What I Learned from Reading “Gamma: Exploring Euler’s Constant” by Julian Havil: Part 10

Suppose $p_n$ is the $n$ th prime number, so that $p_{n+1} - p_n$ is the size of the $n$ th gap between successive prime numbers. It turns out (Gamma, page 115) that there’s an incredible theorem for the lower bound of this number:

$\displaystyle \limsup_{n \to \infty} \frac{(p_{n+1}-p_n) (\ln \ln \ln p_n)^2}{(\ln p_n)(\ln \ln p_n)(\ln \ln \ln \ln p_n)} \ge \displaystyle \frac{4 e^{\gamma}}{c}$ ,

where $\gamma$ is the Euler-Mascheroni constant and $c$ is the solution of $c = 3 + e^{-c}$ .

Holy cow, what a formula. Let’s take a look at just a small part of it.

Let’s look at the amazing function $f(x) = \ln \ln \ln \ln x$ , iterating the natural logarithm function four times. This function has a way of converting really large inputs into unimpressive outputs. For example, the canonical “big number” in popular culture is the googolplex, defined as $10^{10^{100}}$ . Well, it takes some work just to rearrange $\displaystyle f \left(10^{10^{100}} \right)$ in a form suitable for plugging into a calculator:

$\displaystyle f \left(10^{10^{100}} \right) = \displaystyle \ln \ln \ln \left( \ln 10^{10^{100}} \right)$

$= \displaystyle \ln \ln \ln \left( 10^{100} \ln 10 \right)$

$= \displaystyle \ln \ln \left[ \ln \left(10^{100} \right) + \ln \ln 10 \right]$

$= \displaystyle \ln \ln \left[ 100 \ln 10 + \ln \ln 10 \right]$

$= \displaystyle \ln \ln \left[ 100 \ln 10 \left( 1 + \frac{\ln \ln 10}{100 \ln 10} \right) \right]$

$= \displaystyle \ln \left( \ln [ 100 \ln 10] + \ln \left( 1 + \frac{\ln \ln 10}{100 \ln 10} \right)\right)$

$\approx 1.6943$

after using a calculator.

This function grows extremely slowly. What value of $x$ gives an output of $0$ ? Well:

$\ln \ln \ln \ln x = 0$

$\ln \ln \ln x = 1$

$\ln \ln x = e$

$\ln x = e^e$

$x = e^{e^e} \approx 3,814,279.1$

What value of $x$ gives an output of $1$ ? Well:

$\ln \ln \ln \ln x = 1$

$\ln \ln \ln x = e$

$\ln \ln x = e^e$

$\ln x = e^{e^e}$

$x = e^{e^{e^e}}$

$\approx e^{3,814,279.1}$

$\approx 10^{3,814,279.1 \log_{10} e}$

$\approx 10^{1,656,420.367636}$

$\approx 2.3315 \times 10^{1,656,420}$

That’s a number with 1,656,421 digits! At the rapid rate of 5 digits per second, it would take over 92 hours (nearly 4 days) just to write out the answer by hand!

Finally, how large does $x$ have to be for the output to be 2? As we’ve already seen, it’s going to be larger than a googolplex:

$\displaystyle f \left(10^{10^{x}} \right) = 2$

$\displaystyle \ln \ln \ln \left( \ln 10^{10^{x}} \right) = 2$

$\displaystyle \ln \ln \ln \left( 10^{x} \ln 10 \right) = 2$

$\displaystyle \ln \ln \left[ \ln \left(10^{x} \right) + \ln \ln 10 \right] = 2$

$\displaystyle \ln \ln \left[ x\ln 10 + \ln \ln 10 \right] = 2$

$\displaystyle \ln \ln \left[ x\ln 10 \left( 1 + \frac{\ln \ln 10}{x\ln 10} \right) \right] = 2$

$\displaystyle \ln \left( \ln [ x\ln 10] + \ln \left( 1 + \frac{\ln \ln 10}{x \ln 10} \right)\right) = 2$

Let’s simplify things slightly by letting $y = x \ln 10$ :

$\displaystyle \ln \left( \ln y + \ln \left( 1 + \frac{\ln \ln 10}{y} \right)\right) = 2$

$\displaystyle \ln y + \ln \left( 1 + \frac{\ln \ln 10}{y} \right) = e^2$

This is a transcendental equation in $y$ ; however, we can estimate that the solution will approximately solve $\ln y = e^2$ since the second term on the left-hand side is small compared to $\ln y$ . This gives the approximation $y = e^{e^2} \approx 1618.18$ . Using either Newton’s method or else graphing the left-hand side yields the more precise solution $y \approx 1617.57$ .

