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I have a question regarding the natural logarithm $e$.

Simply, why is $e$ special enough to:

  1. Have its own special notation $\ln$?
  2. Be used in derivatives?
  3. Have Wikipedia pages dedicated to it?

From my current understanding, it is simply a non-reoccuring irrational number. So what makes it apart from other irrational numbers such as pi?

David G. Stork
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Mildwood
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    $e$ is natural enough, because by virtue of being the series sum of the inverse factorial sequence allows beautiful formulas such as $\frac{d}{dx} \ln x = \frac{1}{x}$ to emerge. Similarly, $\frac{d}{dx} e^x = e^x$. With which other constant can you get this relationship? Furthermore, $e$ has many interesting expansions, and the focal Euler's equation $e^{i\theta} = i\sin \theta + \cos \theta$ forms the backbone of complex analysis. Do you need more justification? – Sarvesh Ravichandran Iyer Oct 10 '16 at 08:20
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    Because of this ODE: $$ y' = y $$ – ILoveMath Oct 10 '16 at 08:20
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    I remember that years ago, when I was first introduced to the concept of logarithm, I asked the same question and I didn't get a satisfying answer. But as years passed, I realized that I should change the question to its contrary: why do we even bother to define logarithm in base 10? – polfosol Oct 10 '16 at 08:21
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    -1. The OP could have done a bit of research on this. – N.S.JOHN Oct 10 '16 at 08:33
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    to continue on what actoh said : since $\log_a(x) = \frac{\ln x}{\ln a}$ we have $\frac{d}{dx}\log_a x = \frac{1}{x \ln a},\ \frac{d}{dx} a^{x} = a^x \ln a$ and $a^{i \theta} = \cos(\theta \ln a)+i \sin(\theta \ln a)$, $\ a^{ \frac{2i \pi }{\ln a}} = 1$ so that choosing $\ln a = 1$ is not a bad idea – reuns Oct 10 '16 at 08:38
  • @polfosol - Log base 10 was favoured for practical use because you could divide by factors of 10 then add integers onto the value you looked up - so the log10 of 1,900,000 was found by looking up 1.9 - that was before pocket calculators etc, it was a way of approximating large engineering and scientific calculations - for example. – Cato Oct 10 '16 at 09:11
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    Suppose that $f$ and $g$ are real functions such that $f=f'$ and $g=g'$. Then $(\frac{f}{g})'=\frac{f'g-g'f}{fg}=\frac{fg-gf}{fg}=0$. Hence $\frac{f}{g}=C$ for some number $C$. Thus $f=Cg$. Hence there is only one function $f$ (up to multiples) such that $f'=f$. Since $f(x)=e^x$ satisfies this equation, one sees that $e^x$ is special. – Mathematician 42 Oct 10 '16 at 09:22
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    @N.S.JOHN, Asking the question here is a sort of research, doesn't it? – Alex Silva Oct 10 '16 at 11:02
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    @AlexSilva I agree. And since you are more experienced than me, I stand corrected – N.S.JOHN Oct 10 '16 at 11:04
  • Thank you for all your explanations! Can someone post these in an answer so I may close the question? – Mildwood Oct 12 '16 at 05:51
  • $e$ also occurs in probabilities concerning a poisson-distributed random variable. You cannot close your own question (you can delete it) , but with an accept of an answer you can indicate that you are not interested in more answers. – Peter Aug 27 '23 at 16:13

2 Answers2

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The function $f(x) = e^x$ is the only function such that $f'(x)=f(x)$. Moreover, the formula $$ e^{i \theta} = \cos \theta + i \sin \theta $$ is widely used in complex analysis. And then of course you have Euler's identity $$ e^{\pi i} + 1 = 0, $$ which elegantly connects the constants $0$, $1$, the imaginary unit $i=\sqrt {-1}$, $\pi$ and Euler's constant $e$.

Klangen
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Suppose we have a function like below. Just call it $F(x)$ for now rather than any fancy name.

$$ y = F(x) = \int_1^x\frac{1}{t}dt $$

From the fundamental laws of calculus, we can easily conclude that:

$$ \frac{dy}{dx} = F'(x) = \frac{1}{x} --- (1) $$

And with a bit effort we can conclude:

$$ F(x^r) = rF(x) --- (2) $$

Now let's say the reverse function of $F(x)$ is $G(x)$:

$$ x = G(y) = F^{-1}(y) $$

We can easily conclude that the derivative of $G(y)$ is:

$$ G{'}(y) = \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}} = \frac{1}{\frac{1}{x}} = x = G(y) --- (3) $$

We can write $G(y)$ to $G(x)$ and it has a unique behaviour that:

$$ G'(x) = G(x) $$

Now let's do some calculation about $a^x$. Since $G$ and $F$ are reverse function to each other. We have:

$$ a^x = G(F(a^x)) $$

Then let's continue the calculation with the help of (2):

$$ a^x = G(F(a^x)) = G(xF(a)) $$

Only when $F(a) = 1$ can we have:

$$ a^x = G(x) --- (4) $$

To make $F(a) = 1$, $a$ must be $2.71828...$. Euler called this constant $e$.

And if we call $F(x)$ as $ln(x)$. And $G(x)$ as $exp(x)$, we have:

(1) => $ln'(x) = \frac{1}{x}$

(2) => $ln(x^r) = rln(x)$

(3) => $exp'(x) = exp (x)$

(4) => $e^x = exp(x)$

and (3), (4) =>

(5) $(e^x)' = e^x$

So from the above construction and deduction, we can see that $e$ is a special value that is determined by $F(x) = 1$.

And with such a value we can define some nice functions that have nice behaviors. i.e. $ln(x)$ and $exp(x)$ or $e^x$.

We construct these functions through the calculus operations. Besides the nice differential behaviors, these functions also have some similar behaviors to the conventional logarithm/exponent functions, which I didn't list here. I think that's why they are given the fancy names $ln$ and $exp$.

smwikipedia
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  • For any downvote, please give a reason so I can improve on this answer. – smwikipedia Aug 28 '23 at 12:56
  • Not my downvote. But it may be because this is a long (almost longwinded) answer to an old question with an already accepted answer. Moreover, I don't think it actually answers the OP's request for some intuition about why "$e$"? – Ethan Bolker Aug 28 '23 at 13:03
  • Even the question is answered, it shouldn't prevent some further discussion. And I am not satisified with my answer either because it is not convincing enough at least to myself. But I haven't found a better explanation when I wrote this post. I made it long because I want to make the understanding curve gentle. – smwikipedia Aug 29 '23 at 08:52