April 13

Hi!

There is a line integral ∫_L yds, where L is y²=2x and from (1/2, -1) to (2, 2). My answer is (5√5−2√2)/3, but L is a quadratic curve (instead of a line) and the answer seems odd, therefore I am not sure my calculation is correct or not.

How can I calculate a line integral result by computer (by Wolfram Alpha, python etc.)? James King 2009 (talk) 11:49, 13 April 2024 (UTC)[reply]

Swapping ${\displaystyle x}$ and ${\displaystyle y}$, this is the arc length of the segment of the parabola given by the equation ${\displaystyle y={\frac {1}{2))x^{2))$ from ${\displaystyle x=-1}$ to ${\displaystyle x=2}$. It can be turned into an ordinary integral of the form

${\displaystyle L=\int _{a}^{b}{\sqrt {1+\left({\frac {dy}{dx))\right)^{2))}dx.}$

In view of the square root in this formula, the appearance in the result should not be odd, but I find a different result, involving not only square roots but also logarithms (or arcsines). Obviously, the length of the arc should exceed that of the straight line segment connecting the end points, which is ${\displaystyle {\tfrac {3}{2)){\sqrt {5))>3.35,}$ but your result evaluates numerically to less than ${\displaystyle 2.79.}$ To check your result: its numerical value should be close to ${\displaystyle 4.10568.}$ --Lambiam 14:19, 13 April 2024 (UTC)[reply]

The question calls for the integral of yds, so the first moment wrt the x-axis, not arc length. So basically you're integrating ${\displaystyle \int y{\sqrt {1+y^{2))}dy}$. --RDBury (talk) 15:37, 13 April 2024 (UTC)[reply]

PS. if you just want a ballpark estimate to check your result, break the curve up and approximate by line segments. In this case the section from y=-1 to y=1 cancels itself out, so you really just have to do the segment from (1/2, 1) to (2, 2). If you replace the curve with a line segment, its center of mass is the midpoint, so (5/4, 3/2). The length is √13/2, so the moment is (3/2)(√13/2) ≈ 2.7. That's off from the original result by about .1 so I'd count it as verified. --RDBury (talk) 15:54, 13 April 2024 (UTC)[reply]

Symbolic integration by Maxima confirms it as well.  --Lambiam 20:54, 13 April 2024 (UTC)[reply]

Actually I should try that out, so thanks. There are limits to what you can do with pencil & paper (and maybe a calculator). I've found you can do wonders with an off-the-shelf spreadsheet program, and if you already know how to use one that would work as well. As with many things, the trade-off is the time it takes to learn how to use the tool in question. You can write a Python program when the task is too complex for a spreadsheet, but if you don't know Python already that's a lot of effort for solving one problem. The OP mentioned Wolfram Alpha as well, and I assume that would tell you the answer too if you know the right input string. I use Alpha occasionally, but mostly for things where the right command is obvious, "solve x^3-x-1=0" for example. --RDBury (talk) 02:17, 14 April 2024 (UTC)[reply]

For this specific integral it isn't too hard to do the integration purely mentally by change of variables ${\displaystyle ({\sqrt {1{+}y^{2))}\leftrightarrow u,}$ so ${\displaystyle d(1{+}y^{2})=d(u^{2})}$ and thus ${\displaystyle y\,dy=u\,du).}$  --Lambiam 10:09, 14 April 2024 (UTC)[reply]

April 15

I'm trying to read this article translated from a 1934 article in Swedish by B. Berggren. The word "catheter" is mentioned several times, and I gather it should be "cathetus" meaning a leg, or side other than the hypotenuse in a right triangle. I don't think I've heard of a cathetus before, nor a catheter with reference to a triangle. Is this term still used by anyone now? --RDBury (talk) 20:25, 15 April 2024 (UTC)[reply]

It makes it into wiktionary: cathetus, but there are no quotations. This says it's fairly rare, but some prefer it over the term "leg". -- Jack of Oz ^{[pleasantries]} 20:47, 15 April 2024 (UTC)[reply]

Thanks. I've mostly heard and used "side", but that's wrong if you think about it; triangles have three sides. I didn't think to look at NGram Viewer before, but I did just now and it seems that "cathetus" had about a 10 year window of popularity between the mid-1920's and mid-1930's, but still only about 3% as common as "hypotenuse". Perhaps it's more common in Swedish. --RDBury (talk) 06:44, 16 April 2024 (UTC)[reply]

The Swedish Wikipedia uses it in its article on the Pythagorean theorem. It did not help the translator that in Swedish the form kateter is both the word for "catheter" and the plural of katet, which means "cathetus".  --Lambiam 06:14, 17 April 2024 (UTC)[reply]

There's a room with 1,000 people, each with a computer. Their task is to access Wikipedia, and click Random article, editing whatever they see needs fixing each time, before repeating the same process endlessly. They're well paid. They do this for 8 hours a day, 5 days a week, until they either quit out of sheer boredom, kill themselves, or ascend to some higher plane of consciousness.

What are the chances that 2 or more of the people could be editing the same article simultaneously, causing an edit conflict when the second or later ones press Publish? And how long might it take for the first such conflict to occur? -- Jack of Oz ^{[pleasantries]} 23:20, 15 April 2024 (UTC)[reply]

This is essentially the same as the birthday problem for a set of ${\displaystyle n=1000}$ randomly chosen people on a planet in which a year takes ${\displaystyle d}$ days, where ${\displaystyle d}$ is equal to the number of Wikipedia articles (currently 6,817,521). Assuming all are in the same time zone, working 9 to 5, the probability that any article is being edited simultaneously by multiple persons, using the square approximation, equals approximately

${\displaystyle {\frac {n^{2)){2d))\approx {\frac {10^{6)){2\times 6.813{\times }10^{6))}\approx 0.073\,.}$

A slightly better approximation is given by the formula

${\displaystyle 1-\exp \left({\frac {n^{2)){2d))\right)\approx 0.071\,.}$

How long it will take for a conflict to arise (assuming that all simultaneous edits are conflictual) depends on how long editors stay within the same article, which may be characterized by individual-dependent probability distributions and hard to impossible to tackle analytically. If they are synchronized, each editing one article per time slot in equal time slots, the expected number of time slots until conflict is approximately

${\displaystyle {\frac {1}{0.071))\approx 14.1\,.}$

 --Lambiam 03:31, 16 April 2024 (UTC)[reply]

Thanks. The first answer is about what I expected. But the time till the first conflict is rather counter-intuitive. If the time slots were 5 minutes, that means a conflict would occur after only a bit over an hour. I don't question your maths, but that is amazingly short for such a huge database. Thanks again.

But then, I guess the number of editors does make a rather significant difference (* cough), since the number of possible conflicts with 1,000 editors in each time slot is half factorial 999, a number with c. 2,565 digits, and about 10²⁴⁸⁵ times the number of atoms in the universe. Compared to this, a mere 6-odd million articles is infinitesimally small, so having to wait an hour for a conflict does seems a bit tardy, on reflection. If it were only 4 editors, there are only 6 possible conflicts and the expected time would increase massively, to about 500 days. Very nice. -- Jack of Oz ^{[pleasantries]} 23:56, 16 April 2024 (UTC)[reply]

Yes, at first glance it is counterintuitive, which is why the problem in its birthday suit setting is also known as the "birthday paradox".  --Lambiam 05:41, 17 April 2024 (UTC)[reply]

April 18

Interest compounds daily. The effective annual interest rate is 5.15%. Is there some formula to calculate what is the actual daily interest rate? Using a lot of trial and error on spreadsheets, I estimated that the daily interest rate is 5.02235% (or so). But, that was through a lot of trial and error … and it only comes “close to” an estimate of the actual daily interest rate. Is there some mathematical formula that will give me the exact figure? Thanks. 32.209.69.24 (talk) 02:04, 18 April 2024 (UTC)[reply]

To clarify, when I estimated the daily interest rate, I came up with 5.02235% (divided by 365) = 0.01375986301370%. Thanks. 32.209.69.24 (talk) 02:14, 18 April 2024 (UTC)[reply]

