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April 2

Is a stretched circular arc similar to a circular arc of a different circle?

Consider a circular segment, such as the one bounded by circular arc S and chord C in the diagram below. Assume that θ is "small". If we stretch (scale) the circular segment horizontally, will the "stretched" circular arc still be (approximately?) similar to a circular arc on a circle with a different R? If that's the case, will the (approximate?) similarity get worse as the scaling factor increases? Is there a simple formula that can characterize how similar the stretched circular arc is to a true circular arc?

134.242.92.97 (talk) 03:26, 2 April 2024 (UTC)

By 'stretching', do you mean a non-uniform scaling, such that the size is enlarged along the chord while preserved along sagitta...? --CiaPan (talk) 14:48, 2 April 2024 (UTC)

Yes (scaling along the direction of the chord but not along the direction of the sagitta). --134.242.92.97 (talk) 16:25, 2 April 2024 (UTC)

When you scale a circle, the resulting figure is an ellipse. An elliptical segment is not similar to a circular one in geometric sense. Of course the elliptic segment can be approximated with a circular one, so in a common speech they can be called 'similar', but I suppose that's not an appropriate use of the word here, at math ref.desk. --CiaPan (talk) 17:36, 2 April 2024 (UTC)

Note that the question assumes θ is "small" and the "similarity" may be approximate. Is there a simple answer to the question in terms of some intuitive similarity metric? (A similarity metric can be defined, in different ways, in terms of the deviation of the stretched arc relative to the best-fitting circular arc.) --134.242.92.97 (talk) 19:06, 2 April 2024 (UTC)

The area between the two curves (the stretched circle and the circle approximating it) could be used as a measure of the error. If the stretching factor is $f_{s}$ the area of the stretched segment is

${\begin{aligned}A_{s}&={\frac {1}{2}}f_{s}(\theta R^{2}-dc)\\&=f_{s}\left(\arcsin \left({\frac {c}{2R}}\right)R^{2}-{\frac {(R-h)c}{2}}\right)\\&={\frac {f_{s}}{4}}\left(\arcsin \left({\frac {4h}{4h^{2}+c^{2}}}c\right)\left({\frac {4h^{2}+c^{2}}{4h}}\right)^{2}+{\frac {4h^{2}-c^{2}}{4h}}c\right)\end{aligned}}$

If the approximating circle is chosen to go through the end-points and the mid-point off the stretched arc, then its radius $R_{a}$ satisfies

$R_{a}^{2}=(R_{a}-h)^{2}+\left({\frac {f_{s}c}{2}}\right)^{2}$

so

$R_{a}={\frac {4h^{2}+f_{s}^{2}c^{2}}{8h}}$

and the area of the approximating segment is

${\begin{aligned}A_{a}&=\arcsin \left({\frac {f_{s}c}{2R_{a}}}\right)R_{a}^{2}-{\frac {(R_{a}-h)f_{s}c}{2}}\\&={\frac {1}{4}}\left(\arcsin \left({\frac {4h}{4h^{2}+f_{s}^{2}c^{2}}}f_{s}c\right)\left({\frac {4h^{2}+f_{s}^{2}c^{2}}{4h}}\right)^{2}+{\frac {4h^{2}-f_{s}^{2}c^{2}}{4h}}f_{s}c\right)\end{aligned}}$

and the area between the curves is

$A_{a}-A_{s}={\frac {1}{4}}\left(\arcsin \left({\frac {4h}{4h^{2}+f_{s}^{2}c^{2}}}f_{s}c\right)\left({\frac {4h^{2}+f_{s}^{2}c^{2}}{4h}}\right)^{2}-f_{s}\arcsin \left({\frac {4h}{4h^{2}+c^{2}}}c\right)\left({\frac {4h^{2}+c^{2}}{4h}}\right)^{2}-{\frac {c^{3}f_{s}(f_{s}^{2}-1)}{4h}}\right)$

or given that the angle is small, approximating $\arcsin(x)\approx x+{\frac {x^{3}}{6}}$

