Talk:e (mathematical constant)/Archive 2

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Early questions

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! —The preceding unsigned comment was added by 62.255.0.7 (talk • contribs) . 12:06, January 1, 2005 (UTC)

It's too bad nobody has meanignfully responded this perfectly reasonable requests that some-one made. Kdammers 10:43, 3 February 2006 (UTC)

This is probably way too late in response to the question above, but the following might be a possibility: If you look at various graphs of functions like y=2^x and y=3^x, you will see that they cross the y-axis at the point (0,1) and have varying gradients. y=2^x has gradient less than 1 at that point, and y=3^x has gradient greater than 1. The base you have to use to get a gradient of exactly 1 is the number 2.718281828... i.e. 'e' (Clearly, this is only a thinly disguised way of defining e^x as a solution of f'(x) = f(x), but it might be slightly more intuitive to a 'layman') Madmath789 20:53, 10 June 2006 (UTC)

I guess because it is not possible. e is a very important number in mathematics, actually one of the most important, but that's where its importance is, as the base of e^x and of log_e x, known as the natural log. Unless you care about math or its applications (like engineering, physics, statistics, etc), you won't have any need for this number, and neither is there a way of explaining it beyond the math way. Oleg Alexandrov (talk) 17:53, 3 February 2006 (UTC)

Also, questions are more likely to be seen when they get proper headings and go to the bottom of the page, which this did not. -lethe ^{talk +} 19:53, 3 February 2006 (UTC)

Not only are e^x and the slope of e^x equal to 1 at x=0, but EVERY DERIVATIVE of e^x evaluated at x=0 is 1. Another way of looking at it is this: Imagine if there was a function that at every point, the height (y-value) of the function was also the slope of the function at that point. f(x)=f '(x). If you tried to do this free hand you'd see some interesting things: first start out by drawing in little dash marks representing the slope that you'd need to have at each point(the dash marks will all be parallel to each other for a given value of y - thus all along the x-axis the dashs are horizontal y=0, at y=1 the dash marks all have a slope of 1, at y=-1 they all have a slope of -1). You should be able to see these things: 1) the zero-function works out, you can draw it and you will always be moving in the directions the hash marks indicate. 2) if you are negative, you are heading further negative & going that way more & more quickly, 3) same with positive, & 4) You should see this can be done starting with any point! It turns out any function that does this has the form: A*e^x.

That is the easiest way I know to illustrate the neatness of e. —The preceding unsigned comment was added by 64.122.234.42 (talk) 18:04, 4 April 2007 (UTC).

One more way of looking at it is. To calculate interest... if we have an annual interest rate like 5%, and you only got the interest paid to you at the end of the year & started with some amount A, then at the end of the first year you'd have A*(1.05) or A*(1+r) where r is the rate 5%=0.05. At the end of the second year the amount you had at the end of the first year [A*(1+r)] would get interest resulting in A*(1+r)*(1+r) = A*(1+r)^2. Similarly after n years you'd have A*(1+r)^n. Now what happens when you get interest every month!?! In that case we divide the interest rate by 12 and now use months (m) to do the compounding: A(m) = A*(1+r/12)^(m), so that after 1 month you now have A*(1+r/12) in the bank. Next month you have A*(1+r/12)^2, and so on. We can compound this even further and say we wanted to compound interest x times a year. (then use y as years, and allow fractions), then A(y) = A*(1+r/x)^(y*x), that is every (1/x) of a year (ie every quarter if x=4, every month if x=12), your new balance will be (1+r/x) higher than before. If we take the limit as x->infinity (to compound continuously, more than every second) we get: lim (x->inf) of (1+r/x)^(y*x) = e^(y*r). This is the result that is mentioned elsewhere in this article. —The preceding unsigned comment was added by 64.122.234.42 (talk) 18:36, 4 April 2007 (UTC).

The last infinite series

The last infinite series listed in this article is

$e={\frac {-12}{\pi ^{2}}}\left[\sum _{k=0}^{\infty }{\frac {1}{k^{2}}}\ \cos \left({\frac {9}{k\pi +{\sqrt {k^{2}\pi ^{2}-9}}}}\right)\right]^{-1/3}$.

But for k=0 we have division by zero (and also the square root of a negative number). I think the sum should start from k=1. Can anyone confirm? Eric119 04:08, 2 January 2006 (UTC)

Hey thanks for being on the lookout, Eric; the sum does indeed start at 1 (I contributed this sum) and I fixed it.--Hypergeometric2F1[a,b,c,x] 07:11, 2 January 2006 (UTC)

Title in lower case

A friend of mine wrote a little computer program that can make an article title begin with a lower case letter. Is anyone interested in changing the title of this article from E (mathematical constant) to e (mathematical constant)? Uncle Ed 13:16, 10 January 2006 (UTC)

How ugly are the hacks involved? MediaWiki 1.6 will be case-insensitive and nativly support names which start with a lower case letter. —Ruud 13:20, 10 January 2006 (UTC)

It probably involves using some of the "fullwidth" letters in Unicode. æle ✆ 20:35, 17 January 2006 (UTC)

I don't think it is worth the trouble. Oleg Alexandrov (talk) 01:37, 18 January 2006 (UTC)

Most important numbers...

