Talk:Ordered Bell number

A paper reviewed by...
Why does the article refer to "A paper reviewed by Ellison and Klein" and not "A paper by Archangeli and Langendoen" or whomever wrote the paper in this anthology? That would make more sense. --WiseWoman (talk) 23:04, 3 March 2015 (UTC)
 * Because that paragraph is about what Ellison & Klein say in their review, not so much about what Archangeli and Langendoen say. —David Eppstein (talk) 23:28, 3 March 2015 (UTC)
 * Oh, it's clear that the paragraph is about Ellison & Klein's review, and their contributions in sentences 1, 2, and 4 of the paragraph. I was just curious to know who wrote the paper which Ellison & Klein refer to in their review that you refer to in your third sentence. --WiseWoman (talk) 23:58, 3 March 2015 (UTC)

Where is this name from?
This article has a nice historical introduction, but gives no or few explanation about the (curious, indeed) choice of naming this sequence of numbers after E.T.Bell (apparently, it's from a 2005 paper). Isn't the case to go back to the 1974 introduction by Louis Comtet, or to look for some more dignified choice? (btw, "Bell numbers" for the Dobiński's numbers of set partitions seems already quite a local attribution). pm a 08:57, 28 January 2021 (UTC)

Logarithm notation
David Eppstein has reverted my edits about the choice of logarithm notation. I would like to explain why I think $$\ln$$ is better than $$\log$$ for this article.

Firstly, I don't know anywhere where it says $$\ln$$ is not used in mathematical writing, unless you're Paul Halmos. I point out that the articles for Bell number and Stirling's approximation, use the notation $$\ln$$.

Secondly, David also says that articles should be written for the widest audience. The widest audience would therefore include non-expert mathematicians as well. In that case, $$\ln$$ is recognised more widely than $$\log$$ as the natural logarithm. The fact that you have to write "here, $$\log$$ denotes the natural logarithm" is evidence.

On the question about whether the base should be $$\frac 1 {\ln 2}$$ or $$\log_2 e$$: you've already explained later that it is approximately 1.4427. In my opinion, $$\log_2 e$$ is more confusing. It could easily be misconstrued as $$\log_e 2$$, since $$e$$ is far more commonly the base than 2. It takes a few seconds to realise this number is greater than one, whereas $$\ln 2$$ is a well known to be about 0.693 (off the top of my head).

Could we come to some agreement about this? IntGrah (talk) 20:33, 30 April 2024 (UTC)


 * Also, $$\frac 1 {\ln 2}$$ follows from the previous formula much more clearer; otherwise, the reader has to use the change of base formula themselves as well as converting the negative exponent to a reciprocal. IntGrah (talk) 20:51, 30 April 2024 (UTC)
 * This is a mathematics articles. Mathematics articles use log for the natural logarithm. To do otherwise would be to violate the standard conventions of mathematics writing, both off-wiki and on.
 * As for $$\log_2 e = \tfrac{1}{\log 2}$$: Yes, obviously. BUT, when trying to read the formula $$(\tfrac{1}{\log 2})^n$$ one has to first realize that the unwritten base e is larger than 2, then that the unusual backwards ordering of larger base and smaller argument means that the logarithm is less than one, then that taking one over that produces a number greater than one, to eventually maybe conclude that you're actually looking at an exponentially-growing thing rather than an exponentially-shrinking thing. Is it obvious at a glance to you, without working through the steps and carefully keeping track of signs, that $$(\tfrac1{\log_4 2})^n$$ is really just $$2^n$$? It isn't obvious to me.
