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December 11

Unique normal ultrafilter

So I'm supposed to know the answer to this, I suppose, but I don't seem to :-)

"Everyone knows" that, in $L[U]$ , Gödel's constructible universe relative to an ultrafilter $U$ on some measurable cardinal $\kappa$ , there is only a single normal ultrafilter, namely $U$ itself. See for example John R. Steel's monograph here, at Theorem 1.7.

So I guess that must mean that the product measure $U\times U$ , meaning you fix some identification between $\kappa \times \kappa$ and $\kappa$ and then say a set has measure 1 if measure 1 many of its vertical sections have measure 1, must not be normal. (Unless it's somehow just equal to $U$ but I don't think it is.)

But is there some direct way to see that? Say, a ~~continuous~~ function $f:\kappa \to \kappa$ with $\forall \alpha f(\alpha )<\alpha$ such that ~~the set of fixed points of $f$ is not in the ultrafilter~~ no singleton has a preimage under $f$ that's in the ultrafilter? I haven't been able to come up with it. --Trovatore (talk) 06:01, 11 December 2024 (UTC)[reply]

December 15

What is the cause of this paradox?

I recently completed a calculus term, in which one of the last units involving how much one aspect of an object was changing in relation to time at a certain point, given the rate of change of another aspect. Many specific questions could be analyzed as a right triangle with one leg (the x) remaining constant and the other leg (the y) growing at a specified rate. When it came time to solve for the value of the dz/dt (the rate of the hypotenuse’s growth with respect to time) at a certain point, it ended up as less than the provided dy/dt. Here’s an illustration:

The x is the distance from me to a tower. This remains constant.

The y is the distance from the tower to a flying bird.

The dy/dt is the speed at which the bird is flying from the tower.

The z is my distance from the bird.

In this illustration, the distance between me and the bird is increasing at a slower rate than the speed at which the bird itself is flying. What is the cause of this paradox? Primal Groudon (talk) 19:43, 15 December 2024 (UTC)[reply]

I do not see any paradox here. Ruslik_Zero 20:30, 15 December 2024 (UTC)[reply]

If the bird is between you and the tower (

0 \leq y < x

), the distance between you and the bird is even decreasing:

dz / dt < 0.

By the time it flies right overhead (

y = x

), the distance is momentarily stationary:

dz / dt = 0.

After that, it increases:

dz / dt > 0.

This rate of increase will asymptotically approach

dy / dt

from below as the bird flies off into an infinite distance. --Lambiam 00:34, 16 December 2024 (UTC)[reply]

I think the issue here is that even though the rate of change of z is less than the rate of change of y, z never actually becomes less than y. You can see this graphically, for example, by comparing the graphs of y=x and y=√(x²+1). The second graph is always above the first graph, but the slope of the first graph is x/√(x²+1), which is always less than 1, the slope of the first graph. But this is typical behavior when a graph has an asymptote. As a simpler example, the slope of 1/x is negative, but the value never goes below 0 (at least for x>0). Similarly, the slope of x+1/x is always less than 1, but the value of x+1/x is always greater than x (again, for x>0). The graph of y=√(x²+1) is one branch of a hyperbola having y=x as an asymptote, and this looks very much like the x>0 part of y=x+1/x. In general the difference in rates of change can imply that that two quantities get closer and closer to each other, but this does not mean they ever become equal. This phenomenon is, perhaps, counterintuitive for many people, but the math says it can happen anyway. I don't know if this rises to the level of a paradox, but I can see that it might be concerning for some. --RDBury (talk) 09:39, 16 December 2024 (UTC)[reply]

For x > 0, the graph of y=√(x²+1) looks even more like that of y=x+1/(2x). For example, when x = 5, √(x²+1) = √26 ≈ 5.09902 is approximated much more closely by 5.1 than by 5.2. --Lambiam 18:35, 16 December 2024 (UTC)[reply]

December 19

Who is the following unknown?

