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What's so “natural” about the base of natural logarithms?

Why the number e(=2.71828) was chosen as the natural base for logarithm functions ? Mainly I am interested in knowing why is it called "natural " . The number "2" could instead have been chosen as the most natural base.

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Geek
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    Just to be clear $e$ is not $2.71828$. $e = \lim_{n \rightarrow \infty} (1 + \frac{1}{n})^n$. You can prove that $e$ is irrational. – William Jul 29 '12 at 07:53
  • @William for this question let us stick to e being a finite value and lets not get into the infinite sequence . Technically. though you are correct. – Geek Jul 29 '12 at 07:55
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    @Geek If you use the finite value of $e$, you lose the "naturalness" of $e$. Read the wikipedia article on $e$: http://en.wikipedia.org/wiki/E_(mathematical_constant). There is a section about $e$ and continuously compounded interest. – William Jul 29 '12 at 07:57
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    @Geek: $e = \lim_{n \rightarrow \infty} (1 + \frac{1}{n})^n$ is also a finite value. :-) – ShreevatsaR Jul 29 '12 at 08:13
  • I don't know why no one except me up-voted this question. – Michael Hardy Dec 09 '12 at 16:06
  • I posted some reasons in the comments here. – Akiva Weinberger Dec 09 '14 at 01:16

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The simplest answer is this:

If you draw the graphs of $y=a^x$ for varying values of $a$, you find that they all pass through the point$(0,1)$ on the $y$-axis. There is exactly one of these curves that passes through that point with a gradient of exactly 1, and that value is obtained by taking $a=2.718281828459 \dots$.

In more analytical terms, this means that this is the value of $a$ which makes the derivative of $a^x$ equal to $a^x$, rather than a constant multiple of $a^x$.

Old John
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  • can you put the graph in your answer somehow ? – Geek Jul 29 '12 at 07:57
  • @Geek Probably not worth it, as it looks like this question might get quickly closed as a duplicate. – Old John Jul 29 '12 at 07:58
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    @Geek But there is a good example of the graph here – Old John Jul 29 '12 at 08:32
  • that doesn't show the tangent though . However its okay . – Geek Jul 29 '12 at 08:43
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    "There is exactly one of these curves that passes through that point with a gradient of exactly 1..." - not so coincidentally, truncating the series for the exponential function at the linear term yields that particular tangent line with a slope of $1$. ;) – J. M. ain't a mathematician Jul 29 '12 at 09:50
  • @J.M., but the truncated series is not $a^x$ for any $a$, so it doesn't count. – hmakholm left over Monica Jul 29 '12 at 14:27
  • @Henning, by "exponential function", I meant $\exp,x$, and certainly if you truncate its Maclaurin series at the linear term, you get the tangent line at $x=0$... – J. M. ain't a mathematician Jul 29 '12 at 14:30
  • @JM: Sure you do, but when Old John says "exactly one of these curves", "these curves" are the ones of the form $a^x$, and the truncated Maclaurin series for $\exp$ is not one of these curves. Therefore it is not a counterexample to Old John's claim. – hmakholm left over Monica Jul 29 '12 at 14:36
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    @HenningMakholm: I'm not sure I understand. $a^x = e^{x\ln a} = 1 + (x\ln a) + (x\ln a)^2/2! + \dots$, and the coefficient of the linear term gives the slope of the tangent line at $x = 0$, namely $\ln a$. This is $1$ precisely when $\ln a = 1$, namely $a = e$. And truncating at the linear term does give the tangent line at $x = 0$, namely the line $y = 1 + x\ln a$ as J. M. said, which has a slope of $1$ for $a = e$. – ShreevatsaR Jul 30 '12 at 08:28
  • @ShreevatsaR: I don't understand what the point you're trying to make is. The truncated series is not of the form Old John is speaking about, so it is not a counterexample to the claim that the only functions of that form that has a tangent with slope 1 at (0,1) is $e^x$. – hmakholm left over Monica Jul 30 '12 at 11:13
  • @HenningMakholm: The only function of the form $a^x$ that has a tangent with slope 1 at x=0 is $e^x$, corresponding to $a = e$. That this function $e^x$ indeed has a tangent with slope 1 at $x=0$ can be seen by examining its tangent line at $x=0$, which is got by truncating its series at the linear term (as one would for any power series in general), and indeed the tangent line $y = 1 + x$ has slope 1 (the coefficient of the "x" term). No one has claimed that $1 + x$ is of the form $a^x$. Nor has anyone claimed the existence of a counterexample (obviously!), only corroborating evidence/proof. – ShreevatsaR Jul 30 '12 at 11:21
  • @ShreevatsaR: If you agree that 1+x is not of the form Old John is speaking about, then I don't understand at all what it is you and J.M. object to about the claim "There is exactly one of these curves that passes through that point with a gradient of exactly 1...". – hmakholm left over Monica Jul 30 '12 at 11:23
  • @HenningMakholm: No one objects to that claim! It is obviously true! J.M. only pointed out evidence (or even a proof) for that claim, and I reiterated what he said. What makes you think he (or I) was objecting to it? – ShreevatsaR Jul 30 '12 at 11:26
  • @ShreevatsaR: J.M.s first comment in this very thread seems to be a disagreement (on grounds that make no sense to me, but a disagreement nevertheless) to the sentence from the answer he quotes in that comment. Since you seem to think that my objection to that disagreement is wrong, I concluded that you must agree with the disagreement. – hmakholm left over Monica Jul 30 '12 at 11:29
  • @HenningMakholm: I still don't understand why it seems to you to be a disagreement? He quotes the statement (about the gradient of the exponential function being 1) and says "Not coincidentally, ... the exponential function... yields ... tangent line ... slope of 1". This is to provide evidence for the claim, not disagree with it. Anyway, the confusion seems cleared up now (I hope). – ShreevatsaR Jul 30 '12 at 11:53