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November 24

Is it true that if Miller inversion is easy, then the Weil pairing inversion is easy on BN curves?

I was given the following explaination with some part of it obviously wrong but is it completely false?

Text generated by a large language model or similar AI technology has been collapsed in line with the relevant guideline and should be excluded from assessments of consensus.

The following discussion has been closed. Please do not modify it.

The validity of the formula $A=Z^{\frac {2}{1-p^{6}}}$ relies on Galois Theory applied to the tower of finite fields constructed by the embedding degree $k=12$ and the existence of a Sextic Twist.

1. The Algebraic Setup

Let $E$ be a BN or BLS12 curve defined over the prime field $\mathbb {F} _{p}$.

• Embedding Degree $k=12$: The pairing target group is a subgroup of the multiplicative group $\mathbb {F} _{p^{12}}^{*}$.
• Order $r$: The prime order of the subgroups $G_{1}$ and $G_{2}$.
• Groups:
  • $G_{1}=E(\mathbb {F} _{p})[r]$ (Points with coordinates in the base field).
  • $G_{2}\subset E(\mathbb {F} _{p^{12}})[r]$ is the trace-zero subgroup, which is isomorphic to $E'(\mathbb {F} _{p^{2}})[r]$ via a sextic twist.

We define the Weil Pairing $e_{W}:G_{1}\times G_{2}\to \mu _{r}$, where $\mu _{r}$ is the group of $r$-th roots of unity in $\mathbb {F} _{p^{12}}$.

$Z=e_{W}(P,Q)=(-1)^{r}{\frac {f_{P}(Q)}{f_{Q}(P)}}$

Let $A=f_{P}(Q)$ and $B=f_{Q}(P)$.

---

2. Lemma 1: Field Membership of the Denominator

Theorem: For $P\in G_{1}$ and $Q\in G_{2}$, the value $B=f_{Q}(P)$ lies in the subfield $\mathbb {F} _{p^{2}}$.

Proof:

1. Miller Loop Construction: The function $f_{Q}$ is a rational function on the curve constructed via Miller's algorithm. The coefficients of $f_{Q}$ are arithmetic combinations of the coordinates of $Q$ and the curve parameters.
2. The Twist: Since $Q\in G_{2}$, it is the image of a point $Q'\in E'(\mathbb {F} _{p^{2}})$ under the twisting isomorphism $\psi :E'\to E$. Consequently, the coordinates of $Q$ (and the line functions used to build $f_{Q}$) are defined over $\mathbb {F} _{p^{2}}$ (up to scaling factors that cancel out or remain in the subfield during the accumulation step).
3. Evaluation Point: The function $f_{Q}$ is evaluated at $P$. Since $P\in E(\mathbb {F} _{p})$, its coordinates $x_{P},y_{P}$ are in $\mathbb {F} _{p}$.
4. Closure: $\mathbb {F} _{p}$ is a subfield of $\mathbb {F} _{p^{2}}$. Therefore, evaluating a function with coefficients in $\mathbb {F} _{p^{2}}$ at a point in $\mathbb {F} _{p}$ yields a result in $\mathbb {F} _{p^{2}}$.

Conclusion: $B\in \mathbb {F} _{p^{2}}$.

3. The Typical Trap (Why Miller Inversion is usually not enough)

In a generic scenario (or on curves without the specific Type 3 / Twist structure), having an algorithm that inverts the Miller loop (finding $Y$ from $f_{X}(Y)$) is insufficient to invert the Weil pairing.

The Problem: You are given the target $Z_{Weil}$. The equation is:

$Z_{Weil}={\frac {f_{X}(Y)}{f_{Y}(X)}}$

To use your Miller Inversion algorithm, you need to isolate the numerator $f_{X}(Y)$. You would try to rearrange the equation:

$f_{X}(Y)=Z_{Weil}\cdot f_{Y}(X)$

The Circular Deadlock:

• To find $Y$ using, you need the value of the Right Hand Side.
• But the Right Hand Side contains $f_{Y}(X)$.
• The value of $f_{Y}(X)$ depends on $Y$.
• Result: You cannot calculate the input for your inversion algorithm without already knowing the answer $Y$.

