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:
${\displaystyle f(\lambda )=\int _{\lambda =0}^{\infty }{\frac {1}{\lambda ^{5}}}d\lambda }$
I notice that ${\displaystyle \lambda }$ is here both a free and bounded variable.
So how can I do this, replacing an integral:
"${\displaystyle \int _{\lambda =0}^{\infty }...d\lambda }$"
by a summation:
"${\displaystyle \sum _{\lambda =10}^{90}...\Delta \lambda }$".
Where ${\displaystyle \Delta \lambda }$ 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 ${\displaystyle \Sigma }$ notation is given below the ${\displaystyle \Sigma ,}$ followed by an equals sign. The bound variable of an integral is not given below the ${\displaystyle \textstyle \int }$ 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:

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

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

A (not very sophisticated) way to compute an approximation of ${\displaystyle \textstyle \int _{a}^{b}\,f(v)\,dv~}$ 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 ${\displaystyle a_{i}=a+ih,b_{i}=a+(i{+}1)h,m_{i}=a+(i+{\tfrac {1}{2}})h.}$ 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 using Simpson's rule.  ​‑‑Lambiam 20:55, 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:

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

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

Where ${\displaystyle \Delta v}$ 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 ${\displaystyle \Delta v}$, 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 ${\displaystyle \Delta v}$ divides evenly into ${\displaystyle b-a,}$ so that ${\displaystyle n={\frac {b-a}{\Delta v}}}$ 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:

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

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

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

 ​‑‑Lambiam 00:53, 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 the floor function with the symbolic writing ⌊n⌋ and the ceiling 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 π 41.122.66.11 (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 ${\displaystyle 42.}$  ​‑‑Lambiam 16:21, 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 of original 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]

I just found a category mistake. Long is the way (talk) 14:58, 27 October 2025 (UTC)[reply]

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

${\displaystyle {\begin{aligned}&A^{-1}=\{a^{-1}:a\in A\}\\[3pt]&A\circledast B=\{a*b:a\in A,b\in B\}\end{aligned}}}$

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

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

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

Using this we proceed,

${\displaystyle x\in (A\circledast B)^{{-}1}\Leftrightarrow }$ ${\displaystyle x^{{-}1}\in A\circledast B\Leftrightarrow }$ ${\displaystyle (\exists \,a\in A,b\in B:x^{{-}1}=a*b)\Leftrightarrow }$${\displaystyle (\exists \,b\in B,a\in A:x=b^{{-}1}*a^{{-}1})\Leftrightarrow }$ ${\displaystyle x\in B^{{-}1}\circledast A^{{-}1}.}$

 ​‑‑Lambiam 18:16, 25 October 2025 (UTC)[reply]

Crans (2000) wrote, "There is no coherence theorem for tetracategories yet" and the reference list of that paper included "T. Trimble, The definition of tetracategory, handwritten diagrams." Do the handwritten diagrams refer to Trimble's Notes on Tetracategories, published in 2006? Is it correct to think that the statement "There is no coherence theorem for tetracategories yet" refers to the coherence condition explained in Trimble (2006)? I am wondering whether to add that there is currently no coherence theorem for tetracategories to Coherency (homotopy theory). Also, are there any newer references to the coherence theorem for tetracategories?

• Crans, S. (2000). "On braiding, syllapses and symmetries" (PDF). Cahiers de Topologie et Géométrie Différentielle Catégoriques. 41 (1): 2–74.
• Trimble, Todd (2006). "Notes on tetracategories".

--SilverMatsu (talk) 02:11, 1 November 2025 (UTC)[reply]

Apparently, Trimble created the handwritten sheets containing the diagrams in 1995, copies of which were in the hands of several mathematicians. They were digitized and made available as pdf files through Baez's website in 2006. Crans's earlier mention refers to these handwritten sheets, one of which, numbered ①, bears the heading "Tetracategories" and starts with the definition, "A tetracategory is given by the following data: ...", followed by six items subject to four conditions expressed in diagrams on further sheets. Taken together, these four sets of in total 51 handwritten sheets constitute the definition.

There have been several papers since then referring to objects by the name of "tetracategory".^[1] Unless they refer to Trimble's definition, it may be non-trivial to verify whether they are talking about the same species.

A coherence condition is something else than a coherence theorem. The former is a requirement that forms part of a definition (as seen, for example, in the formal definition of a monoidal category). The latter is a result of the form that certain coherencies imply further coherencies.  ​‑‑Lambiam 09:09, 1 November 2025 (UTC)[reply]

Thank you for your reply. It seems that the handwritten diagrams refer to Trimble (2006). Thank you for the link, I'll check out the paper you linked to. The question was phrased incorrectly, I should have asked: when Crans's said "There is no coherence theorem for tetracategories yet" did he mean that there is no way to prove Trimble's coherence condition?--SilverMatsu (talk) 14:21, 1 November 2025 (UTC)[reply]

The use of the adverb "yet" implies that Crans does not deem it impossible that someone will formulate and prove such a theorem in the future, comparable to the existing coherence theorem for tricategories.^[2] He cannot have meant that there is no way to prove Trimble's four coherence conditions, because it is a trivial (but boring) execise to construct objects that have six components as required by the definition, which, however, fail each of the conditions. If a definition contains a condition that can be proved for the general case, the condition is unnecessary baggage.  ​‑‑Lambiam 15:42, 1 November 2025 (UTC)[reply]

I see. Thank you so much for explaining everything so thoroughly.--SilverMatsu (talk) 16:02, 1 November 2025 (UTC)[reply]

I found one reference from the link you provided. The reference appears to have Trimble's definition typed in LaTeX. I previously merged tetracategory into higher category theory, but since I can find references, it might be a good idea to split the article.

• Hoffnung, Alexander E. (2011). "Spans in 2-Categories: A monoidal tricategory". arXiv:1112.0560 [math.CT].

--SilverMatsu (talk) 00:17, 2 November 2025 (UTC)[reply]

The reference is self-published, which may raise some doubts about verifiability. Yet the author was a student of Baez^[3][4] and has published a peer-reviewed article in this area of research ([5]). His contribution is described here as follows:

Despite the complexity of tri-categories, Todd Trimble managed to go beyond and write down the definition of a tetra-category, i.e. a weak 4-category, the result has later been polished [Hof13] by Alexander E. Hoffnung.

([Hof 13] is the arXiv paper). So I guess we may consider this a reliable source.  ​‑‑Lambiam 01:03, 2 November 2025 (UTC)[reply]

Thank you for checking if it is a reliable source. We may be able to add the reference to §.External link. I think the previous merge was to merge (strict) 4-category and tetracategory. For example, strict 2-category and weak 2-category are merged into 2-category. There is no coherence theorem for tetracategories yet, so I think (strict) 4-category and tetracategory cannot be merged. I will use Crans (2000) as a reference. --SilverMatsu (talk) 01:41, 2 November 2025 (UTC)[reply]

Consider two people who walk at distinct steady speeds starting at the same time from one end of a straight track. Without any measurement or communication, they can come back to the starting point together if the faster turns round immediately on reaching the far end of the track then, on the pair meeting, both turn round and redo their routes to date in reverse. Can three people with arbitrary distinct steady speeds achieve the same simultaneous arrival back by a similar procedure? It's possible in special cases, e.g. speeds of 1/2/3, but what about in general? If possible, how can it be achieved; if not, how can this be shown? 2A00:23C6:AA0B:3401:DC5D:BE:981F:7BE1 (talk) 14:56, 1 November 2025 (UTC)[reply]

Does exactly the same procedure (any time two people meet, they both reverse direction) not work? 173.79.19.248 (talk) 21:09, 1 November 2025 (UTC)[reply]

Recently, Miller inversions algorithms got more generic in 2025 in a way they can be applied to arbitrary pairing friendly curves.

But the hard part that prevents breaking the q strong Diffie Hellman problem is exponentation inversion. The ate pairing doesn t use final exponentiation, but as far I understand it then breaks this criteria ${\displaystyle \forall a,b\in \mathbb {F} _{q}^{*},P\in G_{1},Q\in G_{2}:\ e\left(aP,bQ\right)=e\left(P,Q\right)^{ab}}$. Several peoples told me I might find how to use Miller inversion for inverting the Weil pairing, but without knowing a point, I fails to see how this can be done.

Now, which type of construction for bn or bls curve is the most likely to find a pairing that respect ${\displaystyle \forall a,b\in \mathbb {F} _{q}^{*},P\in G_{1},Q\in G_{2}:\ e\left(aP,bQ\right)=e\left(P,Q\right)^{ab}}$ with the step after the Miller algorithm remaining easy to inverse even without knowing 1 of the point involved? 82.67.45.113 (talk) 16:16, 1 November 2025 (UTC)[reply]