If I'm interpreting this correctly, then if you are given the annual interest rate ${\displaystyle A}$, then you want a daily interest rate ${\displaystyle D}$ such that ${\displaystyle (1+{\frac {D}{365)))^{365t}=(1+A)^{t))$, where ${\displaystyle t}$ is the number of years. Since both sides are exponentials and the only way they can always match is for the bases to match, we can just remove the ${\displaystyle t}$ to get ${\displaystyle (1+{\frac {D}{365)))^{365}=1+A}$. Taking the root on each side, you get ${\displaystyle 1+{\frac {D}{365))={\sqrt[{365}]{1+A))}$. Rearranging yields ${\displaystyle D=365({\sqrt[{365}]{1+A))-1)}$. For your value of ${\displaystyle A=0.0515}$, this yields ${\displaystyle D=0.05022117}$, which is close to but slightly off from your value. GalacticShoe (talk) 02:15, 18 April 2024 (UTC)[reply]

Thank you so much ... for such a quick reply and for such a detailed explanation. Much appreciated. A follow-up question, if I may. That final formula that you cite contains a "root index" of 365. Is there a comparable formula that can be entered into an Excel spreadsheet? (I have no idea of how to -- or even if -- one can express a root index of 365 in Excel.) Thanks! 32.209.69.24 (talk) 02:29, 18 April 2024 (UTC)[reply]

Some preliminary searching indicates that you can just do "value^(1/365)", so for example "365*((1+A)^(1/365)-1)" GalacticShoe (talk) 02:44, 18 April 2024 (UTC)[reply]

Thanks a million! Very much appreciated! 32.209.69.24 (talk) 05:21, 18 April 2024 (UTC)[reply]

...and not pentillion? Someone who's wrong on the internet (talk) 05:43, 18 April 2024 (UTC)[reply]

All those number names are based on Latin (in this case quinque) rather than Greek (πέντε). --Wrongfilter (talk) 06:14, 18 April 2024 (UTC)[reply]

Then why is it pentagon and not quintagon? Someone who's wrong on the internet (talk) 09:07, 18 April 2024 (UTC)[reply]

Because -gon is Greek (γωνία = angle, from γόνυ = knee), hence the prefix is Greek, too. Some terms are Greek, others are Latin, mixtures are to be avoided. --Wrongfilter (talk) 09:21, 18 April 2024 (UTC)[reply]

German makes a bit more sense when it comes polygons, Dreieck=Triangle, Viereck=Quadrilateral, Fünfeck=Pentagon, ... . Literally (the number in German)+"Eck" = corner or angle. I imagine other languages not so heavily influenced by Latin and Greek are similar, none of this "If it's Tuesday we must be using Greek roots." The word for "polygon" is "Polygon" in German, breaking the pattern, though "Vieleck"="many"+"angle" is used as well. As for the numbers, people still don't seem to agree on what these names even mean (see Long and short scales). When in doubt, use 10¹⁸. --RDBury (talk) 14:21, 18 April 2024 (UTC)[reply]

My favourite shape is the octangular quadragon. -- Jack of Oz ^{[pleasantries]} 21:14, 18 April 2024 (UTC) [reply]

Like these? —Tamfang (talk) 18:45, 25 April 2024 (UTC)[reply]

April 20

Imagine one tinder gender has a max 100 right swipes per day and the other have X max right swipes per day. Would it be possible to find the value of X that would be needed to make sure they have equal amount of matches per day at average with just that information I am presenting here or you would need internal data to solve this mathematical problem?

75% of tinder is male, 25% is female.

Woman swipe right 7% of time while man swipe 40%

Woman match with 33% of man they swiped right while men match with 2.5% of person they swiped right.

Woman vote at 200 profiles per day while man do with 137. 179.134.97.227 (talk) 17:34, 20 April 2024 (UTC)[reply]

I'm not familiar with tinder, but it seems that the question is about a selection process for strictly binary and straight users, in which each of the two genders is only presented candidates of the other gender. Each candidate presented is a match for female users with probability 0.07 × 0.33 = 0.0231 and for male users with probability 0.40 × 0.025 = 0.0100 .