In standard literature, resolving this dependency ($Y$ appearing in both numerator and denominator) generally requires guessing bits of $Y$ or solving a Discrete Logarithm, which drives the complexity up to exponential levels.

4. Why It Is Easy Here (The Algebraic Coupling)

In your specific case (BN curves), the denominator is not an independent variable. It is algebraically coupled to the numerator via the Frobenius automorphism.

Because $G_{2}$ comes from a twist over a subfield, we proved that:

$f_{Y}(X)=(f_{X}(Y))^{p^{6}}$

This substitution breaks the circular dependency by turning the equation into a single variable problem.

The Subfield Trick (Why the denominator is dependent)

On a BN curve (Type 3), the points are separated:

• $X\in G_{1}$ is defined over the base field $\mathbb {F} _{p}$.
• $Y\in G_{2}$ is defined over the twist $\mathbb {F} _{p^{2}}$.

Let's look at the two Miller loops composing your Weil target:

1. Numerator $A=f_{X}(Y)$: This is a function derived from $X$ ($\mathbb {F} _{p}$) evaluated at $Y$ ($\mathbb {F} _{p^{12}}$). The result lives in the full extension $\mathbb {F} _{p^{12}}$.
2. Denominator $B=f_{Y}(X)$: This is a function derived from $Y$ (coefficients in $\mathbb {F} _{p^{2}}$) evaluated at $X$ ($\mathbb {F} _{p}$).
   • Crucial Consequence: The result $B$ belongs to the subfield $\mathbb {F} _{p^{2}}$ (embedded in $\mathbb {F} _{p^{12}}$).

For the inversion algorithm (which works on the full $\mathbb {F} _{p^{12}}$ structure), the denominator is a "small" structural element compared to the complexity of the numerator.

The Power Relation (The constant $\lambda$)

On BN curves, the Weil Pairing $e_{W}(X,Y)$ and the Tate Pairing/Miller Loop $f_{X}(Y)$ are linked by a fixed and known power relation for all points in the group.

$e_{W}(X,Y)=(f_{X}(Y))^{\lambda }$

• Why? Because the denominator $f_{Y}(X)$ is linked to the numerator via the Frobenius endomorphism (anti-trace).
• The value of $\lambda$: This is not a secret. It is an integer that depends solely on the curve parameters. On many BN curves, this is linked to $(1-p^{6})$.

The Polynomial Reduction: Instead of: :$Z_{Weil}={\frac {{\text{Unknown}}_{1}(Y)}{{\text{Unknown}}_{2}(Y)}}$ You have: :$Z_{Weil}={\frac {{\text{Unknown}}_{1}(Y)}{({\text{Unknown}}_{1}(Y))^{p^{6}}}}={\text{Unknown}}_{1}(Y)^{1-p^{6}}$ The Solution Path: Since $(1-p^{6})$ is a known constant integer, you can simply invert this exponent modulo $r$: :$f_{X}(Y)=(Z_{Weil})^{(1-p^{6})^{-1}{\pmod {r}}}$ ==== The $p^{6}$ Frobenius Trick (Conjugation) ==== In the field $\mathbb {F} _{p^{12}}$, the operation $x\mapsto x^{p^{6}}$ acts as a complex conjugation relative to the quadratic extension. * Property of the Result $Z$ (Pairing): : $Z$ is an $r$-th root of unity. On BN curves, modulo the order $r$, we have $p^{6}\equiv -1$. : Therefore, for any valid pairing result $Z$: :$Z^{p^{6}}=Z^{-1}$ * Property of the Denominator $B$: : Since $B\in \mathbb {F} _{p^{2}}$, it is invariant under $x\mapsto x^{p^{2}}$. A fortiori, it is invariant under $x\mapsto x^{p^{6}}$ (since $p^{6}$ is a multiple of $p^{2}$). :$B^{p^{6}}=B$