When asked "WHO IS YOUR X?" (X still being unknown to me but is known to the respondents), here are the answers I get:

A answers: "A"

B answers: "C"

C answers: "C"

D answers: "F"

E answers: "F"

F answers: "F"

To sum up, the special phenomenon here is that, everybody has their own X (usually), and if any respondent points at another respondent as the first respondent's X, then the other respondent must point at themself as their X.

I wonder who the unknown X may be, when I only know that X is a natural example from everyday life. I thought about a couple of examples, but none of them are satisfactory, as follows:

X is the leader of one's political party, or X is one's mayor, and the like, but all of these examples attribute some kind of leadership or superiority to X, whereas I'm not interested in this kind of solution - involving any superiority of X.

Here is a second solution I thought about: X is the first (or last) person born in the year/month the respondent was born, and the like. But this solution involves some kind of order (in which there is a "first person" and a "last person"), whereas I'm not interested in this kind of solution - involving any order.

Btw, I've published this question also at the Miscellaneous desk, because this question is about everyday life, but now I decide to publish this question also here, because it's indirectly related to a well known topic in Math. 79.177.151.182 (talk) 13:27, 19 December 2024 (UTC)[reply]

Head of household comes to mind as a fairly natural one. The colours then correspond to different households which can be just one person. One objection is that "head of household" is a fairy traditional concept. With marriage equality now being the norm it's perhaps outdated. --2A04:4A43:909F:F9FF:397E:BBF9:E80B:CB36 (talk) 15:11, 19 December 2024 (UTC)[reply]

I have already referred to this kind of solution, in the example of "my mayor", see above why this solution is not satisfactory. 79.177.151.182 (talk) 15:31, 19 December 2024 (UTC)[reply]

The question has been resolved at the Miscellaneous reference desk.

Resolved

79.177.151.182 (talk) 15:48, 19 December 2024 (UTC)[reply]

X may well be 'the oldest living person of your ancestry'. --CiaPan (talk) 20:46, 19 December 2024 (UTC)[reply]

Resolved or not, let's try to analyze this mathematically. Given is some set

S

and some function

f:S\to S.

For the example,

S=\{{\mathsf {A}},{\mathsf {B}},{\mathsf {C}},{\mathsf {D}},{\mathsf {E}},{\mathsf {F}}\},

with

f({\mathsf {A}})={\mathsf {A}},

f({\mathsf {B}})=f({\mathsf {C}})={\mathsf {C}},

f({\mathsf {D}})=f({\mathsf {E}})=f({\mathsf {F}})={\mathsf {F}}.

Knowing that "everybody has their own X (usually)", we can normalize the unusual situation that function

f

might not be total in two ways. The first is to restrict the set

S

to the domain of

f,

that is, the set of elements on which

f

is defined. This is possible because of the condition that

f(u)=v

implies

f(v)=v,

so this does not introduce an undue limitation of the range of

f.

The second approach is to postulate that

f(u)=u

whenever

f(u)

might otherwise be undefined. Which of these two approaches is chosen makes no essential difference.

Let

T=f(S)

be the range of

f

, given by:

T=\{v\in S:(\exists u\in S:f(u)=v)\}.

Clearly, if

f(v)=v,

we have

v\in T.

We know, conversely, that

v\in T

implies

f(v)=v.

Let us also consider the inverse image of

f

, given by:

f^{-1}(v)=\{u\in S:f(u)=v\}.

Suppose that

f^{-1}(v)\neq \emptyset .

This means that there exists some

u\in f^{-1}(v),

which in turns means that

f(u)=v.

But then we know that

f(v)=v.

Combining this, we have,

f(v)=v\Leftrightarrow v\in T\Leftrightarrow f^{-1}(v)\neq \emptyset .

The inverse-image function restricted to

T,

to which we assign the typing

f^{-1}:T\to \mathbb {P} (S),

now induces a partitioning of

S

into non-empty, mutually disjoint subsets, which means they are the classes of an equivalence relation. Each class has its own unique representative, which is the single element of the class that is also a member of

T

. The equivalence relation can be expressed formally by

u\sim v\Leftrightarrow f(u)=f(v),

and the representatives are the fixed points of

f.

Applying this to the original example,

T=\{{\mathsf {A}},{\mathsf {C}},{\mathsf {F}}\},

and the equivalence classes are:

$\{{\mathsf {A}}\},$ with representative ${\mathsf {A}},$
$\{{\mathsf {B}},{\mathsf {C}}\},$ with representative ${\mathsf {C}},$ and
$\{{\mathsf {D}},{\mathsf {E}},{\mathsf {F}}\},$ with representative ${\mathsf {F}}.$

Conversely, any partitioning of a set defines an equivalence relation; together with the selection of a representative for each equivalence class, this gives an instance of the situation defined in the question. --Lambiam 20:47, 19 December 2024 (UTC)[reply]

FWIW, the number of such objects on a set of size n is given by OEIS: A000248, and that page has a number of other combinatorial interpretations. If you ignore the selection of a representative for each class, you get the Bell numbers. --RDBury (talk) 00:35, 21 December 2024 (UTC)[reply]

December 20

Give a base b and two base b digits x and z, must there be a base b digits y such that the 3-digit number xyz in base b is prime?

Give a base b and two base b digits x and z (x is not 0, z is coprime to b), must there be a base b digits y such that the 3-digit number xyz in base b is prime? 1.165.207.39 (talk) 02:10, 20 December 2024 (UTC)[reply]

In base 5,

3Y1_{5}=76+5Y

is composite for all base-5 Y. GalacticShoe (talk) 03:39, 20 December 2024 (UTC)[reply]

3Y1_{7}

also offers a counterexample. While there are many counterexamples for most odd bases, I did not find any for even bases. --Lambiam 09:58, 20 December 2024 (UTC)[reply]

December 23

Is it possible to make Twisted Edwards curve birationally equivalent to twisted weirestrass curves ?

Is there an equation fo converting a twisted Edwards curve into a tiwsted weierstrass form ? 2A01:E0A:401:A7C0:6D06:298B:1495:F479 (talk) 04:12, 23 December 2024 (UTC)[reply]

According to Montgomery curve § Equivalence with twisted Edwards curves, every twisted Edwards curve is birationally equivalent to a Montgomery curve, while Montgomery curve § Equivalence with Weierstrass curves gives a way to transform a Montgomery curve to an elliptic curve in Weierstrass form. I don't see a definition of "twisted Weierstrass", so I don't know if you can give an extra twist in the process. Perhaps this paper, "Efficient Pairing Computation on Twisted Weierstrass Curves" provides the answer; its abstract promises: "In this paper, we construct the twists of twisted Edwards curves in Weierstrass form." --Lambiam 10:35, 23 December 2024 (UTC)[reply]

December 24

How did the Romans do engineering calculations?

The Romans did some impressive engineering. Engineers today use a lot of mathematical calculations when designing stuff. Calculations using Roman numerals strike me as being close to impossible. What did the Romans do? HiLo48 (talk) 05:50, 24 December 2024 (UTC)[reply]

The kind of engineering calculations that might have been relevant would mostly have been about statics – specifically the equilibrium of forces acting on a construction, and the ability of the design to withstand these forces, given its dimensions and the mechanical properties of the materials used in the construction, such as density, modulus of elasticity, shear modulus, Young modulus, fracture strength and ultimate tensile strength. In Roman times, only the simplest aspects of all this were understood mathematically, namely the statics of a construction in which all forces work in the same plane, without torque, and the components are perfectly rigid. The notion of assigning a numerical magnitude to these moduli and strengths did not exist, which anyway did not correspond to precisely defined, well-understood concepts. Therefore, engineering was not a science but an art, mainly based on experience in combination with testing on physical models. Any calculations would mostly have been for the amounts and dimensions of construction materials (and the cost thereof), requiring a relatively small number of additions and multiplications. --Lambiam 19:03, 24 December 2024 (UTC)[reply]

Calculating with Roman Numerals might seem impossible, but in some ways it's simpler than our positional system; there are only so many symbols commonly used, and only so many ways to add and multiply them. Once you know all those ways you efficiently can do calculations with them, up to the limits imposed by the system.

