught in schools and other places as well as the circle area non-rigorous method? Until I found that website page all I could find were rigorous calculus proofs. יהודה שמחה ולדמן (talk) 20:57, 19 October 2016 (UTC)
I feel like there's really not much difference between stretching and calculus here. I mean, to "stretch" a circle on one dimension means that for any given rectangular element with a large dimension y and a small dimension delta-x, the length of delta-x is altered accordingly. In the limit as delta-x approaches zero, the rectangular elements cover the entire ellipse, while retaining the same degree of stretching; hence, the area of the ellipse is altered accordingly. I feel like there must be some general theorem to this effect, regarding any shape, which any real mathematician could name. Wnt (talk) 21:21, 19 October 2016 (UTC)
(edit conflict) It is taught here in the USA. It's just that the ellipse itself has much less pedagogical value than the circle (yes it's a conic section, but the circle is best for teaching trigonometric functions). And at least here, non-rigorous methods are not actually mentioned that much in general.
Strictly speaking, the formula wasn't derived until it was first rigorously proven. The method of circum- and in-scribing polygons implicitly assumes that the circumference of the circle lies between those of the polygons, and the least upper bound property of the reals - assumptions that the Greeks did not have the tools to address rigorously.--Jasper Deng (talk) 21:22, 19 October 2016 (UTC)
I thought Eudoxus of Cnidus had proved the circle area formula by contradiction, no? יהודה שמחה ולדמן (talk) 22:20, 19 October 2016 (UTC)
As far as I know, he only established that the area is proportional to r2 (which can now be easily proven using Fubini's theorem), and did not establish the constant of proportionality. Given that pi most certainly is an "incommensurable" quantity, and that he didn't like dealing with those, I doubt he would have proven the whole formula. Even if he did, the assumptions I mentioned above are still implicit in his method.--Jasper Deng (talk) 23:06, 19 October 2016 (UTC)

# October 20

## Sum of the reciprocals of the subfactorials

What is the sum of the reciprocals of the subfactorials of the integers greater than 1?? (1 subfactorial is 0 and 1/0 is undefined.) Georgia guy (talk) 15:04, 20 October 2016 (UTC)

• Going by derangement, the question is to compute ${\displaystyle \sum _{n=2}^{+\infty }{\frac {1}{!n}}=\sum _{n=2}^{+\infty }{\frac {1}{n!\sum _{i=0}^{n}{\frac {(-1)^{i}}{i!}}}}}$. I see no obvious way to attack this. TigraanClick here to contact me 15:58, 20 October 2016 (UTC)
If I understand derangement rightly, it says that ${\displaystyle !n=\left[{\frac {n!}{e}}\right]}$, where that is, bizarrely, a round off to the nearest integer function in brackets. And factorial says that ${\displaystyle e^{x}=\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}.}$ Now I know that we can't say that ${\displaystyle \sum _{n=2}^{+\infty }{\frac {e}{n!}}}$ = e * e1 minus the first two points, because there's that round off to be considered. But can this be a start, if you understand why this round-off happens and can calculate just the remainders to offset them from this sum??? Wnt (talk) 17:40, 20 October 2016 (UTC)
I just looked into this a bit further and found [1] which says that the subfactorial is equal to the "uppercase incomplete gamma function" gamma(n+1, -1)/e. Which is really neat, except that I wish they'da named the function something like Blarf234385(tm), because nothing short of at threat of prosecution seems capable of getting two programs to mean the same thing by a function name with a gamma in it. (Well, alright, that's being facetious; blasphemy is the most satisfying form of profanity) I installed R package 'gsl' and used gamma_inc() and variants, but none of them work for negative values of the second parameter. This was kind of off the topic anyway though, since what seems most relevant is that this has the same /e as above. The factorial n! is equal to the complete gamma function, and so it is simply gamma(n+1) = gamma(n+1, 0)... Wnt (talk) 04:15, 21 October 2016 (UTC)
It converges quickly: I get 1.6382270745053706475428931141511226610635932496444... at n=40. No results found using inverse symbolic calculators. 24.255.17.182 (talk) 04:13, 21 October 2016 (UTC)

# October 21

## The union of lines joining points on a variety to a fixed point on it

Let ${\displaystyle Y\subset {\mathbb {P}}^{n}}$ be a projective variety, and fix a point ${\displaystyle p\in Y}$. Define X to be the closure of the union of all lines (pq) where ${\displaystyle q\in Y\setminus \{p\}}$. Assume ${\displaystyle Y\neq X}$. How can I find a birational morphism ${\displaystyle Y\times {\mathbb {P}}^{1}\rightarrow X}$? trying to parameterize the line (pq) by P^1 and send (q,t) to the point corresponding to the parameter t is not one-to-one. This is exercise I7.7 (a) in Hartshorne.--46.117.104.173 (talk) 09:56, 21 October 2016 (UTC)