3. Solving the Equation

We have the following system: # $Z=A/B$ # $Z^{-1}=Z^{p^{6}}$ (Target group property) # $B^{p^{6}}=B$ (Subfield property) Apply the Frobenius $p^{6}$ to equation (1): :$Z^{p^{6}}={\frac {A^{p^{6}}}{B^{p^{6}}}}$ Substitute using properties (2) and (3): :$Z^{-1}={\frac {A^{p^{6}}}{B}}$ We now have two expressions involving $B$: * From (1): $B=A/Z$ * From derived equation: $B=Z\cdot A^{p^{6}}$ Equating the two expressions for $B$: :$A/Z=Z\cdot A^{p^{6}}$ :$Z^{2}=A/A^{p^{6}}=A^{1-p^{6}}$

4. The Practical Formula for Your Algorithm

You are looking for $A=f_{P}(Q)$ (the input for inversion) given $Z$ (your Weil target). The final equation is: :$A=Z^{\frac {2}{1-p^{6}}}$ *((The exponent is calculated modulo the group order $r$. Division by $1-p^{6}$ is multiplication by the modular inverse).)* Why does this work without knowing the denominator? Because the denominator $B$ is "eaten" by the operation. By squaring and playing with the conjugate, we exploit the fact that $B$ is "real" (invariant under $p^{6}$) while $A$ is "complex". This eliminates $B$ from the equation.

Summary

You have effectively reduced the "Weil Inversion Problem" to the "Miller Inversion Problem" using simple arithmetic. If your algorithm solves Miller Inversion in polynomial time, the entire Weil inversion becomes polynomial.

~2025-35962-07 (talk) 09:13, 24 November 2025 (UTC)

Why would you expect anything but garbage to come out of ChatGPT? ~2025-31850-11 (talk) 11:59, 24 November 2025 (UTC)

Because some actual experts suggested a similar thing https://crypto.stackexchange.com/questions/117412/can-this-algorithm-about-pairing-inversion-work-in-case-of-pairings-that-don-t-u#comment247437_117430?

Also the advandced model in addition to answer correctly on some mathematical terms did quoted papers on this discussion that achieve bilinear pairings wihtout final exponentiation but on hyperelliptic curves.

The quailty improved from garbage to half truths. ~2025-35962-07 (talk) 19:56, 24 November 2025 (UTC)

A mathematical argument with half-truths is garbage. ~2025-31850-11 (talk) 11:14, 25 November 2025 (UTC)

Since LLMs generate texts that statistically resemble human-produced texts, one should expect, given an LLM-generated text, to find human-produced texts that statistically resemble it The fact that a text statistically resembles human-produced accurate texts does not make it particularly more likely that it is accurate, unless there are many of such human texts, which, moreover, use similar wordings. This is the case if you ask for the capital of New Caledonia, or how to say "Merry Christmas" in Polish. I doubt that it is the case for your conjectured implication.  ​‑‑Lambiam 21:26, 25 November 2025 (UTC)

November 25

Euler necklaces

I wrote a program to find Euler circuits on the complete directed graph of n vertices. Now I want to eliminate isomorphisms. Such a circuit is a kind of bracelet (where bead colors represent vertices of the graph), and there probably exist algorithms to detect isomorphism between bracelets. I wonder whether it would be easier to generate all bracelets (of n-1 beads of each of n colors) and filter the list for the desired quality of including all sequences of two colors. —Tamfang (talk) 02:47, 25 November 2025 (UTC)

Doesn’t sound easier; why would you want to enlarge your set of objects to classify? ~2025-31850-11 (talk) 11:23, 25 November 2025 (UTC)

It could save time if the bracelet-generating algo inherently excludes isomorphs, somehow. —Tamfang (talk) 00:36, 26 November 2025 (UTC)

There exist, rather obviously, algorithms to detect that two Euler circuits on a given complete digraph are isomorphic (enumerate all $n!$ automorphisms of the digraph), so I assume the purpose is to find a faster method.