... would generally include "0" and "1". As per the "famous five" of Euler's identity, indeed. I won't edit this directly in case anyone's upset, y'all just having had a poll, n'stuff. :) Alai 08:54, 30 January 2006 (UTC)

I agree that they are important. I am not sure it is worth the trouble adding them to this article, as it is primarily about e. But, whatever. :) Oleg Alexandrov (talk) 04:56, 31 January 2006 (UTC)

I make the point as at present, it's a little counter-intuitive to the non-mathshead, as we start off with a comparison with the "unusual" numbers pi and i, which may be puzzling to people not clear what the importance of those actually is. Juxtaposing with the more familiar numbers makes the point more strongly, I feel. Alai 08:26, 31 January 2006 (UTC)

I am persuaded by your argument and Oleg's lack of objection. -lethe ^{talk +} 09:41, 31 January 2006 (UTC)

I'd say 1 is important because it a generator of Z, not because it is the multiplicative identity. More importantly though, do people really care that 0 and 1 are the additive/multiplicative identity (do non-mathheads know what this means?) when they are reading an article about e? —Ruud 19:23, 31 January 2006 (UTC)

So the 1 that appears in Euler's identity seems to have more to do with multiplicative identities than generators of integers to me. e⁰ is e times itself no times, which must be the multiplicative identity. The function e^x does have period 2iπ, so the one there in Euler's identity is a multiplicative identity. I think if you viewed the exponent as 1*iπ, then the 1 in the exponent would be the additive generator of the integers, but that's suppressed. Now, as for whether mentioning identities is good for the intro, well if we think they're important, we have to say why. We already say that i is the imaginary unit. We should say what 0 and 1 are as well. I mean, everyone knows the numbers 0 and 1, but might not have the foggiest why someone should consider them as important as π and e. If I knew a 2 word description of why π was important, I would include that as well. -lethe ^{talk +} 19:35, 31 January 2006 (UTC)

Circumference-to-diameter ratio? Paul August ☎ 23:18, 31 January 2006 (UTC)

That's cheating :) —Ruud 23:22, 31 January 2006 (UTC)

Well, it's half the rotational identity, or "semicircular period", or "trigonometric period"? How about "pandisciplinary constant"? ... Well, I suppose that applies to all of them. Damn, I give up. Confusing Manifestation 09:15, 1 February 2006 (UTC)

I'm going to try out Paul's suggestion. I think it's a bit too bulky for the sentence, but you guys take a look and tell me what you think. I'm counting on you, R Koot, to not like it. -lethe ^{talk +} 09:41, 1 February 2006 (UTC)

Given that nobody can come up with something better, it will have to do. —Ruud 12:24, 1 February 2006 (UTC)

Clearly defining e

There are a number of more complicated concepts in math that have been explained in laymen's terms. The problem with the e discussion is that no attempt is made to explain what "natural" means. It is either arrogant or wrong to say that no one si or will be interested in e if they are not concerned with pure math, physics, etc.

("Also, questions are more likely to be seen when they get proper headings and go to the bottom of the page, which this did not. -lethe ^{talk +} 19:53, 3 February 2006 (UTC)"

perhaps so, but the thread is lost this way.) Kdammers 11:21, 6 February 2006 (UTC)

Is wikipedia suppost to teach people math? It is your responsibility to learn e in the context of a higher math class, or in a historical setting such as the book E: Story of a Number.--Hypergeometric2F1[a,b,c,x] 19:27, 6 February 2006 (UTC)

No, you are wrong. Wik is an encyclopedia, not a dictionary or a discussion board. Its obligation is to provide clear explanations (and many if not most Wikians seem to think these should be for lay-people, since there are - especially in math entries - statements llike "We don't need togo into the technical stuff since experts have their own journals and books and won't be looking here"). Kdammers 01:28, 7 February 2006 (UTC)

Hypergeometric is not wrong. Wikipedia is not a textbook or a classroom (but see wikibooks and wikiversity). We are a reference work. You are certainly wrong that experts do not use this encyclopedia. Now if someone has some constructive comments, I'm sure we'd be happy to address them. But complaining and making broad generalizations about the nature of the pedia and what experts do with their time is helping nothing. -lethe ^{talk +} 03:58, 7 February 2006 (UTC)

Are you guys replying to me?--Hypergeometric2F1[a,b,c,x] 05:16, 7 February 2006 (UTC)

Well, I was replying to Kdammers, but I was talking about you, but for some reason I called you linas. That was really weird, I'm sorry for the mixup! (I fix it). -lethe ^{talk +} 21:06, 7 February 2006 (UTC)

e is the unique number, such that the derivative of e^x at x=0 equals 1.

Thanks for the image. A suggestion. I think the fonts on the axes should be made much bigger, the lines a bit thicker, and a few colors would not hurt. I would think of something of the style at subgradient (yeah, my own picture, so I am bragging here :) Oleg Alexandrov (talk) 04:02, 7 February 2006 (UTC)

PS Renaming the picture to something like E-math_constant.png from E-derivative.svg might be good also. :) Oleg Alexandrov (talk) 04:02, 7 February 2006 (UTC)

Sorry :) it was a crude sketch in mathemtica so see if people liked it, still have to learn how gnuplot works. —Ruud 14:58, 7 February 2006 (UTC)

E; the unique number such that 14242+e^(-23) = 14242 +e^(-23) --Hypergeometric2F1[a,b,c,x] 05:21, 7 February 2006 (UTC)

So e = 4 and e = 3 and ..? That doesn't sound very unique to me. —Ruud 15:04, 7 February 2006 (UTC)

I was being sarcastic...n/m lol..--Hypergeometric2F1[a,b,c,x] 21:17, 7 February 2006 (UTC)

E is also the second most found word on google.

