The Riemann Hypothesis (see here, here, and here) is perhaps the most famous (and also most important) unsolved problems in mathematics. Gamma (page 207) provides a way of writing down this conjecture in a form that only uses notation that is commonly taught in high school:

If $\displaystyle \sum_{r=1}^\infty \frac{(-1)^r}{r^a} \cos(b \ln r) = 0$ and $\displaystyle \sum_{r=1}^\infty \frac{(-1)^r}{r^a} \sin(b \ln r) = 0$ for some pair of real numbers $a$ and $b$ , then $a = \frac{1}{2}$ .

As noted in the book, “It seems extraordinary that the most famous unsolved problem in the whole of mathematics can be phrased so that it involves the simplest of mathematical ideas: summation, trigonometry, logarithms, and [square roots].”

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 17

Let $\pi(n)$ denote the number of positive prime numbers that are less than or equal to $n$ . The prime number theorem, one of the most celebrated results in analytic number theory, states that

$\pi(x) \approx \displaystyle \frac{x}{\ln x}$ .

This is a very difficult result to prove. However, Gamma (page 172) provides a heuristic argument that suggests that this answer might be halfway reasonable.

Consider all of the integers between $1$ and $x$ .

About half of these numbers won’t be divisible by 2.
Of those that aren’t divisible by 2, about two-thirds won’t be divisible by 3. (This isn’t exactly correct, but it’s good enough for heuristics.)
Of those that aren’t divisible by 2 and 3, about four-fifths won’t be divisible by 5.
And so on.

If we repeat for all primes less than or equal to $\sqrt{x}$ , we can conclude that the number of prime numbers less than or equal to $x$ is approximately

$\pi(x) \approx \displaystyle x \prod_{p \le \sqrt{x}} \left(1 - \frac{1}{p} \right)$ .

From this point, we can use Mertens product formula

$\displaystyle \lim_{n \to \infty} \frac{1}{\ln n} \prod_{p \le n} \left(1 - \frac{1}{p} \right)^{-1} = e^\gamma$

to conclude that

$\displaystyle \frac{1}{\ln n} \prod_{p \le n} \left(1 - \frac{1}{p} \right) \approx \displaystyle \frac{e^{-\gamma}}{\ln n}$

if $n$ is large. Therefore,

$\pi(x) \approx x \displaystyle \frac{e^{-\gamma}}{\ln \sqrt{x}} = 2 e^{-\gamma} \displaystyle \frac{x}{\ln x}$ .

Though not a formal proof, it’s a fast way to convince students that the unusual fraction $\displaystyle \frac{x}{\ln x}$ ought to appear someplace in the prime number theorem.

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 16

Let $\pi(n)$ denote the number of positive prime numbers that are less than or equal to $n$ . The prime number theorem, one of the most celebrated results in analytic number theory, states that

$\pi(x) \approx \displaystyle \frac{x}{\ln x}$ .

This is a very difficult result to prove. However, Gamma (page 172) provides a heuristic argument that suggests that this answer might be halfway reasonable.

Consider all of the integers between $1$ and $x$ .

About half of these numbers won’t be divisible by 2.
Of those that aren’t divisible by 2, about two-thirds won’t be divisible by 3. (This isn’t exactly correct, but it’s good enough for heuristics.)
Of those that aren’t divisible by 2 and 3, about four-fifths won’t be divisible by 5.
And so on.

If we repeat for all primes less than or equal to $\sqrt{x}$ , we can conclude that the number of prime numbers less than or equal to $x$ is approximately

$\pi(x) \approx \displaystyle x \prod_{p \le \sqrt{x}} \left(1 - \frac{1}{p} \right)$ .

From this point, we can use Mertens product formula

$\displaystyle \lim_{n \to \infty} \frac{1}{\ln n} \prod_{p \le n} \left(1 - \frac{1}{p} \right)^{-1} = e^\gamma$

to conclude that

$\displaystyle \frac{1}{\ln n} \prod_{p \le n} \left(1 - \frac{1}{p} \right) \approx \displaystyle \frac{e^{-\gamma}}{\ln n}$

if $n$ is large. Therefore,

$\pi(x) \approx x \displaystyle \frac{e^{-\gamma}}{\ln \sqrt{x}} = 2 e^{-\gamma} \displaystyle \frac{x}{\ln x}$ .

Though not a formal proof, it’s a fast way to convince students that the unusual fraction $\displaystyle \frac{x}{\ln x}$ ought to appear someplace in the prime number theorem.

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 15

I did not know — until I read Gamma (page 168) — that there actually is a formula for generating $n$ th prime number by directly plugging in $n$ . The catch is that it’s a mess:

$p_n = 1 + \displaystyle \sum_{m=1}^{2^n} \left[ n^{1/n} \left( \sum_{i=1}^m \cos^2 \left( \pi \frac{(i-1)!+1}{i} \right) \right)^{-1/n} \right]$ ,

where the outer brackets $[~ ]$ represent the floor function.

This mathematical curiosity has no practical value, as determining the 10th prime number would require computing $1 + 2 + 3 + \dots + 2^{10} = 524,800$ different terms!