Therefore, $x \approx 1617.57 \ln 10 \approx 3725.99$ , so that

$f \left(10^{10^{3725.99}} \right) \approx 2$ .

One final note: despite what’s typically taught in high school, mathematicians typically use $\log$ to represent natural logarithms (as opposed to base-10 logarithms), so the above formula is more properly written as

$\displaystyle \limsup_{n \to \infty} \frac{(p_{n+1}-p_n) (\log \log \log p_n)^2}{(\log p_n)(\log \log p_n)(\log \log \log \log p_n)} \ge \displaystyle \frac{4 e^{\gamma}}{c}$ .

And this sets up a standard joke, also printed in Gamma:

Q: What noise does a drowning analytic number theorist make?

A: Log… log… log… log…

When I researching for my series of posts on conditional convergence, especially examples related to the constant $\gamma$ , the reference Gamma: Exploring Euler’s Constant by Julian Havil kept popping up. Finally, I decided to splurge for the book, expecting a decent popular account of this number. After all, I’m a professional mathematician, and I took a graduate level class in analytic number theory. In short, I don’t expect to learn a whole lot when reading a popular science book other than perhaps some new pedagogical insights.

Boy, was I wrong. As I turned every page, it seemed I hit a new factoid that I had not known before.

In this series, I’d like to compile some of my favorites — while giving the book a very high recommendation.

What I Learned from Reading “Gamma: Exploring Euler’s Constant” by Julian Havil: Part 9

When teaching students mathematical induction, the following series (well, at least the first two or three) are used as typical examples:

$1 + 2 + 3 + \dots + n = \displaystyle \frac{n(n+1)}{2}$

$1^2 + 2^2 + 3^2 + \dots + n^2 = \displaystyle \frac{n(n+1)(2n+1)}{6}$

$1^3 + 2^3 + 3^3 + \dots + n^3 = \displaystyle \frac{n^2(n+1)^2}{4}$

$1^4 + 2^4 + 3^4 + \dots + n^4 = \displaystyle \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$

What I didn’t know (Gamma, page 81) is that Johann Faulhaber published the following cute result in 1631 (see also Wikipedia): If $k$ is odd, then

$1^k + 2^k + 3^k + \dots + n^k = f_k(n(n+1))$ ,

where $f_k$ is a polynomial. For example, to match the above examples, $f_1(x) = x/2$ and $f_3(x) = x^2/4$ . Furthermore, if $k$ is even, then

$1^k + 2^k + 3^k + \dots + n^k = (2n+1) f_k(n(n+1))$ ,

where again $f_k$ is a polynomial. For example, to match the above examples, $f_2(x) = x/6$ and $f_3(x) = x(3x-1)/30$ .

Boy, was I wrong. As I turned every page, it seemed I hit a new factoid that I had not known before.

In this series, I’d like to compile some of my favorites — while giving the book a very high recommendation.

What I Learned from Reading “Gamma: Exploring Euler’s Constant” by Julian Havil: Part 8

I had always wondered how the constant $\gamma$ can be computed to high precision. I probably should have known this already, but here’s one way that it can be computed (Gamma, page 89):

$\gamma = \displaystyle \sum_{k=1}^n \frac{1}{k} - \ln n - \sum_{k=1}^{\infty} \frac{B_{2k}}{2k \cdot n^{2k}}$ ,

where $B_{2k}$ is the $2k$ th Bernoulli number.

Boy, was I wrong. As I turned every page, it seemed I hit a new factoid that I had not known before.

In this series, I’d like to compile some of my favorites — while giving the book a very high recommendation.

What I Learned from Reading “Gamma: Exploring Euler’s Constant” by Julian Havil: Part 7

Suppose that two positive integers are chosen at random. What is the probability that they are relatively prime (that is, have no common factors except 1)?

The answer is exactly what you’d expect it be (Gamma, p. 68): $6/\pi^2$ , or about 60.8%.

Yes, that was a joke.

Indeed, if $k$ positive integers are random, the probability that they are relatively prime is $1/\zeta(k)$ , where Riemann’s zeta function arises once again.

Even more, the probability that $k$ random positive integers lack a $n$ th power common divisor is $1/\zeta(nk)$ .

I’ll refer the interested reader to Gamma and also to Mathworld (and references therein) for more details.

Boy, was I wrong. As I turned every page, it seemed I hit a new factoid that I had not known before.

In this series, I’d like to compile some of my favorites — while giving the book a very high recommendation.

What I Learned from Reading “Gamma: Exploring Euler’s Constant” by Julian Havil: Part 6

As noted in Gamma (page 65), mathematician meant astrologer in context. Still, what a terrific quote.

Boy, was I wrong. As I turned every page, it seemed I hit a new factoid that I had not known before.

In this series, I’d like to compile some of my favorites — while giving the book a very high recommendation.

What I Learned from Reading “Gamma: Exploring Euler’s Constant” by Julian Havil: Part 5

Check out this lovely integral, dubbed the Sophomore’s Dream, found by Johann Bernoulli in 1697 (Gamma, page 44):

$\displaystyle \int_0^1 \frac{dx}{x^x} = \displaystyle \frac{1}{1^1} + \frac{1}{2^2} + \frac{1}{3^3} + \frac{1}{4^4} + \dots$ .

I’ll refer to either Wikipedia or Mathworld for the derivation.

Boy, was I wrong. As I turned every page, it seemed I hit a new factoid that I had not known before.

In this series, I’d like to compile some of my favorites — while giving the book a very high recommendation.

What I Learned from Reading “Gamma: Exploring Euler’s Constant” by Julian Havil: Part 4

For $s > 1$ , Riemann’s famous zeta function is defined by

$\zeta(s) = \displaystyle \sum_{n=1}^{\infty} \frac{1}{n^s}$ .

This is also called a p-series in calculus.

What I didn’t know (Gamma, page 41) is that, in 1748, Leonhard Euler exactly computed this infinite series for $s = 26$ without a calculator! Here’s the answer:

$\displaystyle 1 + \frac{1}{2^{26}} + \frac{1}{3^{26}} + \frac{1}{4^{26}} + \dots = \frac{1,315,862 \pi^{26}}{11,094,481,976,030,578,125}$ .

I knew that Euler was an amazing human calculator, but I didn’t know he was that amazing.

Boy, was I wrong. As I turned every page, it seemed I hit a new factoid that I had not known before.

In this series, I’d like to compile some of my favorites — while giving the book a very high recommendation.

What I Learned from Reading “Gamma: Exploring Euler’s Constant” by Julian Havil: Part 3

At the time of this writing, it is unknown if there are infinitely many twin primes, which are prime numbers that differ by 2 (like 3 and 5, 5 and 7, 11 and 13, 17 and 19, etc.) However, significant progress has been made in recent years. However, it is known (Gamma, page 30) the sum of the reciprocals of the twin primes converges:

$\displaystyle \left( \frac{1}{3} + \frac{1}{5} \right) + \left( \frac{1}{5} + \frac{1}{7} \right) + \left( \frac{1}{11} + \frac{1}{13} \right) + \left( \frac{1}{17} + \frac{1}{19} \right) = 1.9021605824\dots$ .

This constant is known as Brun’s constant (see also Mathworld). In the process of computing this number, the infamous 1994 Pentium bug was found.

Although this sum is finite, it’s still unknown if there are infinitely many twin primes since it’s possible for an infinite sum to converge (like a geometric series).

Boy, was I wrong. As I turned every page, it seemed I hit a new factoid that I had not known before.

In this series, I’d like to compile some of my favorites — while giving the book a very high recommendation.

What I Learned from Reading “Gamma: Exploring Euler’s Constant” by Julian Havil: Part 2

Let’s define partial sums of the harmonic series as follows:

$H(m,n) = \displaystyle \frac{1}{m} + \frac{1}{m+1} + \frac{1}{m+2} + \dots + \frac{1}{n-1} + \frac{1}{n}$ ,

where $m < n$ are positive integers. Here are a couple of facts that I didn’t know before reading Gamma (pages 24-25):

$H(m,n)$ is never equal to an integer.
The only values of $n$ for which $H(1,n)$ is an integer are $n = 2$ and $n=6$ .

Boy, was I wrong. As I turned every page, it seemed I hit a new factoid that I had not known before.

In this series, I’d like to compile some of my favorites — while giving the book a very high recommendation.

What I Learned from Reading “Gamma: Exploring Euler’s Constant” by Julian Havil: Part 1

Boy, was I wrong. As I turned every page, it seemed I hit a new factoid that I had not known before.

In this series, I’d like to compile some of my favorites — while giving the book a very high recommendation.

It is well known the harmonic series diverges:

$\displaystyle 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots = \infty$ .

This means that, no matter what number $N$ you choose, I can find a number $n$ so that

$\displaystyle 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{n} > N$ .

What I didn’t know (p. 23 of Gamma) is that, in 1968, somebody actually figured out the precise number of terms that are needed for the sum on the left hand side to exceed 100. Here’s the answer:

15,092,688,622,113,788,323,693,563,264,538,101,449,859,497.

With one fewer term, the sum is a little less than 100.