To enumerate the Euler circuits one would use a depth-first algorithm to search the tree of partial Euler paths, backtracking when no Eulerian continuation is open while the cycle is not yet complete. (The problem can also be viewed as finding Hamiltonian circuits in a directed graph whose nodes are formed by edges $(a,b)$ of the original graph and whose edges are the pairs of the form $((a,b),(b,c)),$ and then the search tree is a spanning tree of the new graph.) In enumerating the bracelets, one would, I presume, use a functionally equivalent algorithm, not waiting with filtering till completion of a bracelet, but not attempting to extend a partial bracelet ...AB...A with another B and backtracking when no lawful continuation is possible. So I think there is no gain. And in either case you still have to test for isomorphism.  ​‑‑Lambiam 21:10, 25 November 2025 (UTC)

It's not merely n! but 2n(n-1)n!, because the path can begin anywhere and I want to consider reversal equivalent. (My program reduces by n!, by requiring that the vertices' first appearances be in order.) —Tamfang (talk) 04:53, 28 November 2025 (UTC)

@Tamfang: May I check my understanding of the setting of your problem? Concretely, I suppose that here a "complete directed graph" refers to a tournament on a complete graph. Is that correct?

If so, indeed, by Rédei's theorem, there is a directed Hamilton path in the tournament (or complete directed graph); and any such path is naturally and uniquely divided into a sequence $T_{1},\ldots ,T_{r}$ of constituents of two kinds of induced subtournaments, namely, either consisting of just one vertex, of containing a directed Hamilton cycle; and such that for any vertices $s\in T_{i}$ and $t\in T_{j}$ with $i<j$ the edge between s and t is directed from s to t. Moreover, it is fairly easy to find these T_i, together with one Hamilton cycle for each (for the nontrivial ones). The main trouble I perceive would be that any such T_i (except the smallest ones) could be expected to contain a large number of different directed Hamilton cycles (even disregarding those which just are rotations, obtained by choosing a new 'first vertex'). Are these T_i your bracelets?

If i guessed right about the setting, so far, I think an important point would be whether or not you consider it probable to have many small bracelets (i.& nbsp;e., small T_i) in your application. My ten cent opinion is that in 'sufficiently random' situations you would rarely have more than one bracelet (i. e., you would have r=1, i. e., the graph itself would contain a directed Hamilton cycle); simply since there are fairly few ordered bipartitions of an n-set (actually $2^{n}-1$ many), and fairly low probability for any given one of them to have all edges between them directed 'to the right'. If you seldom or never accounts neclaces with some smaller size 'bracelets', you would gain little in preloading a list of these.

But perhaps you have reasons to believe different in your sought application (or I misunderstood your question). Regards, JoergenB (talk) 23:20, 28 November 2025 (UTC)

Since a directed graph is also called a digraph, I assume that "complete directed graph" means the same as "complete digraph", with twice the number of edges of the tournament for the same number of vertices.  ​‑‑Lambiam 00:23, 29 November 2025 (UTC)

Lambiam reads me right. —(the user formerly known as Tamfang) —Antonissimo (talk) 01:13, 29 November 2025 (UTC)

Seems to me that most tournaments would have no Euler path. —Antonissimo (talk) 01:15, 29 November 2025 (UTC)

Eulerian, not Hamiltonian. ~2025-31850-11 (talk) 12:16, 29 November 2025 (UTC)

Thank you for pointing out that distinction! —Antonissimo (talk) 03:50, 30 November 2025 (UTC)

November 26

Busy Beaver numbers, Rayo's number, inaccessible cardinals

Aren't Rayo numbers(using arbitrary arguments instead of a googol),conceptually the same as Busy Beaver numbers, except using logical symbols instead of theoretical computer science? Also, it seems that although infinite instead of finite, inaccessible cardinals are conceptually similar in that they use the idea of defining them as what one cannot "get to" using all previously used techniques.Rich (talk) 20:33, 26 November 2025 (UTC)

There is a conceptual similarity. You may find the essay "Who Can Name the Bigger Number?" interesting. It was penned years before Rayo offered his solution. See also List of numbers/Uncomputable numbers on the (sometimes somewhat cryptic) Googology Wiki.  ​‑‑Lambiam 23:54, 26 November 2025 (UTC)

November 27

Hasse diagram of the "is a divisor of" relation

1. For which positive integer n, the Hasse diagram of all positive divisors of n ordered by the "is a divisor of" relation is a planer graph?
2. If n is a positive integer, and the Hasse diagram of all numbers >=1 and <=n ordered by the "is a divisor of" relation is a planer graph, what is the largest possible value of n?