Then the expected number of matches for a female user when presented 200 candidates equals 200 × 0.0231 = 4.62, while that for a male user when presented 137 candidates equals 137 × 0.0100 = 1.37 .

So far so good, but where does the maximum of 100 come in if a woman can do 200 swipes per day?

Given match rates p_f and p_m for the two genders, the expected numbers of matches for users of these genders equal n_f × p_f and n_m × p_m, in which n_f and n_m stand for the numbers of candidates presented to the respectively gendered users. To make these expected numbers equal requires achieving a ratio between these numbers of candidates such that

n_f : n_m = p_m : p_f.

For the data supplied, this means,

n_f : n_m = 0.0100 : 0.0231 = 100 : 231 .

So if female users are presented 100 candidates, male users need to get presented 231 candidates to achieve the same number of matches, to wit:

100 × 0.0231 = 231 × 0.0100 = 2.31 .

 --Lambiam 20:33, 20 April 2024 (UTC)[reply]

At tinder, people are presented with a "random" (not exactly), person, they can swipe left (not like) or swipe right (like), if male A like female B and female B like male A they match and can start to talk. The 200 swipes means at average, the woman rates at this an day (like OR dislike) 200 people. Max amount of right swipes would be the amount of likes they would be able to give before not being able to give likes for that day. Woman receive way more matches than man at this apps for various reasons, the question here is if that information presented at the question, would be possible to find what would need to be the limit at the amount of likes a man can do (assuming woman can do 100 per day) to make sure the amount of matches is the same per day at average.179.134.97.227 (talk) 21:09, 20 April 2024 (UTC)[reply]

A question about your terminology. If male A likes female B and female B likes male A, which of the two "receives" the match? It seems to me that if every match is between two users of different genders, then each gender will always have the same number of matches as the other gender. A difference, if any, can only be in who initiated the process that led to the match being established.  --Lambiam 21:18, 20 April 2024 (UTC)[reply]

"If male A likes female B and female B likes male A, which of the two "receives" the match? It seems to me that if every match is between two users of different genders, then each gender will always have the same number of matches as the other gender."

Man A and female B receive the match. A gender can have different numbers of matches at average, one example, if man 1 likes woman 1 and 2, man 2 likes woman 1 and 2, man 3 likes no one and woman 1 likes man 1 and woman 2 likes man 1. Thats man 1 having 2 matches and other man having no matches, thats 0.66 matches at male side and 1 match at average at the woman side.179.134.97.227 (talk) 21:30, 20 April 2024 (UTC)[reply]

Let M stand for the number of matches under some procedure. In your example M = 2, since we have the two matches man 1 ⇆ woman 1 and man 1 ⇆ woman 2. Let, furthermore, u_f and u_m denote the number of female and male users. In the example, u_f = 2 and u_m = 3. Then the average number of matches for the two genders are M⁄u_f and M⁄u_m. In the example, we get 2⁄2 and 2⁄3. One way to get the two averages equal is to make sure that M = 0, for example by not allowing any right swaps at all. The only other way does not involve the process for indicating preferences: make sure that u_f = u_m.  --Lambiam 18:22, 21 April 2024 (UTC)[reply]

April 22

With spheres there's only 1 right answer, there's also formulas that can be used to find points 0.5π radians from another point, is there a formula for the line with half the ellipsoid surface area on each side? Is there a formula for the line where an infinitely far Star of Bethlehem and an anti-Star of Bethlehem at the other end of the ellipsoid normal line would have equal zenith distances ignoring refraction and geoids? Are these lines the same? How far apart can they be? Sagittarian Milky Way (talk) 18:30, 22 April 2024 (UTC)[reply]

Unless an ellipsoid is a sphere (which the WGS 84 reference ellipsoid is not), no portion of it is a mathematical hemisphere. Any plane through its centre divides it though into two equal (congruent) parts. Usually the plane will be a meridional or the equatorial plane. In more general geodetic systems the equator and meridians, although not ellipses, also lie in a plane and can be used for a fairly fair cutting into two parts, which however will normally not be congruent. Calling the two parts "hemispheres", although not correct in a strictly mathematical sense, is nevertheless conventional.  --Lambiam 19:11, 22 April 2024 (UTC)[reply]

Yes but that's the easy way out, hemispheres centered on the equator or pole are exactly zero percent of all possible centers. Sagittarian Milky Way (talk) 21:56, 22 April 2024 (UTC)[reply]

Is division by any plane through the centre not general enough?

