The number $e$ (and the exponentiation function $e^x$) appears in so many places in mathematics and engineering. There seem to be a multitude of applications of it. I want to know why.
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Why is e so special ?
Because very many of humanity's age-old mathematical interests ultimately converged towards it:
Arithmetic:
Addition begets multiplication; multiplication begets exponentiation; exponentiation begets e, since by studying it we inevitably arrive at the conclusion that this number is its most natural base.
$\sqrt[\large^n]{\text{LCM}(1,2,3,\ldots,n)}~$ tends to e as n tends towards infinity.
Geometry:
- Circles and hyperbolas have been studied since ancient times; e is to the latter what $\pi$ is to the former.
Finance:
- Examining the way in which banking interests are computed leads us to discovering the same quantity.
Calculus:
The harmonic series has been studied since ancient times; its continuous equivalent is $\displaystyle\int\frac1x~dx$ $=\log_ex$.
The solution to $f(x)=f'(x)$ is $a~e^x$, meaning that the exponential function is immune to the operations of differentiation and integration.
Lucian
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1Why do you say "we inevitably arrive at the conclusion that this number is its most natural base"? The only reasons I could give for this are analytical, not arithmetic... – A.P. Jun 09 '15 at 08:43
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@A.P.: Because, when we differentiate all basic operations and elementary functions, we notice that $\big(a^x\big)' ~=~ a^x\cdot\displaystyle\lim_{h\to0}\frac{a^h-1}h$, which naturally makes us wonder for what value of a might that limit be $1$, so as to have a “nice” exponentiation base which yields us no “leftovers”. And that limit is nothing else than the natural logarithm of a, as shown here. – Lucian Jun 09 '15 at 09:22
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That is precisely my point. This reason is analytic and it is precisely the same as the second one you cited under the "Calculus" heading. – A.P. Jun 09 '15 at 10:24
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@A.P.: Wanting to simplify an expression, and hunting for a super-special function, are two completely different things altogether. That, for some mysterious reason, they happen to point in the same direction, is mere coincidence, so to say. We would still have discovered e even if the derivative of $a^x$ would have been $f(x)\cdot\displaystyle\lim_{h\to0}\frac{a^h-1}h$ with $f(x)\neq a^x$. – Lucian Jun 09 '15 at 10:30
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I'm sorry: how is looking for a value of $a$ for which that limit is $1$ different from finding a value of $a$ for which $(a^x)' = a^x$ (which would then be a particular solution of $f'(x) = f(x)$)? – A.P. Jun 09 '15 at 10:45
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@A.P.: Because it's just a coincidence that one of the basic arithmetic operations also happens to be the answer to a very simple and very beautiful differential equation. The former has nothing directly to do with the latter. Simplifying the expression of $\big(a^x\big)'$ would still have been done, even if its expression would not have had any remarkable properties. And the solution $f(x)=f'(x)$ would still have been studied, even it would not have been a combination of known or elementary functions. – Lucian Jun 09 '15 at 11:05
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Exponentials pop up in linear differential equations, which govern innumerable physical phenomena. Such equations appear every time the variation of a physical quantity is proportional to that same physical quantity.
Let us take the discharge of a capacitor in a resistor: due to the resistor, the current is proportional to the tension, and due to the capacitor the tension is proportional to the load. Lastly, the capacitor empties at a speed proportional to the current.
This is summarized by the archetypal equation below:
$$\frac{du}{dt}=-u.$$
This equation is solved as
$$\frac{du}u=-t,$$ then $$\int\frac{du}u=-\int dt,$$ then $$\log_e(u)=-t$$ then $$u=e^{-t}.$$
It expresses that the capacitor will discharge exponentially, i.e. after $1$ unit of time, only $1/e=37\%$ of the initial load remains, after two units of time, $1/e^2=14\%$, after three units of time, $1/e^3=5\%$... The decrease is fast and follows a geometric progression.
This is a typical behavior of systems, that move from one state to another following a transient stage with an exponential decay. The differential equation shows you how the exponential function appears. It also shows you that $e$ is a natural base of exponentials and logarithms, as it is the only base such that
$$(\log_b(x))'=\frac1x$$and similary$$(b^x)'=b^x.$$
$e$ is a natural "unit" in the same sense that $2\pi$ is a natural angular unit for the trigonometric functions, as
$$(\sin(x))'=\cos(x)$$
is valid in radians, i.e. when a full turn is $2\pi$.
