## Curious Constants (part 1 of 4)

I’m giving a talk tomorrow about a handful of mathematical constants. I’ve been working on digging through references, finding the constants I want to talk about, and seeing how much I can make sense of, and I think I’m getting on towards good shape. I want to not focus on more common constants, like $e,\pi,\phi$, though, of course, they have a tendency of sneaking up, sometimes apparently out of nowhere. I’d also like not to pick specific integers to talk about, though I will end up talking about sequences and subsets of integers.

Without further ado, here’s my hand-picked collection of curious constants. I’ve truncated (not even bothered rounding) the decimals off at about 3 places, because the number itself doesn’t strike me as the interesting part. Resources are mentioned in the last post in this series.

## The First Batch

### 1.291… & 0.704…

These are the values of the integrals $\int_0^1 \frac{1}{x^x}dx$ and $\int_1^{\infty} \frac{1}{x^x}dx$. What’s more interesting is some series representations of these integrals. We can write the first as

$\begin{array}{rcl}\displaystyle\int_0^1\frac{1}{x^x}dx &=& \displaystyle\int_0^1 e^{-x\ln x}dx\\ &=& \displaystyle\int_0^1 \sum_{n=0}^{\infty} \frac{(-x\ln x)^n}{n!}dx \\ &=& \displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}\int_0^1 x^n(\ln x)^n dx \\ &=& \displaystyle1+\sum_{n=1}^{\infty}\frac{(-1)^n}{n!}\cdot \frac{(-1)^nn!}{(n+1)^{(n+1)}}\\ &=&\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^n}\end{array}$

Notice that if we just integrated $x^x$, instead of it’s reciprocal, the minus signs wouldn’t cancel, and we’d just get the alternating series. By the way, Ramanujan rewrote the second integral as $\int_1^{\infty} \frac{1}{x^x}dx=\sum_{n=0}^{\infty}(-1)^nn^n$, using a technique known as Borel summation. This won’t be the first divergent series we’ll see.

### 1.444… & 0.0659…

While we’re on iterated exponentials (tetrations), let’s make an infinite tower. But first, some notation. Define ${}^1x=x$, ${}^2x=x^x$, and in general ${}^nx=x^{({}^{n-1}x)}$. So ${}^nx$ is a tower of exponentials, with $n$ (count ’em) $x$s. Then let $h(x)=\lim_{n\to \infty} {}^nx$, if the limit exists. That is, $h(x)=x^{x^{x^{.^{.^.}}}}$. Suppose that $a$ is a point for which this limit exists. Then $a=x^a$, so $a^{1/a}=x$. What this means is that if we define $g(x)=x^{1/x}$, then $g$ and $h$ are inverse functions. This helps us identify the domain of $h(x)$, because $g(x)$ has a maximum value of $e^{1/e}\approx 1.444$. So this forms the upper limit of the domain of $h(x)$. It turns out that the lower limit is $e^{-e}\approx 0.0659$. I, personally, like the number $e^{1/e}$, because it is the unique value $a$ so that $a^x$ and $\log_a x$ have a common tangent.

### 16.177… & 23.103…

Let me write $\sum_n'\frac{1}{n}$ for the sum of the reciprocals of those natural numbers $n$ whose decimal expansion contains no 1. Let’s first convince ourselves that this series converges. To do so, notice that there are $8\cdot 9^{k-1}$ values for $n$ that have $k$ digits, none of which is a 1. For such an $n$, the reciprocal is never greater than $1/10^{k-1}$. Therefore, $\sum_n'\frac{1}{n}$ is no greater than the convergent geometric series $\sum_{k=1}^{\infty} 8 \left(\frac{9}{10}\right)^{k-1}$. In fact, the sum is around 16.177. There was nothing particularly special about choosing to exclude the digit 1, and excluding any digit still gives a convergent series. The greatest of these sub-harmonic series is when 0 is excluded, with a sum around 23.103. The others lie between 16.177 and 23.103. This series is known as the Kempner Series, for his paper on the topic in 1914.

### .577…

Perhaps the most well-known constant on the lineup, Euler’s (shared with Mascheroni) constant, $\gamma$, given by $\lim_{n\to \infty} \left(\sum_{i=1}^{n}\frac{1}{n}-\ln n\right)$. I mention this mostly as a reminder for when it shows up shortly.