The move

I don't like the current ℮ (mathematical constant) title. I would think moving it back to E (mathematical constant) would be better. Comments? Oleg Alexandrov (talk) 23:31, 26 February 2006 (UTC)

Agreed. ℮ is the "estimated symbol". It has nothing to do with the constant e. I have moved the article back, accordingly. Nohat 00:10, 27 February 2006 (UTC)

Mnemonic

I think Martin Gardner mentioned some mnemonics for recalling the first several digits of e. The only one that springs to mind is "I'm forming a mnemonic to remember a constant in analysis", the words of which have 2, 7, 1, 8, 2, 8, 1, 8, 2 and 8 letters. I don't know if it's worth including in the article, but I thought it was kind of cool. Kensson 16:09, 8 March 2006 (UTC)Kensson

Another mnemonic canbe used through the birth date of Leo Tolstoy. --BorisFromStockdale 21:13, 9 March 2006 (UTC)

Or you could just remember the first 8 digits of the number, which are by far the easiest to remember compared to other important irrational numbers.--Hypergeometric2F1(a,b,c,x) 22:41, 10 March 2006 (UTC)

Another mnenonic is the story of Andrew Jackson (though it uses incorrect information):

Andrew Jackson was the seventh President of the United States and served two terms (2.7)

He was elected in 1828, and since he served two terms, we'll write that twice (2.718281828)

Rumor has it he lived to the ripe old age of 90 and carred a .45 on each hip (2.718281828459045)

The age thing and the .45 thing are convenient made-up "facts," but they do help one memorize many digits.Tristan 01:14, 30 April 2006 (UTC)

Natural occurences of e

In my oppinion, the natural occurences of e section of the article sould be creaded and expanded. What are your oppinions on this topic?--BorisFromStockdale 21:09, 9 March 2006 (UTC)

1^infinity is undefined

$\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}$

Should we show that L'Hopital's Rule is required in order to actually find the limit of this expression? As it is now, it would just come out to 1^infinity. - Christopher 07:47, 10 March 2006 (UTC)

Strictly speaking, L'Hopital isn't meant to be used on limits of sequences, only limits of differentiable functions., though you can probably replace this limit with a corresponding limit of real functions with which you could use L'Hopital. Certainly L'Hopital is not required to show that such a limit exists. In fact, to use L'Hopital here would be circular; to find the limit using L'Hopital, you have to know the derivative of the logarithm, which is predicated on a knowledge of e. The right way to understand this limit, as described at Characterizations of the exponential function, is to recognize that the sequence is bounded above, and therefore the limit exists. Define e to be that limit, and then calculate the derivative of the exponential function in terms of that definition, and then you can use logarithmic differentiation to calculate similar limits. Does any of this need mention in the article to explain that limit? I'm not sure. -lethe ^{talk +} 10:50, 10 March 2006 (UTC)

But is Infinity^0 undefined? What about Infinity/Infinity..all of which equal e? ANSWER ME THAT kind SIR! Things that make ya say hmm.....--Hypergeometric2F1(a,b,c,x) 22:44, 10 March 2006 (UTC)

Actually, those are indeterminate (and so not equal to e). -lethe ^{talk +} 22:53, 10 March 2006 (UTC)

Sarcasm.--Hypergeometric2F1(a,b,c,x) 18:53, 12 March 2006 (UTC)

So, where thrn did $\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}$ come from? Surely, Bernoulli didn't one day randomly decide to find the limit of this function. - Christopher 18:43, 14 March 2006 (UTC)

It arises naturally when you are trying to tabulate logarithms for the purposes of calculation. You start by choosing a number like 1.01, and then you calculate powers of 1.01 by repeated multiplication. (Napier actually chose 0.999999 and worked backward, but it comes to the same thing.) Then you can look up some number in the table and find its log to the base 1.01, and you can multiply numbers by looking up the logs in the table, adding the logs, and then looking up the sum in the table the other way. THis was the original application of logarithms.

Now it might not strike you as unusual that entry #100 in your table happens to be for 2.704. But the following year, when you have finished your more accurate table, which is a table of powers of 1.001 instead of powers of 1.01, you might well notice that entry #1000 is almost the same: 2.716. And in fact this is the context in which e was first noticed: if you construct a table of powers of 1+10^-n, then the 10ⁿ entry in the table is more or less independent of n.

