Talk:Pascal's triangle/Archive 3

Addition of a section on Morton's discovery
Hi, I'm considering adding the following section and would like any feedback on whether it is worth mentioning, and if so, whether the description is up to standards. This description would replace the fourth bullet point (The value of a row) under Rows and be appended to the Extensions section. Any pointers to related demonstrations of this would be also be greatly appreciated:

NOTE: Please see my sandbox for the latest revision. — Preceding undated comment added 23:47, 3 June 2023 (UTC)

Arbitrary Bases
In 1964, Dr. Robert L. Morton presented the argument that each row $$n$$ can be read as a radix $$a$$ numeral, where $\lim_{n \to \infty} 11^{n}_{a}$ is the hypothetical terminal row, or limit, of the triangle, and the rows are its partial products. He proved the entries of row $$n$$, when interpreted directly as a place-value numeral, correspond to the binomial expansion of $$(a + 1)^n = 11^{n}_{a}$$. To better understand the principle behind this interpretation, here are some things to recall about binomials:
 * A radix $$a$$ numeral in positional notation (e.g. $$14641_{a}$$) is a univariate polynomial in the variable $$a$$, where the degree of the variable of the $$i$$th term (starting with $$i = 0$$) is $$i$$. For example, $$14641_{a} = 1 \cdot a^{4} + 4 \cdot a^{3} + 6 \cdot a^{2} + 4 \cdot a^{1} + 1 \cdot a^{0}$$.
 * A row corresponds to the binomial expansion of $$(a + b)^{n}$$. We can eliminate $$b$$ from the expansion by setting $$b = 1$$. The expansion now typifies the expanded form of a radix $$a$$ numeral, as demonstrated above. Thus, when the entries of the row are concatenated and read in radix $$a$$ they form the numerical equivalent of $$(a + 1)^{n} = 11^{n}_{a}$$. For example, $$14641_{a}$$ is $$11^{4}_{7} = 8^{4}$$ in radix $$7$$, $$11^{4}_{8} = 9^{4}$$ in radix $$8$$, and so on (in general, if $$c = a + 1$$ for $$c < 0$$, then  let $$a = \{c - 1, -(c + 1)\} \;\mathrm{mod}\; 2c$$, with odd values of $$n$$ yielding negative row products).

The frequently cited row products $$(a + 1)^{n} = 11^{n}_{1} = 2^{n}$$ and $$(a + 1)^{n} = 11^{n}_{10} = 11^{n}$$ are obtained by setting the rows' radix (the variable $$a$$) equal to one and ten, respectively. As another example, setting $$a = n$$ yields the row product $$n^n \left( 1 + \frac{1}{n} \right)^{n} = 11^{n}_{n}$$. The numeric representation of $11^{n}_{n}$ is formed by concatenating the entries of row $$n$$. In the image above, the twelfth row denotes the product:

$$11^{12}_{12} = \boldsymbol{1}:\boldsymbol{1}0:56:164:353:560:650:560:353:164:56:1\boldsymbol{0}:\boldsymbol{1}_{12} = \boldsymbol{2}7433a96997\boldsymbol{01}_{12} = 2613035290224..._{10}$$

with compound digits (delimited by ":") in radix $$12$$ and a priori values in bold. The digits from $$k = n - 1$$ through $$k = 1$$ are compound because these row entries compute to values greater than or equal to $$12$$. To normalize the numeral, simply carry the first compound entry's prefix, that is, remove the prefix of the coefficient $${n \choose n - 1}$$ from its leftmost digit up to, but excluding, its rightmost digit, and use radix-$$12$$ arithmetic to sum the removed prefix with the entry on its immediate left, then repeat this process, proceeding leftward, until the leftmost entry is reached. In this example, the normalized string ends with $$01$$ for all $$n$$. The leftmost digit is $$2$$ for $$n > 2$$, which is obtained by carrying the $$1$$ of $$10_{n}$$ at entry $$k = 1$$. It follows that the length of the normalized value of $$11^{n}_{n}$$ is equal to the row length, $$n + 1$$. The integral part of $$1.1^{n}_{n}$$ contains exactly one digit because $$n$$ (the number of places to the left the decimal has moved) is one less than the row length. Below is the normalized value of $$11^{1234}_{1234}$$. Compound digits remain in the value because they are radix $$1234$$ residues represented in radix ten:

$$11^{1234}_{1234} = 2:885:2:35:977:696:\overbrace{...}^\text{1227 digits}:0:1_{1234} = 2717181235..._{10}$$

[Removed summary]
Twoxili (talk) 04:09, 30 May 2023 (UTC)


 * Hi, as written this has the appearance of being the product of your own original research. Wikipedia is an encyclopedia; the articles here are based on published sources, as described in our policy WP:OR (and other related content policies, like WP:V and WP:DUE).  I would be opposed to adding this to our article without evidence that it has been described in reliable published sources.  --JBL (talk) 18:15, 30 May 2023 (UTC)
 * You've now added a reference, but since the cited paper does not mention the number e, it does not improve the situation. --JBL (talk) 20:46, 30 May 2023 (UTC)
 * Hi JBL. Thanks for providing links to the relevant guidance documents. Yes, I added a source that gives a more thorough treatment to all the points I made except Step 4 (the identity involving the row binomial and the limit expression of e). Otherwise the steps I have listed are a paraphrase of the cited paper. As for the identity in Step 4, it is a trivial observation, meaning, anyone who is farmilar with the cited paper, specifically the attention that is given to the special case of b = 1 in the binomial (a + b)^n, and is also familiar with the limit expression of e, can clearly detect that they are identical, that is, each row is (a + 1) multiplied by itself n many times. I was not able to find much more discussion on the conclusion reached by the cited paper (and I assume this is because it is common knowledge in this domain), but my aim is to present the findings of that paper in a way that helps convey how Morton interprets the rows. My question is, does the cited paper help to resolve your concern? Twoxili (talk) 20:58, 30 May 2023 (UTC)
 * The paper mentions (a + 1)^n. The limit of this value as n -> ∞ is e scaled to a whole number for a = n. This is what I mean by "trivial observation". Does this help mitigate your concerns? Twoxili (talk) 21:20, 30 May 2023 (UTC)
 * I've included a summary in the introductory paragraph. I do agree that, based on the Wikipedia guidelines, and given the state of available material, "a = n" is treading into original territory, due to the sparsity of discourse on the topic. As I mentioned, the formula product of which e is a limit (though not mentioned by name) is stated in the published material, and should seem obvious to any student of math if they are familiar with the paper. The cited paper includes a version of Pascal's triangle in base 12 and explicitly states that row n denotes 13^n. It seems Prof. Morton couldn't be bothered to mention the significance of the 12th (or nth $$a$$th) power of 13 specifically [Edit (not even as a footnote)], nor can I find any discussions about row n of the radix n triangle. As for why that is, my thoughts are that hardly anyone has read the paper (as of today it is cited only 8 times according to Google Scholar), or that "a = n" doesn't add anything to the discourse. To say that the structure of the triangle is e given the definition of the entries as outlined in Morton's paper is to merely restate that definition [Edit: in terms of e]. At any rate, I thank you for your feedback. My understanding of this topic and the issues involved in editing Wikipedia articles are greatly improved by your comments. Twoxili (talk) 12:07, 31 May 2023 (UTC)
 * To me personally, this seems far from passing the bar set by WP:OR and WP:DUE, but it's possible that other editors will disagree. --JBL (talk) 17:24, 31 May 2023 (UTC)
 * I made some more edits to remove any original insights or ideas, with the aim of distilling it to the prior result obtained by the cited paper. The discussion of this result is convoluted, that's for sure. The result is alluded to multiple times in the Wikipedia page, but never explicitly asserted. The use of arbitrary bases, and the fact that the subject of the discussion (the row entries) have multiple mathematical objects associated with them (the index, the polynomial term, it's solution, the degree, the coefficient, etc) doesn't help with communicating the mechanism at work, so I focused on clarifying the discussion. As for whether "a = n" is original, at this point I feel it's "prior art" in need of more discussion and clearer framing. [Edit: My only contribution is nomenclature.] Even the Wikipedia page applies the b = 1 case and changes base a to elucidate certain properties, (in the Binomial Theorem section) setting a = 1 to obtain 2^n and (in the Rows section) setting "a = 10" to obtain 11^n. The paper states "The entries of Pascal's triangle can be expressed in any place-value numeral system which we may choose..." to obtain (a + 1)^n = 11^n in the chosen base a. The results 2^n and 11^2 are due to changing the base a. The problem is that the Wikipedia page never mentions this treats the two (the row sum and the base change) as separate phenomena. This property is the thesis of the cited article, and it's something I feel the Wikipedia page is missing. Twoxili (talk) 22:18, 31 May 2023 (UTC)
 * I've revised the writeup to better highlight the connection between Morton's paper and e. Basically, the sequence 11 (the topic of Morton's paper) is the root of e. This wasn't made clear in the paper, perhaps because it's unremarkable at first glance. In the comments below, in response to Mgnbar's comments, I presents some arguments for why I feel this is important to explicitly capture in the discussion about Pascal's triangle. Twoxili (talk) 17:01, 2 June 2023 (UTC)