 * When the formula is expressed in the form $$(\log_2 e)^n$$, one doesn't have to do that confusing double negation to reach the same conclusion. The base and argument of the log are in the more commonly seen order of a smaller base and larger argment. Therefore the log gives a number greater than one. Therefore it's exponentially growing. I think it is MUCH MUCH CLEARER this way. Specifically, I found the older formulation to be very confusing, myself, to the point where I had to work through it step by step to understand the growth rate. The current formulation expresses the point I am trying to express (exponential growth) more clearly, in the sense that I don't personally find it confusing. I do find the version you want to reinstate confusing. The change of base is a a bigger step from the summation formula but I think that is less important. —David Eppstein (talk) 22:54, 30 April 2024 (UTC)
 * Thanks for your reply. However, I still don't agree; the exponential growth is already highlighted by the succeeding sentence, and frankly it's stupid to write $$\log_2 e$$. And the change of base step is still an issue. I'll leave it for now until someone else agrees with me. IntGrah (talk) 23:31, 30 April 2024 (UTC)
 * The expression $$\log_2 e$$ appears in multiple other articles. To pick one that has little or no computer science content, and little or no connection to this article, it appears in clique game. What makes you think writing this expression in this form, there or here, is "stupid"? It is more concise than $$\tfrac1{\log 2}$$, in its TeX coding, in the number of symbols in the rendered expression, and especially in the vertical extent of the rendered expression. It uses logs in a way that I find more natural (base smaller than argument). It is just a logarithm of two numbers that happens to produce a value seen in some mathematical topics. What is it about that pair of numbers that makes it "stupid"? Which other pairs of numbers produce logarithms that are "stupid"? Is there a unique non-stupid way of writing each possible logarithm that one might see? What is your method for canonicalizing a logarithm into a non-stupid form? —David Eppstein (talk) 00:24, 1 May 2024 (UTC)
 * Writing $$\ln x$$ does not violate the conventions of the wiki; the contrary is true. Thousands of Wikipedia mathematical articles use $$\ln x$$. It is the ISO standard. The off-wiki conventions do not apply to this article, because the wiki is not a scientific journal, and the wiki is also written for readers who may not know these "conventions". It was already mentioned by Bilorv that this is ambiguous notation. There is quite literally zero gain by writing $$\log$$ and having to explain its meaning in prose.
 * For one, the article clique game uses base 2 logarithms consistently, unlike this article. I expect an article to stick to one base, and not assume the reader can follow with the an arbitrary change of base formula.
 * It's not "more concise" than $$\frac1{\ln 2}$$. They use exactly the same number of symbols ( and  ). Is vertical space really getting expensive nowadays? Should Stirling's formula be written as $$\sqrt{2\pi n}n^n e^{-n}$$? Should $$\frac12$$ also be chopped down to 0.5?
 * I could not find any style guide which says logarithm bases should be ideally smaller than their argument. How would you write $$\ln 2$$ then? Is there a unique 'natural' way of writing every possible logarithm that one might see?
 * What is it about that pair of numbers that makes it "stupid"? As I already said, it could be misconstrued as $$\log_e 2$$, which you failed to address.
 * Which other pairs of numbers produce logarithms that are "stupid"?
 * This is about the numbers 2 and $$e$$, which are two of the most common bases.
 * What is your method for canonicalizing a logarithm into a non-stupid form? I never said there was a canonical form for logarithms. The goal is readability, which entails consistency and familiarity but does not entail "golfing" the expression in a way which might suggest the number is greater than one, but is already fully explained in the sentence after anyway. IntGrah (talk) 11:09, 1 May 2024 (UTC)
 * Ok, you lost all credibility with me when you invoked ISO standards on how to write logs. ISO does write standards for this, it is true. Nobody follows them. See Binary logarithm. —David Eppstein (talk) 18:26, 1 May 2024 (UTC)
 * And how exactly is their recommendation for binary logarithm related to this? Do you completely ignore the thousands of Wikipedia articles that use $$\ln$$? Do you assume everyone who reads mathematics articles assumes base $$e$$? IntGrah (talk) 18:44, 1 May 2024 (UTC)
 * @IntGrah No skin in this game, but a heads up that there's further discussion going on over at WT:WPM. GalacticShoe (talk) 18:59, 1 May 2024 (UTC)
 * No, I do not assume base e. The article explicitly states that it defaults to the natural logarithm, a statement you tried to remove. —David Eppstein (talk) 19:00, 1 May 2024 (UTC)
 * ...and you wouldn't need that statement if you'd just used $$\ln$$ in the first place. IntGrah (talk) 19:03, 1 May 2024 (UTC)