So no, nobody randomly tried to find the limit of that function, but they were looking at related matters and e cropped up naturally. Hope this helps. -- Dominus 05:04, 23 March 2006 (UTC)

Good question. Actually that limit had been around for quite some time since Bernoulli's day. I believe it was Napier that first approximated e. Anyway, according to the book, "E: The story of a number" by Eli Maor, e first arose as a means to calculate logarithms. It turns out that if you build the theory of logarithms from scratch, then it becomes appearant that e is the best number to choose as a common base, or, it is the most "natural" base for logarithms. This is basically because the of the infinite series for e being elegant, but if you want to know more I suggest you read that book. Its a shame that in school they never really explain how in the hell e was "chosen" as the common base for logs, but then again, most math in school is taught in a completely arbitrary way which is part of the reason I dropped out.--Hypergeometric2F1(a,b,c,x) 12:00, 21 March 2006 (UTC)

You say that "if you build the theory of logarithms from scrath, it becomes apparent that e is the best base". I'm curious, is there any way to recognize e as the best base without calculus? I certainly don't know of any. -lethe ^{talk +} 14:37, 21 March 2006 (UTC)

It depends what you mean by "calculus", but the short answer is "yes". If by calculus you mean infinitely large and infinitely small numbers than the answer is "no". Basically e has been known to be the best base to use for logarithms because so called "natural logarithms" were much easier to approximate compared to logs in other bases. This comes from ln's infinite series, which you could argue has to do with "calculus" but the notion of e and logarithms at that time had nothing to do with functions (the notion of logarithms being a function was introduced by Euler and others) ; logarithms were considered not functions but numbers that were used to aid multiplying large numbers. Back then the sole concern was mainly approximating logariths to aid their use, hence the developement of a natural base, not developing any systematic theory..the calculus changed all that. A brilliant non-calculus derivation for e^x and ln can be found in Euler's Analysis of the Infinities, although Euler uses infinitely small numbers in his argument. --Hypergeometric2F1(a,b,c,x) 16:38, 21 March 2006 (UTC)

move to Mathematical constant e

This page was moved to mathematical constant e. I don't support such a move, so I've moved it back to e (mathematical constant). If and when a consensus should arise for another name, then I won't oppose. -lethe ^{talk +} 09:04, 15 March 2006 (UTC)

I agree; the name Mathematical constant e is also bad-sounding. Oleg Alexandrov (talk) 19:05, 15 March 2006 (UTC)

LAYMAN Terms

Is the discussion about the lack of "layman terms" for e one that can be solved by adding an example to the article text? A friend taught me e by using an (imagined) drum that contained 1 liter of water, trying to add water to the container at an initial rate of 1 l per second, for one second. Now, impossibilities arise when, in infinitely small increments, more water is added to the rate as time goes by. After one second, you have what can be rounded off to 2,7 liters of water, but which in maths will always be an endless number. For physics: Should one assume there is a definite particle that all is made of, and that the observer was in posession of instruments to count every single added particle, it still wouldn't limit the end value e. For that, you need (granted I understood this right - I may have made a huge fool out of myself just now) a definite timeunit, say a hundredth of a second. As there is none, e is considered endless. Sorry if this is redundant, just trying to be constructive and such.

one more digit

Someone named Hiiiiiiii recently added one more digit. The text claimed 29 digits, so adding another made the text incorrect. I fixed the text, and clarified that it's truncating, not rounding (though maybe we should prefer to round up?). Furthermore, I added the digit back, becaues I seem to remember my old fashioned tables of numbers format things in blocks of five digits aligned to the decimal point. I could be misremembering though. -lethe ^{talk +} 18:08, 7 April 2006 (UTC)

page move

The page was moved a second time to ℮ (mathematical constant). I guess there is some support for the idea, but I'm sure also some opposition. I've moved back. -lethe ^{talk +} 20:31, 11 April 2006 (UTC)

My unicode browser tells me that ℮ is the "estimated symbol". Can someone defend the stance that it denotes the natural logarithm base? -lethe ^{talk +} 20:34, 11 April 2006 (UTC)

If the sole reason is that this symbol looks like a miniscule e, then I find this wholly unconvincing. If that were a convincing reason, I would prefer the unicode ℯ, but it's not convincing, so I don't. -lethe ^{talk +} 20:36, 11 April 2006 (UTC)

I've protected the page from being moved. These semi-clever unicode-hacks are starting to become a bit tiring by now and it really shouldn't be moved without a discussion first. —Ruud 16:55, 12 April 2006 (UTC)

From the edit summary, I infer that this semi-clever (heh) idea originated on the Hebrew wikipedia. I notice that the Italian article also uses it, but no other language does. -lethe ^{talk +} 21:18, 12 April 2006 (UTC)

e in the real world

Isn't the curvature of e used in erecting structures? I remember learning that a parabolic shape wouldn't be as good as an e shape. Anyone have any information on this? 71.250.59.37 20:22, 3 May 2006 (UTC)

The inverted catenary ${\frac {e^{-x}+e^{x}}{2}}$ is one of the most stable structures if that is what you are thinking of. The St Louis arch is an inverted centenary. --Hypergeometric2F1(a,b,c,x) 17:24, 13 December 2006 (UTC)

truncated AND rounded

Not sure about others, but this sounds a bit odd to me - how can a number be both truncated and rounded? I would suggest "or" instead, or preferably replace with "correct to 20 decimal places"? Madmath789 13:16, 30 May 2006 (UTC)

To answer your question, for example, a number could be rounded to the nearest third (e would be 2.666666..etc) and then truncated to two decimal places (yielding 2.66) or rounded to two decimal places (yielding 2.67), neither being equal to each other or a multiple of one third.

