Talk:e (mathematical constant)

Recent Stochastic Representations

You may ignore this, but I think the recent changes to Stochastic representations are a funny misinterpretation of the mathematics in general and the pheneomena in particular, and that it was correct before those recent changes, but only almost. My opinions are laughable. I mean that literally, for I am not a mathematician.

The nature of e in chance theory:

• It's not a representation, its a property. (Resist the temptation to use it as a competitive algorithm.)
• It's not "representations", its a unique stochastic property that might be called "exceed me".
• U(0,1) is "a closed standard realm" for random variables, the equivalent of some function random(X), as simple as the simplest operation we might call "pick a card". (Please don't say "U=min<anything>", say V or something.)
• One sample is usually three random variables, after that it's a simple average of a secondary population of n. There is no more sampling done after the simple ones that makes the n's happen.

What it should return to, almost:

In addition to analytical techniques and expressions involving e, there is a unique stochastic process that acertains e. If random variables X₁, X₂, ..., X_n of sample size n from the standard uniform distribution are limited such that

$V= \min { \left \{ n \mid X_1+X_2+\cdots+X_n > 1 \right \} },$

then the expected value of V is e, or E(V) = e.

I say all this because I'm working hard on the discussion section above this one, and I really want to get lambasted by an understanding mathematicians so my proposal for this section will become correct. But in all respect, I am not wholy unqualified to speak, or even revert the edit, just (perhaps overly) curious about the resolution of these recent discussions. — CpiralCpiral 07:52, 18 November 2011 (UTC)

Agree calling the section something like 'Stochastic estimation' would be better and U was a bad choice for the sum. Otherwise the English bit there at the moment is better, we need to remember the target audience. The method is simple not unique. We should be pretty explicit rather than just referring to 'standard'. I would go for something like

There is simple stochastic process that estimates e. The number of random variables with a uniform distribution on [0, 1] that need to be added up to first exceed 1 is on average e. In other words if X_i ~ U(0, 1) and

$V= \min { \left \{ n \mid X_1+X_2+\cdots+X_n > 1 \right \} },$

then the expected value of V is e, or E(V) = e.

and if it can be simplified a biut more then well and good. Dmcq (talk) 10:17, 18 November 2011 (UTC)

Looks good, though there's no need to write 'simple' as readers can decide that for themselves, per WP:EDITORIAL.--JohnBlackburne^words_deeds 10:43, 18 November 2011 (UTC)

No. It is simple to anyone, for who can count using natural numbers? even most crows. The arithmetic part is taking the mean. What you are refering to is more complex, is generally misunderstood, is the nomenclature and naming conventions the equation packs in. It's too formal. Please review the proposal without bias. Four days ago in our discussion, starting at "Encyclopedia" near "Heuristic" I was doubly sure of it. As always I have you to thank. Thank you. — CpiralCpiral 16:39, 18 November 2011 (UTC)

No. The stochastic process does not "estimate e", the mean does.

The uniform distribution is complex conceptually, but I have not seen its equivelent, some unit line segment, to refer to as I have seen unit square and unit cube. We could omit mention of the thing except for the formal need of this section to use "random variable X", and say a much wordier "a random selection n over the closed, real interval [0,1]", but... the section is only a compact, formal explication. The uniform distribution is not a number line, its a equiprobability space that may contain much more mathematical tooling than a simple output of one number. The standard uniform distribution is "like" the missing "unit line segment" except that a random variable may "return" an interval, not just a number. Where is the unit line segment when you need it to build a unit square or cube? Nowhere that I know of.

Unit interval — CpiralCpiral 11:06, 22 November 2011 (UTC) Embarrassed red. (Was green)

Furthermore about the U: the original version was mis-copied from the less-than stellar reference I just now studied, where they decided to use U. (They were high school teachers?) The copy left out the difference between n and N, which I pointed out 12 days ago. The other reference bar graphic from some fancy math-suite app is inferior to to the numeric output the C program (above section) performed for me: ten steps, three numbers (illuminated me 20 days ago. BTW, 4 days ago, I finally got the terminology and nomenclature. Ya know... I am the wider audience I write for in the proposal above. What's 20-4 days to me? A lot of effort for widening the audience per WP:CLASSES. )

About the section title. It should be Stochastic under Properties. — CpiralCpiral 16:39, 18 November 2011 (UTC)

No no no. These remarks all miss the point of this section. The article has now been saying for 12 hours that there are an infinite number of X's. You should watch more carefully what the words are saying everywhere: in your watchlist, in my proposal, and in these talk page sections here and the one above, and in the references. Now I'm becoming bolder to add my proposal outright and get some more eyes. Let us conclude. Let us resolve, not the courts please. That's inefficient. — CpiralCpiral 16:39, 18 November 2011 (UTC)

