Wikipedia:Reference desk/Mathematics#Heat equation on the ice rink

For an unbalanced channel, where the probability of 0 is p₀, and the probability of 1 is p₁, with p₀+p₁=1, the Entropy (aka Shannon Entropy) is given by:

:$-p\_0\; \backslash ln(p\_0)-p\_1\; \backslash ln(p\_1)\; .$

(Here the units are nats, but you can multiply by the appropriate constant to get bits or dits.) This is covered in the article. Now suppose that instead of getting a clear signal each time, you only get one with probability p_S. If there is no signal then there is no information received, and so the entropy is 0. Given that there is a signal, the probabilities of getting is 0 and 1 are

:$p\_\{0|S\}=\{p\_0\; \backslash over\; p\_S\},\; \backslash \; p\_\{1|S\}=\{p\_1\; \backslash over\; p\_S\}$

respectively, where the probability of 0 is p₀, and probability of 1 is p₁, with p₀+p₁=p_S. This gives a revised entropy of

:$p\_S\; (-p\_\{0|S\}\; \backslash ln(p\_\{0|S\})-p\_\{1|S\}\; \backslash ln(p\_\{0|S\}))\; .$

With a bit of algebra, this is

:$=\; -p\_0\; \backslash ln(\{p\_0\; \backslash over\; p\_S\})-p\_1\; \backslash ln(\{p\_1\; \backslash over\; p\_S\})$

:$=\; -p\_0\; \backslash ln(p\_0)\; +\; p\_0\; \backslash ln(p\_S)\; -\; p\_1\; \backslash ln((p\_1)\; +\; p\_1\; \backslash ln(p\_S)))$

:$=\; -p\_0\; \backslash ln(p\_0)\; -\; p\_1\; \backslash ln(p\_1)\; +\; p\_S\; \backslash ln(p\_S).$

This agrees with some numerical experiments I've done, but the last term seems counterintuitive to me. In terms of a weighted average, the weights p₀ and p₁ are already less that the original values by a factor of p_S < 1. But on top of this seeming reduction, you have to reduce the value again by the last term, and it seems like it shouldn't be included. I don't know anything about information theory other than what's included in the article, so is the entropy of a partial or lossy channel even a thing that's studied? Does this formula make sense and does it already exist in the literature? --RDBury (talk) 00:06, 9 June 2025 (UTC)

::When I do the maths, I get $\{\}-p\_0\backslash ln(p\_0)-p\_1\backslash ln(p\_1)+\backslash ln(p\_S)$ for the "revised entropy", which is not bounded from below.  ​‑‑Lambiam 06:49, 9 June 2025 (UTC)

:::I'm pretty sure the last term is $p\_S\; \backslash ln(p\_S)$; remember p₀+p₁=p_S here. In the usual case p_S = 1, you get a value of 0 for $p\_S\; \backslash ln(p\_S).$ --RDBury (talk) 17:40, 9 June 2025 (UTC)

:The signal on the lossy channel is the bitwise $\backslash mathsf\{AND\}$ of the original channel and that of the interruptor. The situation is in fact symmetrical: the interruptor could be sending a message through their channel, so the interruptions also carry some entropy.

:The probabilities of $\backslash mathsf\; 0$ and $\backslash mathsf\; 1$ on the lossy channel are $q\_0=1\{-\}q\_1,\; q\_\{\backslash mathsf\; 1\}=p\_1p\_S.$ I'd say that the Shannon entropy is then

::$\{\}-q\_0\backslash log(q\_0)-q\_1\backslash log(q\_1).$

: ​‑‑Lambiam 06:37, 9 June 2025 (UTC)

::I'm struggling with this myself, and perhaps I'm oversimplifying the situation because the actual situation is pretty involved. But if what you're saying is true, then look at the case where p_0|S = 0, p_1|S = 1 and p_S = 1/2. So the sender is just transmitting an endless stream of 1's which has 0 information, and the difference is that the receiver is only getting the 1's half the time. So the information being sent is 0, and the loss of signal certainly doesn't add information to it, so the received information should be 0. According to your formula, q₁ = 1/2, q₀ = 1/2, and the received information works out to ln 2 or 1 bit per symbol received. In the way I've worked it out, you get p₀ = 0, p₁ = 1/2, p_S = 1/2, and the received information is ln 2 - ln 2 = 0. You're assuming that the loss of signal is really a second sender transmitting a stream of information that's combined with the original stream.