 * A literature search finds hundreds (speaking conservatively) of examples where the authors say where ln is the natural logarithm. So, color me doubtful that the notation is so ubiquitously understood that it would not also need just as much clarification. XOR&#39;easter (talk) 19:11, 1 May 2024 (UTC)
 * There are also examples where authors write "where log is the natural logarithm" too, so what's the advantage? Wikipedia isn't a research journal. Even if it was not understood by everyone, it is still understood by more people than "log" — the article was in the DYK page, and is of interest to amateur combinatorists too. IntGrah (talk) 19:37, 1 May 2024 (UTC)
 * If an amateur combinatorist can't understand the plain text explanation "where log is the natural logarithm", they're probably not a very good amateur combinatorist. XOR&#39;easter (talk) 20:19, 1 May 2024 (UTC)
 * That's unrelated to the point that $$\ln$$ is more widely known. IntGrah (talk) 20:36, 1 May 2024 (UTC)
 * It is more widely used by engineers and physical scientists. It is very infrequently used by pure mathematicians. This is not an article in engineering or physical science. We should follow the conventions of the fields our articles belong to. —David Eppstein (talk) 23:22, 1 May 2024 (UTC)
 * Using "log" for the natural logarithm is indeed standard for mathematicians, and writing $$\frac{1}{\ln 2}$$ instead of $$\log_2 e$$ is a mite peculiar. The point about making clear that the final functional form is an exponential growth rather than an exponential decay is a good one. In this case, $$\log_2 e$$ is the more clear way to express the meaning the formula is supposed to convey, while also fitting more nicely into the following text. On the other hand, $$\frac{1}{\ln 2}$$ is an ungainly fraction that disrupts the prose flow. Changing the base between one formula and the next isn't a problem in this style of writing, since the reader is told that steps are being skipped over ("This leads to an approximation for..."). XOR&#39;easter (talk) 19:08, 1 May 2024 (UTC)
 * How does it disrupt the prose flow? I don't think $$\log_2 e$$ is clear. It's base two, which is unexpected. It's a different base from the one used beforehand. And $$e$$ is the argument for some reason. As I said, the fact that it is 1.443 is already explained clearly; there is no need for such a convoluted expression. IntGrah (talk) 19:31, 1 May 2024 (UTC)
 * It disrupts the prose flow by being a big honkin' fraction sticking up and down in the middle of a sentence. It's more convoluted than $$\log_2 e$$, because it requires taking a logarithm and then dividing by it, rather than just taking a logarithm. XOR&#39;easter (talk) 20:22, 1 May 2024 (UTC)
 * What's so scary about a fraction? There is already a fraction in the expression. IntGrah (talk) 20:40, 1 May 2024 (UTC)
 * I don't understand the objections to $$\log_2 e$$ or any other base-2 logarithm. I don't understand how the number 2 is "unexpected". And it seems simpler to me than $$1 / \log 2$$, $$1 / \log_e 2$$, or $$1 / \ln 2$$, for XOR'easter's reason.
 * In advanced math, $$\log$$ (unsubscripted, meaning the inverse of $$\exp$$) is more common than $$\ln$$. And this article is advanced math. So I prefer $$\log$$ here. Mgnbar (talk) 20:42, 1 May 2024 (UTC)
 * It's not 'advanced math'. It's the number of outcomes of a horse race. It's readable for a high school student and understandable by an undergrad. There is no point in defining the meaning of the logarithm for the [two] times it is mentioned, especially when you override it by making it base 2. IntGrah (talk) 20:50, 1 May 2024 (UTC)
 * When base 2 is used in mathematics, I expect it to be meaningful, suggesting a relationship with the number two. In this scenario, it is not being used to convey twoness. Rather, it's a pointless application of change of base. IntGrah (talk) 21:12, 1 May 2024 (UTC)
 * Whether you write it as $$\tfrac{1}{\log 2}$$ or $$\log_2 e$$, the same number two is there both times. There is an inherent meaningful relation with the number two in this formula, regardless of whether you put it in the base or the argument. —David Eppstein (talk) 23:02, 1 May 2024 (UTC)
 * It's one more operation. Taking a log, which happens to be smaller than one, and then taking a reciprocal to get a number bigger than one, takes multiple steps of reasoning. Just taking a log and seeing that is bigger than one is fewer steps. It is precisely this issue that caused me to be confused by the earlier version of the approximation formula in the article, the one that you want to revert to, and it was my confusion over this formula that led me to the present form. —David Eppstein (talk) 20:43, 1 May 2024 (UTC)
 * Then what about the number of steps between the previous summation and the approximation? The working is omitted, making it even more important to highlight the parallels between the two expressions. I fail to see how the 'number of function calls' is more important than the readability of the expression. Taking the logarithm of an irrational number with an integer base raises more questions and is certainly more work.