And I've reverted it as it made things much worse. In particular: it's not unique; 'acertains' apart from being misspelt does not make sense; there are not only n variables, as n is undetermined; the standard uniform distribution is not well known and should be specified; 'limited' like 'acertains' is poor English and also does not make sense.--JohnBlackburne^words_deeds 17:00, 18 November 2011 (UTC)

Remarkably, you restored the version with infinite X's, reverting my reversion. (I made an honest mistake between "::" and ":" and interpreted your "Looks good" as a go ahead, when in actuallity it was your comment to Dmcq. My apologies. Additionally, thanks for correcting the discussion's order.) More importantly do you mean that n is infinite?. I thought it was usually 3. — CpiralCpiral 21:10, 18 November 2011 (UTC)

Yes, "ascertain". But reconsider "limited". In fact the infinite X's error is an easy misunderstanding without it. The standard uniform distribution is well known. That's why they call it a standard. I am still emboldened, even without "proper English", and maintain that the section needs to be renamed and moved (see above), and re-written (see previous). — CpiralCpiral 17:13, 18 November 2011 (UTC)

Question by an outsider: is it really unique? I would think that there are many stochastic techniques that approximate e. CRGreathouse (t | c) 20:10, 18 November 2011 (UTC)

I'd guess there's a way of transforming the secretary problem for instance into one so I doubt one would have to look far. I think they must have had a different idea of unique like saying 'very unique' to mean something specially interesting. Saying very unique grates on me so please don't! :) Dmcq (talk) 20:53, 18 November 2011 (UTC)

Looks better as it stands now. And, as CRGreathouse and Dmcq suggest, there are other stochastic techniques that approximate e. -- 203.171.197.251 (talk) 23:35, 18 November 2011 (UTC)

Pray tell, most honorable and welcome non-parameterized statistic, why does not the most recent version

$U= \min { \left \{ n \mid X_1+X_2+\cdots+X_n > 1 \right \} }$

not say

$U= { \left \{ \min { n \mid X_1+X_2+\cdots+X_n > 1 } \right \} }$

More to the point I ask you hopefully and expectantly, what, recently, is "infinite" about X, a sequence or a series? (Does "+" mean "," i.e. the Complex plane?) According to the recent wording its an "infinite sequence". Assuming standard notation is used, its a finite series. All this assumes the awesome E(U)=e, and therefore that my personal role as (Cpiral) has an an unstated function domain U = {2,3,4,3,2,3,2,5,3,3...} — CpiralCpiral 01:55, 19 November 2011 (UTC)

Because this

$U= \min { \left \{ n \mid X_1+X_2+\cdots+X_n > 1 \right \} }$

is correct while the other is meaningless. As for infinite that's needed for the problem. U could be any integer: with decreasing probability it could be 2, 3, 4, ... but every integer has a finite probability. Therefore the range of U is infinite and you have to have an infinite number of X_i to test it. If there are only finite X_i, 1000 say, then E(U) will not be e but slightly less.--JohnBlackburne^words_deeds 02:20, 19 November 2011 (UTC)

Thank you so very very much. — CpiralCpiral 21:07, 19 November 2011 (UTC)

I think it's worth pointing out that U (which I've just changed to V in the article) is a random variable. It's a random variable defined in terms of the infinite (yes, infinite) sequence of random variables X₁, X₂, X₃, .... It's because V is a random variable that it makes sense to take its expectation. Ozob (talk) 03:21, 19 November 2011 (UTC)

Thanks! But about the infinite X's part: the equation has metastasized. The heralded section has gone beyond "too formal" being one step worse for me. The English use of the symbols are wrong because they are not about the equation, but rather they are, for me, now biased and intuitively comfortable, in the advanced algebraic equation, which homes sui generis, multi-purposed symbols. I am also for homophonics in English, but not when it is, as it should be here, about a formal math equation. I've acquired a bias for this equation, except the new "infinite X's" part, the one putting what I will now call Me, the prototypical "next step audience", N_e, back into confusion from a curiosly driven talkpage breakthrough. It is for the turn of Me now as JohnBlackburne says "worse". (Given my authorship of the previous proposal for the section as the only step forward, I am actually not confused myself, but Me is now two steps backward.) Let me explain My position, analyzing the one, too obvious confusion from the several angles offered therein, and I quote.

