Benford's Generalized Law

First of all, apologies for my lack of posts recently; school started for me, so I don't really have time for much. I have been working on an article regarding the Hollow Earth theory in which I debunked it, but, after writing the article, I realized that I made a mistake somewhere along the way, and, frankly speaking, I rage-quited. I still wanted to write an article, so I took out my 'notebook with notable problems' (I have one of those, I'm a full-time nerd), and I came across the Newcomb-Benford law, which states that the numerical values of the vast majority of social phenomena (the majority modeled by exponential evolutions in rapport to time) start with 1. The law takes its name from its two discoverers, Frank Benford, who wrote about it in 1938, and Simon Newcomb, who discovered it in 1881. Both discovered it independently from one another.  This problem was also proposed in a Gazeta Matematica back in February 2017.

Let \(f(t)=a^t\) be the function modeling a specific social event. We are looking at the values of t for which $$10^n<f(t)<2*10^n$$. By taking logarithms, we get that \( n<t log(a)<lg2 + n\). Thus, the probability that the first digit of f is 1 is log2, which is roughly equal to 0.301. For the general case, $$lg(d)+n<t lg(a)<lg(d+1)+n$$,where d is the first digit of f, so the probability that the first digit is d is equal to \(log(d+1)-log(d)\). The function \(log(x+1)-log(x)\) is decreasing for all positive x, so the probability of each digit appearing is decreasing. Obviously, the law works for all numerical bases.

We can also do a quick generalization of this result. How about we derive the n-th digit of an exponential function? Let us presume that the exponential function has m digits, with m bigger than n. Now, we know that \((10M+d)*10^{m+1-n}<a^t<(10M+d+1)*10^{m+1-n}\), where M is the number composed by the first n-1 digits. Thus, by taking logs again, we get that $$(m+1-n)log(10M+d)<tloga<(m+1-n)log(10M+d+1)$$, so the required probability, for a certain M, is \(log(1+ \frac {1}{10M+d})\). Since M is the number made by the first n-1 digits, it takes values from \( 10^{n-2}\) to \(10^{n-1}-1\), so the required probability is $$P(d)= \sum_{k=10^{n-2}}^{10^{n-1}-1} log(1+ \frac {1}{10k+d})$$.

After doing a quick search on Wikipedia, I found out that this result has some practical uses. Apparently, price digits and election results can be somehow checked by this law, and it appears to have applications in molecular genetics, too. Perhaps some old results like this are still revolutionary to this day.

Cris.