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May 25

The spherical cow of stadiums

If a point is x decibels cause 1,000 randomly distributed mouths in radius r then how loud is 10r (10⁵ mouths)? What if it's 3D (10⁶ mouths)? What if the inner ⅓r & 3⅓r respectively have no mouths? Sagittarian Milky Way (talk) 00:21, 25 May 2025 (UTC)[reply]

First, note that a reading of

x

decibels is equal to a sound pressure of

p_{0}10^{x/20}

pascals or a sound intensity of

I_{0}10^{x/10}

watts per square meter. I'll call these

p_{x}

and

I_{x}

for short.

Second, note that standard wave laws mean that the sound intensity given a spherical source will be

I(r)={\frac {P}{4\pi r^{2}}}

so multiplying the distance by

10

gives an intensity

1/100

the original intensity for a single mouth. Sound pressure is proportional to the square root of sound intensity and thus to

1/r

.

However, we're considering a sphere with the same density, so

100

times as many mouths. To combine different sound sources, we assume the different mouths are not coordinated enough in their frequency and phase to give strongly constructive or destructive interference, in which case the intensity will be additive and the pressure will be square-root-of-summed-square. (This relates to the additivity of variance in the central limit theorem.) This means that with a constant density of mouths on the surface of a sphere, we multiply the intensity at the reader by

r^{2}

, and get a constant that does not depend on the radius; this will carry back to the decibel reading, which will continue to be

x

.

For the case where the mouths are evenly distributed in space, one can consider it as an integral/sum of spherical shells, and will get a result proportional to

r

for intensity,

{\sqrt {r}}

for pressure, and a reading of

x+10\,\log _{10}(r_{2}/r_{1})

. Sesquilinear (talk) 18:45, 25 May 2025 (UTC)[reply]

May 29

Trying to understand the guy who wrote the deleted article…

Hello, I tried to ask him but fails to get an answer. Although it doesn’t seems important does, someone understand his logic here in this section ?

He talks about the equation should have the form RSA260^4+18*RSA260^2+1 but then use ((2 (RSA260^2 - 1^2 ))^2 + 1 (RSA260^2 - 1^2)^2 + 5 (((2 1 RSA260)^2)))/(5) 2A01:E0A:ACF:90B0:0:0:A03F:E788 (talk) 21:49, 29 May 2025 (UTC)[reply]

The article was deleted for a reason... byhill (talk) 23:41, 29 May 2025 (UTC)[reply]

Yep. And it was mainly WP:OR, not because it was that much wrong.

What he did works for finding remainders equal to 5, I need to understand how to use it for other remainders. 2A01:E0A:ACF:90B0:0:0:A03F:E788 (talk) 08:40, 30 May 2025 (UTC)[reply]

Perhaps this will help: User:Endo999 § Adolf Kunerth's 1878 Modular Square Root Algorithm. ‑‑Lambiam 08:49, 30 May 2025 (UTC)[reply]

Sorry, but I'll push back on the claim "not because it was that much wrong". Before I proposed the article's deletion, I spent a lot of time trying to understand what the article's creator was doing, as it looked promising to what I was trying to do. I can now say with confidence that whatever the article's creator was doing does NOT match up to Kunerth's original paper. You can go through the paper and verify this yourself, even if you can't read German. For one, Kunerth's original paper only works when the moduli is prime. In addition, the article (including the "algorithm") as written was completely nonsensical, and they were making false claims about the algorithm, claims that directly contradict well known facts about the hardness of finding modular square roots and the hardness of integer factorization. While WP:OP was enough of a reason for it to get deleted, I wouldn't have gone to the effort of deleting the article if the mathematics was sound. That is why I proposed its deletion.

Maybe the article's creator is doing some interesting mathematics, but it is not what Kunerth was doing. byhill (talk) 11:43, 30 May 2025 (UTC)[reply]

Yep. He did his own math, hence WP:OR. But I managed to get most of his equations working. 78.245.7.54 (talk) 16:58, 31 May 2025 (UTC)[reply]

Yay! Great! byhill (talk) 19:32, 31 May 2025 (UTC)[reply]

But I need help for modifying that part… 2A01:E0A:ACF:90B0:0:0:A03F:E788 (talk) 19:52, 2 June 2025 (UTC)[reply]

The reasons for which the article was proposed for deletion can be found here. The vast majority of the participants in the ensuing discussion agreed that the article should be deleted. The person who closed the discussion and deleted the article was not a participant in the discussion but followed the instructions for closing such discussions. ‑‑Lambiam 08:42, 30 May 2025 (UTC)[reply]

In the end of the discussion, the reason is WP:OR isn’t Ok. It can work in part for what it claims (non generic algorithm) but it’s not Ok for wikipedia. 78.245.7.54 (talk) 17:02, 31 May 2025 (UTC)[reply]

June 2

Adding matrices of different dimensions

Is there any reason we have to leave this undefined?? That is, it's not something we just choose to leave undefined because we keep definitions of things to where they are easiest to define. Georgia guy (talk) 22:08, 2 June 2025 (UTC)[reply]