::Look at a special case where the loss of signal is independent of what is being sent. Let's say that the sender is sending 0 with probability s₀ and 1 with probability s₁ with s₀ + s₁=1. And, as before, the probability of the signal being received is p_S. Then p₀ = s₀p_S and p₁ = s₁p_S, and if you plug these into the formula given above, the result comes out as:

:::$p\_S(\; -s\_0\; \backslash ln(s\_0)-s\_1\; \backslash ln(s\_1))\; .$

::I think this is to be expected; if the signal only gets through 80% of the time then the rate of information will be reduced by a factor of .8. I'm trying to allow for more general possibilities, for example if a loss of signal is more likely if it's a 1 than if it's a 0. Or it may be more complicated, say if the loss of signal only occurs on the bit after a 1 is sent. I can imagine even more complex scenarios, and I'm trying to get a formula that covers most of them. --RDBury (talk) 18:45, 9 June 2025 (UTC)

:::I assumed that if the signal on the original channel is {{mono|01110010}} and the on-off of the connection is {{mono|01001011}}, that the result was {{mono|01000010}}, the same length as the orginal but with some information-carrying bits blinded by being replaced by {{mono|0}}'s. I think my assumption may have been wrong and that the intention of the question was not just that the information carried by the censored bits was lost, but that the bits themselves were dropped, resulting in the shorter signal {{mono|1010}}.

:::Regardless, the "loss of signal" could be a steganographic signal, possibly a covert channel, disguised as randomly unreliable hardware. For example, if the original message is plain text and losses are relatively infrequent, the original can be recovered from the lossy signal with good probability, and so an interceptor can determine the interval lengths between losses, which can be used to encode a hidden message. Therefore we cannot say that the loss of signal doesn't convey information.

:::If the original signal was one bit long, the signal received can be {{mono|ε}} (zero bits), {{mono|0}} or {{mono|1}}, with probabilities:

:::: {{mono|ε}}: {{quad}}$1\{-\}p\_S$

:::: {{mono|0}}: {{quad}}$p\_0p\_S$

:::: {{mono|1}}: {{quad}}$p\_1p\_S$

:::Assuming the receiver knows the original length, the entropy is then

::::$\{\}-(1\{-\}p\_S)\backslash log(1\{-\}p\_S)-p\_0p\_S\backslash log(p\_0p\_S)-p\_1p\_S\backslash log(p\_1p\_S).$

:::This can be rewritten as:

::::$p\_S(\{\}-p\_0\backslash log(p\_0)-q\_0\backslash log(q\_0))-p\_S\backslash log(p\_S)-(1\{-\}p\_S)\backslash log(1\{-\}p\_S).$

:::Note that this exceeds the original entropy $\{\}-p\_0\backslash log(p\_0)-q\_0\backslash log(q\_0)$ if

::::

:::To get the average per bit received, divide by $p\_S.$ This is always as high as or higher than the original entropy.

:::Similarly, if the original signal was two bits long, the signal received can be {{mono|ε}} (zero bits), one bit or two bits, with probabilities:

:::: {{mono|ε}}: {{quad}}$(1\{-\}p\_S)^2$

:::: {{mono|0}}: {{quad}}$2p\_0p\_S(1\{-\}p\_S)$

:::: {{mono|1}}: {{quad}}$2p\_1p\_S(1\{-\}p\_S)$

::::{{mono|00}}:{{quad}}$p\_0^2p\_S^2$

::::{{mono|01}}:{{quad}}$p\_0p\_1p\_S^2$

::::{{mono|10}}:{{quad}}$p\_0p\_1p\_S^2$

::::{{mono|11}}:{{quad}}$p\_1^2p\_S^2$

:::The entropy is then $\backslash textstyle\backslash sum-p\backslash log(p),$ with $p$ ranging over these 7 probabilities (which add up to $1$). To get the average per bit sent, divide by $2,$ and per bit received, divide by $2p\_S.$

:::The same analysis can be done for longer signals, and I expect that the entropy per bit, if not already constant, tends to a definite limit, but I don't see a ready method for deriving it.  ​‑‑Lambiam 14:43, 11 June 2025 (UTC)

::::I think you're trying to compute something different from what I'm trying to compute. If the entire signal is lost then you're getting none of the original signal so the information received is 0. In any case, I'm getting different probabilities from what you're getting, for example I compute the probability of 01 to be p₀p₁; I think you may be using p₀ for what I'm calling p_0|S. If you work out the sums you get p₀ ln(p₀) + p₁ ln(p₁) + q_S ln(q_S) per bit, which is the entropy you'd get if it was a signal with a third possible value occurring with probability q_S = 1- p_S). But again, you're counting the loss of signal as an new signal with information, but it only represents the lack of information. I'm not trying to compute the average per bits received, but the average per bits sent.