~2025-36798-45 (talk) 20:21, 27 November 2025 (UTC)

The Hasse diagram based on the divisors of a positive integer is constructed as follows:

1. For a prime power $p^{a}$, the diagram is effectively a line with $a+1$ vertices, e.g. 1-2-4-8 having 4 vertices.

2. For a product of $n$ prime powers $p_{1}^{a_{1}}...p_{n}^{a_{n}}$, it is the Cartesian product of the $n$ line graphs corresponding to the contained factors.

If your number has two prime power factors, then this graph is pretty much just a grid, and obviously planar. If your number has four prime factors, then it must admit the non-planar hypercube graph $Q_{4}$ as a subgraph. If your number has exactly 3 prime factors then it seems to be a bit complicated, and I will have to reflect on this a bit. GalacticShoe (talk) 23:14, 27 November 2025 (UTC)

Doesn't $Q_{3}$ contain the "utility graph" $K_{3,3}$ as a subgraph?  ​‑‑Lambiam 00:22, 28 November 2025 (UTC)

$Q_{3}$ should just be the graph of a 3D cube, which is planar (e.g. File:Cube skeleton.svg) GalacticShoe (talk) 00:57, 28 November 2025 (UTC)

Okay, so when there are three prime factors, if at most one of them has a power above 1, then it is planar; this can be seen as drawing the graph is a series of nested squares. In the article for Hasse diagrams, the example given for factors of 60 is a graph of this form that can be rearranged into a planar graph. Meanwhile, I think the graph corresponding to a number $p_{1}^{2}p_{2}^{2}p_{3}$ (essentially four cubes pasted together in a 2 by 2 grid) is nonplanar, but I can't find a convenient online tool to test planarity, and I'm having an annoying time trying to find a $K_{3,3}$ or $K_{5}$ minor a la Kuratowski's theorem. If it is the case that this is nonplanar, then the criterion would be as follows:

The Hasse diagram of the divisors of a positive integer is planar if and only if the integer either has at most 2 prime factors, or it has exactly 3 prime factors of which only one may have a power greater than or equal to 2.

Or more succinctly,

The Hasse diagram of the divisors of a positive integer is planar if and only if the integer has at most 4 semiprime divisors.

GalacticShoe (talk) 00:51, 28 November 2025 (UTC)

This online SageMath program shows that indeed the graph is nonplanar, confirming the condition mentioned. The list of positive integers with nonplanar divisors Hasse diagram, i.e. the list of positive integers with at least 5 semiprime divisors, starts 180, 210, 252, 300, 330, 360, 390, 396, 420, 450, 462, 468... GalacticShoe (talk) 02:44, 28 November 2025 (UTC)

As for the latter question, the largest value for which the Hasse diagram is still planar is 27. When you add in edges for (4,28) and (14,28), it breaks down, as per this online SageMath program. GalacticShoe (talk) 08:05, 28 November 2025 (UTC)

December 2

I need help here

Hi all, I am stuck on this math problem:

Consider the function

$f(x)={\frac {3.14^{x}\cdot x^{5/2}}{x^{2}+0.5}}$

A space curve $C$ in $\mathbb {R} ^{3}$ is parameterized by

$\mathbf {r} (t)=(x(t),\ y(t),\ z(t))=(t,\ t^{3},\ {\sqrt {t+1}}),\quad t\geq 0.$

The density along this curve at parameter $t$ is given by $\rho (t)=f(t)$.

Compute the exact value of the limit

$\lim _{b\to +\infty }{\frac {1}{b}}\int _{0}^{b}{\frac {3.14^{t}\cdot t^{5/2}}{t^{2}+0.5}}\,dt,$

which represents the long-term average density along the curve as $b\to +\infty$.

Can I just use the average value theorem for integrals? Like, isn’t the limit of ${\frac {1}{b}}\int _{0}^{b}{\frac {3.14^{t}\cdot t^{5/2}}{t^{2}+0.5}}\,dt,$ equal to something in the middle? I would greately appreciate your help here. Thank you. ExtraTerrestrial120 (talk) 14:21, 2 December 2025 (UTC)

The mean value theorem is not going to help. It tells you that some point exists, but not how to find it.