 * With $$\frac1{\ln 2}$$, the reader sees $$\ln$$, then 2, and knows this is some transcendental constant, which the reader does not need to evaluate. The reader can simply glance over it as 'some random number'. And, it obviously comes from the previous expression. The following sentence immediately explains it is exponential growth.
 * Compare with $$\log_2 e$$: the reader sees a base 2 logarithm. Does this imply some relationship with the number 2? Why e? They then realise that this probably came from an application of the change of base formula. They recall the formula to verify what they are seeing: $$\log_2 e = \frac{\ln e}{\ln 2} = \frac1{\ln 2}$$. Only then is it clear where it comes from. Meanwhile, they have not even begun to realise this number is greater than one; it would not cross their mind as to why greater-than-one-ness is important. They are simply baffled at why someone would write that, instead of one-over-ln-2. IntGrah (talk) 21:09, 1 May 2024 (UTC)
 * You're overthinking it. $$\log 2$$ is the natural logarithm of some number, which happens to be 2. $$\log_2 e$$ is the binary logarithm of some number, which happens to be e. They have exactly the same descriptive complexity. —David Eppstein (talk) 22:59, 1 May 2024 (UTC)
 * I don't have any issue with using log instead of ln, and in principle $$\log_2 e$$ is totally fine, but I have to admit that in context I found $$\log_2 e$$ in the following a little jarring, and it took me a moment to figure out where it came from:
 * Based on a contour integration of this generating function, the ordered Bell numbers can be expressed by the infinite sum
 * $$a(n)=\frac{n!}{2}\sum_{k=-\infty}^{\infty}(\log 2+2\pi ik)^{-(n+1)},\qquad n\ge1.$$
 * Here, $$\log$$ stands for the natural logarithm, whose base is $$e\approx 2.718$$. This leads to an approximation for the ordered Bell numbers, obtained by using only the term for $$k=0$$ in this sum and discarding the remaining terms:
 * $$a(n)\sim \frac12 {n!}(\log_2 e)^{n+1}.$$
 * Thus, the ordered Bell numbers are larger than the factorials by an exponential factor, whose base $$\log_2 e$$ is approximately $$1.4427$$.
 * Wouldn't it be clearer to just say instead " ... and discarding the remaining terms:
 * $$a(n)\sim \frac{n!}{2}(\log 2)^{-(n+1)}\approx 1.4427^{n+1}\frac{n!}{2}.$$
 * Thus, the ordered Bell numbers are larger than the factorials by an exponential factor."? It both makes the point of exponential growth immediately, and also makes comparison to the previous formula completely direct. Gumshoe2 (talk) 21:17, 1 May 2024 (UTC)
 * I am happy with this solution. I can concede $$\ln$$ versus $$\log$$ on the basis that $$\log_2$$ is removed.