(By contrast, $(2^x)'=0.69314718\cdots 2^x$, and $(\sin_°(x))'=0.01745329\cdots\cos_°(x)$ in degrees.)
Actually, all the elementary functions that you find on a calculator are closely related to the exponential: the exponential $e^x$ itself, its inverse $\ln(x)=\log_e(x)$, the exponential and logarithms in other bases, $b^x$ and $\log_b(x)$, the powers $x^y$, computed from $x^y=e^{y\ln(x)}$; then the trigonometric functions constructed from the imaginary exponential $e^{ix}=\cos(x)+i\sin(x)$, the functions derived from them $\tan,\sec,\csc,\cot$ and their inverses $\arccos,\arcsin,\arctan,\cdots$. Ditto for the hyperbolic functions.
As logarithms and antilogarithms can be used to perform multiplies and divides, a great deal of the maths can be done with $+, -, \ln(z),e^z$ only ! ($z$ to denote complex numbers.)
For example,
$$\arccos(x)=-i\ln\left(x+ie^{e^{\ln\left(\ln\left(1-e^{e^{ln(ln(x))+ln(2)}}\right)\right)-\ln(2)}}\right).$$
(This section is informal.)
To compute the value of $e$, let us use the differential equation. As $$(e^x)'\Big|_{x=0}=\lim_{x\to0}\frac{e^x-1}x=e^0=1,$$ we have $$e^x\approx1+x$$ for small $x$.
Then $$e=(e^{1/n})^n\approx\left(1+\frac1n\right)^n,$$ and $$e=\lim_{n\to\infty}\left(1+\frac1n\right)^n.$$
Also, by the binomial theorem,
$$\left(1+\frac1n\right)^n=1+\frac nn+\frac{n(n-1)}{2n^2}+\frac{n(n-1)(n-2)}{3!n^3}\cdots\approx1+1+\frac12+\frac1{3!}\cdots$$ and in the limit, $$e=\sum_{k=0}^\infty\frac1{k!}.$$
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Technically, you only need $-,\ln(z),e^z$ (and $1$), because $a+b=a-((1-1)-b)$. (And, in case anyone thinks it's cheating for him to use $i$: $i=\exp(\exp(-\ln2)\ln(-1))$. And the $\Im$ isn't necessary, either.) – Akiva Weinberger Jun 09 '15 at 10:00
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@columbus8myhw: you can also consider $i$ as immaterial, when defining a complex number $z$ as a couple of reals $(x,y)$; this avoids the "multiplies" by $i$. – Jun 09 '15 at 10:12
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True. I was thinking more along the lines of "use only the symbols $1$, $\exp$, $\ln$, and $-$ (and parentheses) to get as many numbers as possible" (with the principle branch of log, though it doesn't matter). – Akiva Weinberger Jun 09 '15 at 11:01
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$e$ is the positive real number which fulfils differential equation \begin{equation} \frac{d}{dt} e^t = e^t \end{equation} Also, in engineering problems, the number $e$ appears in the solutions of systems of differential equations (see e.g. the wikipedia article about Linear differential equations).
smo
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The idea of exponential growth or decay is natural: certain continuous quantities increase $10\%$ every year, other quantities get halved in the course of $10\,000$ years, etcetera. Given a unit of time there are still "slow" and "fast" exponential increases: a quantity can double in one unit of time, or it can increase by a factor of $100$ in one unit of time. Therefore we need a unit for the "speed" of exponential growth. We arrive at it in the following way: Among all exponential growth curves $t\mapsto f(t)$ with $f(0)=1$ there is exactly one which has slope $1$ at $(0,1)$. The base for this particular $f$ is a certain number $>1$. This number is called $e$, its value turns out to be about $2.718$. In this way the standard exponential function is $f(t)=e^t$.
Christian Blatter
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This too: http://math.stackexchange.com/questions/272507/how-would-you-explain-why-e-is-important-and-when-it-applies?rq=1
– hjhjhj57 Jun 09 '15 at 06:54