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Misc. Math

Landau Notation

$$ f(x) = O(g(x)) \iff \limsup_{x \to \infty} \frac{f(x)}{g(x)} < \infty $$

$$ f(x) = o(g(x)) \iff \lim_{x \to \infty} \frac{f(x)}{g(x)} = 0 $$

$$ f(x) = \Omega(g(x)) \iff \liminf_{x \to \infty} \frac{f(x)}{g(x)} > 0 $$

$$ f(x) = \omega(g(x)) \iff \liminf_{x \to \infty} \frac{f(x)}{g(x)} = \infty $$

$$ f(x) = \Theta(g(x)) \iff f(x) = O(g(x)) \ \& \ f(x) = \Omega(g(x)) $$


$e$ is Irrational

Proof by contradiction.

Assume $e$ is rational. This means $e = \frac{p}{v}$.

We use the identity:

$$ e = \sum_{k=0}^{\infty} \frac{1}{k!} $$

With the usual convention that $0!=1$.

$$ \begin{array}{lcl} & e & = \frac{p}{v} \\ & & = \sum_{k=0}^{v} \frac{1}{k!} + \sum_{k=v+1}^{\infty} \frac{1}{k!} \\ \to & (v!) e & = (v-1)! p \\ & & = \sum_{k=0}^{v} \frac{v!}{k!} + ( \frac{1}{v+1} + \frac{1}{(v+1)(v+2)} + \cdots ) \end{array} $$

By assumption, $(v!) e = (v-1)! p \in \mathbb{Z}$. We also have the first sum $\sum_{k=0}^{v} \frac{v!}{k!} \in \mathbb{Z}$.

The second sum is greater than 0 and we can get bounds:

$$ \begin{array}{lrcl} & 0 & < \frac{1}{v+1} + \frac{1}{(v+1)(v+2)} + \cdots & \le \frac{1}{v+1} \sum_{k=0}^{\infty} \frac{1}{(v+1)^k} \\ \to & 0 & < \frac{1}{v+1} + \frac{1}{(v+1)(v+2)} + \cdots & \le \frac{1}{v} \end{array} $$

Choose $v > 1$ and we have the second sum as non-integral, contradicting the assumption of rationality.


2020-10-31