::::Perhaps it would be better to consider the following scenario, closer to the situation I'm trying to model. The sender sends 0's and 1's with certain probabilities, say s₀ and s₁ as above, but when the receiver receives a 1 it overheats and does not function again until the sender sends another 0. During this time it receives no information, and the receiver records a 0, and keeps recording 0's until the sender sends a 0 followed by another 1. To compute the, let's call it for sample information, of a string of n bits, you have to include the previous bit. So for the sample information given by a single bit you have to consider four combinations of bits sent. The only way a 1 can be received is for the bits sent to be 01, and the entropy per bit works out to:

:::::$-(s\_0\; s\_0+s\_0\; s\_1+s\_1\; s\_1)\backslash ln(s\_0\; s\_0+s\_0\; s\_1+s\_1\; s\_1)-s\_0\; s\_1\; \backslash ln(s\_0\; s\_1).$

::::When you increase the sample size to 2 you have to sum over 8 sent bit combinations, and the result is:

:::::$(-(s\_0\; s\_0\; s\_0+s\_0\; s\_0\; s\_1+s\_0\; s\_1\; s\_1+s\_1\; s\_1\; s\_1)\backslash ln(s\_0\; s\_0\; s\_0+s\_0\; s\_0\; s\_1\; s\_1+s\_0\; s\_1\; s\_1+s\_1\; s\_1\; s\_1)-2\; (s\_0\; s\_0\; s\_1+s\_0\; s\_1\; s\_1)\backslash ln(s\_0\; s\_0\; s\_1+s\_0\; s\_1\; s\_1)\; )/2.$

::::This turns out to be slightly less than the previous expression.

::::When you increase the size to 3, the result is again slightly smaller but seems to be converging to a limit:

:::::$(-(s\_0\; s\_0\; s\_0\; s\_0+s\_0\; s\_0\; s\_0\; s\_1+s\_0\; s\_0\; s\_1\; s\_1+s\_0\; s\_1\; s\_1\; s\_1+s\_1\; s\_1\; s\_1\; s\_1)\; \backslash ln(s\_0\; s\_0\; s\_0\; s\_0+s\_0\; s\_0\; s\_0\; s\_1+s\_0\; s\_0\; s\_1\; s\_1+s\_0\; s\_1\; s\_1\; s\_1+s\_1\; s\_1\; s\_1\; s\_1)$

::::::$-2(s\_0\; s\_0\; s\_0\; s\_1+s\_0\; s\_0\; s\_1\; s\_1+s\_0\; s\_1\; s\_1\; s\_1)\; \backslash ln(s\_0\; s\_0\; s\_0\; s\_1+s\_0\; s\_0\; s\_1\; s\_1+s\_0\; s\_1\; s\_1\; s\_1)$

::::::$-(s\_0\; s\_0\; s\_0\; s\_1+2s\_0\; s\_0\; s\_1\; s\_1+s\_0\; s\_1\; s\_1\; s\_1)\; \backslash ln(s\_0\; s\_0\; s\_0\; s\_1+2s\_0\; s\_0\; s\_1\; s\_1+s\_0\; s\_1\; s\_1\; s\_1)$

::::::$-s\_0\; s\_0\; s\_1\; s\_1\; \backslash ln(s\_0\; s\_0\; s\_1\; s\_1))/3.$

::::Using the ideas in the original post I get that the limiting value should be

::::::$-(s\_0\; s\_0+s\_1\; s\_1)\; \backslash ln(s\_0\; s\_0+s\_1\; s\_1)-s\_0\; s\_1\; \backslash ln(s\_0\; s\_1)+(s\_0\; s\_0+s\_0\; s\_1+s\_1\; s\_1)\; \backslash ln(s\_0\; s\_0+s\_0\; s\_1+s\_1\; s\_1).$

::::I used a spreadsheet to plot all four curves, and the plots do seems consistent with last one being the limit of the first three. In this case

::::::$p\_0\; =\; s\_0\; s\_0+s\_1\; s\_1,\; p\_1\; =\; s\_0\; s\_1,\; p\_S\; =\; s\_0\; s\_0+s\_0\; s\_1+s\_1\; s\_1.$