### .261…

Before getting to what this constant is, I’d like to present Euler’s proof that the sum of the reciprocals of the primes diverges (an easy corollary being that there are infinitely many primes). Set

$M=\sum_{n}\frac{1}{n}=\prod_{p}\left(1-(1/p)\right)^{-1}=\prod_{p}\left(1+\frac{1}{p}+\frac{1}{p^2}+\frac{1}{p^3}+\cdots\right)$

Here we have expanded each factor $\left(1-(1/p)\right)^{-1}$ using the usual expansion $1/(1-x)=\sum_{n=0}^{\infty} x^n$, and the products are taken over all primes. This helps see why the product is the same as the harmonic series, if you stare at it for a few minutes. Now, taking the natural log on both sides of $M=\prod_p \left(1-(1/p)\right)^{-1}$, and using the Taylor expansion $\ln(1-x)=-\sum_{n=1}^{\infty} \frac{x^n}{n}$, we get

$\begin{array}{rcl}\ln M &=& -\ln(1-1/2)-\ln(1-1/3)-\ln(1-1/4)-\cdots \\ &=& \phantom{+}\frac{1}{2}+\frac{1}{2}\left(\frac{1}{2}\right)^2+\frac{1}{3}\left(\frac{1}{2}\right)^3+\cdots \\ &{} & +\frac{1}{3}+\frac{1}{2}\left(\frac{1}{3}\right)^2+\frac{1}{3}\left(\frac{1}{3}\right)^3+\cdots\\ &{}& + \frac{1}{5}+\frac{1}{2}\left(\frac{1}{5}\right)^2+\frac{1}{3}\left(\frac{1}{5}\right)^3+\cdots\\ &{}& + \ddots\end{array}$

Here, the rows are indexed by the primes, and the columns by the natural numbers. Next, sum down columns to obtain

$\ln M=\sum_p \frac{1}{p}+\frac{1}{2}\sum_p \frac{1}{p^2}+\frac{1}{3}\sum_p \frac{1}{p^3}+\frac{1}{4}\sum_p \frac{1}{p^4}+\cdots$

Thinking about the integral test, if $s>1$ then $\sum_{k=2}^{\infty} \frac{1}{k^s}\leq \frac{1}{s-1}$, so if we only sum over primes, the inequality still holds. Therefore

$\begin{array}{rcl}\ln M &=& \sum_p \frac{1}{p}+\sum_{n=2}^{\infty}\left( \frac{1}{n}\sum_p \frac{1}{p^n}\right) \\ &\leq & \sum_p \frac{1}{p}+\sum_{n=2}^{\infty} \frac{1}{n(n-1)}\end{array}$

Since the final sum on the right is convergent, and we know the harmonic series diverges (so $\ln M$ “diverges”) we must determine that $\sum_p \frac{1}{p}$ diverges.

In 1874, German mathematician Franz Mertens established the following two results ($p$ indicates we index over primes):

1. $\sum_{p\leq x}\frac{\ln p}{p}=\ln(x)+O(1)$
2. $\prod_{p\leq x}\left(1-\frac{1}{p}\right)^{-1}=e^{\gamma}\ln(x)+O(1)$

He also determined that $\lim_{n\rightarrow \infty}\left(\sum_{p\leq n}\frac{1}{p}-\ln(\ln(n))\right)\approx .261$. He was beaten to this result, though he obtained it independantly, by Ernst Meissel, who determined the same limit in 1866.

### 1.902… & 0.660…

Though the sum of the reciprocals of all primes diverges, Brun, in 1919, showed that the sum of all prime pairs converges. In fact,

$\left(\frac{1}{3}+\frac{1}{5}\right)+\left(\frac{1}{5}+\frac{1}{7}\right)+\left(\frac{1}{11}+\frac{1}{13}\right)+\cdots\approx 1.902$

This value is known as Brun’s constant. Let $\pi(x)$ denote the number of primes less than $x$ and $\pi_2(x)$ denote the number of primes, $p$, less than $x$, for which $p+2$ is also a prime. One way to state the prime number theorem is to say $\pi(x)/x\sim 1/\ln(x)$. Similarly, it is reasonably conjectured that $\pi_2(x)/x\sim A/(\ln x)^2$ for a constant $A$. If the events “$p$ is prime” and “$p+2$ is prime” were independant, we would expect $A=1$. However, this seems to not be the case, and in fact $A=2C_2$ where $C_2=\prod_{p\geq 3} \left(1-\frac{1}{(p-1)^2}\right)\approx 0.660$ is the “twin primes constant.” Before discussing where this constant comes from, it is worth mentioning that computer work with twin primes in the mid ’90s led to the discovery of the Pentium bug.

Several papers have aimed at motivating the appearance of $C_2$ in the asymptotic estimates for $\pi_2(x)$. The one that finally made things click for me is this one, which I found here, and whose methods I discuss here. Suppose $f_1,\ldots f_k$ are polynomials with integer coefficients, and let’s see what we can say about the probability that they are simultaneously prime when evaluated at $n$. If they were independant, the probability would be the product of the probabilities that the $f_i(n)$ were prime, which is $1/\ln(f_i(n))$, by the prime number theorem. However, this is, in general, too much to ask for, and we introduce a scaling factor.