In this case (since the digits in the truncation to 20th decimal place happen to round down anyway) rounded is correct, specifically to the nearest 100 quintillionth which implies truncation. I agree with you that specifying both rounded and truncated is redundant since rounded decimals are typically not printed with their infinity of trailing zeros. I would support a move to remove the truncated part from that sentence. Soltras 13:44, 30 May 2006 (UTC)

Yes, that is obvious - what I was actually questioning was whether it was sensible to talk about rounding and truncating to the same number of places. Never mind, it has been changed now :-) Madmath789 13:52, 30 May 2006 (UTC)

The image

Is there any good reason why the horizontal and vertical axes of the Image:E-ruud.png are not drawn to the same scale? Evercat 20:12, 9 June 2006 (UTC)

More effort, I was lazy. —Ruud 20:31, 10 June 2006 (UTC)

Continued fraction

The text cites A005131 as the simple continued fraction of e, but OEIS:A003417 says it is. The latter looks more reasonable to me. Then the question is, what is OEIS:A005131?Lee S. Svoboda tɑk 22:11, 26 August 2006 (UTC)

The article is wrong. Both are continued fraction expansions of e; the only difference is that one begins "2,..." and the other begins "1,0,1,...". These have the same value. (In general, any continued fraction expansion that contains "...m, 0, n, ..." is equal to one that contains "...m+n, ..." at the same position.) But only the first of these is a simple continued fraction expansion, because simple continued fractions are required to have positive terms. I will correct the article. -- Dominus 14:06, 27 August 2006 (UTC)

Thank you. Lee S. Svoboda tɑk 17:18, 30 August 2006 (UTC)

========

I added a third generalized continued fraction that converges far more quickly than the first two. It corresponds to the series

1/1, 3/1, 19/7, 193/71, 2721/1001, ...

Each convergent adds 2-3 decimal digits of precision to its predcessor.

I also added a general formula for e^(2m/n) for |m|,n = 1,2,3,...; however, convergence slows as |m| increases in value. When m=1 and n=2 you get the formula for e.

Glenn L

Thank you, Glenn L. I am amazed – someone else actually cares about these things! DavidCBryant 12:19, 17 January 2007 (UTC)

Graph

The graph comparing f(x)=e^x to f'(x) is nice, but a little confusing, if only because x=.7 (whatever it is, don't have calculator handy), which gives f(x) = 2. Which is nice, but not entirely evident unless you know to look for it. Perhaps showing x=1 would help, and showing the equation "e^(.7whatever) = 2, which is the slope at that point, and e^1 = 1, which is the slope there." Perhaps even more explicitly labelling the slope of 2 would help (either with a direct label, or an old middleschool "rise over run" triangle, showing that it goes up 2 for every 1 over). And, I may be splitting hairs, but it seems like the line for e^x should be bolded, or that the f' line should not be given equal weight. I say this just because it is the first (and only) graph on the page, so it is doing double duty: showing the function e^x (very important) and showing that that derivative of e^x equals e^x (important, but less important than showing that function itself). Sir Isaac Lime 20:13, 19 September 2006 (UTC)

APR

Reinstated my paragraph on APR (interest rates) after it was completely deleted by RandomP on the grounds that he didn't know what "percentage rate" meant. Expanded the text slightly to clarify, but a really detailed explanation of APR/interest probably belongs in an article on that subject. --DarelRex 13:34, 26 September 2006 (UTC)

No personal attacks, please, and assume good faith.

See the annual percentage rate article for the definition overwhelmingly (for good reasons) in use. Saying that an APR of 100 should result in more than a doubling of debt or credit after a year is simply not what APR is.

If you want the section back in, fix it. Using APR is inappropriate, and the other phrase is, at best, incomprehensible.

$1 deposited at APR 100 becomes $2 after a year. That's the whole point of using APR numbers.

Reverting for now.

RandomP 13:44, 26 September 2006 (UTC)

Oh, by the way: I wouldn't mind the information in the article: if you go back and forth between compound and "simple" interest, e units is indeed the maximum you can make out of the "1 unit doubling once" scenario. But implying that that is somehow related to the concept of APR is unacceptable.

RandomP 13:47, 26 September 2006 (UTC)

This is not a personal attack; please do not repeatedly delete my entire post on the grounds that it needs a tiny fix. I have now reposted it (again) with an extra sentence that hopefully alleviates the problem as best as I understand it from your above criticism. If further modification is necessary, please advise or revise instead of axing the whole thing again and again. Thanks. --DarelRex 13:56, 26 September 2006 (UTC)

It doesn't need a tiny fix; if I knew how to fix it, I would. It's just plain wrong.

The point is that while going back and forth between simple and compound interest might be interesting to do in theory, it has nothing to do with either finance or the term APR, which has a very specific definition that excludes exactly that kind of trick.

My suggestion would be to just write that if the doubling period of an exponential growth process is estimated using a linear approximation, the actual factor will be e, not 2.

Again, I do not see an easy way to salvage the section: working with years, then months, then weeks, simply appears too awkward for me to include that information, and adding to the confusion about compound interest is simply unnecessary.

I know it's crazy, but there are people out there who will quote your section to explain how an APR of 5 should mean they should get $5.13 for every $100 invested for a year. Unacceptable, and not what an APR is.

