Euler's number, $e$, is a very important irrational constant, it has wide uses across maths, science and economics. In particular it's useful in modelling the natural processes of growth and decay, compound interest as well as having a number of useful properties in calculus. We define $e$ as the exponential function that has a gradient of $1$ when it crosses the coordinate axis $x=0$, this can be represented in functions as:
\[ \frac{\mathrm{d}}{\mathrm{d}x} e^x = 1, \ \text{when} \ x = 0\]
Or graphically. The graph shows the plots of $e^x$, $3^x$ and the line $y = x+1$. The line also has a gradient of 1 for comparison:
But what is the value of $e$? Lets suppose there exists an exponential function $a^x$ where $a$ is an arbitrary constant, the derivative of this function is: \[\begin{align} \frac{\mathrm{d}}{\mathrm{d}x} a^x &= \lim_{\delta x \to 0} \ \frac{a^{x+ \delta x} - a^x}{\delta x} \\ \\ &= \lim_{\delta x \to 0} \ \frac{a^{x} ( a^{\delta x} - 1)}{\delta x} \end{align} \] $e$ has the unique property that its gradient is 1 when it crosses the coordinate axis. So make the derivative 1 and set $x=0$. This is: \[\begin{align} &\lim_{\delta x \to 0} \ \frac{a^{0} ( a^{\delta x} - 1)}{\delta x} = 1 \\ \\ &\lim_{\delta x \to 0} \ \frac{( a^{\delta x} - 1)}{\delta x} = 1 \\ \\ &\lim_{\delta x \to 0} \ a^{\delta x} = 1 + \delta x \\ \\ &\lim_{\delta x \to 0} \ 1 + \delta x = a^{\delta x} \\ \\ &\lim_{\delta x \to 0} \ (1 + \delta x)^{\frac{1}{\delta x}} = (a^{\delta x})^{\frac{1}{\delta x}} \\ \\ &\lim_{\delta x \to 0} \ (1 + \delta x)^{\frac{1}{\delta x}} = a \end{align}\] When this limit is evaluated the arbitrary constant $a$ is found to be $2.71828...$ This is Euler's number!
Euler's number can be expressed as another limit as well, by making the substitution $\delta x = \frac{1}{\xi}$. It can be seen that for $\frac{1}{\xi} \to 0$, $\xi \to \infty$, this can replace the limit of the final expression, so:
\[ \lim_{\xi \to \infty} \ \Big(1+ \frac{1}{\xi} \Big)^{\xi} = e \]
I have graphed this limit over a small range of $\xi$ values to demonstrate that it converges onto $e$.
Lets say I wanted to find the derivative of $a^x$: \[\begin{align} \frac{\mathrm{d}}{\mathrm{d}x} a^x &= \lim_{\delta x \to 0} \ \frac{a^{x+ \delta x} - a^x}{\delta x} \\ \\ &= \lim_{\delta x \to 0} \ \frac{a^{x} ( a^{\delta x} - 1)}{\delta x} \\ \\ &= \lim_{\delta x \to 0} \ a^{x} \Big( \frac{a^{\delta x} - 1}{\delta x} \Big) \end{align} \] Now that derivative isn't very pleasant. However we saw earlier that in the case where the arbitrary constant $a$ is equal to $e$ the limit evaluates to 1, so: \[ \lim_{\delta x \to 0} \ \frac{ (e^{\delta x} - 1) }{\delta x} = 1\] So it can be seen that: \[ \frac{\mathrm{d}}{\mathrm{d}x} e^x = \lim_{\delta x \to 0} \ e^{x} \Big( \frac{e^{\delta x} - 1}{\delta x} \Big) = e^x \cdot 1 = e^x \] So not only does $e$ make finding the derivative of exponents easier it has a rather unique property, it is its own derivative!
We can also find the antiderivative of $e^x$ using the the fact that $\frac{\mathrm{d}}{\mathrm{d}x} e^x = e^x$: \[ \int e^x \ \mathrm{d}x = \int \frac{\mathrm{d}}{\mathrm{d}x} (e^x) \ \mathrm{d}x \] From here it is simple enough to apply the fundamental theorem of calculus to find the antiderivative: \[ \int e^x \ \mathrm{d}x = \int \frac{\mathrm{d}}{\mathrm{d}x} (e^x) \ \mathrm{d}x = e^x + C \]
$e$ is also very useful when dealing with logarithms. Suppose I find the differential of an arbitrary logarithm in base $a$: \[ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x} \log_{a} x &= \lim_{\delta x \to 0} \ \frac{\log_{a} (x + \delta x) - \log_{a} x }{\delta x} \\ \\ &= \lim_{\delta x \to 0} \ \frac{ \log_{a} \Big( \frac{x + \delta x} {x} \Big) } {\delta x} \\ \\ &= \lim_{\delta x \to 0} \ \frac{ \log_{a} \Big( 1 + \frac{\delta x}{x} \Big) } {\delta x} \end{align} \] Making the substitution $\xi = \frac{\delta x}{x}$, the limit is now becomes $\xi \to 0$ \[ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x} \log_{a} x &= \lim_{\xi \to 0} \ \frac{ \log_{a} (1 + \xi) } {x \xi} \\ \\ &= \lim_{\xi \to 0} \ \frac{1}{x} \Big( \frac{1}{\xi} \log_{a} (1 + \xi) \Big) \end{align} \] Now it isn't a particularly pleasant derivative, but what if $ \lim_{\xi \to 0} \frac{1}{\xi} \log_{a} (1+ \xi) = 1$, then the derivative would just be $\frac{1}{x}$. This can be solved fairly easily using algebra: \[ \begin{align} &\lim_{\xi \to 0} \ \frac{1}{\xi} \log_{a} (1+ \xi) = 1 \\ \\ &\lim_{\xi \to 0} \ \log_{a} (1+ \xi) = \xi \\ \\ &\lim_{\xi \to 0} \ 1 + \xi = a^{\xi}\end{align} \] Raising both these expressions to the power $\frac{1}{\xi}$: \[ \lim_{\xi \to 0} \ (1+\xi)^{\frac{1}{\xi}} = a\] Now that limit should look familiar, we derived it earlier, it is $e$. The logarithm to base $e$ is very useful as it is easy to find derivatives of logarithms. The logarithm to base $e$ is given a special symbol $\log_{e} x = \ln x$
We established that $e$ is its own derivative, this means it can be differentiated infinitely, so it can be expressed as a Taylor expansion: \[ \exp(x) = 1 + x + \frac{1}{2}x^2 + \frac{1}{6} x^3 + \cdots \] To find $e$ we must set $x=1$, this simplifies the expansion down to: \[ e = 1 + 1 + \frac{1}{2} + \frac{1}{6} + \cdots \] or in sigma notation: \[ e = \sum_{k=0}^{\infty} \frac{1}{k!} \] The function $e^x$ is defined by the area under the curve $e^x$ so $e^x$ can be represented by the integral: \[ e^x = \int_{-\infty}^{x} e^x \ \mathrm{d}x \] So an expression of $e$ is: \[e = \int_{-\infty}^{1} e^x \ \mathrm{d}x\] The function $\ln x$ is defined by the area under the curve $\frac{1}{x}$. This can be used to form an interesting expression for 1: \[ \int_{1}^{e} \frac{1}{x} \ \mathrm{d}x = 1 \]
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