X_n: The X_n is either a series of trials "X₁ + X₂ + X₃, ...." in a sample limited by "> 1", or a sequence of samples "X₁, X₂, X₃, ...." not limited at all. X_n is portrayed as both a trial — what constitutes the continuous random variable X, and is what the binary operator + on X_n require — and a sample of trials — what the discrete random variable V accumulates for E — each denoted X_n. The same symbol, X_n is now heralded in the English Wikipedia to mean, both at once, the continuous random variable trials and the discrete random variable, whose samples comprises the number V. Which is it?

'n' or 'N':The succesor to the last n < 1, i.e. the first n > 1, i.e. the min n, let us denote as N, as the referenced source to the equation does. The article says "Let V be the least [discrete random variable] n", so "n" is now associate with both X and V. Is n multipurposed here?. Isn't that situation unacceptable by Wikipedia standards? Granted, both n and N are on [2,∞), but n is stochastic, and N is a blip at the ending in an infinite-loving mean operation on E(V(N1, N2, ...N∞)) To Me, n seems finite, limited by "< 1" with an X that seems finite by association. Is there an unstated-in-English, but implied step with N = the successor to the nth term, and a similarly implied step that V is multivalued? I think these are so, for I use implied steps plentifully in my proposal (previous section). Let it stand, then, but without "ininite X's". Let it be stated though, in a way that differentiates n from N, not so error-frought in it's implications, as does my proposal (in the previous section).

sum(X) and avg(V): Each X_n is a trial, and each V_N is a sample. The article says "sum of the first n samples", similar to where historical versions have said "sample averages of U". The new version confuses "sample" with "trial". The older version was more accurately stated.

Random variable: The article says "independent random variable" X, while the wikilink says only that there is either a "discrete random variable" (multivalued, like V) or a "continuous random variables". Why say an "independent random variable"?

Please help Me if you can. — CpiralCpiral 21:07, 19 November 2011 (UTC)

No, the current version is clear and correct. You are simply seeing problems that aren't there that seem to arise from your own misunderstanding of the techniques being used. As such, because this article is not about stochastic methods, you probably need to look elsewhere for help with this, to other articles or more probably sources other than Wikipedia, as Wikipedia serves poorly as a textbook on topics you don't understand.--JohnBlackburne^words_deeds 22:12, 19 November 2011 (UTC)

I do not understand your first paragraph at all.

In the present context, a trial is the same thing as a sample.

You should not denote the minimum n by N. It already has a name, V.

You seem to be misunderstanding some of the terminology used in the article. I believe that it is within your reach, but you will have to study more probability first. Ozob (talk) 22:44, 19 November 2011 (UTC)

My pained style is discomforting, and yours is an admirable example.

• We should differentiate "trial" from "sample" precisely as is done in Sample (statistics) because, as it says in the sourced reference [11], "Students will generally be comfortable with the notion of the sum of a fixed number of random variables, but the sum of a random number of random variables may cause some difficulty." N is a sample, and n is a trial. Now make both variables random variables. Yes?
• X_n: Still, the X's are each presented, indistinguishable symbolically, as two different entities differing in cardinality. Are you saying that's not confusing? Yes infinitely? — CpiralCpiral 00:11, 20 November 2011 (UTC)

It's perfectly clear. Again, if you don't understand what is a straightforward example you need to look elsewhere for explanation. --JohnBlackburne^words_deeds 00:28, 20 November 2011 (UTC)

I'm sure "it" is, but this does not seem about you or me in the previous section. What is "it"? "It" remains unspecified.— CpiralCpiral 01:11, 20 November 2011 (UTC)

By 'it' I mean the current version, the subject of my previous comment. It's clear and correct and if you don't understand it you need to look elsewhere for such understanding.--JohnBlackburne^words_deeds 01:29, 20 November 2011 (UT

Specifically which part of X and "sample" is clear, and how am I to be satisfied by more than mere strokes of keyboards? You did ask the WikiProject:Mathematics to help us. Now I'm going out, to help someone I love. Goodbye for now. — CpiralCpiral 02:27, 20 November 2011 (UTC)

┌────────────────────────────────────────────────────────────────────────────────────────────────────┘The whole of the subsection is clear: if you do not understand it then you won't find the information here: the articles linked in that section would be a good place to start. I posted to the WikiProject Mathematics talk page as you and I were at a standstill. It is normal in such circumstances to ask for a third opinion. There is a semi-formal mechanism for this at WP:3 but as it was a mathematical disagreement it made sense to ask mathematicians for help resolving the dispute.--JohnBlackburne^words_deeds 02:40, 20 November 2011 (UTC)

OK, thank you for your perseverence. Then the three links and two refs in the section should work, and

1. the n of X_n really is both limited (the section rightly says "the least number n") and infinite (the section strongly implies "an infinite sequence" with n terms), making that
2. trial and sample really are the same word because V is a function that generates one sample with infinite "trials" in a single probability space (not a multiset): all possible values of V and similarly all possible values of each X_n at once, making
3. a "+" operator with really unusual meaning that is
   1. not unlike the sum of probability spaces I document in my proposal, and also
   2. not unlike the "+" in a complex number
4. and an unusual set notation really has extensional definitions that go outside the curly bracket.