Before we can try to help you, could you tell us where the original problem came from and how it was formulated? This specific problem looks utterly implausible, both as an exercise and as a practical problem.  ​‑‑Lambiam 15:42, 2 December 2025 (UTC)

Is not this limit just zero? Ruslik_Zero 20:42, 2 December 2025 (UTC)

I think it is a bit more. If $t>1,~f(t)>2^{\,t{-}1}.$  ​‑‑Lambiam 23:22, 2 December 2025 (UTC)

The limit is in fact quite a bit more. In the interest of not spoiling the answer to what I presume is a homework question, I'll simply mention that when $t>{\sqrt {1/2}}$, $2t^{2}>t^{2}+0.5$, and when $t>1$, ${\sqrt {t}}>1$. GalacticShoe (talk) 23:53, 2 December 2025 (UTC)

My "a bit more" was meant as an understatement. This may originate from a homework question, but if so, and if this is the original homework question, it is a very strange one, bizarrely convoluted to drive a simple point home. Note that the form of the space curve, and in fact everything before and after the imperative "Compute ... $dt$", is irrelevant to the problem as posed.  ​‑‑Lambiam 06:59, 3 December 2025 (UTC)

It’s also not true that the given integral computes the average density along the curve. ~2025-31850-11 (talk) 11:22, 3 December 2025 (UTC)

I was wondering about that myself. Why is the information about the curve even included in the problem? --RDBury (talk) 17:07, 3 December 2025 (UTC)

I just misread the limit. I thought it b->0. 20:33, 3 December 2025 (UTC)

If b were going to 0, this would be a better question (about the definition of the derivative and the fundamental theorem of calculus—still no meaningful connection to density on a curve, though). ~2025-31850-11 (talk) 02:42, 6 December 2025 (UTC)

December 6

prime divisors of 2^(n+1) choose 2^n, also known as C(2^(n+1),2^n)

when n=13, wolfram alpha says there are 1293 prime distinct prime divisors of N=C(16384,8192). According to the OEIS, the number of primes less than 16384 is 1900, and the number of primes less than 8192 is 1028. So there are 607 primes less than 16384 that don't divide N. But it's known that all primes between 8192 and 16384 will divide N. So the 607 missing primes are from the 1028 primes less than 8192. Now, 1028/(1028-607) is about 2.45. This ratio has been inching upward from smaller values of n. As n->oo, does the ratio go to e~2.71828182846? Rich (talk) 21:46, 6 December 2025 (UTC)

Determining whether a given prime p divides C(m,n) is not too hard: Write out m and n base p, then p|C(m,n) iff each digit in the expression for m is ≥ the corresponding digit in the expression for n. For example does 11 divide C(16384,8192)? 16384 base 11 is 11345₁₁ and 8192 base 11 is 6178₁₁. In the last digit, 8>5 so 11 divides C(16384,8192). The smallest prime that does not divide C(16384,8192) is 61. Note that if p is between 8192 and 16384 then 8192 base p is the single digit 8192 and 16384 base p is 1,16384-p, and 8192>16384-p so p will divide C(16384,8192). On the other hand, if p is between 16384/3 and 8192 then p does not divide C(16384,8192).

I think finding the limit as P→∞ of the proportion of p's < 2P where p divides C(2P,P) would be difficult. You could start with π(2P)-π(P) as first approximation for the number of primes, where π is the Prime-counting function. But you're interested in the relative error of this approximation. You could get additional terms for a better approximation by ignoring primes < √2P (assuming π(√(2P)) is small compared to π(2P)), and looking at the first digit of 2P base p. Then p is included if this digit is odd and excluded if it's even. We then get an alternating series

S = π(2P)-π(2P/2)+π(2P/3)-π(2P/4)+π(2P/5)-π(2P/6)...

and would like the limit of S/π(2P). But I don't know enough about the asymptotics of π to answer this. In any case, I'd be very surprised if the limit is something nice like e. There are many numbers > 2.45, and there are functions that increase very slowly but which don't converge to a limit at all. --RDBury (talk) 10:16, 7 December 2025 (UTC)