Let me see whether I can summarize this argument concisely. The nth row of the triangle is
 * $${n \choose 0}, {n \choose 1}, \cdots, {n \choose n}.$$

If we interpret those integers as digits in a base-n expansion of another integer &mdash; let's call it k &mdash; then
 * $$k = {n \choose 0} n^n + {n \choose 1} n^{n - 1} + \cdots + {n \choose n} = (n + 1)^n = n^n \left(1 + \frac{1}{n}\right)^n.$$

As $$n \to \infty$$, the $$(1 + 1 / n)^n$$ on the far right limits to e, while the $$n^n$$ on the far right limits to infinity. So k limits to infinity, and $$k n^{-n}$$ limits to e.

First question: Am I correctly summarizing the argument?

Second question: If so, then is the result $$k n^{-n} \to e$$ notable? Mgnbar (talk) 19:00, 1 June 2023 (UTC)


 * Hi Mgnbar. You are correct in summarizing the argument. The value k is infinitely large (it has no decimal places). The statement $$k n^{-n} \to e$$ says "the hypothetical bottom row of Pascal's triangle in base $$n$$ is, on the first order of magnitude, $$e$$". As to your question of whether this is notable, this is also my question. Here are my thoughts. Firstly, when I search for "e in Pascal's triangle", the ratio of the row products turn up, but that's it. I've never heard anyone mention this explicitly, though I suspect it's probably taken for granted by most. While the result may not be very surprising, I think it deserves to be said for anyone just wanting to know all the ways e is connected to the triangle. What the argument says to me is that, as you calculate the rows from top to bottom, you're computing e. The triangle computes e. The casual student might find that quite interesting. This convergence is what the cited paper was getting at, but it was written without the benefit of the better framework we have today for discussing the triangle. At least for me, seeing the rows as powers of the sequence 11, and knowing that the root of e is the sequence 1.1 gives me a certain satisfaction with regards to wanting to understand what, if any, connections there are between e and the triangle. I'd be very interested to hear your thoughts about this. * This has been edited to clarify the meaning of the original comment. Twoxili (talk) 20:15, 1 June 2023 (UTC)
 * For posterity, the statement "the hypothetical bottom row of Pascal's triangle in base n is, on the first order of magnitude, e" is just a fancy way of stating the limit definition of e, which is why its noteworthiness in the below discussion is dubious. Twoxili (talk) 14:11, 5 June 2023 (UTC)


 * I'll try adding that parse of your statement: "the hypothetical bottom row of Pascal's triangle in base n is, on the first order of magnitude, e" if it helps clarify the point of interest. Twoxili (talk) 23:24, 1 June 2023 (UTC)


 * I guess my opinion right now is that this limit is not notable. It's a nice math fact, but not all nice math facts need to be on Wikipedia. (For example, no theorem that I have professionally proved is on Wikipedia.) I could be convinced otherwise, if a few Reliable sources were marshaled, to confirm that this limit is important, commonly used, or especially interesting. Mgnbar (talk) 23:32, 1 June 2023 (UTC)
 * I'll look at what's out there, and if I can't find anything, then I'll have to concur. Twoxili (talk) 23:43, 1 June 2023 (UTC)
 * Does the broad interest in the $$2^{n}$$ row sums and the $$11^{n}$$ row values not imply a similar level of interest in the shared mechanism behind these two phenomena? What I mean is, the reason why these two cases are generalizable across bases is due to the fact that the rows converge to $$e$$ when $$a = n$$. The base of the natural log is the mathematical "glue" that connects the bases and gives their powers identical numerical expressions. This was Prof. Morton's point, but he didn't say it in these terms (apart from including "11" in the title of his article), and for that reason I think the point has been overlooked. We've been discussing the implications of this principle but not the principle itself. I've updated the intro of this thread to propose this description replace the The value of a row bullet point under the Rows section. * This has been edited Twoxili (talk) 10:35, 2 June 2023 (UTC)
 * The divisibility rule for 11 is generalizable to $$11_{n}$$. This is a corollary of Morton's paper and a couple of follow-up papers by author's other than Morton were written on this topic (I'll add these citations [Edit: the total is 6 now compared to the 1 citation in the earlier revision]). That further establishes the relevance of $$11$$ to discussions about the triangle or binomials in general. When we make statements about $$11$$, we're making statements about the root of $$e$$, which is the common thread through the bases that makes this generalization possible. Twoxili (talk) 14:44, 2 June 2023 (UTC)
 * When we make statements about 11, we're making statements about the root of e You keep repeating versions of this statement, and I am perfectly happy to accept that you believe it strongly. But Wikipedia is based on what reliable sources write, and until there is evidence that reliable sources explicitly make this connection, I don't think you're going to find a lot of support to include it in the article.  (Stackexchange sites, which you mention below, are WP:USERGEN and so generally unacceptable.)  --JBL (talk) 17:17, 2 June 2023 (UTC)
 * I also agree that the connection is not explicitly asserted in the paper. And I do realize the importance of keeping original content out of the articles. I agree Morton and the other authors who wrote follow up papers might not have realized they were talking about $$e$$. I'm not claiming they did. My question, though, is if an author uses the Leibniz formula for $$\pi$$, or the Euler formula, rather than the symbol $$\pi$$, and makes statements about those formulas, is it not okay for the Wikipedia article on $$\pi$$ to reference the paper, and replace those formulas with the symbol $$\pi$$ when paraphrasing the paper, if it helps get the points of the paper across to an audience interested in $$\pi$$? Alternatively, if I remove all instances of the symbol $$e$$, and replaced them with expressions involving $$1.1_{n}$$, will the edit be acceptable? because the content would not change as a result of that refactoring. The limit of $$1.1^{n}_{n} = (1 + \frac{1}{n})^{n}$$ is the constant $$e$$ as $$n$$ approaches $$\infty$$. This is a natural language description of the limit equation of $$e$$. There are many more descriptions of $$e$$. The limit of $$11^{n}_{n}$$ is the last row of the base-$$n$$ triangle as $$n$$ approaches $$\infty$$. That is the binomial theorem as applied by Morton. The former description is on the first order of magnitude of base $$n$$, and the latter is on the last order of magnitude of base $$n$$. Morton's paper, by choosing to discuss 11 in arbitrary bases (radix $$n$$) and its powers as rows of Pascal's triangle (which is unbounded at its base), is tantamount to a discussion about $$e$$. * This has been edited to correct several errors in describing e and Morton's use of the binomial theorem Twoxili (talk) 17:33, 2 June 2023 (UTC)
 * I do not think your analogy is even slightly apt. Rather than try to force your (idiosyncratic, to say the least) views into this article, you should try to improve Wikipedia by making sure it reflects accurately what the best-quality sources have to say on the subject.  Otherwise, you run a significant risk of becoming disruptive.  --JBL (talk) 17:51, 2 June 2023 (UTC)
 * I greatly appreciate your feedback. Twoxili (talk) 17:53, 2 June 2023 (UTC)
 * I've removed all instances of e, and changed the language to focus on explaining the paper faithfully and succinctly. Does the revised version come any closer to resolving your concerns? Twoxili (talk) 22:19, 2 June 2023 (UTC)
 * So far I was able to find this math.stackexchange.com post. I realize this may not be a standard source, but it is an example of what I think is a big problem about this topic. Here, the OP discovered this property in reverse, but never got an answer as to what he discovered. He appears to have started by looking at $$(1.1)^{n}$$, and realized the numbers match the Pascal rows with decimals, and asks about the relationship of $$e$$ to the triangle (he even cites the row product ratio connection as something he wanted to know more about, just as I did). He offers a conjecture that turns out to be incorrect. The accepted answer correctly explains that the binomial $$a^n \left( 1 + \frac{1}{a} \right)^{n} = 11^{n}_{a}$$ is the culprit, but not in those terms. The answer fails to explain why $$a^n$$ is essential for the convergence to work, and instead demonstrates arithmetically why OP's approach doesn't work. According to my writeup, the OP's product is off because $$e$$ cross-cuts the sequence of radix-$$a$$ triangles at row $$n$$, and moving off that plane/triangle will move you away from $$e$$: "If it drops faster or slower, the limit will be smaller than or larger than $$e$$" the answer explains in describing the asymmetry of OP's scaling, but what I wish to draw attention to is how it goes about explaining it. The answer makes no mention at all of radices. It didn't come up in the comments either. And no one comments that OP is partially correct in seeking a formula for $$e$$ in the Pascal rows: "Is there any way to make this series converge to $$e$$ with the adding of zeroes suitably as needed?" Yes, but you need to think of $$e$$ as moving across bases forever towards its limit. The OP posits "$$n$$-many $$0$$s" between $$a = 1$$ and $$b = 1/n$$, so he intuits the scaling of $$n$$ to the power $$n$$ (he neglects to scale the other term, $$a$$, of the binomial), but the answer simply responds in binary. The answer mentions another indication that you're off the plane $$e$$: "... then even in the limit the first digit (the only digit before the decimal point) is always $$1$$, which is the left edge of the triangle", but does not mention that a first digit of $$2$$ nevertheless holds for all rows $$a = n$$ for $$n > 2$$ in base $$n$$ (because these rows are indeed converging to $$e$$). In my writeup, I note this also. The last digit of OP's product is $$1$$ because the penultimate $$1$$ will not carry in the normalized value of the row in OP's scheme. That 1 is one of the four a priori values in the limit. If you don't know this, then the triangle's leftmost diagonal will mislead you to believe the convergence is not possible. I myself nearly dismissed the idea because of this, and OP actually does: "Yeah I guess that was a blunder on my side. I guess there is indeed no way." I also show that one application of this fact is that we know the exact length of the partial product's numerical value. The reply appears to do a nice job of explaining why the OP's conjecture doesn't converge to $$e$$ (not all of which I understand after my cursor look at it, I admit), but fails to mention the useful fact that the rows are, in fact, partial products of $$e$$. More to the point, I believe the OP would have appreciated and might have been enriched by a more thorough explanation of the intricate relationship between $$e$$ and the hypothetical bottom row. I think my writeup is what the OP might have been looking for, or would have at least found interesting. The post has 779 views, so it seems he and I are not the only ones interested in this relationship. I'll continue searching (scholarly searches are turning up nothing so far), but do you think this post lends any credit to why it's appropriate to add this to the Wikipedia page? * This has been edited Twoxili (talk) 01:29, 2 June 2023 (UTC)
 * Here is another post that get's the row formula's structure right (but fatally fails to equate $$n$$ and $$k$$). I can't tell by "make's sense at all" whether OP asks how to normalize the the $$k$$th row numeral or if the question is whether the presented formula describes the $$k$$th row, but his second question seems to ask whether the rows have a limit. As stated, his conjecture does not converge. The post has no answers, but over 100 views. Twoxili (talk) 11:51, 2 June 2023 (UTC)
 * Since we know that the ratio of the ratio of successive row (coefficient) products converge to $$e$$, and we also have a definition of the row that converges to $$e$$, the rows are partial products of the product ratio growth. That connection could be useful in understanding the product ratio growth, a topic that is widely of interest. Twoxili (talk) 03:04, 2 June 2023 (UTC)