 * Although, telling the reader that $$e \approx 2.718$$ is a minor insult to intelligence, and it feels pointless to define the base when it is only used twice. That was the reason I preferred $$\ln$$. IntGrah (talk) 21:23, 1 May 2024 (UTC)
 * I don't think it's necessary to say e is 2.718, it's enough just to say "Here log stands for the natural logarithm." Gumshoe2 (talk) 21:26, 1 May 2024 (UTC)
 * Okay, that is fine. IntGrah (talk) 21:27, 1 May 2024 (UTC)
 * The reason for saying what $$e$$ is was to explain its later explicit appearance in $$\log_2 e$$. There are so many other uses of the letter e that, if it is to be used in a formula, it needs to be disambiguated somewhere, just like log without an explicit base always means natural log in mathematics but still could use disambiguation. —David Eppstein (talk) 21:45, 1 May 2024 (UTC)
 * Yes, but if we are not to write $$\log_2 e$$, then there is no need to highlight that 2.718 > 2 anymore. IntGrah (talk) 21:47, 1 May 2024 (UTC)
 * I am unhappy with this solution. I considered it and discarded it. The reason is that I am unsure what a formula combining both $$\sim$$ and $$\approx$$ is supposed to mean. Here, $$\sim$$ has a precise technical meaning that works for formulas with a variable $$n$$ in them (the ratio goes to one as $$n$$ goes to infinity) and $$\approx$$ has a meaning that has not been formally defined but informally means something like "this specific number is numerically close to this other specific number", without dependence on $$n$$. It is not true that the left and right sides of the combined formula have ratios that go to 1 as $$n$$ goes to infinity. Instead, the ratio blows up to something that is itself exponentially large because of the numerical approximation. So we get a mismatch between what we are telling the readers (this approximation formula gets more accurate for larger $$n$$) and what actually happens in the formula (it gets much less accurate). It is also not even true that the middle formula and the right hand side are numerically close for all $$n$$ (they are exponentially far apart for large $$n$$). —David Eppstein (talk) 21:36, 1 May 2024 (UTC)
 * That hadn't occurred to me, it's a fair objection. I don't have a better suggestion.
 * So then my only input is that for me as a reader it's definitely clearer to stick with a single choice of log base. Gumshoe2 (talk) 21:49, 1 May 2024 (UTC)
 * It seems to me that using "ln" instead of "log" would be clearer here, and 1/ln 2 instead of log_2 e. Tito Omburo (talk) 21:58, 1 May 2024 (UTC)
 * To be clear, how obvious is it to you whether the expression $$\bigl(\tfrac1{\ln2}\bigr)^n$$ produces a big value or a small value? Can you tell that instantly just looking at the formula, or do you have to think about it for a while? That is the main thing I want this part of the formula to convey. —David Eppstein (talk) 22:27, 1 May 2024 (UTC)
 * The fact that it isn't instantly clear (to me) can be addressed by immediately subsequent text: "Since $$\textstyle\frac{1}{\ln 2}\approx 1.44,$$ this shows that the ordered Bell numbers are larger than the factorials by an exponential factor." (Or analogous.) I think that would read really easily and clearly. To be honest, I don't see the issue with it. Gumshoe2 (talk) 22:36, 1 May 2024 (UTC)
 * Next question: Since your preferred version fails to make this point clear, what is your reason for formatting it in this way rather than $$(\ln 2)^{-n-1}$$ (putting the +1 back into the exponent) or the even more abbreviated but often-seen form $$\ln^{-n-1}2$$? Why separate out the inversion and exponentiation when they can be jammed into a single operation? After all, if we no longer are attempting to make the formula understandable, maybe we should at least make it concise? And next readability question: in this form, can you quickly tell whether the +1 makes the formula bigger or smaller? Readability is important. In case you hadn't noticed, this article is up for a GA nomination (although I suspect IntGrah may have endangered that), and WP:TECHNICAL is an explicit part of the GA requirements. —David Eppstein (talk) 22:50, 1 May 2024 (UTC)
 * I think the following is completely clear and readable (and as I said, I don't see the issue with it):
 * ... and discarding the remaining terms:
 * $$a(n)\sim\frac{n!}{2}\left(\frac{1}{\log 2}\right)^{n+1}.$$
 * Since $$\textstyle\frac{1}{\log 2}\approx 1.44,$$ this shows that the ordered Bell numbers are larger than the factorials by an exponential factor.
 * Answers to your questions: if you use $$(\log 2)^{-n-1}$$ instead, the followup sentence would be clunkier. The expression $$\log^{-n-1}2$$ is potentially confusing, since $$f^{-n}$$ sometimes represents n-fold iteration of $$f^{-1}$$. I don't see the benefit of trying to jam everything into a single operation. The premise of your second to last question is incorrect. The answer to the last question is yes (and immediately so), at least in the way I just wrote it, where the formula is immediately followed by saying $$\textstyle\frac{1}{\log 2}\approx 1.44.$$ Gumshoe2 (talk) 23:08, 1 May 2024 (UTC)
 * From a practical point of view, that formula is going to require huge amounts of long-term on-going policing to stop gnomes from jamming the two fractions back into one fraction.