$A_{a}-A_{s}\approx {\frac {2c^{3}f_{s}h^{3}(f_{s}^{2}-1)}{3(4h^{2}+f_{s}^{2}c^{2})(4h^{2}+c^{2})}}={\frac {f_{s}c^{3}h(f_{s}^{2}-1)}{96RR_{a}}}$

(barring mistakes) catslash (talk) 00:01, 3 April 2024 (UTC)

Thanks for the reply. It seems that your formula is for the difference in area between the two segments, but not the area between the two curves. --134.242.92.97 (talk) 17:12, 3 April 2024 (UTC)

That is true, but the the two segments have the same lower boundary, namely the stretched arc, and so the difference in their areas is area between their upper boundaries, i.e. the area between the two curves. But in any case, Lambian's answer is better. catslash (talk) 22:30, 4 April 2024 (UTC)

A suitable measure for the dissimilarity of two curves is the Fréchet distance.  --Lambiam 21:04, 3 April 2024 (UTC)

Then if the approximating circle is again chosen to go through the end-points and the mid-point off the stretched arc

$F=f_{s}{\sqrt {R^{2}+{\frac {(f_{s}^{2}-1)(2R-h)^{2}}{4}}}}+{\frac {h(f_{s}^{2}-1)-2Rf_{s}^{2}}{2}}$

(this gives $\left.F\right|_{f_{s}=1}=0$ and $\left.F\right|_{f_{s}=0}=-{\frac {h}{2}}$ as expected, so could be correct). However, it is possible to get a smaller $F$ by choosing an approximation which passes above/below ($f_{s}\gtrless 1$)the mid-point of the stretched arc rather than through it. catslash (talk) 22:22, 4 April 2024 (UTC)

One way someone could define how noncircular a curve is, is by taking the minimum, over all possible circles and all positions and orientations of the curve, of the cumulative distance between curve and circle. At the same time, one doesn't need to specify the positions of both circle and curve; one can just assume the circle is centered at the origin, and use the curve's offset instead. Similarly, because of circular symmetry, one can assume that the curve is one specific orientation. This means that we can parametrize a circle purely on radius, and the curve on offset (in addition to whatever else is used to parametrize the curve.) We essentially want:

$\min _{(X,Y,R)\in \mathbb {R} ^{3}}\int _{{\mathcal {C}}(X,Y)}D(\mathbf {r} ,proj_{R}(\mathbf {r} ))ds$

Where we take the line integral over the curve ${\mathcal {C}}(X,Y)$ parametrized with offset $(x,y)$, of the distance between each point $\mathbf {r}$ and its projection $proj_{R}(\mathbf {r} )$ onto the circle of radius $R$.

Note that although $proj_{R}(\mathbf {r} )$ doesn't make sense at the origin, the distance to the circle still does. For all values of $\mathbf {r}$, it can be easily seen that the distance from $\mathbf {r}$ to the circle of radius $R$ is $|\|\mathbf {r} \|-R|$, so we can rewrite this as:

$\min _{(X,Y,R)\in \mathbb {R} ^{3}}\int _{{\mathcal {C}}(X,Y)}|\|\mathbf {r} \|-R|ds$

Now, ${\mathcal {C}}$ here is a stretched circle segment, so it can be further parametrized with angle $\theta$ and stretch factor $S$. We write $\mathbf {r} (t)=(\cos(t)+X,S\sin(t)+Y)$ for $t\in [-\theta /2,\theta /2]$. We now have our final formulation for the problem:

$\min _{(X,Y,R)\in \mathbb {R} ^{3}}\int _{-\theta /2}^{\theta /2}|{\sqrt {(\cos(t)+X)^{2}+(S\sin(t)+Y)^{2}}}-R|{\sqrt {\sin(t)^{2}+S^{2}\cos(t)^{2}}}dt$

Or, using nonnegativity to bring the last term into the absolute value:

$\min _{(X,Y,R)\in \mathbb {R} ^{3}}\int _{-\theta /2}^{\theta /2}|({\sqrt {(\cos(t)+X)^{2}+(S\sin(t)+Y)^{2}}}-R){\sqrt {\sin(t)^{2}+S^{2}\cos(t)^{2}}}|dt$

Although I have no idea how to solve this analytically, I imagine there's a way to solve it numerically, and also I suspect that for intuitive reasons the minimum $(X,Y,R)$ occurs with $Y=0$. If someone else could weigh in, that would be greatly appreciated. GalacticShoe (talk) 20:35, 10 April 2024 (UTC)

For which n, A269252(n) is -1?

[1] proves that A269254(110) and A269252(34) are both -1, and there is a sequence for the n such that A269254(n) = -1: A333859, but there seems to be no OEIS sequence for the n such that A269252(n) = -1, and A269254 and A269252 are similar sequence, so for which n, A269252(n) is -1? 118.170.19.90 (talk) 09:41, 2 April 2024 (UTC)

A269252(n) = -1 for n = 1, 2, 34, 53., but there may not be a sequence 1, 2, 34, 53, ... Bubba73 ^{You talkin' to me?} 20:04, 2 April 2024 (UTC)

Could you compute the terms <= 1000? Thanks. A333859 has the terms <= 1500. 49.217.123.95 (talk) 08:21, 9 April 2024 (UTC)

April 4

K-triviality using conditional complexity?

To account for length, the definition of K-trivial set makes the complexity of the length part of the upper-bound on the complexity of the string. Has anyone determined what happens if one instead asks for a bound on the length-conditional complexity? This could be done with either plain or prefix-free Kolmogorov complexity. JumpDiscont (talk) 00:12, 4 April 2024 (UTC)

If there's a constant bound on the length-conditioned complexity, then the set is computable. For a bound that tends to infinity (maybe $K(n)$?), this is similar to c.e.-traceability, so I suspect it won't line up exactly with $K$-triviality.

For research level math like this, you'll probably have better luck on mathoverflow.--Antendren (talk) 10:04, 6 April 2024 (UTC)

April 8

Constructing Wieferich numbers according to Agoh, Dilcher and Skula

I am trying to understand how to use the method in https://doi.org/10.1006/jnth.1997.2162 to construct composite base-a Wieferich numbers from a given integer a > 1 and a base-a Wieferich prime p. In section 6. EXAMPLES the authors use Theorem 5.5 to construct the list of base-2 Wieferich numbers based on the two known Wieferich primes 1093 and 3511. They compute the prime factorizations of 1092 and 3510 and apply Theorem 5.5 to obtain the 104 known base-2 Wieferich numbers. How does this work for arbitrary Wieferich primes p and bases a? For example, suppose I have the prime number p = 5209 which is a Wieferich prime to base a = 1359624 (see https://oeis.org/A143548 comment no. 3 by T. D. Noe). 5208 = 2³ * 3 * 7 * 31. How does one obtain the base-1359624 Wieferich numbers generated by 5209 with Theorem 5.5? I also find the notation in the paper a bit confusing: q(a, m) is an Euler quotient per definition 1.2, but I don't really understand the definition of σ(a, p) at the top of p. 46. Is this simply the multiplicative order of a (mod p)? Toshio Yamaguchi (talk) 11:40, 8 April 2024 (UTC)

Is your question related to [2]? Which is the largest known Wieferich numbers to bases b<=25 2402:7500:916:7991:1CB3:D7F2:BD28:3F3A (talk) 08:11, 9 April 2024 (UTC)

Mhm, I guess it is related. If I understand correctly, Theorem 2 in that paper could be used to generate a Wieferich number w based on a given Wieferich prime p, although I have yet to understand how exactly that would work. Toshio Yamaguchi (talk) 09:22, 9 April 2024 (UTC)

A further question: On p. 2 of the Akbari, Siavashi paper there is a following statement I do not understand: "For a prime p and positive integer n, we denote the largest power of p in n by v_p(n)". What does "power of p in n" mean? Toshio Yamaguchi (talk) 09:38, 9 April 2024 (UTC)

April 9

What is the smallest positive integer which is known to not divide any (even or odd) perfect number?