The sight lines to a point on the celestial sphere and to its celestial opposite are parallel. So are the directions to the respective zeniths from a given place on the ellipsoid and its antipodal place. Therefore the angular distances are the same.  --Lambiam 11:03, 23 April 2024 (UTC)[reply]

I don't know how to do the center plane. Either finding points of surface tangency from the point of surface perpendicularity or the point of surface perpendicularity from 2 surface points on the plane. There seems to be a u and a v involved I keep seeing u and v but don't know what that is, or if that's needed when the plane is not arbitrary but has 1 of 3 defining points fixed to the ellipsoid center. I stupidly dropped out before learning u, v and pseudo-delta swirl. Sagittarian Milky Way (talk) 22:31, 23 April 2024 (UTC)[reply]

If they are boldfaced u and v, these variables probably stand for some 3D-vector (x, y, z).  --Lambiam 16:51, 24 April 2024 (UTC)[reply]

I must admit to being intrigued by the idea that the Star of Bethlehem might have been arranged by angels using WGS 84! However as far as I can see there's just ocean at the antipodal point for the birth of Damien Thorn. Pehaps we're safe for a while yet ;) NadVolum (talk) 19:42, 22 April 2024 (UTC)[reply]

The normal on the ellipsoidal through Bethlehem won't go through the center of the Earth and so won't go though the antipodal point. NadVolum (talk) 20:04, 22 April 2024 (UTC)[reply]

It doesn't really matter though cause the idealized celestial sphere/astronomical coordinate system is infinitely far, the lines to the star from anywhere on Earth would be parallel. It would matter for the "ranking all points by distance and picking the nearer half" way as an extremely flattened ellipse could have the (geographic, not geocentric) latitude minus 90 be only a few miles away (plus 90 in the southern hemisphere) Sagittarian Milky Way (talk) 22:15, 22 April 2024 (UTC)[reply]

April 23

The article on Fibonacci numbers mentions the pineapple, but I am unable to discover when, in the literature, this was first discussed. Later in the aforementioned article it states, "In 1830, K. F. Schimper and A. Braun discovered that the parastichies (spiral phyllotaxis) of plants were frequently expressed as fractions involving Fibonacci numbers." Given that the discussion of pineapples was extremely popular at this time in the early 19th century, one would expect it to be found within that time frame. However, I cannot find anything until the mid to late 20th century, possibly starting with Onderdonk 1970. Does anyone know when Fibonacci numbers were first discussed in reference to the pineapple, and if it was before the 20th century? As it stands, 1830 would fit absolutely perfectly into the pineapple timeline I'm working on, but I can find no supporting evidence for this idea. Viriditas (talk) 23:17, 23 April 2024 (UTC)[reply]

Pineapples are mentioned as an example in a long list of diverse plant species. In other words there's nothing specifically notable about pineapples in relation to Fibonacci numbers. Further on the article also mentions daises and the image shows a chamomile. I think the reason pineapples are mentioned is because someone was able to find a citation for them, and perhaps also because many people can find them in their local supermarket. I remember counting rows on a teasel flower head and coming up with Fibonacci related numbers, but I doubt it's mentioned a lot in the literature. It seems to be a general property of Phyllotaxis, or the way plants grow, though there are exceptions, and the "Repeating spiral" section of that article mentions more about Fibonacci numbers. The article mentions Kepler having pointed out the presence of Fibonacci numbers in nature; that was 400 years ago. I don't know if Kepler ever saw a pineapple though. --RDBury (talk) 02:02, 24 April 2024 (UTC)[reply]

