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.

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

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