I don't think in general there is a birational map between these two varieties. Sławomir Biały (talk) 10:18, 21 October 2016 (UTC)
The exercise asks to prove dim(X)=dim(Y)+1, and an online hint is to find a birational map ${\displaystyle Y\times {\mathbb {A}}^{1}\rightarrow X}$. I believe this to be equivalent to the existence of a (perhaps more natural) birational map ${\displaystyle Y\times {\mathbb {P}}^{1}\rightarrow X}$, since ${\displaystyle Y\times {\mathbb {A}}^{1}}$ is open in ${\displaystyle Y\times {\mathbb {P}}^{1}}$.--46.117.104.173 (talk) 11:14, 21 October 2016 (UTC)
You don't actually need a birational map to prove that the dimension is the same. The "obvious" mapping from ${\displaystyle Y\times {\mathbb {A}}^{1}\to X}$ is not birational when Y is a hypersurface, but it generically has maximal rank. You can use this to give affine charts on X. Sławomir Biały (talk) 13:57, 21 October 2016 (UTC)
I think that a truly birational map can be defined by taking an affine set ${\displaystyle U_{i}}$ containing the point p, and send (q,t) to the point on the affine line (pq) that corresponds to the parameter t. Then if the line (pq) intersects Y in an additional point r, the map can't decide which parameter to use, so we take the parameterization with unit speed ${\displaystyle (q,t)\mapsto p+t{\frac {(q-p)}{d(p,q)}}}$. Is that all right?--46.117.104.173 (talk) 21:52, 25 October 2016 (UTC)

## Rare math symbol unicode

I am trying to quote an old mathematics book. It is using a three-dot notation that I've never seen, so I don't know the name, so I cannot find the unicode for it. I need the unicode so I can simply reprint the original. The two symbols used have three dots. One symbol has a high dot on the left, a mid-dot in the middle and a mid-dot on the right. The other symbol has a low dot on the left, a mid-dot in the middle, and a mid-dot on the right. I've been searching through unicode charts, but I don't know how to efficiently find rarely used symbols. 209.149.113.4 (talk) 13:09, 21 October 2016 (UTC)

Try http://shapecatcher.com. PrimeHunter (talk) 13:14, 21 October 2016 (UTC)
Thanks. That helped me realize that I could use Braille to fake it. 209.149.113.4 (talk) 13:55, 21 October 2016 (UTC)
I tried shapecatcher and it said it was "Down right diagonal ellipsis, Unicode hexadecimal: 0x22f1, In block: Mathematical Operators". That's &#8945; = ⋱ So I'm duly impressed... especially considering the quality of my scrawled circles and the fact that I don't think I've ever seen this symbol before. Oh, and as you might have guessed, the other is next to it at &#8944; = ⋰ Wnt (talk) 12:58, 22 October 2016 (UTC)
But the OP's descriptions have two dots on the same level. —Tamfang (talk) 17:43, 22 October 2016 (UTC)
Erm.... ooops! (This kind of thing happens way too often when I start playing with a new tool... too busy admiring it to remember what I was trying to do) Wnt (talk) 11:22, 24 October 2016 (UTC)

# October 22

## 0.01825 as fraction and percentage

How do I convert 0.01825 into either a fraction or percentage? Uncle dan is home (talk) 00:37, 22 October 2016 (UTC)

0.01825 = 001825/100000 = 1825/100000 110.22.20.252 (talk) 01:41, 22 October 2016 (UTC)
and
0.01825 = 1.825/100 = 1.825% Rojomoke (talk) 04:18, 22 October 2016 (UTC)
Many folks would assume (perhaps wrongly!) that the first question is a request specifically for a reduced fraction. In this case this starts off easy because you see a "5" at the end and know you can divide by 5 to get 365, and divide that by 5 to get 73. When you get to 73 though, what do you divide by? That's tougher - a problem called factorization really that is key to a lot of advanced math. At a basic level it might be easiest to look up 73 (number), which says it is a prime number, which means you can't divide it by anything else. Otherwise you probably do not just have to try dividing by 2, 3, 5, 7, 11 (i.e. prime numbers) until they see 11*11 is more than 73. This is not that hard for a small number but with a big one it gets to be really tough.
On the bottom, you can easily divide 100000 / 25 = 4000 for a reduced fraction of 73/4000. Wnt (talk) 13:08, 22 October 2016 (UTC)
• The extended Euclidean algorithm gives a sequence (${\displaystyle \pm s_{i}:t_{i}}$) of rational approximations to any real ratio; if the true ratio is rational, it terminates with that rational, in lowest terms. (To use it you don't need to know the prime factors.) —Tamfang (talk) 17:49, 22 October 2016 (UTC)
@Tamfang: Obviously I don't know enough mathematics. Even the Euclidean algorithm, if I knew it, I forgot it. Just take the difference between the numbers, or the remainder. Genius! From 300 B.C. yet. So to go back to elementary school for a minute (though I doubt my teacher ever read this book) that means take 100000 modulo 1825, hmmm, try 100000 - 91250 (took half, multipled by 1000) = 8750, now take 8750 mod 1825 = 8750 - 9125 = -375, take 1825 mod -375 = 1825 - 1500 = 325, take 375-325 = 50, take 325 mod 50 = 25, take 50 mod 25 = 25. Seems pretty robust even for head-calculation. Scary thing is that it seems like no matter how basic the math question there's always something to be learned if you look at it. Now I'll have to look at the extended version... Wnt (talk) 13:59, 26 October 2016 (UTC)
Hmmm, to apply the extended algorithm I need the genuine qi sequence, so no fumbling around with wrong divisors:
 subscript r q s t ~ -s/t 0 100000 ? 1 0 infinite 1 1825 54 0 1 0 2 1450 1 1 -54 0.01851852 3 375 3 -1 55 0.01818182 4 325 1 4 -219 0.01826484 5 50 6 -5 274 0.01824818 6 25 2 34 -1863 0.01825013 7 0 0 -73 4000 0.01825
Wnt (talk) 14:42, 26 October 2016 (UTC)

# October 23

## What is a Zwicky Search?

In reading Robert Heinlein's Time Enough for Love I came across the term "Zwicky Search which the author has the hero mention in the context of discusssions of genetics and interstellar travel. I assume therefore this has something to do with the astronomer Fritz Zwicky, who discovered neutron stars and (the evidence for) dark matter. Heinlein does not explain what is meant by the term. Can anyone elucidate? Thanks. μηδείς (talk) 20:22, 23 October 2016 (UTC)

Just a possibility, but he proposed the existence of gravitational lenses, which have since been confirmed, and you could use one of those, say with a black hole (not currently consuming matter) as the lens, with a spaceship that moves around it to change the focus, as a "super telescope" to see distant objects more clearly than with any other telescope. See Fritz_Zwicky#Gravitational_lenses. StuRat (talk) 21:57, 23 October 2016 (UTC)
See Morphological box. -- ToE 22:03, 23 October 2016 (UTC)
Yep, Morphological box is it. I'd never have figured that one out. I am also surprised that I have never heard of Fritz Zwicky before searching for this term. The last time I read Time Enough for Love, Wikipedia did not exist. Thanks. μηδείς (talk) 00:35, 24 October 2016 (UTC)

# October 27

## Hash functions based on quadratic residues

Initially, I wrote this simple Javascript implementation (you can download the raw HTML file here) as an exercise in creating a hash function based on the Quadratic residuosity problem. The algorithm uses several large safe primes (including the modulus) on the order of roughly ten thousand digits and generates hashes of comparable size which seem to be of the highest quality. At first I was convinced that the fact that the primes in question are "exposed" (and thus known variables) shouldn't matter, given their size (Quadratic_residue#Prime_or_prime_power_modulus seems to support that conclusion), but now I'm not so sure. Was my assumption correct, or am I missing something here? Earl of Arundel (talk) 01:11, 27 October 2016 (UTC)

Perhaps there's something I'm not understanding, but if the modulus is prime then the quadratic residuosity problem would not apply since it depends on the modulus being composite with unknown factors. The section you pointed to in the quadratic residue article talks about finding the actual square root once you know you have a residue, a different problem. The article suggests that there are algorithms which are efficient in practice that find square roots for prime moduli, and from there you could get square roots for moduli whose factorization is known. In fact it seems like pretty much everything depends on the factors of the modulus being unknown. --RDBury (talk) 12:43, 27 October 2016 (UTC)
Yes, I was afraid that might be the case! Fortunately, I think there is a very simple solution. The basic idea is this: generate two hashes using a pair of distinct safe-prime triplets and then just add them together. That should be more than sufficient, considering that it is unfeasible to deduce which two addends have been used to produce any given sum (provided, of course, that the numbers are large enough to preclude a brute-force search). I've updated the implementation to further demonstrate the concept here (also at pastebin). Earl of Arundel (talk) 16:50, 27 October 2016 (UTC)

# October 28

## if ANOVA fails for a given population X, (and/or Y), under what circumstances can you still cross compare X and Y?

(Assume that the data has been or will be de-identified in the process and HIPPA laws are not an issue)

Suppose we have patient data for six hospitals in two cities:

NYC: