# A fun surprise with Euler’s identity coming from Manjul Bhargava’s generalized factorials

Earlier this week I wrote about a really neat paper by one of the 2014 Fields Medalists Manjul Bhargava. My original post is here:

Fibonacci Factorials

and Bhargava’s fascinating paper is here:

The Factorial Function and Generlizations

I was thinking about the paper today because one of the questions Bhargava poses in the paper was stuck in my mind.  Near the end Bhargava asks about the analogue of the exponential function: $e^x = \sum_{k = 0}^{\infty} \frac{x^k}{k!}$

for the generalized factorial function.  It certainly seems that for the generalized factorials, we’ll see many different values for “e”.

I also thought that it might be fun to look at the Taylor expansions for cos(x) and sin(x) and see if there was any analogue to the formula $e^{\pi i} = -1$ for the generalized factorials.  I got a fun little surprise when I got home and played around a little.

My idea during the day was to pretend that first positive root of sin(x) was the analog to $\pi$ for generalized factorials and that evaluating the exponential function above at x = 1 was the analog for e.     When I tried to estimate these values for the Fibonacci Factorials that I’d found previously, I didn’t notice anything interesting.  I then moved on to looking at the example that Bhargava gives in the paper – generalized factorials over the prime numbers.    For the generalized factorial function over the primes we get the following values for the first 10 factorials:

0! = 1
1! = 1

2! = 2

3! = 24

4! = 48

5! = 5760

6! = 11,520

7! = 2,903,040

8! = 5,806,080

9! = 1,393,459,200

10! = 2,786,918,400

I calculated up to 19! in order to use 10 terms each in the series for Sin(x) and Cos(x).    Here’s what the graphs for Sin(x) and Cos(x) looked like for x between 0 and 6 when we use the first 10 terms of the Taylor series for both functions:

y = Sin(x) = $x - x^3 / 3! + x^5 / 5! - x^7 / 7! . . .$ and here’s what y = Cos(x) = $1 - x^2 / 2! + x^4 / 4! - x^6 / 6! + . . .$ looked like over the same interval: As you can see from the two graphs, the usual identity $Sin(x)^2 + Cos(x)^2 = 1$ does not hold in this setting!

You can also see from the graph that the first positive root of Sin(x) is a little bit larger than 5.  According to Mathematica, the root is approximately x = 5.1819247.    So in this setting the analogy we have is that $\pi \approx 5.18. . .$

Now for the surprise.  Cos(x) evaluated at that root is equal to 1 to quite a high precision.  A high enough precision, in fact, that Mathematica simply returns the value 1.  I have not done enough work even to know how to calculate the remaining (infinitely many, ha!) factorials and see if that result holds in general.  Actually, I doubt that calculation is even within the realm of something that I could do.  However, if the result does hold in the general case rather than just when we approximate the various Taylor series with 10 terms, it would mean that when you evaluate Bhargava’s generalized factorials over the primes, and use the analogies for e and $\pi$ that I mention above, you get the amazing and quite surprising identity that $e^{\pi i} = 1$

[ Further – the proof uses Dirchlet’s Theorem on primes and arithmetic progressions. I will try to write up the proof more carefully when I have a little more time, but the idea is that for a given prime p, Dirchlet’s theorem tells us that there are infinitely many primes with remainder 1,2,3, . . . , and p – 1 when divided by p. This means that Bhargava’s factorials over the primes will add new powers of p in steps of p – 1 after the $p^{th}$ step. But for odd primes, since p – 1 is even it will be only in the odd numbered factorials where the powers of odd primes increase. Furthermore , the even steps will increase the previous odd step by multiplying by 2. You can see the beginning of these patterns in the list above.
What this all means is that Bhargava’s prime factorials have some interesting relations. One in particular is that 2* (2n – 1)! = (2n)!. When you look at the power series for $e^x, Cos(x),$ and $Sin(x)$ you can see that this simple relation implies that $Cos(x) = 1 - (x / 2) * Sin(x).$  Thus, when $Sin(x) = 0, Cos(x) = 1$. Since my analogy for $\pi$ in this setting was the first positive root of $Sin(x)$, this all shows that for factorials over the primes it seems that the the analogy for Euler’s formula is: $e^{\pi i} = 1.$ Fun!
1. Renan says: