Another Math Blog

Today, we’ll be looking at $p$-adic valuation defined on the integers, with a specific look at a special formula for finding the $p$-adic valuation of $n!$ for any given $n.$ We won’t be as concerned about optimizations this time, since … Continue reading →

While looking through some of my older programming projects, I found my old Egyptian Form project, which implements a greedy algorithm described by Solomon W. Golomb [1] for turning any fraction of positive integers into a sum of reciprocals of … Continue reading →

mathematicsnumber-theory

One of my favorite facts from combinatorics is that $\sum\limits_{0\leq k\leq n}\binom{n}{k} = 2^{n}$. To prove it is simple: Note that $2 = 1 + 1$, and that $\binom{n}{k} = \binom{n}{k}\cdot 1^{k}\cdot 1^{n-k}$, and appeal to binomial theorem: $$\sum_{0\leq k\leq … Continue reading →

combinatoricsmathematics

For each real number $t>1$, one can define a probability distribution $P_{t} = \left\{\mathrm{pr}_{t}\left(n\right)\right\}_{n\in\mathbb{N}}$ by $\mathrm{pr}_{t}\left(n\right) = \displaystyle{\frac{n^{-t}}{\zeta\left(t\right)}}$. This class of probability functions was studied by Golomb in [1]. In this post, we will prove a theorem about relatively $r$-prime … Continue reading →

mathematicsprobability

I was recently introduced to an interesting polynomial problem. Problem. Determine all polynomials $p\left(x\right)\in \mathbb{R}\left[x\right]$ such that $\left(x – 16\right)p\left(2x\right) = 16\left(x – 1\right)p\left(x\right)$ for all $x\in \mathbb{R}$. The way I solved this was to essentially use this equation as … Continue reading →

algebramathematics

While reading some course notes from MIT 18.703 (Modern Algebra), I came across the following statement on page 3: Lemma 22.3. The product of all primes $r$ between $N$ and $2N$ is greater than $2^{N}$. However, this can quickly be … Continue reading →

mathematicsnumber-theory

Recall that the harmonic series $\displaystyle{\sum_{n=1}^{\infty}\frac{1}{n}}$ diverges. This is because we may bound the partial sums below, like so: $$1+\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+\cdots >$$ $$\frac{1}{2}+\frac{1}{2}+\left(\frac{1}{4}+\frac{1}{4}\right)+\left(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\right…

divergencemathematicsseries