 * It is also not clear to me how "since (some number in the middle of a complicated formula has a value)" provides any explanation for why "larger than the factorials by an exponential factor". How would your sentence be any more meaningful if you wrote "since $$\pi\approx 3.1416$$, ... are larger than the factorials by an exponential factor"? In what way does providing a numerical approximation to part of the formula explain the exponential behavior of a different part of the formula? —David Eppstein (talk) 23:19, 1 May 2024 (UTC)
 * What is this comment supposed to even mean? It clearly explains that the base is greater than one and is thus exponential growth. IntGrah (talk) 23:24, 1 May 2024 (UTC)
 * Unfortunately, this time I don't understand your questions (even a little), since I think it's crystal clear how $$\textstyle\frac{1}{\log 2}\approx 1.44$$ is relevant in a way which $$\pi\approx 3.1416$$ isn't. I also don't understand what you mean by a "different part of the formula." Gumshoe2 (talk) 23:26, 1 May 2024 (UTC)
 * The numerical value of the logarithm, buried two levels below the exponentiation under a division operation, is not relevant to the fact that the exponentiation is an exponentiation. Your sentence is worded in a way that makes it sound like, if the log had a different value, the exponentiation would not be an exponentiation, which is nonsensical. The actual explanation, that the base of the exponentiation being larger than one causes the factor to be exponentially large rather than exponentially small, has been almost entirely skipped over in your wording. The result is a sentence that contains the two things we want to connect (numeric value ... exponentially larger) but where the path from one to the other has been muddied. We're missing the intermediate step. Because 2 < e, $$\ln 2 < 1$$. Because $$\ln 2 < 1$$, $$\frac{1}{\ln 2} > 1$$. Because the base of the exponential is > 1, the whole formula is exponentially larger. Look at all those steps when we put them in. I was hoping that writing $$\log_2 e$$ would make this clearer because fewer steps but instead it's leading to all this "I don't understand binary logarithms! Let's do everything we possibly can to contort things to avoid binary logarithms!" fuss. What's wrong with binary logarithms? —David Eppstein (talk) 23:31, 1 May 2024 (UTC)
 * Ok, I understand now. That makes sense, but it's really easy to fix: change the sentence to "Since $$\textstyle\frac{1}{\log 2}>1$$, this shows that the ordered Bell numbers are larger than the factorials by an exponential factor" or "Since $$\textstyle\frac{1}{\log 2}\approx 1.44$$ is larger than one, this shows ..." or, etc. Gumshoe2 (talk) 23:37, 1 May 2024 (UTC)
 * This was essentially how my edit looked. IntGrah (talk) 23:55, 1 May 2024 (UTC)

Another possibility
The following variation avoids all the horrible confusion of using binary logarithms (oh the humanity) at the expense of introducing a completely unnecessary new variable. In exchange we also never need to define or use $$e$$. Maybe some might like it better:


 * This leads to an approximation for the ordered Bell numbers, obtained by using only the term for $$k=0$$ in this sum and discarding the remaining terms:
 * $$\displaystyle a(n)\sim \frac12 {n!}\,t^{n+1},$$
 * where $$t=\tfrac{1}{\log 2}\approx 1.4427$$. Thus, the ordered Bell numbers are larger than the factorials by an exponential factor.

What think you? —David Eppstein (talk) 23:44, 1 May 2024 (UTC)


 * I had this in mind as a possible solution, but thought it was unnecessary to introduce a single-use variable, so probably not. IntGrah (talk) 23:53, 1 May 2024 (UTC)
 * I don't understand why you think not using $$\log_2$$ means you have to introduce new variables? As I asked above, what's wrong with
 * ... and discarding the remaining terms:
 * $$a(n)\sim\frac{n!}{2}\left(\frac{1}{\log 2}\right)^{n+1}.$$
 * Since $$\textstyle\frac{1}{\log 2}\approx 1.44$$ is larger than one, this shows that the ordered Bell numbers are larger than the factorials by an exponential factor.
