Motivation for taylor series of $\mathrm{e}$ A classical way to introduce "Euler's number" $\mathrm{e}$ is via compound interest which leads to 
$$
\mathrm{e} = \lim_{n\to \infty} \left (1 + \frac{1}{n} \right )^n 
$$
Is there a also such a real world motivation for 
$$
\mathrm{e} = \sum_{n = 0}^{\infty} \frac{1}{n!}
$$
 A: My favorite way is to
look for solutions to
the functional equation
$f(x+y) = f(x)f(y)$.
Setting $y=0$,
we get
$f(0) = 1$.
If we assume that they are differentiable,
$f(x+h)= f(x)f(h)
$,
so
$\begin{array}\\
f(x+h)-f(x)
&= f(x)f(h)-f(x)\\
&= f(x)(f(h)-1)\\
\text{so}\\
\dfrac{f(x+h)-f(x)}{h}
&= \dfrac{f(x)(f(h)-1)}{h}\\
&= f(x)\dfrac{f(h)-1}{h}\\
\end{array}
$
Letting $h \to 0$,
this gives
$f'(x) = f'(0)f(x)$,
so that
$f(x) = a^x$
where
$f'(0) = \ln(a)$.
$e$ is the value
that makes
$f'(0) = 1$.
A: One important property of $e$ (some would say the important property) is that $\frac{d}{dx} e^x = e^x$. Since we know that $a^0 = 1$ for any nonzero $a$, this property can be used to find the Taylor expansion about 0, since for $f(x) = e^x$, $f^{(n)}(0)=1 \forall n$. 
$$e^x = \sum_{n=0}^{\infty} \frac{x^n f^{(n)}(0)}{n!} = \sum_{n=0}^{\infty} \frac{x^n}{n!}$$
Since $1^n = 1 \forall n$,
$$e^1 = e = \sum_{n=0}^{\infty} \frac{1}{n!}$$
Another way to look at this is to examine what happens when you take the derivative of the Taylor series:
$$\frac{d}{dx} e^x = \frac{d}{dx}\sum_{n=0}^{\infty} \frac{x^n}{n!} = \frac{d}{dx}\left(1 + \sum_{n=1}^{\infty} \frac{x^n}{n!}\right) = \sum_{n=1}^{\infty} \frac{d}{dx}\frac{x^n}{n!}= \sum_{n=1}^{\infty} \frac{nx^{n-1}}{n!} = \sum_{n=1}^{\infty} \frac{x^{n-1}}{(n-1)!} = \sum_{n=0}^{\infty} \frac{x^n}{n!} = e^x$$
Similarly, when you integrate the Taylor series, you also get the same Taylor series back (plus a constant of integration). 