::::The last equation reflects that the bit received following a 1 must be 0, so this information is lost every time a 1 appears, and this happens with frequency s₀s₁. (If this seems confusing, I did say above that I was trying to simplify since the actual situation is pretty involved, and I was trying to avoid going into the gory details, but perhaps I didn't explain it well. And yes the reasoning seems pretty sketchy, but my previous attempts, following what seemed like more intuitive methods, didn't come close to the apparent limiting value. That's why I posted the question. We're supposed to be using logic, not induction. Perhaps the general situation might be described as a Turing machine which can only move left to right. It receives an input signal as a string of 0's and 1's placed on the tape, and the output signal is what is left behind after the machine passes. The scenario just described can be modeled with a 2 state machine.) --RDBury (talk) 23:10, 11 June 2025 (UTC)

:::::PS. After a third and fourth look at the reasoning I've come to the conclusion that the formula given above is, in fact, wrong in the previous example. I'm still trying to simplify expressions, but it appears that the actual value is an infinite sum which can't be reduced to closed form. And I think that the error was, in essence, that I was taking E(ln(X)) = ln(E(X)). That might be a decent approximation in some cases, which is why the result seemed to fit the computed approximate values, but it's not equality. Anyway, thank you {{ping|Lambiam}} for the lively discussion; sometimes arguing your point of view leads to the realization that your point of view isn't correct after all. --RDBury (talk) 02:18, 16 June 2025 (UTC)

Do the interval signatures introduced in Orthogonal coordinates#Table of three-dimensional orthogonal coordinates have any significance? Are they OR? They seem to be badly defined. Thanks. catslash (talk) 21:48, 9 June 2025 (UTC)

:I think the meaning is, well not really clear, but something you can figure out from the example. It uses C or O for each end of an interval, with C meaning the interval is closed at that endpoint, and O if the interval is open at the endpoint. So [0, 1] is CC, [0, ∞) is CO, and (−∞, ∞) is OO. If the domain of the coordination system is given as the Cartesian product of three intervals then the signature uses a letter pair for each interval. So COxCCxCO for spherical, COxCOxCO for parabolic, and COxOOxOO for bipolar cylindrical coordinates. I think the intent was to organize the table in a way that's a bit more logical than alphabetical by name. And since it's just being used as a way to organize existing information rather than being included as new information, I don't know if it should really count as OR. The term is certainly unsourced, which is sketchy, and not well explained, also sketchy, and I haven't checked to see if people are actually following this scheme when adding new entries, though this seems unlikely. But the person who inserted this scheme, {{ping|DGoncz}}, does seem to be active editor for this page so I'd do some consensus building before just removing it. In any case, this isn't really the forum for discussing whether content should be added or removed; that's the kind of thing WT:WPM covers. I will say though, that in general I'm not a fan schemes where only a few people understand the "proper" way to add new information to an article. As it is, to add a new entry to that table you'd need working knowledge of Wiki's version of TeX and table markup, we don't need to make it even more difficult. --RDBury (talk) 05:39, 10 June 2025 (UTC)

::Thank you for responding. Yes, the meaning is clear enough to me. Also, the matter has already been raised on the article talk page Talk:Orthogonal coordinates#explanation of notation not clear, without any response from DGoncz. Those are separate issues from what I am asking about here. What I am asking here is: are interval signatures of coordinate systems a recognized concept or just something that was made up for the purpose of the WP article? Also, if they are a recognized concept, then what is their mathematical significance? catslash (talk) 10:24, 10 June 2025 (UTC)

:::All I can say I that it's unsourced in the article, I've never heard of it, and it seems that Google has never heard of it. You can't really prove a negative when it comes whether something is a recognized concept, that's why WP presumes OR until a source is provided. --RDBury (talk) 00:04, 11 June 2025 (UTC)

::::@RDBury Thanks for that input. catslash (talk) 18:41, 16 June 2025 (UTC)

::If the answers is perpendicular to the plane of the circle, translating produces cylindrical coordinates.

::Giankoplis explains the four regions or intervals within which the confocal elliptical coordinates are not degenerate, and within those four regions of course there are three point at which the solutions are degenerate. Four intervals times three affine transfornms is 12. 13th is Cartesian.

::Combinatorially, guided by the work of, and by some conversations with Dr Giancarlo Rota, we can start by saying since these are 3D orthogonal coordinates that each has three variables. And so we ask are they rotational or translational variables. And we ask is the associated interval, quite aside from any concern for generating a finite volume, a closed, left closed, right clothed, or open interval. 4 interval types times 3 variable is 12. Again, Cartesian as re the 13th.

::Math world has some interesting reading on scale factors. Sure, we can translate to and from Cartesian coordinates. But what about the 144 other possibilities? The reason I put the ordering there, the way it was that day, is it so that someone else... Don't think it would be me... Can fill in the other 144 possibilities a partition of the group of 12 at a time. You see if you order them a certain way, the variables naturally fall into a certain few patterns. They are not ordered that way already. Imm inently that is the next step. But then we need to draw the partitions. Then we need to say out of 144 theoretical translations, how many have we found matching intervals of similar nature to intervals of similar nature in an entirely different coordinate system?

