Many years ago I got interested in minimum spanning trees and, as an extension, independently conceived of theDelaunay triangulation (and thence of its dual theVoronoi tessellation), before I knew of those names. Now theminimum spanning tree article inspires me to ask whether there is an analogous extension of theSteiner tree; perhaps that is not even meaningful.—Tamfang (talk)00:11, 16 October 2025 (UTC)[reply]

How do I write Zipf’s law as a sequence in the abbreviated form that’s expressed as a function of an index variable (not the expanded form where several terms are fully written out)? Is this complicated by Zipf’s law terms being defined in relation to both their ordinal position and the first term? The best I can think of is something like a_n = 1/n * a₁, but I think even this looks odd.Primal Groudon (talk)02:41, 19 October 2025 (UTC)[reply]

You could use

${\displaystyle a_n=\frac{a}{n}.}$

Compare the formula we give for ageometric progression,

${\displaystyle a_n=a{r}^{n-1},}$

in which also${\displaystyle a_1}$ equals the constant parameter${\displaystyle a}$.  ​‑‑Lambiam06:41, 19 October 2025 (UTC)[reply]

Given a group${\displaystyle (\text{G},\ast )}$ and${\displaystyle \text{N}⊴\text{G}}$ a normal subgroup, for all${\displaystyle a\in \text{G}}$ we define a coset to be of the form

${\displaystyle a\text{N}=\{a\ast n:n\in \text{N}\}}$

and for all${\displaystyle a,b\in \text{G}}$ we show that coset multiplication is closed. The common proof is so:

The problem? the symbols represent different mathematical objects – sets and elements.
Using associativity and commutativity like clay here is a misuse or anabuse of notation.
Only after reading carefully in an old group theory book, it hit me:
${\displaystyle a\text{N}}$ is actually an abbreviated notation ofset multiplication, with the formal notation being

${\displaystyle a\text{N}\triangleq \{a\}\cdot \text{N}=\{a\ast n:a\in \{a\},n\in \text{N}\}}$.

The proof above becomes by definition:

My point is simply this:
I am NOT saying we must use formal notation everywhere. Not at all. But just one line of formal notation in the main definition could have cleared this all up for me in advance. Instead, I had to struggle quite a lot learning this topic.
Unfortunately, most sources I saw until now tend to skip this tiny little important detail completely.יהודה שמחה ולדמן (talk)19:34, 19 October 2025 (UTC)[reply]

For a normal subroup, all of these equalities are true.Tito Omburo (talk)19:49, 19 October 2025 (UTC)[reply]

Seems clear to me. Though I've not looked at a group textbook for decades and not used group theory directly for as long the patternaN is just the obvious product, as it is in algebra I'm more familiar with. It's the set formed by multiplyinga by everything inN. The proof follows from the properties that group multiplications are associative, and multiplication withN is commutative from the definition of a normal subgroup. --2A04:4A43:903F:F2A9:B144:FD8B:1D1F:3F13 (talk)00:25, 20 October 2025 (UTC)[reply]

I do not expect the rebracketing step

${\displaystyle (aN)(bN)=(a(Nb))N}$

to be obvious to a learner who is somewhat new to abstract algebra. You really need to unfold the definitions to see that this is valid. Otherwise, why not simply write

${\displaystyle (aN)(bN)=aNbN=abNN=abN}$?

I don't know the text this is from, but it may help the reader if, where set product is introduced, the text sets an exercise to prove that this product inherits associativity from the group's product.  ​‑‑Lambiam09:05, 20 October 2025 (UTC)[reply]

The old book I read does give such an exercise.

But once again: without both the formal definition and notation in advance, we simply cannot justify rebracketing the symbols only by set multiplication properties.יהודה שמחה ולדמן (talk)13:22, 20 October 2025 (UTC)[reply]

You're right, expressions like aS, Sa, and ST have to be defined, and the generalizations of the associative law, (aS)b = a(Sb), (Sa)b = S(ab), (ab)S = a(Sb), (Sa)T = S(aT), (ST)a = S(Ta), a(ST) = (aS)T, (ST)U = S(TU), are theorems that must be proven. But they hold for any sets S, T, U, not just subgroups or cosets, and the proofs are straightforward enough so most authors leave them as exercises. (Moral: Do at least read the exercises in a math text because sometimes they contain material you will need later.) Also, if H is a subgroup then HH = H by closure, another exercise. Note that your definition of coset is questionable. Aleft coset of a group H is a set of the from aH, and aright coset of H is a set of the from Ha. It's only when H is a normal subgroup that you can assume that left coset is a right coset of H and you can just say coset. (Fun fact, a left coset of aH of H can be written (aH)a^-1a = (aHa^-1)a which is a right coset of of aHa^-1. So every left coset is also a right coset, but the corresponding subgroups may not be the same.) There is also an equivalent equivalence class definition of left and right cosets, where you define the left cosets of H to be equivalence classes under the relation a ~ b when a^-1b ∈ H. That this is indeed an equivalence relation and that the equivalence classes are indeed the sets of the form aH are yet more exercises. --RDBury (talk)16:25, 20 October 2025 (UTC)[reply]

I realise that an integral is an infinite sum of a variable function, for example${\displaystyle \lambda }$, with an infinitesimal parameter as a step value, for example${\displaystyle d\lambda }$, as follow:
${\displaystyle f(\lambda )={\int }_{\lambda =0}^{\mathrm{\infty }}\frac{1}{{\lambda }^5}d\lambda }$
I notice that${\displaystyle \lambda }$ is here both a free and bounded variable.
So how can I do this, replacing an integral:
"${\displaystyle {\int }_{\lambda =0}^{\mathrm{\infty }}...d\lambda }$"
by a summation:
"${\displaystyle \sum _{\lambda =10}^{90}...\mathrm{\Delta }\lambda }$".
Where${\displaystyle \mathrm{\Delta }\lambda }$ is a finitesimal parameter acting as the step parameter?
Malypaet (talk)12:29, 22 October 2025 (UTC)[reply]

The bound variable of a summation in${\displaystyle \mathrm{\Sigma }}$ notation is given below the${\displaystyle \mathrm{\Sigma },}$ followed by an equals sign. The bound variable of an integral is not given below the${\displaystyle \int }$ sign but instead after the symbol${\displaystyle d.}$ (Physics texts often use a Roman${\displaystyle \text{d}.}$) Using${\displaystyle v}$ as the name of the bound variable and${\displaystyle a}$ and${\displaystyle b}$ as the bounds, a summation looks like

${\displaystyle \sum _{v=a}^{b}f(v),}$

whereas a (definite) integral looks like

${\displaystyle {\int }_a^{b}f(v)dv.}$

Your definition of${\displaystyle f(\lambda )}$ mixes these two notations. In any case,${\displaystyle \lambda }$ does not occur free in the right hand side. (And "bounded variable" has a different meaning than "bound variable"). Also, the integral diverges:

${\displaystyle {\int }_{\epsilon }^{\mathrm{\infty }}\frac{1}{{\lambda }^5}d\lambda =\frac{1}{4{\epsilon }^4}}$ if${\displaystyle \epsilon >0,}$

but for${\displaystyle \epsilon =0}$ this is undefined.

A (not very sophisticated) way to compute an approximation of${\displaystyle {\int }_a^{b}f(v)dv}$ is to divide the interval${\displaystyle [a,b]}$ in${\displaystyle n}$ small segments${\displaystyle [a_i,b_i],i=0,...,n-1,}$ where${\displaystyle a_0=a,a_{i+1}=b_i,b_{n-1}=b.}$ The function can be sampled in the midpoints of each segment${\displaystyle m_i=\frac{a_i+b_i}{2}.}$ By summing the function values, weighted by the width${\displaystyle h_i=b_i-a_i}$ of each segment, you get an approximation of the integral. In a formula, this amounts to

${\displaystyle \sum _{i=0}^{n-1}f(m_i)\cdot h_i.}$

If all segments are given equal width${\displaystyle h=\frac{b-a}{n},}$ we get${\displaystyle a_i=a+ih,b_i=a+(i+1)h,m_i=a+(i+\frac{1}{2})h.}$ The integral is then approximated by

${\displaystyle h\cdot \sum _{i=0}^{n-1}f(m_i).}$

As I have already indicated elsewhere, you get a much better approximation for the same computational effort by usingSimpson's rule.  ​‑‑Lambiam20:55, 22 October 2025 (UTC)[reply]

Thanks, I am reading Simpson's rule.

But I am trying to begin with a simple formula.

From yours:${\displaystyle {\int }_a^{b}f(v)dv.}$

That I traduct by:

${\displaystyle f(a)\cdot dv+f(a+1)\cdot dv+f(a+2)\cdot dv+...+f(b)\cdot dv}$

So, an infinite sum of surfaces of rectangles of height${\displaystyle f(v)}$ and infinitesimal width${\displaystyle dv}$.

And then, trying to make a first simple adaptation in discrete mathematics like that:

${\displaystyle v_i=a+((i-a)\cdot \mathrm{\Delta }v)}$

${\displaystyle \sum _{i=a}^{b}f(v_i)\cdot \mathrm{\Delta }v}$

Where${\displaystyle \mathrm{\Delta }v}$ is a parameter of approximately${\displaystyle a/100}$ and is the segment width, and also the step value.

Then if ok, I continue with better approximation based on your examples or Simpson's rule.

