e

The number e is the unique number a > 0 satisfying 1 + x \le a^x for all x.

If y = a^x for some a > 0 then its graph meets the line y = 1 + x exactly once if and only if a = e.

We can show e\in[2,3] with simple calculations.

Exercise. Show 2^{1/2} > 1 + 1/2.

Since 1 + x \le e^x we have 1 - x \le e^x so 1+x \le e^x \le 1/(1 - x) when 1 - x > 0.

Exercise. Show (1 + x/n)^n \le e^x for all x and e^x \le (1 - x/n)^{-n} if x < 1.

Newton’s binomial formula is \begin{aligned} (1 + x)^a &= 1 + ax + a(a - 1) x^2/2 + \dots \\ &= \sum_{k = 0}^\infty (a)_n x^n/n! \\ \end{aligned} when |x| < 1 and (a)_n = a(a - 1)\cdots(a - n + 1) is the (decreasing) Pochhammer symbol.

Exercise. _Show (n)_n = n!.

Exercise. _Show $(m)n = 0 if m > n.

This gives a lower bound for computing e^x: \begin{aligned} e^x &\ge \sum_{k=0}^n n(n - 1)\cdots(n - k + 1)x^k/k! \\ &= \sum_{k=0}^n C(n,k) x^k \end{aligned} where C(n,k) = n!/(n - k)!k! is the number of ways to choose k items from a collection of n items.