::Lastly I should mention I have a transcript of a chat GPT log. GPT and I it started with extraordinary symmetry 64 theoretical coordinate systems. Working together when we identified 13. Then we looked at the 14th and the 15th all the way up to 27 I think it was. I mean I'm just typing this in for memory and using my voice by the way. would the Wikipedia system for comments... even parse such a file?

::Good night for now sorry I didn't check in more often. Doug Goncz (talk) 03:58, 16 June 2025 (UTC)

= June 11 =

I just encountered the concept of the Hamming distance for the first time. I see that it is the number of positions at which the corresponding symbols are different, and the minimum number of substitutions required to change one string into the other, and the minimum number of errors that could have transformed one string into the other.

What is the Hamming distance between 0010 and 0100? Is it two, since a digit-by-digit replacement forces us to change first to 0000 or 0110, and second to 0100? Or is it one, since each one has three 0s and a 1, and the only difference is a single error of sequence? Nyttend (talk) 00:29, 11 June 2025 (UTC)

:It's two. There are two positions at which the strings differ (the second and third positions).--Antendren (talk) 05:29, 11 June 2025 (UTC)

= June 12 =

I know they are better algorithms. But I want to solve a discrete logarithm in a finite field having a finite field of several Kb long and where the discrete logarithm solution lies into a 200bits subgroup.

The problem of such finite fields is there’s no birational equivalence to finite rings : such finite field element are polynomials. In such a case, what does it means for a finite field element to be smooth ? How do you achieve factorization into prime elements in such a case ? 2A01:E0A:ACF:90B0:0:0:A03F:E788 (talk) 21:27, 12 June 2025 (UTC)

= June 13 =

For the decimal expansion of n² to contain no digits other than 1, 4 and/or 7, what constraints are there on n? Thanks, cmɢʟee⎆τaʟκ 15:56, 13 June 2025 (UTC)

:The sequence of such $n$ is OEIS:A053898, while the sequence of their corresponding $n^\{2\}$ is OEIS:A053899. I'm searching for a pattern/constraints on $n$. GalacticShoe (talk) 17:04, 13 June 2025 (UTC)

::Thanks very much. Amazing it's already in the OEIS! cmɢʟee⎆τaʟκ 22:51, 13 June 2025 (UTC)

:::I think part of the appeal is the rarity. There are no squares containing only one digit other than the trivial single-digit squares, and the sequence of squares containing at most two digits, OEIS:A018885, seems to eventually become all multiples of powers of 10 by 1, 4, and 9 after $6661661161\; =\; 81619^\{2\}$. Squares containing exactly three distinct digits would be a natural place to continue. GalacticShoe (talk) 01:32, 15 June 2025 (UTC)

::::It is currently unknown, though, whether A018885 contains only a finite number of members without 0 tail.

::::An element of the ring of n-adic numbers (a field when n is prime) can be viewed as an infinite path on a rooted tree, in which the nodes of the path correspond to the tail segments of the element. One can do a breadth-first tree unfolding of the 10-adic numbers whose squares (or cubes etc.) use only a small finite number of digits. For example, for the case of at most two distinct digits, the square of ···1619 is ···1161, whose tail at the same length uses just two digits. So 1619 is a node in the tree, at depth level 4. For which d do the last five digits of ···d 1619 use only digits 1 and 6? Well, ···31619² = ···61161, while also ···81619² = ···61161. All other choices for d break the just-1-and-6-ness. So the tree node 1161 sprouts two branches, to nodes 31619 and 81619. Continuing, we find that ···331619² = ···161161 and ···831619² = ···161161, so 31619 sprouts branches to 331619 and 831619. Likewise, node 81619 sprouts branches to 081619 and 581619.

::::A node may fail to sprout branches. For the one-digit case, ···12² = ···44, but no ···d 12 squares to ···444, so it is a dead end. In fact, here all paths terminate except for ···00000000000. Some nodes are bingo nodes, like for the two-digit case the node 81619, since 81619² = 6661661161 in ordinary arithmetic.  ​‑‑Lambiam 08:45, 15 June 2025 (UTC)

= June 16 =

As far I understand, when it comes to finite fields, Pollard rho and Pollard’s lambda are still the best algorithm for solving discrete logarithms in a multiplicative subgroup/suborder…

Index calculus methods like ɴꜰꜱ can be made to target multiplicative subgroups, but they don’t work in multiplicative subgroups : given a finite fields with a very large characteristic which is enough larger than the target subgroup, they can have an higher complexity than using a Pollard’s method.