 * ? Exactly the same content but no extra variables. You said "that formula is going to require huge amounts of long-term on-going policing to stop gnomes from jamming the two fractions back into one fraction" (which I don't really follow), but is there any other reason? Gumshoe2 (talk) 23:54, 1 May 2024 (UTC)
 * To me, a formula that is only an exponential, $$t^{n+1}$$ (or whatever other variable; I have no particular preference for t) is more obviously exponential than a formula that combines a logarithm, a reciprocal, and an exponentiation and leaves it to the reader to figure out which of those operations to focus on. Especially so when the notation for the reciprocal and the parens around it are so much more heavyweight and visually in your face than the notation for the other operations. Writing it with only an exponential, and no other operations, emphasizes that the exponential is the important operation in the formula, and all the rest is just a number whose precise calculation isn't so important. —David Eppstein (talk) 00:00, 2 May 2024 (UTC)
 * ... and leaves it to the reader to figure out which of those operations to focus on ...
 * It's not left to the reader, because it's explained in the following sentence. IntGrah (talk) 00:06, 2 May 2024 (UTC)
 * To take Stirling's approximation as an example, do you see any issues with
 * $$\sqrt{2\pi n}\left(\frac ne \right)^n$$?
 * There's a fraction, which you despise for some reason. It doesn't take away from the fact that there is exponentiation going on. IntGrah (talk) 00:08, 2 May 2024 (UTC)
 * There's a "divided by a number" and an "exponentiation" from which I infer "exponentially smaller". That's exactly why putting both a division and an exponentiation into a formula that has the opposite outcome, "exponentially larger" confuses me so much. —David Eppstein (talk) 00:10, 2 May 2024 (UTC)
 * I see. It seems like you're looking at the formula (to some extent) entirely in and of itself, which is not how IntGrah and I (and, I believe, most readers) are viewing it. I think there's not too much more to discuss on it (at least on my part), maybe we can wait and see what other people think. Gumshoe2 (talk) 00:10, 2 May 2024 (UTC)
 * Well, for those readers who look at a big formula and think "big formula! let's just skip over it and read the text" rather than hoping for the formula to be readable and informative on its own, and not merely a calculation recipe, I wonder why the details of the formula even matter. Those readers are just going to see the "exponentially larger" part and not need any explanation of how that relates to the formula that they skipped over. —David Eppstein (talk) 00:12, 2 May 2024 (UTC)
 * I think it's kind of silly of you to describe looking for context or explanation in a followup sentence as "big formula! let's just skip over it and read the text." For further discussion I'll wait for some others to chime in. I should say though that I think the version you've proposed here is fine despite being (to my eyes) needlessly strange. Gumshoe2 (talk) 00:23, 2 May 2024 (UTC)
 * I think that version is permissible but not ideal, in the sense that I wouldn't edit it if I saw it. IntGrah (talk) 00:28, 2 May 2024 (UTC)
 * Writing it with only an exponential, and no other operations, emphasizes that the exponential is the important operation in the formula, and all the rest is just a number whose precise calculation isn't so important. I tend to agree with this. It's a subtle effect, but a noticeable one. XOR&#39;easter (talk) 02:28, 2 May 2024 (UTC)
 * I'm not really sure what all the fuss is about, but it strikes me as reasonable to use $$\log$$ with the base spelled out, reasonable to use $$\ln$$ with or without the base spelled out, reasonable to use $$(\log_2 e)^{n+1}$$, passable to use $$(\log 2)^{-(n+1)}$$ and unnatural to use $$(1/\ln 2)^{n+1}$$ or $$t^{n+1}$$. (On the other hand, $$\sqrt{2\pi n}\left (\frac{n}{e}\right)^n$$ is ghastly in merging $$n^{1/2}$$ with the constant $$\sqrt{2\pi}$$ rather than the $$n^n$$ term, but there's no nice way to write Stirling's approximation.) — Bilorv ( talk ) 20:56, 2 May 2024 (UTC)