In general, concepts are defined if they're useful in some way that makes sense logically, and if they're left undefined is mean they aren't. One can define fine a matrix sum of matrices with different sizes, but there's no way to do it that's useful and makes sense logically, no obvious one at least. --RDBury (talk) 00:22, 3 June 2025 (UTC)[reply]

Adding matrices is defined elementwise. If the matrices aren't the same height and width, some elements that exist in one matrix don't exist in another, so it doesn't really make sense. Duckmather (talk) Duckmather (talk) 04:43, 3 June 2025 (UTC)[reply]

Given a matrix ring (or semiring)

M_{n}(R)

with entries from

R=(R,+,0,\,\cdot \,,1),

we can view its elements as functions from

I_{n}^{2}

to

R,

where

I_{n}=\{1,...,n\}.

We can assign a "rarified" version to each element

M\in M_{n}(R)

by "removing" all entries that are equal to

0,

meaning that we take the partial function from

\mathbb {N} ^{2}

to

R\setminus \{0\}

whose domain is the set

\{(i,j)\,|\,M_{i,j}\neq 0\}

and which agrees with

M

on its domain. These rarified matrices can be restored to fullness in an obvious way. We can use this to make them also into a semiring: to perform an operation on rarified matrices, make the operands full, perform the operation in the (full) matrix semiring, and rarify the result.

Now, in the process of rarifying, it can happen that the original value of

n

is lost. The rarified version of a matrix in

M_{n{+}1}(R)

whose last row and column vectors are all

0

's is the same as that of the matrix in

M_{n}(R)

obtained by removing these vectors. However, this does not matter: the value of

n

is immaterial in our rarified matrix ring.

This gives us a way of adding matrices of different dimensions, but the result is then not a neat, rectangular matrix, unless we pad it out with

0

's. This is effectively the approach taken in the TORRIX system. ‑‑Lambiam 06:18, 3 June 2025 (UTC)[reply]

Previously asked here: Wikipedia:Reference_desk/Archives/Mathematics/2023_May_16 --Amble (talk) 15:04, 4 June 2025 (UTC)[reply]

The identity

(\mathbf {A} +\mathbf {B} )\mathbf {v} =\mathbf {Av} +\mathbf {Bv}

holds if

\mathbf {A}

and

\mathbf {B}

are both

m\times n

matrices while

\mathbf {v}

is a vector of length

n.

The earlier responses raised the issue what remained of this for matrices that are allowed to have different shapes. In the semiring as constructed above, this identity remains to hold in full generality. ‑‑Lambiam 18:54, 4 June 2025 (UTC)[reply]

June 3

When using the cubic sieve for computing discrete logarithms, how to compute the factor base ?

I also have other questions : How exactly to perform the linear algebra step ? How many relations should I collect for a 255 bits safeprime ? What’s the complexity of each phases in O(n) notation ? 2A01:E0A:ACF:90B0:0:0:A03F:E788 (talk) 2A01:E0A:ACF:90B0:0:0:A03F:E788 (talk) 13:17, 3 June 2025 (UTC)[reply]

June 5

When was the last time any of the 60,001 least rare scalp hair counts had only 1 human?

(Google says 90,000 to 150,000's a typical number) Rough statistical estimate of course not actual counting. It should be long after the most recent time world population was 60,001x2 cause protecting 60,001 hair counts for every Planck time of the rest of time with random chance is hard though pigeonhole principle will win eventually. Sagittarian Milky Way (talk) 20:16, 5 June 2025 (UTC)[reply]

June 6

Flatter than logarithmic curve

I have some data in Excel and I made a chart. I wanted to add a trendline. The data fits a logarithmic curve rather well. The problem is that the logarithmic curve starts out close to the real data, but after the hump, it keeps going up. The real data goes up some, but is much more flat. It isn't a straight line, so that option is clearly wrong. It never curves downward. It keep increasing, little by little, so the polynomial doesn't fit on the right side. Power and exponential curve the wrong way. Is there a trendline that is like a logarithmic curve but has a flatter top to it? I don't know how to collapse a data box. If someone knows, please do it here. This is the data that I wanted to make a model from. It shows the total number of times a purchase of resources are checked out. It shouldn't go down (it does sometimes due resources being lost or destroyed and removed from the system, causing the counts for that resource to vanish.). I wanted to see if there was a way to estimate the life-span of purchases by seeing when the checkout count flattens out. Then, it would be time to get more resources and remove the old ones. So, the data: 0 2335 4158 5213 5975 6710 7383 7665 8013 8637 9249 9941 10397 10550 10833 10912 10950 10969 10887 11025 11213 11258 11464 11715 11835 12015 12146 12067 12066 12023 11987 11834 11855 11853 12018 12173 12306 12354 12269 12269 12125 12044 11810 11677 11646 11560 11558 68.187.174.155 (talk) 17:32, 6 June 2025 (UTC)[reply]

Try a logistic function. –jacobolus (t) 18:57, 6 June 2025 (UTC)[reply]

June 7