What is the smallest positive integer which is known to not divide any (even or odd) perfect number? Also, how many positive integers <= 400 are known to not divide any (even or odd) perfect number? 49.217.123.95 (talk) 08:05, 9 April 2024 (UTC)

All positive integers divide a perfect number. Writing $n=2^{a}b,$ in which $b$ is odd, it divides a perfect number of the form $2^{p-1}(2^{p}-1)$ for $p\leq a+b.$  --Lambiam 17:03, 9 April 2024 (UTC)

Are you sure about that, Lambiam? We know of only 51 perfect numbers, and we do not know whether there are infinitely many of them. Assuming there is a highest one, there must be infinitely many numbers greater than that, which, by definition, do not divide any lower numbers. Or, is your formula somehow proof that there are indeed infinitely many? -- Jack of Oz ^{[pleasantries]} 19:22, 9 April 2024 (UTC)

No, thank you for correcting me. All positive integers divide a number (in fact, infinitely many) of the form $2^{p-1}(2^{p}-1),$ but these need not be perfect numbers.  --Lambiam 00:13, 10 April 2024 (UTC)

Check. :) -- Jack of Oz ^{[pleasantries]} 02:27, 10 April 2024 (UTC)

As far as I can tell, from our article on perfect numbers (in particular, the section on odd perfect numbers; even perfect numbers are precisely divisible by all odd primes of the form $2^{p}-1$, and assuming that there are infinitely many such primes, all powers of 2 as well), it would seem that the answer is 105. GalacticShoe (talk) 02:22, 10 April 2024 (UTC)

Obviously this gives a trivial lower bound of 3 on the number of positive integers at most 400 known to not divide any perfect number. GalacticShoe (talk) 02:23, 10 April 2024 (UTC)

Even numbers cannot be factors of odd perfect numbers, thus there should be a larger lower bound. 220.132.230.56 (talk) 21:11, 12 April 2024 (UTC)

Good point. The sequence of even numbers that divide no perfect numbers is $10,12,18,20,22,24,26,30,34,36,38,40,42,44,46,48,50...$, basically even numbers not of the form $2^{k}(2^{p}-1)$ where $2^{p}-1$ is prime and $1\leq k<p$. GalacticShoe (talk) 23:47, 12 April 2024 (UTC)

So how many positive integers <= 400 are known to not divide any (even or odd) perfect number? 220.132.230.56 (talk) 13:35, 13 April 2024 (UTC)

Is it possible to use all n-cubes with n <= 5 assembled into a 2*3*31 rectangular cuboid?

Is it possible to use all n-cubes with n <= 5 (reflections counted as distinct), with totally 186 cubes, assembled into a 2*3*31 rectangular cuboid?

Also, for which positive integer N, it is possible to use all n-cubes with n <= N (reflections counted as distinct), assembled into a rectangular cuboid? 2402:7500:916:7991:1CB3:D7F2:BD28:3F3A (talk) 08:49, 9 April 2024 (UTC)

There are 11-polycubes with holes in the center, and it's impossible to fill this hole with anything but a single cube. There are several such shapes, and you only get one single cube, so N≥11 is impossible. You run into similar problems with holes when you try to tile with heptominos. You could stipulate that the pieces with holes not be included, but that would be a different problem. For N=5, it seems to me that best (or at least the most fun) approach would be to whip up a set with a 3-D printer and experiment. Sometimes a parity argument can give a simple impossibility proof, but that seems unlikely in this case. --RDBury (talk) 12:56, 9 April 2024 (UTC)

April 13

How to calculate a line integral by computer?

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)

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

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

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 ${\tfrac {3}{2}}{\sqrt {5}}>3.35,$ but your result evaluates numerically to less than $2.79.$ To check your result: its numerical value should be close to $4.10568.$ --Lambiam 14:19, 13 April 2024 (UTC)

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

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)

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

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)

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

April 15

Does anyone use the word "cathetus" anymore?

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)

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)

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)

1,000 random editors

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)

This is essentially the same as the birthday problem for a set of $n=1000$ randomly chosen people on a planet in which a year takes $d$ days, where $d$ is equal to the number of Wikipedia articles (currently 6,813,000). 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

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

A slightly better approximation is given by the formula

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

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

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

 --Lambiam 03:31, 16 April 2024 (UTC)