Thanks again.
Malypaet (talk)22:38, 23 October 2025 (UTC)[reply]

Your (symbolic) infinite sum of infinitesimals should have been

${\displaystyle f(a)\cdot dv+f(a+dv)\cdot dv+f(a+2dv)\cdot dv+\cdots +f(b-dv)\cdot dv,}$

If, instead of${\displaystyle dv,}$ you use a finite increment${\displaystyle \mathrm{\Delta }v}$, which I denoted by${\displaystyle h}$ in my preceding reply, you will get from${\displaystyle a}$ to${\displaystyle b}$ in a finite number of steps. But make sure that${\displaystyle \mathrm{\Delta }v}$ divides evenly into${\displaystyle b-a,}$ so that${\displaystyle n=\frac{b-a}{\mathrm{\Delta }v}}$ is an integer, and that the finite sum then has${\displaystyle n}$ terms. Also, you sample the function at the start of each segment. It is better to use the midpoints:

${\displaystyle f(a+\frac{h}{2})\cdot h+f(a+\frac{3h}{2})\cdot h+f(a+\frac{5h}{2})\cdot h+\cdots +f(b-\frac{h}{2})\cdot h.}$

To use Simpson's rule,${\displaystyle n}$ needs to be even. Then the sum becomes

${\displaystyle \frac{1}{3}f(a)\cdot h+\frac{4}{3}f(a+h)\cdot h+\frac{2}{3}f(a+2h)\cdot h+\frac{4}{3}f(a+3h)\cdot h+\frac{2}{3}f(a+4h)\cdot h+\cdots }$

${\displaystyle \cdots +\frac{2}{3}f(b-2h)+\frac{4}{3}f(b-h)+\frac{1}{3}f(b)\cdot h.}$

So the fractional factors alternate between${\displaystyle \frac{2}{3}}$ and${\displaystyle \frac{4}{3},}$ except at the two endpoints, where they are equal to${\displaystyle \frac{1}{3}.}$ This sum has actually${\displaystyle n+1}$ terms, but the weights still add up to${\displaystyle b-a.}$ By rearranging the terms to get those with equal weights together, you can rewrite this as the weighted sum of three plain sums:

${\displaystyle \frac{h}{3}[f(a)+f(b)]+}$

${\displaystyle +\frac{2h}{3}[f(a+2h)+f(a+4h)+f(a+6h)+\cdots +f(b-6h)+f(b-4h)+f(b-2h)]+}$

${\displaystyle +\frac{4h}{3}[f(a+h)+f(a+3h)+f(a+5h)+\cdots +f(b-5h)+f(b-3h)+f(b-h)].}$

 ​‑‑Lambiam00:53, 24 October 2025 (UTC)[reply]

Thanks very well, it is clear to me now.

Sorry that I have made the the good (symbolic) infinite sum of infinitesimals first, but tired I changed it to the bad one.

For the integer problem, I found thefloor function with the symbolic writing ⌊n⌋ and theceiling function ⌊n⌋.

So, all is right for me now.

Thanks again.
Malypaet (talk)07:43, 24 October 2025 (UTC)[reply]

I have found a unique proof of π41.122.66.11 (talk)15:46, 24 October 2025 (UTC)[reply]

The proof of the${\displaystyle \pi }$ is in the eating. Apart from that,${\displaystyle \pi }$ is a number. I myself have three unique proofs of${\displaystyle 2\pi }$, as well as (just) one of${\displaystyle 41.}$ I'm working now on${\displaystyle 42.}$  ​‑‑Lambiam16:21, 24 October 2025 (UTC)[reply]

Congratulations. Not just on your proof, but on being able to demonstrate that it is unique, which seems to me to be an even harder thing to achieve. Seriously though, this isn't a mathematical forum, it is a reference desk, and since Wikipedia isn't a publisher oforiginal research, merely announcing your find here is pointless, and I doubt that you'll find anyone willing to do the sort of rigorous checking on your proof necessary to (a) demonstrate it is both valid and mathematically interesting, and (b) confirm it is unique. So, do you have a question, or are you just looking for applause?AndyTheGrump (talk)16:24, 24 October 2025 (UTC)[reply]

Given a group${\displaystyle (G,\ast )}$ and${\displaystyle A,B\subseteq G}$, let us define WLOG:

Can we prove that${\displaystyle (A⊛B{)}^{-1}=B^{-1}⊛A^{-1}}$?יהודה שמחה ולדמן (talk)17:07, 25 October 2025 (UTC)[reply]

First an easy part, an auxiliary result. We establish that${\displaystyle x\in A^{-1}\iff x^{-1}\in A,}$ as follows:

${\displaystyle x\in A^{-1}\iff }$${\displaystyle (\mathrm{\exists }a\in A:x=a^{-1})\iff }$${\displaystyle (\mathrm{\exists }a\in A:x^{-1}=a)\iff }$${\displaystyle x^{-1}\in A.}$

Using this we proceed,

${\displaystyle x\in (A⊛B{)}^{-1}\iff }$${\displaystyle x^{-1}\in A⊛B\iff }$${\displaystyle (\mathrm{\exists }a\in A,b\in B:x^{-1}=a\ast b)\iff }$${\displaystyle (\mathrm{\exists }b\in B,a\in A:x=b^{-1}\ast a^{-1})\iff }$${\displaystyle x\in B^{-1}⊛A^{-1}.}$

 ​‑‑Lambiam18:16, 25 October 2025 (UTC)[reply]

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