Sorry I'm trigger-happy on taking this out, but wikipedia, unfortunately, does have clear policies on dealing with unverified (and factually incorrect) information: if in doubt, take it out. WP:V isn't applied strictly to mathematical content, but the paragraph in question wasn't that.

RandomP 14:09, 26 September 2006 (UTC)

Proposal

e is also relevant to measuring the difference between linear growth and exponential growth. If the doubling period for a process of exponential growth is estimated by linear approximation, the actual growth factor will be e after that time.

For example, assume that a bank continuously compounds interest, and we start with $1 in the bank, with current interest payments at a rate of $1/year. In order to calculate the annualised percentage rate, we might consider this rate to remain constant, and estimate the doubling period at a year — and thus, the annualised percentage rate at 100.

However, taking compound interest into account, we see that after a year, the account balance is $e, for an annualised percentage rate of approximately 172, much higher than our original estimate.

Admittedly, bad writing, but at least the term APR is used correctly. Feel free to include further approximations for monthly, daily, or whatever interest payments in there, but do not misrepresent "APR" to mean something it doesn't. (The current APR article suggests it's actually 12 * monthly interest rate; I'm pretty sure there's different definitions around, and at the interest rates usually considered, it doesn't make a lot of difference).

RandomP 14:20, 26 September 2006 (UTC)

So if any "crazy" person might lift something out of context and use it to mispresent something else, then it has to be entirely removed? You are obviously determined to keep my paragraph out, so I will not repost it, even though I do not understand your criticisms. Your suggestion of what to replace my paragraph with will surely obfuscate its meaning to all but the most mathematically advanced readers. I give up; you win. --155.188.247.6 14:27, 26 September 2006 (UTC)

• I did not call anyone crazy
• "Taking something out of context" is otherwise known as "citing". It's what we do.
• I'm not determined to keep "your" paragraph out. I just sat down and wrote a paragraph with essentially the same information, and I did not do so in order to keep it out of the article.
• I'm not sure about "mathematically advanced", but there is no non-mathematical meaning to the compound interest comparison. Readers not interested in mathematics should not be interested in that paragraph.

RandomP 14:39, 26 September 2006 (UTC)

I think a lot of people would get more out of a "real world" example, such as e being involved in the difference between simple and compound interest, than all those very abstract equations. --Hugh7 08:52, 24 January 2007 (UTC)

If you want information on something "real world" like compound interest, then go to the "real world" oriented article on Compound interest. This article is for those who are interested in the abstract mathematical properties of e. JRSpriggs 11:52, 24 January 2007 (UTC)

picture

I think that a picture showing the tangent line at 1, instead of 2 would ilustrate the point much more clearly... a slope of one is easier to see. 160.39.168.58 04:25, 11 December 2006 (UTC)

"alongside the additive and multiplicative identities 0 and 1"

Does that mean the additive identity is 0 and the multiplicative identity is 1, or does it mean "the [actions] of addition and multiplication and the numbers 0 and 1"? Could someone please clarify this on the page. --Hugh7 08:44, 24 January 2007 (UTC)

If you do not know that zero is the additive identity and that one is the multiplicative identity, then this article is beyond your level. JRSpriggs 11:57, 24 January 2007 (UTC)

A Possible Problem with the Definition of e

Right now it's listed as the sum of (1/n!) from 0 to infinity; this doesn't seem to be right because 1/0! is 1, or am I mistaken? (I had learned before that 0! is a special case equal to 1) Since 1/0! + 1/1! is already 2, the next few numbers (2 through 6, I think) make the value pass 1... could somebody clear this up for me? Robinson0120 02:04, 28 January 2007 (UTC)

No, 1/n! gets smaller very fast. Look at the first few terms: 1 + 1 + 1/2 + 1/6 + 1/24 + 1/120 + 1/720... that's the 0 through 6 terms, which sum to 2.71805...; the first 3 decimal places are correct, e converges quite fast! –EdC 02:25, 28 January 2007 (UTC)

Graph and definition

How's this?

Ruud's graph showing the tangent to e^x at x=0 was replaced by a nice SVG graph with a tangent drawn at a different place, and the simple definition that e is the unique number such that the slope of the tangent to e^x at 0 is 1 was replaced by the much more scary-sounding (equivalent but over-specified) definition that e is the unique number such that the slope of the tangent to e^x is e^x for all x. Why? Can we modify the SVG and the definition back to the much simpler version, such that the graph actually illustrates the condition? We could also draw some light curves for 2^x and 4^x just to illustrate that they have different slope, not 1, at x=0. Dicklyon 03:17, 28 January 2007 (UTC)

Yeah, like it. –EdC 15:07, 28 January 2007 (UTC)

I think it's a good idea, Dick. I'm not so sure about the other curves, like 4^x, though. They don't show up very well on your prototype. Maybe just one light curve, like 1.1^x, would make the point more clearly. Oh – I searched through the history, and the old picture you referred to got wiped out of this article on 21 Aug, 2006, by an anon. DavidCBryant 16:09, 28 January 2007 (UTC)

I updated it with bolder dashed and dotted curves that show up better and don't confuse with other. Does that work better for you? I think these curves that obviously go through (1,2) and (1,4) will make the point better than curves further away. Dicklyon 17:27, 28 January 2007 (UTC)

That looks good to me, Dick. DavidCBryant 18:53, 28 January 2007 (UTC)

Explanation of e

Dicklyon decided to revert rather than follow Wikipedia rules on reverting. Rather than give the editer the benefit of the doubt, he reverted rather than fix it. This is a violation of the revert policy. Please help if you can. —The preceding unsigned comment was added by 172.163.172.47 (talk) 20:47, 4 February 2007 (UTC).