Therefore, if we discard quantum computers and Oracle based situations, there’s no known sub‑exponential algorithm that can take real advantage from knowledge the solution of a specific discrete logarithm is below a specific integer.

This even lead to cryptographic systems where the publicly known and used finite field’s order is relatively small for a would be sub‑exponential algorithm but the finite field itself is too large for index calculus (still used today).

Now is it because of lack of research or is there a reason in number theory that prevents creating an algorithm having at least a time and space complexity ? 2A01:E0A:ACF:90B0:0:0:A03F:E788 (talk) 23:15, 16 June 2025 (UTC)

:: There is a theorem of Victor Shoup that this is not possible. Tito Omburo (talk) 23:45, 16 June 2025 (UTC)

:::Where to find it ? 2A01:E0A:ACF:90B0:0:0:A03F:E788 (talk) 00:44, 17 June 2025 (UTC)

:::What’s the name of the article from the author ? 2A01:E0A:ACF:90B0:0:0:A03F:E788 (talk) 20:57, 18 June 2025 (UTC)

:::: I do not know the name. Shoup allegedly "proves" that in a cyclic group where multiplication is given by an oracle, it is not possible to achieve better than $O(\backslash sqrt\; p)$ time for the discrete logarithm problem. As far as I understand the question, this was it. Multiplicative groups in finite fields have more structure than this, but that structure ultimately reduces to discrete log in the multiplicative subgroups per Pohlig-Hellman. These Pohlig-Hellman subgroups are too "rarified" to expect anything more than Shoup's result. Caveat: I think Shoup is wrong, and all of the complexity estimates are bunk, but that's just me. Tito Omburo (talk) 21:12, 18 June 2025 (UTC)

:::::Here's the paper from his website: [https://www.shoup.net/papers/dlbounds1.pdf] 173.79.19.248 (talk) 01:58, 21 June 2025 (UTC)

= June 18 =

I was looking at a Circuit of the pentagonal faces of a Dodecahedron, wondering if it could be done with only passing from a side to one of the non-adjacent sides. I determined it can't since any set of three faces in such a circuit would be the ones on three consecutive sides of a fourth pentagon, which could then only enter and exit by adjacent sides.

I was looking at moving that up a level. For a given face of a Dodecahedron, designate a move from the face to one of the five adjacent faces as an Ortho move, a move from a face to an opposite face as a Para move and a move to one of the other five faces as a Meta move (see Arene substitution pattern for the naming convention) Can a circuit through the cells of a 120-Cell be done *only* with Meta moves? Can a circuit be done only with Ortho Moves? If no for either, can it be done allowing both that type of move or Para. (Obviously it can't be done with Para only has that will trace out the "equator" of the polydedron (similar to doing it on the cube or hypercube).Naraht (talk) 01:57, 18 June 2025 (UTC)

I just know it was published in 1968 by

Tables of indices and primitive roots, Royal Society Mathematical Tables, vol 9, Cambridge University Press.

But don’t know the article’s exact name nor can I afford the book… Even if I could afford it, I need it as a ᴘᴅꜰ. 2A01:E0A:ACF:90B0:0:0:A03F:E788 (talk) 20:56, 18 June 2025 (UTC)

:That volume was authored by A. E. Western and J. C. P. Miller. Part of the introduction, preceding the tables, can be viewed [https://oeis.org/A002223/a002223.pdf here].

:I don't see anyone with the surname "Stephen" who has published work on elliptic curves. Could this by any chance refer to Nelson Malcolm Stephens (1941–2024), no article on Wikipedia but author of one of the cited works in Feit–Thompson conjecture? The article [https://www.ams.org/journals/proc/2004-132-04/S0002-9939-03-07311-8/ "Primes generated by elliptic curves"] is by Graham Everest, Victor Miller and Nelson Stephens.  ​‑‑Lambiam 00:07, 19 June 2025 (UTC)

= June 20 =

For y=|x| the x=0 point is said in italian "punto angoloso". Paper and on-line dictionaries fail to translate this term. Note that I mean a generic case, not the peculiar case called cusp. In the english wiki I have not found a definition, but I have found such a point called angle, corner, sharp turn and bend. Which is the most-widely accepted term? thanks 151.29.62.127 (talk) 09:49, 20 June 2025 (UTC)