I just might enjoy trying {{rfc}} on the previous section (style and content) much later. (You are all right. I did not know that, but I have a curious intuition Zeno, Reimann, and Deutche are blah blah blah.) — CpiralCpiral 08:36, 20 November 2011 (UTC)

The "+" and the set notation are used in the same manner that they are used in thousands of first textbooks on probability theory, as is the word "random variable" (as opposed to "sample", "trial", and other terms that have been tossed around). This isn't the place to argue against completely standard conventions in probability theory. Also, there really are infinitely many X_i. Sławomir Biały (talk) 14:25, 20 November 2011 (UTC)

The siren call of any noumenon or textbook arouses my sentiment, but I rather feel a multiply beneficial duty to the wiki, and to the participants here, whose every attention fills me with honor. Most importantly I will take my three weeks of study on it, turn it into three minutes for an audience at my level.

Prime Obsession (2004) p 15, says "Every mathematical statement that contains the word 'infinity' can be reformulated without that word" and "Modern analysis does not admit [the concept of the infinite and the infinitesimal]" because they "created serious problems in math during the early nineteenth century...and eventually they were swept away altogether in a great reform." Ugh. Ugh.

Perhaps something like

Consider the terms of a sequence of partial sums generated by n random selections from some finite, zero-based, continuous or discrete, interval until an N^th term exeeds the size of the interval. The expected value of N is e.

A more visual approach considers the random field of each N-term event, each in its own probability space, and then adds them. Here

$e = \sum_{n=2}^\infty { n (n-1)/(n!)}$.

The temporal version of this is e = E(N) where N is a random variable composed of random variables X₁, X₂, ..., drawn from the uniform distribution on [0, 1] such that

$N = \min { \left \{ n \mid X_1+X_2+\cdots+X_n > 1 \right \} }.$.

Besides, "infinite X" is excess verbiage, like the finest print in the rules of a ball game, since it is implied in the theories employed, and N is usually around 3. — CpiralCpiral 21:52, 21 November 2011 (UTC):

That makes no sense: how is a formula a 'visual approach'? How is derived? How is it related to what's there? This could maybe fixed with the lengthy geometric digression in your original proposal, but we've already determined that has no place here.--JohnBlackburne^words_deeds 22:00, 21 November 2011 (UTC)

Thanks. I'll take "maybe" and the questions you ask, and I'll remember that you only make it seem very much like, really like, you are the consensus. (Others like(d?) it short.) That's a fearsome burden that works to motivate only some few offerings. I'll get it right. Happy editing, Gonzales! — CpiralCpiral 10:09, 22 November 2011 (UTC)

┌────────────────────────────────────────────────────────────────────────────────────────────────────┘Well, it is possible to say it without infinite numbers of random variables, but it would be much more complicated. If we're going to include the iid in the more complicated version, we should do it in the simpler version, even though it may not be necessary.

Let V_N be constructed as follows: Let X₁, X₂, ..., X_N be independent random variables with uniform distribution ~U(0,1). Let V_N be the random variable taking the least number n such that the sum of the first n of the X_i is greater than 1, defining V_N = N + 1 if no such n exists. In symbols,

$V_N = \begin{cases} \min \left \{ n \mid X_1+X_2+\cdots+X_n > 1 \right \} & X_1+X_2+\cdots+X_N > 1 \\ N+1& \text{otherwise} \end{cases}$

Then $\lim_{N \rightarrow \infty} E \left( V_N \right) = e$

I fail to see how this is simpler than:

Let X₁, X₂, ..., be independent random variables with uniform distribution ~U(0,1). Let V be the least number n such that the sum of the first n samples exceeds 1:

$V = \min { \left \{ n \mid X_1+X_2+\cdots+X_n > 1 \right \} }.$

Then the expected value of V is e: E(V) = e.

I might include an additional note that V almost surely exists. — Arthur Rubin (talk) 16:16, 22 November 2011 (UTC)

Based on Cpiral's quotation, he seems to believe that modern mathematics is not supposed to use "infinity" (like "infinite sequence", "infinite set"). This, no doubt, is based on his misunderstanding of a pop-math book that he read. Read a book on modern analysis (like Rudin or Apostol). Flip to the chapter of "Infinite sequences and series". Sławomir Biały (talk) 12:53, 28 November 2011 (UTC)

Confusing Antiderivative Proof/Section

The following section shows a calculation regarding the antiderivative of e^x. It states that the antiderivative of e^x is e^x , but I do not feel that it shows that clearly. The last statement shown does not imply this directly to the layperson (in my opinion.) Can someone review it? Thanks! Sdegan (talk) 17:57, 25 November 2011 (UTC)

Calculus

As in the motivation, the exponential function e^x is important in part because it is the unique nontrivial function (up to multiplication by a constant) which is its own derivative

$\frac{d}{dx}e^x=e^x$

and therefore its own antiderivative as well:

$\begin{align} e^x & = \int_{-\infty}^x e^t\,dt \\[8pt] & = \int_{-\infty}^0 e^t\,dt + \int_0^x e^t\,dt \\[8pt] & = 1 + \int_{0}^x e^t\,dt. \end{align}$

Sdegan (talk) 17:57, 25 November 2011 (UTC)

It doesn't prove it but that's a bit of work to do, and not especially enlightening. It does give two forms for e^x, as a single integral and as 1 + an easier/finite integral, and relates them, which is useful though I'd do it all in one line, perhaps clarifying that it's just a statement:

$e^x = \int_{-\infty}^x e^t\,dt = 1 + \int_{0}^x e^t\,dt.$

--JohnBlackburne^words_deeds 18:35, 25 November 2011 (UTC)

Thanks for the clarification JohnBlackburne. That makes more sense. It's positioning had me feeling that it was intended to be more of a proof. Sdegan (talk) 06:24, 16 December 2011 (UTC)

A consistent style for series notation

I find no recommendation in the MOS, so I'm not surprised we have an inconsistent style for the series notations. Compare:

$p_n = \sum_{k = 0}^n \frac{(-1)^k}{k!} = \frac{1}{0!}-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+-\cdots+\frac{(-1)^n}{n!}.$

$e = \sum_{n = 0}^\infty \frac{1}{n!} = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \cdots$

$e^{x} = 1 + {x \over 1!} + {x^{2} \over 2!} + {x^{3} \over 3!} + \cdots = \sum_{n=0}^{\infty} \frac{x^n}{n!}$

Recommend?:

$p_n = \sum_{k = 0}^n \frac{(-1)^k}{k!} = \frac{1}{0!}-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}\pm\cdots$

$e = \sum_{n = 0}^\infty \frac{1}{n!} = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \cdots$

$e^{x} = \sum_{n=0}^{\infty} \frac{x^n}{n!}= 1 + {x \over 1!} + {x^{2} \over 2!} + {x^{3} \over 3!} + \cdots$

— CpiralCpiral 22:14, 28 November 2011 (UTC)

I would get rid of the 0! in all cases, replacing it with one. I also don't like the recently-added $+-\,$, and would suggest that we get rid of that as well. I prefer the (original) ( − 1)ⁿ to the (proposed) $\pm$, because it more clearly implies that the signs alternate. (Also, your recommendation for p_n is missing its last term.) Sławomir Biały (talk) 22:22, 28 November 2011 (UTC)

plus-minus sign (±) is commonly used "to indicate a value that can be of either sign". We don't need to repeat the same mechanism on both sides of the equation for p_n.

Zero in the denominator is not pretty? 0! is not intutive? Agreed on both. And if aesthetics is not to be over-ruled by mere mechanics, then, we might smooth the compaction's echo:

$p_n = 1+\frac{1}{2!}-\frac{1}{3!}\pm\cdots+\frac{(-1)^n}{n!}$

$e = \sum_{n = 0}^\infty \frac{1}{n!} = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \cdots$

$e^{x} = \sum_{n=0}^{\infty} \frac{x^n}{n!}= 1 + x + {x^{2} \over 2!} + {x^{3} \over 3!} + \cdots$

— CpiralCpiral 23:06, 28 November 2011 (UTC)

This looks just wrong for p_n. I've never seen an alternating series written this way. Sławomir Biały (talk) 23:38, 28 November 2011 (UTC)

Right... Then I conclude here. (Edited 18:10, 30 November 2011 (UTC))

$p_n =1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\cdots+\frac{(-1)^n}{n!}= \sum_{k = 0}^n \frac{(-1)^k}{k!}$

$e =\sum_{n = 0}^\infty \frac{1}{n!} = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \cdots$

$e^{x} = 1 + \frac{x}{1!} + {x^{2} \over 2!} + {x^{3} \over 3!} + \cdots= \sum_{n=0}^{\infty} \frac{x^n}{n!}$

Article consistency "rules", but sectional aesthetics, and practice in the field, matter more.

— CpiralCpiral 00:06, 29 November 2011 (UTC)

Summary: There are two, stylistic edits to make. e^x is the "model equation" .

Exigesis: Having 0! there is not aesthetic, Having "1! + 2! + 3!" is now "the model". 0! is tempting to see when $\textstyle\sum$, is in on the left.

— CpiralCpiral 20:07, 29 November 2011 (UTC)

Continued fraction notation

This notation from the article

$e = [[2; 1, \textbf{2}, 1, 1, \textbf{4}, 1, 1, \textbf{6}, 1, 1, \textbf{8}, 1, 1, \ldots, \textbf{2n}, 1, 1, \ldots]], \,$

is not supported at Continued_fraction#Notations_for_continued_fractions. May I change it to single square brackets here, or should someone update Continued fraction? — CpiralCpiral 02:38, 30 November 2011 (UTC)

Next I would like to change the <;> to a <,>. Shouldn't we show only the changes provided by the borrowing of the one, and write:

or in its more concise form

$e = \,$               $[2, 1, \textbf{2}, 1, 1, \textbf{4}, 1, 1, \textbf{6}, 1, 1, \textbf{8}, 1, 1, \ldots, \textbf{2n}, 1, 1, \ldots], \,$

which can be entirely harmoniously written by inserting the zero:

$e = [1 , \textbf{0} , 1 , 1, \textbf{2}, 1, 1, \textbf{4}, 1 , 1 , \textbf{6}, 1, 1, \textbf{8}, 1, 1, \ldots]. \,$

— CpiralCpiral 09:42, 4 December 2011 (UTC) which can be written

If you're going to follow a standard then you should follow it so why remove the semicolon? Also there is no particular point to putting in the zero, one might as well then make an argument to go back to -2 or something like that. Dmcq (talk) 13:36, 4 December 2011 (UTC)

I thought the point was that there was a clearer pattern if you insert the zero, whereas without it there is an exceptional term. Sławomir Biały (talk) 14:03, 4 December 2011 (UTC)

OK, then let's keep the semicolon in both versions. But the semicolon is only an option:

• "it is customary to replace only the first comma by a semicolon."
• "The semicolon ... is sometimes replaced by a comma." — CpiralCpiral 21:39, 4 December 2011 (UTC)

Showing the "chalkboard" continued fraction when we're mainly interested in representative forms is unecessary, but I encourage it for it's "inviting read" effect. To be consistent with this style, we need to make them both semicolons, for that removes distraction, and keeps the focus on this page. As far as standards go, I have just learned that zero violates the nature of the intention of the concise form, which is have no redundancy; but neither the fact that it is non-standard to insert a zero in the continued fraction's standard nomenclature, nor that one form is more "harmonious" need be said. We want the reader to take in many things on faith while staying on track to finish the article in one sitting. Don't say "permitted" or "harmonious" or throw the nomenclature for loops, just say

or in its more concise form

$e = [2; 1, \textbf{2}, 1, 1, \textbf{4}, 1, 1, \textbf{6}, 1, 1, \textbf{8}, 1, 1, \ldots, \textbf{2n}, 1, 1, \ldots], \,$

which can be written:

$e = [1; \textbf{0} , 1 , 1, \textbf{2}, 1, 1, \textbf{4}, 1 , 1 , \textbf{6}, 1, 1, \textbf{8}, 1, 1, \ldots]. \,$

— CpiralCpiral 21:39, 4 December 2011 (UTC)

I don't think it really matters. Although from the point of view of thinking of the semicolon as the analog of a decimal separator, it doesn't really make sense for the second expansion. So I would just leave it out of both. Sławomir Biały (talk) 13:59, 5 December 2011 (UTC)

But the second expansion has got an integral part 1, just like the first has an integral part 2. Isn't this right:

$1 + \cfrac{1}{0 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{2 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{4 + \ddots}}}}}}} = 2+\cfrac{1} {1+\cfrac{1} {{\mathbf 2}+\cfrac{1} {1+\cfrac{1} {1+\cfrac{1} {{\mathbf 4} +\cfrac{1} {1+\cfrac{1} {1+\cfrac{1} {{\mathbf 6}+\cfrac{1} {1+\cfrac{1} {1+\ddots}}}}}}}}}}$? — CpiralCpiral 07:57, 6 December 2011 (UTC)

Both expressions, being equal to e, have integer part 2. This is why the semicolon is misleading. Sławomir Biały (talk) 14:01, 6 December 2011 (UTC)

Could it be, pray tell, that when zero is allowed, the semicolon is not allowed? In other words, does a generalized fraction have any such a notation? Please see the discussion at talk:Continued fraction#If zero is allowed.— CpiralCpiral 10:27, 7 December 2011 (UTC)

I do not agree that a semicolon implies a specific integer part; it is as Sławomir Biały says: like a decimal separator. A parallel is that a decimal number may be denoted 1.234=−0.766 (typically when using log₁₀ with tables); this uses a relaxation like the one being spoken of. So the semicolon makes perfect sense in both cases. I suggest using the semicolon or comma consistently to avoid the distraction of the difference. Both the regular/harmonious/neat and generalized/nonstandard aspects should be mentioned; the first is the point of the example, the second might confuse if not clarified. Quondum^talk_contr 11:41, 7 December 2011 (UTC)

A more general formula is e^1/n=[1;n-1,1,1,3n-1,1,1,5n-1,1,1,7n-1,...], which is a proper continued fraction when n is an integer greater than 1. Plugging in n=1 to get [1;0,1,1,2,1,1,4,1,1,6,...] seems natural even though it's now a generalized continued fraction. The bracket notation still defines a valid expression which can be evaluated even if the entries are not positive integers, the only problem is you lose some of it's nice properties such as uniqueness and guaranteed convergence.--RDBury (talk) 16:28, 7 December 2011 (UTC)

I'm guessing that this general formula is valid for all real n. It would be a nice addition to Exponential function#Continued fractions for e^x if you have a reference for it. It might also be cooler in this article to mention it as a special case (for n=1) of the general formula, rather than pre-substituting n. I think that the loss of the canonical form is not an issue. Quondum^talk_contr 16:53, 7 December 2011 (UTC)

Apparently it can be found as an exercise in Knuth's Art of Computer Programming, I'm only going by the snippet view in Google Books though. In any case it's fairly straightforward to derive it from Gauss's continued fraction.--RDBury (talk) 20:39, 7 December 2011 (UTC)

I consider myself the target audience for this article. I need a consistent representation between the two bracket notations so I don't have to concern myself with the confusion between a formula and its notation. A student need not know the esoterica of the numeral to do the numbers, nor a reader of e that of the bracket notation. I was confused by the unnecessary inconsistency. Now I'm not, but I want to make the edit reflect the me without my cognitive bias. — CpiralCpiral 19:37, 7 December 2011 (UTC)

I think this is an excellent way to approach it. My immediate reaction is to remove the unfamiliar "concise" notation and to put in the explicit continued fractions you have higher in this talk section. The pattern is still pretty clear; some mention of "regular" would focus the reader's attention on the pattern. Would that achieve the objective? Quondum^talk_contr 19:55, 7 December 2011 (UTC)

The objective has seemed to me to be concise, more precisely in Ozob's terms "I prefer short". Thus I would take your suggestion and propose that we replace the current generalized fraction of e and replace it with the generalized fraction that has the zero in it. Furthermore I would add a citation needed tag on the concise notation that has the zero in it.

$e = 2+\cfrac{1} {1+\cfrac{1} {{\mathbf 2}+\cfrac{1} {1+\cfrac{1} {1+\cfrac{1} {{\mathbf 4} +\cfrac{1} {1+\cfrac{1} {1+\cfrac{1} {{\mathbf 6}+\cfrac{1} {1+\cfrac{1} {1+\ddots}}}}}}}}}} = 1 + \cfrac{1}{{\mathbf 0} + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{{\mathbf 2} + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{{\mathbf 4} + \ddots}}}}}}}$

or in continued fraction notation

$[2; 1, \textbf{2}, 1, 1, \textbf{4}, 1, 1, \textbf{6}, 1, 1, \textbf{8}, 1, 1, \ldots, \textbf{2n}, 1, 1, \ldots]$

which can be written ^{[citation needed]}

$[1; \textbf{0} , 1 , 1, \textbf{2}, 1, 1, \textbf{4}, 1 , 1 , \textbf{6}, 1, 1, \ldots].$

See also Continued fraction#Regular patterns in e.— CpiralCpiral 01:24, 8 December 2011 (UTC)

Q&A out loud:

• The compact notation is only used for generalized continued fractions if there are, as required per the convention of the compact notation, "all ones" in the numerators of the continued fraction represented. Is there any other way to get to a generalized continued fraction with all ones than to add a zero by borrowing from the whole part?
• The semicolon is optional because like a "mixed number" expression 2& 71⁄100, the plus sign is optional. The first number in compact notation is usually the "whole number part" or "integral part" of the number represented, which is "an integer". The semicolon is an operator, while the commas are just separators.
• The semicolon is not like a radix point because the optionally signified "two compartments" have different bases (to say the least).
• The objection to using the semicolon to represent e when the first number is one is understandably like the objection to saying "I have one dozen and fourteen eggs". It involves the "measuring vs counting" aspect of that first number in that representation. Zero was born when people started counting in a positional notation that, previous to zero, measured "nothing yet" in that column or heap. Yes, we can now count on zero to count to zero.
• Yet "If zero is allowed" confuses math (zero) with audience (allowing). It really asks if the writer may take the unusual step of presenting something that does not match at the level of the "whole number part". The math allows it. — CpiralCpiral 23:30, 8 December 2011 (UTC)