You claim to know a surprising amount about Wikipedia policy for someone without a Wikipedia user account. –EdC 21:21, 4 February 2007 (UTC)

Probably he meant to refer to Help:Reverting, which I in fact never read before. According to that, I should have tried to fix it instead of just reverting it. But I got lazy. It looked like too much trouble to figure out where it should go and how to rewrite it, so I just left my edit summary which essentially invites anyone who cares to try again and not make it so lame this time. Dicklyon 22:55, 4 February 2007 (UTC)

By the way, I do agree that the article should include a bit about compound interest; but not so informal, not part of the definition section, and not confused with concepts like APR. Dicklyon 22:58, 4 February 2007 (UTC)

I do have a Wikipedia account; I was merely editing anonymously...but this doesn't concern you, EdC. As for Dicklyon, it's a shame that you perceive yourself above the rule of law. The correct and proper course of action for now would be for you to revert your previous unlawful reversion and place your requests on the talk page for others, including me, perhaps, to further edit. And furthermore, my edit was not a confusion of e with APR, but rather an example of a varied method for reaching e. Hopefully, you do not deny that e can be defined as the maximal limit of principal plus earned interest when said principal x has 100% interest compounded infinitely applied (i.e. (e)(x)). Thus, you should not deny this analogy as a proper and fitting explanation of e to those who may not be able to comprehend its relevance to growth patterns in nature. Your admission of laziness merely "compounds" your otherwise innocent violation of not properly reviewing rules for conduct prior to conducting. —The preceding unsigned comment was added by 172.163.172.47 (talk) 07:56, 6 February 2007 (UTC).

Oh, stop WP:WL. WP:NOT#BUREAUCRACY. Your own grasp of rules for conduct is evidently lacking, or you'd know to WP:SIG. There's myriads of ways e appears in nature; explanations should only be included if they improve the article. –EdC 14:11, 6 February 2007 (UTC)

That's odd. Just about the time EDC was trying to talk some sense into 172.163.172.47, I was trying to cut the section about compound interest down to size. Maybe it doesn't detract from the article quite so much now as it once did. Personally, though, I figure that the motivation for J. Bernoulli's investigation deserves no more than a footnote in the article, so if someone else wants to chop the bit about compound interest clear out of there, it's OK by me. Who knows – I might do it myself in another couple of days.  ;^> DavidCBryant 15:44, 6 February 2007 (UTC)

PS I took the bit about hourly compounding and compounding once a minute out of that section. I thought about adding even more examples – of compounding by the second, by the millisecond, by the microsecond, and by the nanosecond – for about a picosecond!

Integral expression for e

To EJ: I changed the indefinite integral for e to the definite integral form:

$e^{x}=1+\int _{0}^{x}e^{t}\,dt$

because it has a unique solution and it can be used to generate the series expression for e. You replaced it with:

$e^{x}=\int _{-\infty }^{x}e^{t}\,dt$

which appears superficially to be a definite integral, but actually is ambiguous and equivalent to the indefinite integral form from which I was trying to get away. Nor is it simpler as you said it was. It includes an implied limit as the lower bound goes to negative infinity, which makes it more complex than my version. JRSpriggs 08:46, 6 February 2007 (UTC)

I asked EJ a similar question on his talk page yesterday, JR. Here's his response.

Why is the integral improper? AFAICS, it is a usual convergent Lebesgue integral of a nonnegative function, there is no need to go through the limit $\lim _{a\to -\infty }\int _{a}^{x}e^{t}dt$.

Anyway, I'm notoriously bad at estimating the level of mathematical sophistication needed for understanding particular problems, so if you think the original formula is easier, do not hesitate to revert. However, I should point out that the original formula also relies on a convention which might or might not be clear to less sophisticated readers, namely $\int _{0}^{x}e^{t}dt:=-\int _{x}^{0}e^{t}dt$ for negative x. -- EJ 15:22, 5 February 2007 (UTC)

While I have to concede his point (in a measure-theoretic sense, there are no improper integrals), I don't think the typical reader of Wikipedia is up to speed on the fine points that distinguish the many possible definitions of an "integral" from one another. So I'm reinstating JR's version of the formula, because I think it's better for our target audience. DavidCBryant 12:22, 6 February 2007 (UTC)

Compound Interest section

DavidCBryant, your rewrite of this section has at least three problems that I see: 1. You disconnected the reference to the math formula that is involved. You either need to refer to the History or Definition section above, or repeat the formula explicitly. 2. You introduced the informal imperative style "Think about an account...". Not good. 3. You dropped the connection to the illustration. Please work on it some more. Dicklyon 18:11, 6 February 2007 (UTC)

Sorry to be a bit slow with a response, Dick. I've had other fish to fry. I see that you already changed the imperative mood, which is fine by me. I don't agree with your assessment ("not good"), but it's not a big deal as far as I'm concerned.

