Talk:E (mathematical constant)
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Is it possible to explain e in any meaningful sense to a layman, rather than just it being the 'base of the natural logarithm function'? That still doesn't mean a whole lot to the average person!
Why is the page called "E (mathematical constant)"? I've seen this number named "e" (lowercase) almost everywhere, and the only place, other than Wikipedia, where I've seen "E" (uppercase) is Mathematica. So, I think this page should be renamed to "e (mathematical constant)". --Fibonacci 00:47, 27 Apr 2004 (UTC)
- Fiddly technical reasons, I'm afraid. Added this to the page. --Aponar Kestrel 16:27, 2004 Jul 31 (UTC)
In The Book on Numbers by John Conway and Richard Guy, the number e is persistently called Napier's number. I know that John Napier more or less discovered logarithms, but is this really the correct name? -- JanHidders
I don't think that's too common; Weisstein lists it as "Napier's Constant", but the main entry is under "e". Encyclopedia Britannica doesn't list "Napier's Number" or "Napier's Constant" at all. Most people call it "the base of the natural logarithm", I believe.
e is still called Euler's number in many texts too introductory to worry about confusion with &gamma (Euler's constant).
Could somebody explain what `e' is useful for? It was always difficult for me to explain it to mathematical newbies.
And could somebody explain ei*π = -1, and why it is so? -- Taw
See The most remarkable formula in the world (where it is poorly explained to the layman, sorry!) -- drj
It's used mainly because it arises "naturally" in calculus, and is related to useful functions (eg., trigonometric and hyperbolic functions). A connection with pi is inevitable, as pi is related (via polar coordinates) to -1 and the trigonometric functions.
Zundark - as far as I know, you should be entitled to claim that you invented the word "miscorrection" :) Great stuff! - MMGB
- I'd like to take credit for it, but Google finds about 140 pages with this word. --Zundark, 2001 Nov 26
- No. It'd mean "correcting" something you believe to be wrong but it's actually right. In linguistics that's called hypercorrection.
D'oh! I really need to quit editing pages when I'm so tired I can hardly type straight. At least I got the sum notation definition right...--BlackGriffen
Could this be moved to e (base of natural logarithm), to make it more consistent with other disambiguated titles, and to allow for the pipe trick? -- Oliver P. 00:13 Feb 22, 2003 (UTC)
- How about e (number)? -- Tarquin 09:08 Feb 22, 2003 (UTC)
I think that e (number) would be easily confsed with E numbers The Anome
- e (mathematical constant)? -- Oliver P. 15:18 Feb 22, 2003 (UTC)
- Sounds okay. -- Tarquin
- Okay, I've moved it there. -- Oliver P. 00:12 Feb 24, 2003 (UTC)
I think, e should always be spelled e not E...
"The number e is relevant because one can show that the exponential function exp(x) can be written as e^x"
- I'm really confused what this is supposed to mean. Exp(x) is just another notation for e^x. The only thing I can guess is that this is supposed to mean that exp(x) is defined as unique function y' = y and y(0) = 1, and e^x means limit of a^x as a approaches e, where e is given as the limit in the article. But this is a rather convoluted observation to make. Revolver
- It's not just notation. For a^x where x is not rational, you have to define it in terms of exp(x) and natural log. So you need some construction of exp(x) to even define e^x for irrational x. -- Walt Pohl 18:17, 1 Mar 2004 (UTC)
- I don't follow. If you have exp(x) defined, you have e^x, and vice versa. THEY'RE THE SAME FUNCTION. It's just a different notation. To say, "you need some construction of exp(x) to even define e^x for irrational x." is kind of a tautology, like saying, "you need some construction of sine function to define sin(x) for irrational x." Well, sure, that's true for anything. While it's true a^x can be defined in terms of e^x for a &neq; e, it's just as true that e^x is already defined for irrational x as soon as e^x is defined, you don't need "exp(x)", well you do, since it's the same function. I really miss the point. The claim only makes sense if "exp(x)" and "e^x" are actually defined in 2 different ways. I gave an example of this above. You can define exp(x) all sorts of ways (2 in this article, as solution of diff eq, yada yada), to say two notations are equal without saying how they're defined isn't saying anything. Revolver 01:19, 12 Jun 2004 (UTC)
- I see, you're saying a^x = exp(log(a^x)) = exp(x*(log(a)), so e^x = exp(log(e^x)) = exp(x*(log(e)) = exp(x), that's certainly true, it's just not saying anything terribly interesting, IMO, beyond "log is the inverse function of exp" or something like that. So, I still don't get it. Revolver 01:23, 12 Jun 2004 (UTC)
I enjoyed both forms of the expansion (one was removed today). Of course they can be derived from each other, but I still found the patterns that each exhibit interesting. Here are the two forms:
- <math>e = \sum_{n=0}^\infty {1 \over n!} = {1 \over 0!} + {1 \over 1!}
+ {1 \over 2!} + {1 \over 3!}
+ {1 \over 4!} + \cdots<math>
- <math>e = \sum_{n=1}^\infty {n \over n!} = {1 \over 1!} + {2 \over 2!} + {3 \over 3!} + \cdots<math>
Can we put them both back into the article? Bevo 16:39, 21 Feb 2004 (UTC)
- My vote is to leave out the second one. It is trivially equivalent to the first one, and I had been planning to remove it myself. -- Dominus 16:51, 21 Feb 2004 (UTC)
- The reason I removed the second series is that it is completely identical to the first, i.e. if you write out the terms, they are exactly the same, so they are really the same series, not different, just expressed differently. You can see this by cancelling n to get 1/(n − 1)! and then do a shift of index, (drop on the index, up inside the sum). If the series were truly different, I would consider keeping it. Revolver 06:22, 27 Feb 2004 (UTC)
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"Keller's Expression"
The article says:
- In 1975, the Swiss Felix A. Keller discovered the following formula that converges in e ("Keller's Expression" Steven Finch, mathsoft):
- <math>e = \lim_{n\to\infty} \quad {\rm }\frac{n^n}{(n-1)^{(n-1)}} - \frac{(n-1)^{(n-1)}}{(n-2)^{(n-2)}} \quad {\rm for}\quad\left|n\right|>2.<math>
- This formula was published for the first time 1998 on Steven Finch's website www.mathsoft.com/asolve/constant/e/e.html. He refers to it as “Keller's Expression”.</pre>
This is a delightful formula, but I have some problems with the way the description is written. First, the referenced document is not available (404). Second, it seems unlikely that the formula was first discovered in 1975. Even if it was never published before (which I rather doubt) I think that's more likely to be because it is so simple to prove. The formula looks mysterious at first glance, but really it turns out that the left term approaches n·e and the right term approaches (n-1)·e; this can be proved in about two steps of simple algebra, directly from the definition of e. Google search doesn't turn up anything relevant for "Felix A. Keller" or for "Keller's Expression". So we have a formula here which could be discovered in ten minutes of idle tinkering by any bright undergraduate, but it's being credited to Mr. Keller as though it were a big discovery. That seems strange to me. Formulas usually only get names when they are important or at least surprising (Stirling's formula, Euler's identity) and this one is neither. -- Dominus 14:38, 10 Mar 2004 (UTC)
I got rid of this again in October 2004, and again today. -- Dominus 02:09, 20 Mar 2005 (UTC)
Proofs
I think we should add more proofs, eg, that the given continued fraction representation is correct.
Just for the record, I removed the bit about the Pyramids and Greeks. It smells of nonsense, was originally added in bold text, and I can't find any other references to either part of it anyway. (And given that the Greeks were not known for their imprecision in mathematics, I can't imagine they'd mistakenly use 2.72 for e if they knew about it.) --Aponar Kestrel 06:38, 2004 Jul 31 (UTC)
Expansion
I removed the <big> tags surrounding the approximation of e as this caused the number to be breaked at the resolution of 800x600. If there is a need to include longer expantion of e we should break it ahead of time I think. Two possibilities are:
e ≈ 2. 71828 18284 59045 23536 02874 71352 66249 77572 47093 69995 95749 66968
- <math>
\begin{matrix}
e \approx&2.71828\,18284\,59045\,23536\,02874\,71352 \\ &\ \,\,66249\,77572\,47093\,69995\,95749\,66968 \\
\end{matrix} <math> --filu 13:08, 24 Apr 2005 (UTC)
Not terribly important, but with 30 digits shown, and digit 31 being '6' the number shown would be more accurate if rounded up (and is thus misleading), yet wouldn't show the same digits that longers representations do. (It's just that I assumed that this was the motivation for using 64 digits on the pi page, and was mine for using 64 here.) Frencheigh 00:27, 25 Apr 2005 (UTC)
- Thanks for pointing this out. I am not sure I undertand though what you mean by "yet wouldn't show the same digits that longers representations do". I will for now just remove the last digit, to not have issues with rounding. Besides, the number looked kind of unesthetic with the 64 digits split onto two lines. Oleg Alexandrov
- What I meant was if we'd rounded it, our 30th digit wouldn't have been the same as the 30th digit on, for example, the Wikisource page with 10000 digits. (So changing the number of digits, as you did, was precisely what I had in mind.) Frencheigh 01:49, 25 Apr 2005 (UTC)
Google trivia
Just a technical note, currently the article states
- (the first 10-digit prime in e is 7427466391, which surprisingly starts as soon as the 101st digit)
As well as being a subjective statement, it is incorrect to someone who knows about the frequency of 10-digit primes. The prime counting function π which has values listed on Prime number theorem, the number of 10 digit primes is π(1010) - π(109) = 455052511 - 50847534 = 404204977. Out of 9000000000 10-digit numbers, this gives an average density of 404204977/9000000000 which is about 4.5%. So we would expect one in 22 randomly-selected 10-digit strings to be prime. It is surprising the first 10-digit prime starts as late as 101 digits in. I am changng the statement for this reason. Andrew Kepert 03:58, 11 May 2005 (UTC)