:I'm not convinced it has exactly the same meaning as you're going for, but I think vertex comes pretty close. That's the term used in the context of linear programming at least. A direct translation, "corner point" will probably be understood as well. And yes, a cusp is a special case since the tangent line moves continuously there. There are many ways that a function can be non-differentiable though, for example y=x sin (1/x) is not differentiable at 0, and I'd hesitate to call that a corner. As with many language questions, it helps to know the context, can give an example of the word being used in a sentence? --RDBury (talk) 12:10, 20 June 2025 (UTC)

::Thanks very much. I am translating italian lessons for people that know little english and almost no italian, so the context cannot help.151.29.62.127 (talk) 12:34, 20 June 2025 (UTC)

:In old calculus books, you may find two words, "cusp" for where the derivative goes to infinity, and "salient point" for where the derivative has a jump discontinuity. I think the best answer to your question as phrased is "salient point", though that excludes the cusp case. --Trovatore (talk) 20:04, 20 June 2025 (UTC)

:Anticipating a couple of possible objections. First, technically, the first derivative doesn't have a jump discontinuity at the singular point because it isn't defined there. Hopefully you can work out what I mean.

:More importantly, I'm just treating the case of the graph of a function, not more general curves. Again the extension is left as an exercise. --Trovatore (talk) 20:07, 20 June 2025 (UTC)

::I've never personally seen "salient point", and notably Richard Courant's classic Differential and Integral Calculus uses the term "corner" (although as a purely descriptive noun, rather than term of art). It's fair to say that there is no widely accepted term of art for this, and "corner" seems to be common, but even a literal translation of the Italian would be satisfactory. Tito Omburo (talk) 20:12, 20 June 2025 (UTC)

:::It's attested in {{cite book |last= Griffin|first= Frank Loxley|date= 1927|title= Mathematical Analysis Higher Course|publisher= The Riverside Press|location= Cambridge, Massachusetts|page= 390|isbn= }}. --Trovatore (talk) 20:38, 20 June 2025 (UTC)

:::File:SalientPoint.jpg

:::It happens to be PD, so I'm going to go ahead and drop in a shot of it; possibly it will give context, but also I just love these old books. --Trovatore (talk) 20:56, 20 June 2025 (UTC)

::::A book of somewhat dubious provenance, imo. Tito Omburo (talk) 21:07, 20 June 2025 (UTC)

::::: I have to say I find that comment a little reductive. It's actually an excellent book for learning certain techniques. It was my dad's college textbook and I had a great time with it in high school (or was it junior high?). It's true that it's pitched more to the engineering or physics student and less to the mathematician. I did learn some physics from it as well.

::::: This particular part of it is a little frustrating because it's not completely clear what Griffin is talking about. (It might help to know that on the previous page he's discussing curves defined by an equation $f(x,\; y)=0$, and the mysterious equation 61 he mentions is $\backslash frac\{\backslash partial\; f\}\{\backslash partial\; x\}+\backslash frac\{\backslash partial\; f\}\{\backslash partial\; y\}\backslash frac\{dy\}\{dx\}=0$.) In particular the statement that "[a] salient point is possible only when the equation involves transcendental or irrational algebraic functions of special kinds" is a bit obscure. I wondered for years what these cases would be. He does give an example in the exercises, but it's $y=5-\backslash sqrt\{x^2(x+4)\}$, which is obviously just leveraging the convention that the square-root function gives the nonnegative square root if it exists. In any case this is potentially "good trouble" as it might spur the interested student to learn about algebraic varieties.

::::: Griffin was head of the math department at Reed College, its acting president for a while, and vice president of the MAA. We probably ought to have an article about him. There's a starting point [https://www.reed.edu/president/reed_presidents/griffin.html here]. --Trovatore (talk) 00:58, 21 June 2025 (UTC)

::::The term "angular point" is also found: E. J. Townsend, G. A. Goodenough (1910). Essentials of Calculus. New York: Holt. [https://archive.org/details/cu31924000678015/page/n68/mode/1up?q=%22angular+point%22 p. 51]. Or, more recently, [https://books.google.com/books?id=QnpqBQAAQBAJ&pg=PA243&dq=%22angular+point%22&hl=en here] or [https://books.google.com/books?id=WMyZlwiQBNYC&pg=PA98&dq=%22angular+point%22&hl=en here]. Townsend was the translator of the 1902 English edition of Hilbert's Grundlagen der Geometrie.^{[https://archive.org/details/foundationsgeom00towngoog/page/n6/mode/1up]}  ​‑‑Lambiam 21:15, 20 June 2025 (UTC)