I don't understand your first criticism very well. I think that the sentence "Using n as the number of compounding intervals, with interest of 1/n in each interval, the limit for large n is the number that came to be known as e …" says exactly the same thing as

$\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}=e\,$

and that a reader of normal intelligence will make the connection with the formula in the immediately preceding section easily. If you don't agree, then please make the changes you see fit.

Thanks, I might work on it. The thing about "normal intelligence" is always a tough judgement; I prepare to err on the side of a little bit of extra help for the newbies. Dicklyon 21:23, 10 February 2007 (UTC)

The version I redacted contained this sentence: "These simple and compound yields correspond to the red and blue curves, respectively, in the opening figure above, if the horizontal axis represents R (where 100% is R=1)." In my opinion, that's at best confusing, and literally false. If the horizontal axis represents R (does that mean R is the unit on the scale?) then what is left to represent time? Wouldn't the red and blue curves correspond to simple interest of 100% per year, and continuously compounded interest of 100% per year? And wouldn't the horizontal axis then represent time, in units of years? I do not understand how to interpret a 2-dimensional graph of "interest" that does not include "time" on one axis or the other. DavidCBryant 20:05, 10 February 2007 (UTC)

In these expressions, time is constant, or if the time over which the simple interest is R, and what's being compared is simple versus continuously compounded interest as a function of R. It's not that tough, and since we already had a plot of it, it seemed worth referring to. I might work on a less confusing way to say it. Dicklyon 21:23, 10 February 2007 (UTC)

irrational number

I know very little about mathmatical concepts such as this, so I don't feel very comfortable editing the article, but it's my understanding that e is an irrational number, and this doesn't come across very clearly in the article. Does anyone object to this being mentioned in the opening paragraph? I think it would help expalin why the number is not shown in its entirety. -R. fiend 18:44, 10 February 2007 (UTC)

Done. Transcendental, too. Dicklyon 19:10, 10 February 2007 (UTC)

Return of "Keller's Expression"

User:Hypergeometric2F1(a,b,c,x) put in a formula back on 21 December 2005 that had previously been repeatedly inserted by a user who called it "Keller's Expression", presumably after himself. This self-promotion had been removed several times before. Today I removed it yet again. -- Dominus 16:33, 15 February 2007 (UTC)

Let us delete the mnemonics

Let us delete the section on mnemonics, "Remembering the digits of e". It adds nothing of value. JRSpriggs 11:18, 28 February 2007 (UTC)

Yes. Or perhaps we might move them to an article on "mnemonics for digits of real numbers" or something of the sort. -- Dominus 14:38, 28 February 2007 (UTC)

I would disagree. wikipedia guidelines suggest space should be awarded based on interest and importance. remembering the digits of e both allows people to better remember e (which is important)and while it is hard to prove something interesting the methods based upon myself and my math class 100 of people were mildly (or more) interested in the methods for remembering particularly the Andrew Jackson portion Beckboyanch 08:30, 3 March 2007 (UTC)

How precisely is memorizing more than the first two or three digits of e important? CMummert · talk

Inaccurate value?

Hi! Is there a way to calculate some incorrect but approximate value of e using some sort of finite fraction/series? (Like for PI, we have 22/7 or 355/311). If there is, I would really like to add it to the article.--Scheibenzahl 20:23, 9 March 2007 (UTC)

Take one of the infinite series, and truncate it.

For example, e is the sum 1/0! + 1/1! + 1/2! + 1/3! + ... . Truncating at various places gives the rational approximations 2/1, 5/2, 16/6, 65/24, 326/120, 1957/720, etc.

You get the best approximations if you truncate the continued fraction expansion to a finite number of terms. The result will be the best possible rational approximation for e.

The continued fraction expansion begins [2; 1, 2, 1, 1, 4, 1...]. So the first few rational approximations are 2, 3, 8/3, 11/4, 19/7, 87/32, 106/39, 193/71, 1264/465, 1457/536, 2721/1001, 23225/8544, 25946/9545, 49171/18089, 517656/190435, ... 14665106/5394991, etc.

Unfortunately, the terms of the continued fraction expansion for e are relatively small, which means that the rational approximations to e are not particularly accurate. With π, there is a fortunate 292 term early on, which leads to the extremely accurate 355/113 approximation, and there is a 15 term even earlier, resulting in the very accurate 22/7 approximation. With e, there is no such luck; the terms don't get as big as 15 until the 23rd term.

Hope this helps. -- Dominus 21:57, 9 March 2007 (UTC)

Thanks for the information. Since there is no such "number" that one can use promptly, there is no need to add it to the article.--Scheibenzahl 15:32, 10 March 2007 (UTC)

The reason for the difference is that the (simple) continued fraction expression for π [3; 7, 15, 1, 292...] has no obvious pattern, while that for e [2; 1, 2, 1, 1, 4, 1, 1, 6, 1, 1...] *does* have an obvious pattern. The best approximations for e involve truncation after terms 2, 5, 8, 11... yielding 3/1, 19/7, 193/71, 2721/1001... as described in my previous entry in this discussion. Glenn L 15:12, 27 May 2007 (UTC)