Define $w(p)$ to be the number of values $x\in \{0,\ldots,p-1\}$ for which $f_1(x)\cdot f_2(x)\cdots f_k(x)\equiv 0\mod p$. That means $x$ gets counted toward $w(p)$ if its value under any of the polynomials is a multiple of $p$. Now we recognize that the probability that $p$ does not divide any $f_i(n)$, for a randomly chosen $n$, is $(p-w(p))/p$. This gets divided by the probability that $p$ does not divide any $k$ randomly chosen integers, which is $\left(\frac{p-1}{p}\right)^k$. This gives us an appropriate scaling factor for each prime $p$, and we take the product over all of them to obtain $\prod_p \frac{1-w(p)/p}{(1-1/p)^k}$.

Let’s check if any of this made sense, by checking the prime number theorem. We want the probability that a random $n$ is prime to be $1/\ln(n)$, so we want the scaling factor above to be 1. The only polynomial is $f_1(n)=n$, so $k=w(p)=1$ and we see that each term in the product above is simply 1. That’s comforting. To apply these ideas to the case of twin primes, let $f_1(n)=n, f_2(n)=n+2$. So $k=2,w(2)=1$, and $w(p)=2$ for $p>2$. So we write the above product as:

$\begin{array}{rcl} \displaystyle\prod_p \frac{1-w(p)/p}{(1-1/p)^k} &=& \displaystyle\frac{1-1/2}{(1-1/2)^2}\cdot \prod_{p>2}\frac{1-2/p}{(1-1/p)^2} \\ &=& \displaystyle 2\prod_{p>2}\frac{p(p-2)}{(p-1)^2}=2\prod_{p>2}\left(1-\frac{1}{(p-1)^2}\right)\end{array}$

### 0.373.. & 0.247…

For any prime $p$, if $1/p$ is a repeating decimal with period $k$, then we can write $1/p=M/(10^k-1)$ for some integer $M$, which we rearrange to say $10^k\equiv 1\mod p$. Say that 10 has order $k$, mod $p$, if $k$ is the least value for which this equivalence is satisfied. Fermat‘s Little Theorem tells us that $10^{p-1}\equiv 1\mod p$, and so we know that the order (and hence, length of the decimal period) will never be greater than $p-1$. One might call a prime $p$ with maximal decimal period $p-1$ a long prime, an example is 7, whose decimal period is 142857. There’s nothing particularly special about using 10 – let us say that $a$ has order $k$, mod $p$, if $a^k\equiv 1\mod p$ and $k$ is the least (positive integer) value satisfying this condition. Call $a$ a primitive root, mod $p$, if it is not a multiple of $p$ and has order $p-1$, mod $p$.

Define $N_a(x)$ to be the number of primes no larger than $x$ for which $a$ is a primitive root. Artin, in 1927, conjectured that $N_a(x)\sim A(a)\frac{x}{\ln x}$ and, furthermore, that $A(a)$ was independant of $a$, and had a value $\prod_p \left(1-\frac{1}{p(p-1)}\right)\approx 0.373$. This can be interpreted as a sort of probability – if you pick a value for $a$ and then a prime $p$ randomly, the probability that $a$ will be a primitive root for $p$ is about 0.373. The conjecture can also be justified by thinking about how probable it is to be a primitive root.

While we’re talking about probabilities and densities like this, I can hardly resist mentioning abundant numbers. Remember that a number is abundant if the sum of its proper divisors is greater than the number itself. So, for example, 12 is abundant, because 1+2+3+4+6=16, while 7 is not abundant (making it deficient), because it’s only proper divisor is 1. What is the probability that a randomly chosen integer is abundant? Approximately 0.247.

Other posts in this series

1. The First Batch
2. Continued Fractions
3. The Second Batch
4. Resources

### 8 Responses to “Curious Constants (part 1 of 4)”

1. John Says:

The first integration in your post is striking. The integral exactly equals its discrete analog. Of course one often uses something like this to get an approximation, or upper and lower bounds if the integrand is monotone. But it’s surprising that the integral and the sum should be exactly equal.

2. sumidiot Says:

It is, indeed, striking. I’ve been a fan since I first ran across it.

Also, the original version of this post had an error, in that first series. The final answer should have been indexed from 1 to infinity, not 0 to infinity. I’ve corrected it.

3. Yasiru Says:

I encountered the sum integral relation in Gamma: Exploring Euler’s constant by Julian Havil and was enamoured by it immediately. Regarding the Borel summation you might like to take a look at my blog,

http://mathrants.blogspot.com

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