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

Suppose p_n is the nth prime number, so that p_{n+1} - p_n is the size of the nth 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…

green line

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.

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

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.