It's actually Pi(2^n)/(primes less than 2^n that divide C(2^(n+1),2^n)) that I'm interested in.Rich (talk) 17:22, 7 December 2025 (UTC)

Need for a kind of discrete integral

I need a kind of discrete integral, with a sum that progresses in finite steps like in a computer language , whereas an integral progress in infinitesimal steps.
So I have 2 functions:
$\lambda (i)=i\cdot \Delta \lambda +{\frac {\Delta \lambda }{2}}$
and $B(\lambda )$
I would like to replace the following integral:
$\int _{a}^{b}B(\lambda )\cdot d\lambda$
By the following summation:
$\Sigma _{i=a}^{b}B(\lambda (i))\cdot \Delta \lambda$
I suppose I can also write it theoretically:
$\Sigma _{i=0}^{\infty }B(\lambda (i))\cdot \Delta \lambda$
Therefore, I get a summ, not of infinitesimal areas, but of finite areas.
Does this conform to discrete mathematics, or is there a different notation and if so, what is it?
Malypaet (talk) 23:11, 6 December 2025 (UTC)

Isn't this essentially the question you asked on 20 October with a follow-up on 22 October?

It seems that what you want to do is a crude form of numerical integration; see in particular Numerical integration § Quadrature rules based on step functions. This corresponds to the paragraph in my response of 22 October that starts with "A (not very sophisticated) way to compute an approximation of $\textstyle \int _{a}^{b}\,f(v)\,dv~$", especially the part where I give the formula for when all segments are given equal width.  ​‑‑Lambiam 23:42, 6 December 2025 (UTC)

Yes, but your answers don't satisfy me, because I'm studying a case where the step size is finite and cannot correspond to an integral, where the step size is infinitesimal and applicable neither experimentally nor in computer calculations.

Therefore, I'm not looking for an approximate solution based on an integral, but rather a summation of steps of finite size, regardless of whether the number of steps is finite or infinite.

So I'm looking for a formulation based on summation, respecting mathematical notation in terms of function summation.

My question is whether my proposal is acceptable or needs to be modified to comply with the mathematical rules for summation of functions.
Malypaet (talk) 14:55, 7 December 2025 (UTC)

Have you looked at Summation § Capital-sigma notation?

If you'd write the sum as $v(a)+v(a{+}1)+\cdots +v(b),$ using an ellipsis, the capital-sigma notation is $\textstyle {\sum _{i=a}^{b}v(i)}.$ So, the other way around,

$\Sigma _{i=a}^{b}B\left(\lambda (i))\cdot \Delta \lambda =(B(\lambda (a))+B(\lambda (a{+}1))+\cdots +B(\lambda (i)\right)\cdot \Delta \lambda .$

BTW, it is somewhat unusual to use the notation $\Delta \lambda$ for a variable. A more common use is that it is shorthand for a difference between two changeable quantities, as in $\Delta \lambda =\lambda _{1}-\lambda _{0}.$ For a variable used for a stepwise change, as here, the letter $h$ is conventionally used.

Given the definition $\lambda (i)=i\cdot h+{\frac {h}{2}},$ the above sum comes out as

$\left(B(a\cdot h+{\frac {h}{2}})+B((a{+}1)\cdot h+{\frac {h}{2}})+\cdots +B(b\cdot h+{\frac {h}{2}}\right)\cdot h.$

If that is what you want, it is what you get.  ​‑‑Lambiam 17:55, 7 December 2025 (UTC)

Thanks for the link, it's interesting. But unfortunately, Wikipedia often refers back to integrals.

I found a more relevant explanation elsewhere: Discrete Calculus

So my formula actually corresponds to a discrete integral.

And you are right, it is what I want and get.

However, regarding the unusual use of $\Delta \lambda$, my need is ambiguous.

Actually, I'm using it as a parameter, that is, a constant whose value changes depending on the context:

i.e. the function's domain.

But I don't yet know how to introduce this uncommon usage in my text.

Have you an idea ?
Malypaet (talk) 18:32, 7 December 2025 (UTC)

Is indefinite sum relevant (to the original question)? catslash (talk) 18:44, 7 December 2025 (UTC)