You are putting a lot of energy into this. I must point out that this probably belongs in List of representations of e, and that this section is at presently inadvisedly attempting to duplicate a lot of the material there. It is already linked to as the "main article". In view of this, I'd suggest brutally trimming this section and putting the material that is here but not yet there on that page. I'd agree with your "Q&A" points, with the exception of the third bullet. Quondum^talk_contr 09:00, 9 December 2011 (UTC)

Such discussions are brewing. The article should document all of e's characterizations and properties and select a few of the many remarkable representations and applications. The sectioning and titling and inaccessibility and redundancy issues the article has are related to Representations, but my energies to address those issues are moving more towards checking the philosophy of the layout, and it's implications. — CpiralCpiral 04:36, 12 December 2011 (UTC)

Now I'm leaning towards some representations in continued fractions read something like:

$e =\lim_{n \to \infty}[2;1,\mathbf 2,1,1,\mathbf 4,1,1,\mathbf 6,1,1,\mathbf 8,1,1,...,\mathbf {2n},1,1] = [1;\mathbf 0,1,1,\mathbf 2,1,1,\mathbf 4,1,1,...]$

which written out looks like

$2+ \cfrac{1} {1+\cfrac{1} {\mathbf 2 +\cfrac{1} {1+\cfrac{1} {1+\cfrac{1} {\mathbf 4 +\ddots} } } } } = 1+ \cfrac{1} {\mathbf 0 + \cfrac{1} {1 + \cfrac{1} {1 + \cfrac{1} {\mathbf 2 + \cfrac{1} {1 + \cfrac{1} {1 + \cfrac{1} {\mathbf 4 + \ddots} } } } } } }$
— CpiralCpiral 04:50, 12 December 2011 (UTC)

I think this is an improvement on all counts, so I'd suggest replacing what's there for now. Then the question of how much detail to recap from the main article on representations can be considered separately. Quondum^talk_contr 07:25, 12 December 2011 (UTC)

Intro edits

I've edited the intro to better meet the requirements of WP:INTRO. The lede now begins with definitions used in major reference sources, the Oxford English Dictionary and the Encyclopedic Dictionary of Mathematics, among others (I've looked at a bunch), not a random calculus text. A knowledge of calculus should not be necessary to understand the first paragraph of an article like this.--agr (talk) 18:01, 4 December 2011 (UTC)

Looks fine to me. Sławomir Biały (talk) 18:08, 4 December 2011 (UTC)

A recent edit to the intro changed e = 2 + 1/2 + 1/(2×3) + 1/(2×3×4) + 1/(2×3×4×5) + … to e = 2 + 1/2 + 1/(2 × 3) + 1/(2 × 3 × 4) + 1/(2 × 3 × 4 × 5) + …. I think the earlier version is easier to comprehend. Comments?--agr (talk) 17:05, 8 December 2011 (UTC)

Agree. It's easier to parse without the spaces. Sławomir Biały (talk) 18:10, 8 December 2011 (UTC)

Common operators are always spaced in formulae. I don't see why this should be any different. — Edokter (talk) — 18:11, 8 December 2011 (UTC)

The MOS:MATH guideline in WP:MOS#Common mathematical symbols (fifth bullet) is to space operators (but technically that should apply to the division as well). My concern is that this formula maybe does not belong in the lead. If it is kept, then a more compact notation should be sought, e.g. e = ∑∞
n=0 1/n!. Quondum^talk_contr 18:22, 8 December 2011 (UTC)

(Technically, it's a fraction, not a division.) — Edokter (talk) — 19:19, 8 December 2011 (UTC)

Why is my link inappropriate?

It's our only goal to show people how the constant e can emerges using simple math.

• e, the understandable birth of the constant, on paper, on youtube

User: N lasters (talk)

The linked material appears to be a video, (with no sound to give an explanation?), showing calculation of e through what appears at a glance to be successive approximation. That does not explain any underlying principles, so it does not even live up to the byline "the birth of...". In my opinion, it has little value even as a tutorial in any context. To this must be added that in general tutorials and Youtube should not be linked to from Wikipedia; Wikipedia is a reference, not a classroom. — Quondum^☏✎ 07:09, 23 January 2012 (UTC)