i

\(e\)

The derivative of \(e^x\) is itself. That's how \(e\) is (maybe I should say 'can be') defined.

Lets do some math with that. By the above definition, we have \(e^x={e^{x+dx}-e^x\over dx}\) where, of course, \(dx\) goes to 0. We can now simplify this to \(dx=e^{dx}-1\). Hey! We can solve for \(e\)! Lets do that: \(e=(1+dx)^{1\over dx}\). Since \(dx\) is going to 0, we can replace it by, for example, \(1\over n\), where \(n\) goes to infinity. We get the well-known expression\[e=\lim_{n\rightarrow\infty}\left(1+{1\over n}\right)^{n}\]

Just because it's fun to do, I'll also give you another way to approach \(e\). Taylor series are always fun, so consider the following: \[e^x=\sum_{k=0}^{\infty}{a_kx^k}\]Since the \(n\)th derivative on the left at \(x=0\) is always 1, the right side is also 1; \(1=n!a_n\). Thus, \(a_n={1\over n!}\), so we get \[e^x=\sum_{k=0}^{\infty}{x^k\over k!}\]Note that this is not a full proof, but it gives you an idea of what happens. You can also make a Taylor series for \(\sin(x)\) and \(\cos(x)\), and with those, you can see that \(e^{ix}=\cos(x)+i\sin(x)\), where \(i^2=-1\) is the imaginary number. This connects \(\pi\) to \(e\). Awesome, right?

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