And for a lot of things they relied on experience. Romans knew how to build circular arches, but rather than do calculations to build larger arches, or ones with more efficient shapes they used many small circular ones which they knew worked, stacked side by side and sometimes on top of each other. See e.g. any Roman aqueduct or the Colosseum. For materials they probably produce them on site or close by as they're needed.--2A04:4A43:909F:F9FF:FC7B:F1E8:19D6:124C (talk) 20:12, 24 December 2024 (UTC)[reply]

To add to the above, it wasn't really the Romans that were great innovators in science, engineering, etc.. I think more innovation and discovery took place in Ancient Greece and Ancient Egypt. Certainly Greece as we have a record of that. Egypt it's more that they were building on such a monumental scale, as scale no-one came close to repeating until very recently.

Romans were military geniuses. They conquered Greece, and Egypt, and Carthage, and Gaul, and Britannia, and everywhere in between. They then built forts, towns, cities and infrastructure throughout their empire. They built so much so widely that a lot of it still stands. But individually a lot of it isn't technically impressive; instead it's using a few simple patterns over and over again.--2A04:4A43:909F:F9FF:FC7B:F1E8:19D6:124C (talk) 12:09, 25 December 2024 (UTC)[reply]

See Roman abacus. catslash (talk) 22:03, 25 December 2024 (UTC)[reply]

Are these sequences mod any natural number n periodic?

The period of Fibonacci number mod n is the Pisano period of n, but are these sequences mod any natural number n also periodic like Fibonacci number mod n?

42.76.153.22 (talk) 06:06, 24 December 2024 (UTC)[reply]

For 1. through 4., see Pisano period#Generalizations. Although 5. through 8. are not explicitly listed, I'm pretty sure the same argument applies for their periodicity as well. GalacticShoe (talk) 07:05, 24 December 2024 (UTC)[reply]

For 13, I'm not sure, but I think the partition function is not periodic modulo any nontrivial number, to the point that the few congruences that are satisfied by the function are also very notable, e.g. Ramanujan's congruences. GalacticShoe (talk) 07:23, 24 December 2024 (UTC)[reply]

It's possible that a sequences is eventually periodic but not periodic from the start, for example powers of 2 are periodic for any odd n, but for n=2 the sequence is 1, 0, 0, ..., which is only periodic starting with the second entry. In other words a sequence can be become periodic without being pure periodic. A finiteness argument shows that 1-8 are at least eventually periodic, but I don't think it works for the rest. (Pollard's rho algorithm uses this finiteness argument as well.) It says in the article that Bell numbers are periodic mod n for any prime n, but the status for composite n is unclear, at least from the article. Btw, Catalan numbers not on the list?. --RDBury (talk) 08:08, 24 December 2024 (UTC)[reply]

PS. 1-8 are pure periodic. In general, if the recurrence can be written in the form F(k) = (some polynomial in F(k-1), F(k-2), ... F(k-d+1) ) ± F(k-d), then F is pure periodic. The reason is that you can solve for F(k-d) and carry out the recursion backwards starting from where the sequence becomes periodic. Since the previous entries are uniquely determined they must follow the same periodic pattern as the rest of the sequence. If the coefficient of F(k-d) is not ±1 then this argument fails and the sequence can be pre-periodic but not pure periodic, at least when n is not relatively prime to the coefficient. --RDBury (talk) 18:12, 24 December 2024 (UTC)[reply]

Ah, what was tripping me up was showing pure periodicity, recursing backwards completely slipped my mind. Thanks for the writeup! GalacticShoe (talk) 18:31, 24 December 2024 (UTC)[reply]

December 26