:{{ping|151.29.62.127}} I think "singular point" might be the best choice for your needs here. The only thing is it might be more general than you want, in that it at least arguably includes points of discontinuity, but if you specify that the function is continuous, that should be OK. --Trovatore (talk) 21:03, 20 June 2025 (UTC)

:: Singular point could mean anything! That's a bad choice. OP is specifically asking about corners. Tito Omburo (talk) 21:06, 20 June 2025 (UTC)

:::There is, moreover, only a singularity in the derivative.  ​‑‑Lambiam 09:16, 21 June 2025 (UTC)

::::"Singular" doesn't mean "discontinuous". A discontinuity in the derivative is absolutely a singularity, and as I said a discontinuity in the function itself could be ruled out by context. That said, it's true there are various other possibilities. However the heading asks about "non-derivable points" ("non-derivable" presumably meaning "non-differentiable"). In lots of contexts those would be precisely the singular points. --Trovatore (talk) 17:01, 21 June 2025 (UTC)

::File:AngularPoint.jpg

::Wiktionary defines the term salient point solely as: "a prominent feature or detail". This is what English readers will naturally expect the term to mean. Wiktionary lists an obsolete sense "moving by leaps or springs; jumping" for salient; apart from this being obsolete, what is moving by a leap is only the derivative.

::I have uploaded the textbook definition of angular point, which is crisp and clear. Modern texts (refs given above) using the term give equivalent definitions. The term is appropriately descriptive.  ​‑‑Lambiam 09:48, 21 June 2025 (UTC)

:::I don't know how Griffin came to give this sense in his textbook. I had assumed it was standard but perhaps it was not. Searches are confounded by a more recent sense from computer vision and machine learning. --Trovatore (talk) 17:07, 21 June 2025 (UTC)

::::Probably a calque of French point saillant.^{[https://books.google.com/books?id=5foJAAAAIAAJ&pg=PA256&dq=%22point+saillant%22&hl=en][https://books.google.com/books?id=jMsKAAAAYAAJ&pg=PA245&dq=%22point+saillant%22&hl=en][https://books.google.com/books?id=jr8KAAAAYAAJ&pg=PA326&dq=%22point+saillant%22&hl=en]}  ​‑‑Lambiam 18:28, 21 June 2025 (UTC)

= June 21 =

Is James Tanton an Australian mathematician, as stated in the article about him, or an American author, as stated of the article Encyclopedia of Mathematics (James Tanton). None of the references for either of these pages is clear. Possibly there are two people called James Tanton, one of whom is a mathematician and the other wrote an encyclopedia about mathematics. 176.108.139.1 (talk) 01:31, 21 June 2025 (UTC)

:Math genealogy knows about one [https://www.mathgenealogy.org/id.php?id=18868 James Tanton] (actually only one person with the surname Tanton). That seems to be the same person as [https://www.jamestantonmath.com/ this guy], whose CV (linked there) includes degrees from an Australian university and a book he's written called Encyclopedia of Mathematics. All this suggests there is one person, who has lived in both Australia and the US at different times. 173.79.19.248 (talk) 02:08, 21 June 2025 (UTC)

::Since graduating from the University of Adelaide in 1988, Tanton has been living in the US.^{[https://stevementz.com/earth-by-jeffrey-cohen-and-lindy-elkins-tanton-object-lessons/]} He is settled there, with a family – his spouse is planetary scientist Lindy Elkins-Tanton,^{[https://news.berkeley.edu/2025/05/06/planetary-scientist-lindy-elkins-tanton-to-head-space-sciences-laboratory/]} who was born in Ithaca, New York. While not certain, it is reasonable to assume that Tanton is now a naturalized US citizen. If so, he should be referred to as an "Australian–American mathematician", like we do for James Oxley, Terence Tao and Paula Tretkoff.  ​‑‑Lambiam 07:43, 21 June 2025 (UTC)

:::Thanks! I've updated both articles. 176.108.139.1 (talk) 18:16, 21 June 2025 (UTC)

For polynomials over the integers of degree n, one could classify them as Cantor did, by k= sum of the absolute values of the coefficients. For each k,the number of each that are factorable, the number that have Galois group C_n, D_n[or D_2n depending on convention], A_n, S_n could be counted, at least for small n and small k. If the polynomial is factorable, a "quasi" galois group could be noted as the dirct product of the galois groups of the irreducible factors.(not transitive though). So are there results anyone here know about this? Also what books are there where I can I find out about work like this? (I've heard in passing that as n->oo, most polynomials have galois group A_n, but I don't know why or where to find out.Rich (talk) 20:44, 21 June 2025 (UTC)

= June 23 =