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 14

I hadn’t heard of the worm-on-a-rope problem until I read Gamma (page 133). From Cut-The-Knot:

A worm is at one end of a rubber rope that can be stretched indefinitely. Initially the rope is one kilometer long. The worm crawls along the rope toward the other end at a constant rate of one centimeter per second. At the end of each second the rope is instantly stretched another kilometer. Thus, after the first second the worm has traveled one centimeter, and the length of the rope has become two kilometers. After the second second, the worm has crawled another centimeter and the rope has become three kilometers long, and so on. The stretching is uniform, like the stretching of a rubber band. Only the rope stretches. Units of length and time remain constant.

It turns out that, after $n$ seconds, that the fraction of the band that the worm has traveled is $H_n/N$ , where

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

and $N$ is the length of the rope in centimeters. Using the estimate $H_n \approx \ln n + \gamma$ , we see that the worm will reach the end of the rope when

$H_n = N$

$\ln n + \gamma \approx N$

$\ln n \approx N - \gamma$

$n \approx e^{N - \gamma}$ .

If $N = 100,000$ (since the rope is initially a kilometer long), it will take a really long time for the worm to reach its destination!

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 13

I hadn’t heard of the crossing-the-desert problem until I read Gamma (page 127). From Wikipedia:

There are n units of fuel stored at a fixed base. The jeep can carry at most 1 unit of fuel at any time, and can travel 1 unit of distance on 1 unit of fuel (the jeep’s fuel consumption is assumed to be constant). At any point in a trip the jeep may leave any amount of fuel that it is carrying at a fuel dump, or may collect any amount of fuel that was left at a fuel dump on a previous trip, as long as its fuel load never exceeds 1 unit…

The jeep must return to the base at the end of every trip except for the final trip, when the jeep travels as far as it can before running out of fuel…

[T]he objective is to maximize the distance traveled by the jeep on its final trip.

The answer is, if $n$ fuel dumps are used, the jeep can go a distance of

$H_n = \displaystyle 1 + \frac{1}{3} + \frac{1}{5} + \dots + \frac{1}{2n-1}$ .

Since the right-hand side approaches infinity as $n$ gets arbitrarily large, it is possible to cross an arbitrarily long desert according the rules of this problem.

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 12

Let $X_1, X_2, X_3, \dots$ be a sequence of independent and identically distributed random variables, and let $H_n$ be the number of “record highs” upon to and including event $n$ . For example, each $X_i$ can represent the amount of rainfall in a year, where $X_1$ is amount of rainfall recorded the first time that records were kept. As shown in Gamma (page 125), the expected number of record highs is

$H_n = \displaystyle 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}$ .

As noted in Gamma,

Two arbitrary chosen examples are revealing. The Radcliffe Meteorological Station in Oxford has data for rainfall in Oxford between 1767 and 2000 and there are five record years; this is a span of 234 recorded years and $H_{234} = 6.03$ . For Central Park, New York City, between 1835 and 1994 there are six record years over the 160-year period and $H_{160} = 5.65$ , providing good evidence that English weather is that bit more unpredictable.

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 11

The Euler-Mascheroni constant $\gamma$ is defined by

$\gamma = \displaystyle \lim_{n \to \infty} \left( \sum_{r=1}^n \frac{1}{r} - \ln n \right)$ .

What I didn’t know, until reading Gamma (page 117), is that there are at least two ways to generalize this definition.

First, $\gamma$ may be thought of as

$\gamma = \displaystyle \lim_{n \to \infty} \left( \sum_{r=1}^n \frac{1}{\hbox{length of~} [0,r]} - \ln n \right)$ ,

and so this can be generalized to two dimensions as follows:

$\delta = \displaystyle \lim_{n \to \infty} \left( \sum_{r=2}^n \frac{1}{\pi (\rho_r)^2} - \ln n \right)$ ,

where $\rho_r$ is the radius of the smallest disk in the plane containing at least $r$ points $(a,b)$ so that $a$ and $b$ are both integers. This new constant $\delta$ is called the Masser-Gramain constant; like $\gamma$ , the exact value isn’t known.

Second, let $f(x) = \displaystyle \frac{1}{x}$ . Then $\gamma$ may be written as

$\gamma = \displaystyle \lim_{n \to \infty} \left( \sum_{r=1}^n f(r) - \int_1^n f(x) \, dx \right)$ .

Euler (not surprisingly) had the bright idea of changing the function $f(x)$ to any other positive, decreasing function, such as

$f(x) = x^a, \qquad -1 \le a < 0$ ,

producing Euler’s generalized constants. Alternatively (from Stieltjes), we could choose

$f(x) = \displaystyle \frac{ (\ln x)^m }{x}$ .

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 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…

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.

Author: John Quintanilla

Regression

What I Learned from Reading “Gamma: Exploring Euler’s Constant” by Julian Havil: Part 18

What I Learned from Reading “Gamma: Exploring Euler’s Constant” by Julian Havil: Part 17

What I Learned from Reading “Gamma: Exploring Euler’s Constant” by Julian Havil: Part 16

What I Learned from Reading “Gamma: Exploring Euler’s Constant” by Julian Havil: Part 15

What I Learned from Reading “Gamma: Exploring Euler’s Constant” by Julian Havil: Part 14

What I Learned from Reading “Gamma: Exploring Euler’s Constant” by Julian Havil: Part 13

What I Learned from Reading “Gamma: Exploring Euler’s Constant” by Julian Havil: Part 12

What I Learned from Reading “Gamma: Exploring Euler’s Constant” by Julian Havil: Part 11

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