Yes, I’m aware of that. What’s notable in this context is that pineapples were introduced to Europe and it led to a great deal of interest. I’m tying to trace the discussion of pineapples throughout each discipline as it arose within a specific 150 year time frame of interest before cultivation and mechanization led to wider availability of the fruit. One of the reasons so many different disciplines discussed pineapples is because they were considered new, difficult to impossible to grow in cold climates, and didn’t have a previous known history in Europe, giving rise to people in different fields using them as examples in their domain-specific literature. It would be kind of like talking about the Internet in your field of expertise in the 1990s. It was somewhat new and different for the general public and people were trying to apply it to their knowledge base. For example, both Leibniz and Locke wrote about pineapples in the context of philosophy because it was considered unique in taste and unobtainable to the common person due to cost, so it represented an idealized version of an idea that they could use in their work and would attract attention. My post on this topic pertains to pineapple within this timeframe, of which the year 1830 fits. It was at the time, coincidentally, that discussion about Fibonacci numbers and plants arose. My question is whether pineapples were discussed in this context at this time and used as an example, not whether it is of any importance to the math itself. Viriditas (talk) 02:17, 24 April 2024 (UTC)[reply]

The botanists credited with the discovery were Karl Friedrich Schimper and Alexander Braun. Schimper appears to have been the first (in 1830) to describe an observed phyllotactic pattern in term of the Fibonacci sequence (to wit, for the rotational angle between leaves in a stem) and his friend Braun described the next year a Fibonacci pattern in pine cones. I found no evidence these gentlemen or any other 19th-century scientists ever studied the patterns of pineapples. As reported here, the number of spiral rows of fruitlets in pineapples was studied "as early as 1933" by Linford,^[1] however, without referencing the Fibonacci sequence.  --Lambiam 16:45, 24 April 2024 (UTC)[reply]

Do you know where I can read their original work? I assume it is in German somewhere? It turns out that the history of the word "pineapple" has a lot of of confusion. "Pineapple" once referred to pine cones in English, while other languages used variations on "ananas" for pineapple, such as German. Viriditas (talk) 20:36, 24 April 2024 (UTC)[reply]

@Lambiam: I just found something interesting. This is unlikely to be true, but there is an implausible chance that Schimper & Braun were mistranslated: "Two names will exist side by side, and only after a time will one gain the upper hand of the other. Thus when the pineapple was introduced into England, it brought with it the name of 'ananas,' erroneously 'anana,' under which last form it is celebrated by Thomson in his Seasons. This name has been nearly or quite superseded by 'pineapple,' manifestly suggested by the likeness of the new fruit to the cone of the pine. It is not a very happy formation; for it is not likeness, but identity, which 'pineapple' suggests, and it gives some excuse to an error, which up to a very late day ran through all German-English and French-English dictionaries; I know not whether even now it has even disappeared. In all of these 'pineapple' is rendered though it signified not the anana, but this cone of the pine; and not very long ago, the Journal des Débats made some uncomplimentary observations on the voracity of the English, who could wind up a Lord Mayor's banquet with fir-cones for dessert." (On the Study of Words, Richard Chenevix Trench, 1893, p. 254.) Viriditas (talk) 20:51, 24 April 2024 (UTC)[reply]

Schimper 1830^[2] and Braun 1831^[3] are independently written articles. I have not looked at Schimper's article, but as described he only discusses the placement of leaves. Braun being an accomplished botanist, he would not have used the term Tannenzapfen (pine cone) as an ambiguous name for the fruit of Ananas comosus, and his article furthermore identifies specific species or at least genera of conifers (Weisstanne = Abies alba; Lerchen = Larix; Rothtanne = Picea abies) whose cones he studied.  --Lambiam 05:42, 25 April 2024 (UTC)[reply]

Thank you. This is even more confusing given that pine cones were also called pineapples. Viriditas (talk) 20:24, 26 April 2024 (UTC)[reply]

• I just uploaded this photo of the top of a pine cone. Can anyone get the Fibonacci numbers from it? I've seen drawings were they show the Fibonacci numbers, but they may be a little idealized, or maybe they had a better example. Bubba73 ^{You talkin' to me?} 06:36, 27 April 2024 (UTC)[reply]

The photo looks good to me. The two numbers in the sequence are 8 (spirals going right), 13 (spirals going left), from 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55... Viriditas (talk) 07:12, 27 April 2024 (UTC)[reply]

