# Definitions of $e$ Below are various definitions of the natural exponential base $e$. It can be proved mathematically that the definitions are equal, but I don't get the intuition behind why they are equal. It's fascinating that the definitions are equivalent, though its not clear to me 'why' they are. ### Compound Interest According to wikipedia, Bernoulli first encountered the constant $e$ in the context of studying compound interest. * If you have to pay an interest rate of $r$ at the end of the year, the final amount you have to pay is $(1+r)$ times the principal * If you have to pay an interest rate of $r$, compounded twice at 6 months, the amount you pay $(1+r/2)^2$ times the principal * If you have to pay an interest rate of $r$, compounded quarterly, the amount you pay is $(1+r/4)^4$ times the principal * If you have to pay the interest rate, compounded $n$ times, the amount you pay is $(1+r/n)^n$ times. If $r=1$, these values * $(1+1) = 2.0$, * $(1+1/2)^2$ = $2.25$, * $(1+1/4)^4$ = $2.44$, * $(1 + 1/365)^{365}=2.718$ This is where Bernoulli noticed that the value that if you keep extending this series, it converges (each further division increases the overall rate smaller and smaller) to a value that looks that looks like $2.714...$. The limit of this series the natural exponential base $e$. $ \lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n = e $ What if $r \neq 1?$, then $ \lim_{n \to \infty}\left(1+\frac{r}{n}\right)^n = e^r $ ### Probability of Failure in Bernoulli trials Again, taking the wikipedia article as source - if a gambler plays a game $n$ times and has a probability of winning $1/n$ each time, the probability that the gambler never wins once is $(1-1/n)^n$. Similar to the compound interest problem studied above, this approaches $ \lim_{n \to \infty}\left(1-\frac{1}{n}\right)^n=\frac{1}{e} $ Ok, the first time I wrote this I didn't see why the compound interest and this problem were related, but I see it now - If you set $r=-1$ in the compound interest definition of $e$, you get the same result. It's still mind boggling to me that they're related. ### Rate of growth The rate of growth of the $e^x$ is $e^x$. I still have trouble wrapping my head around this concept. It would be interesting to understand the history - [this seems like a good source](https://www.scirp.org/journal/paperinformation?paperid=120075), was this found by accident or from the compound interest equation above? Anyways, from calculus this means: $ \frac{d(e^x)}{dx} = e^x $ The Taylor series expansion of a function can be used to derive a polynomial for $e^x$. To get the intuition behind Taylor Series, refer to [this excellent video by 3Blue1Brown](https://www.youtube.com/watch?v=3d6DsjIBzJ4). The Taylor series expansion for $e^x$ is $ \begin{gather} e^x = 1 + \frac{x}{1} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + ... \\ \\ e^x = \sum_{i=0}^{\infty} \frac{x^i}{i!} \end{gather} $ # Should I just accept the definitions? * Math is all about making abstractions about the real world, and understanding the reasoning or motivation behind making the abstraction can help us internalize the technical details. * Source: 'Is Math Real', Euegenia Cheng [[Is Math Real, Eugenia Cheng]]