First sentence

 * In mathematics, Pascal's triangle is a triangular array of the binomial coefficients that arises in probability theory, combinatorics, and algebra.

or
 * In mathematics, Pascal's triangle is a triangular array of the binomial coefficients that arise in probability theory, combinatorics, and algebra.

i.e., is "arise" plural or singular? User:Maryanne Cunningham recently changed "arises" to "arise", and User:JayBeeEll reverted. I'd say both are possible - the triangle arises, but the coeffficients also arise. I think the "arise" version is marginally easier to read - and, in fact, you could say that it is the coefficients that arise in the fields mentioned, where as the triangle is merely a (slightly quirky) way of displaying them. Nø (talk) 19:09, 19 June 2023 (UTC)


 * The subject of both the verb "arises" and the article is undoubtedly Pascal's triangle. To make the subject of the verb be "binomial coefficients", one would write, "Pascal's triangle is a triangular array of binomial coefficients, which arise in ..." -- but since the article is about Pascal's triangle, it would be odd to have the first sentence be about some other object.  (I agree with you that the sentence structure is slightly difficult to parse, even for native speakers.)  --JBL (talk) 20:18, 19 June 2023 (UTC)
 * How about
 * In mathematics, Pascal's triangle is a triangular array of the binomial coefficients arising in probability theory, combinatorics, and algebra.
 * It circumvents the question, and is shorter. I (a nonnative speaker) find that it is perhaps marginally easier to parse. Nø (talk) 07:57, 23 June 2023 (UTC)
 * Yes that seems reasonable. --JBL (talk) 17:56, 23 June 2023 (UTC)
 * Done. Nø (talk) 12:53, 25 June 2023 (UTC)