April 26

I noticed that Isbell conjugacy and Isbell duality have the same meaning. So, I would like to know the difference in meaning between duality and conjugacy in mathematics. Also, I found Category:Duality theories, but what is the field of mathematics called Duality theory? However, since Baez (2022) said that the Isbell conjugacy is an adjoints rather than a duality of the category, so I changed the category to Category:Adjoint functors. Thank you, SilverMatsu (talk) 03:54, 26 April 2024 (UTC)[reply]

The original duality occurs in projective geometry, see Duality (projective geometry). At some point people noticed that the axioms of the projective plane where the same (or equivalent) if you swapped the undefined terms "point" and "line". So any theorem in projective geometry can be transformed to a dual theorem by changing the roles of points and lines. The new theorem may simply be a restatement of the original theorem as in Desargues's theorem, but sometimes it's not as in Pappus's hexagon theorem. The result is that you often get two theorems for the price of one proof. You can define a dual category for a given category by reversing the arrows, but category theory was invented long after projective geometry so that's not the original meaning. You can also define the dual curve of a plane curve, the dual space of a vector space, the dual polytope of a polytope, etc. As far as I know there is no all-encompassing "theory of duality", just the custom of using "dual" to describe when mathematical objects seem to occur in pairs in some way. Calling something a dual usually implies that the dual of the dual is in some way identifiable with the original object, but this is not always required. For example the dual of a dual vector space is not identifiable with the original vector space unless it's finite dimensional. Duality does not always exist, for example there doesn't seem to be a useful concept for the dual of a finite group, though you can define one for abelian groups. And sometimes there is a duality that's not called that, for example cohomology can be viewed as the dual of homology. I don't think there is a formal distinction between a "dual" and a "conjugate", but usually a conjugate is the the result of applying an automorphism of order two. For example a complex conjugate is the result of applying the automorphism a+bi → a-bi. Again, this is more of a naming custom than a formal mathematical concept, and there is (apparently) some overlap. I'd say a "conjugate" is usually used when the two objects live inside the same structure, and "dual" is used when you're talking about two different structures. For example the dual of a plane curve lives in the dual of the plane in which the original curve lives. Category theory blurs the distinction between an object and a structure so I can see how the distinction is rather meaningless there. --RDBury (talk) 07:12, 26 April 2024 (UTC)[reply]

I thought of Googling orthographic projection ellipsoid and found these, did I interpret everything right?:

ν=a/(1-e²*(sinΦ)²)^1/2 (why not square root?) x=(ν+h)*cosΦ*cosλ y=(ν+h)*cosΦ*sinλ z=(ν*(1-e²)+h)*sinΦ This seems to be a simple spherical to Cartesian converter with latitudes (Φ) "massaged" so it's not slightly wrong (eccentricity²=0.00669437999014 so not much massaging). Then they convert that to topocentric Cartesian with a matrix I can't solve (now I know why galactic Cartesian's UVW!) but it seems like they also say surface points are U=ν*cosφ*sin(λ-λ_O) V=ν*(sinφ*cosφ_O-cosφ*sinφ_O*cos(λ-λ_O))+e²*(ν_O*sinφ_O-ν*sinφ)cosφ_O where _O means "of the topocentric origin". Did I get that right? If so then I can set an initial guess point at or about 0.25 circumference from the W-axis, use the formulae to find its U and V in "W-axis place"-centered coordinates and the test point is of course √(U²+V²) meters from the W-axis and the part of the ellipsoid with the most meters without being too far from the W-axis-test point plane is the limb of the Earth from infinite distance. The worst-case scenario for how spindly a pie the test points have to be in would be looking at ~the 45th parallel limb with the W-axis in the equator plane. The geocenter depth increases roughly quarter mile from 45.5N to 44.5N so ~56 meters poleward shortens the limb to W-axis line segment by 8 inches which is how much Earth curves in a mile. Azimuth accuracy needed increases "exponentially" with limb coordinate accuracy desired though so 10 miles accuracy would be a lot more than 10x easier than 1 mile. Sagittarian Milky Way (talk) 06:24, 26 April 2024 (UTC)[reply]

April 27