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.

Fibonacci convention

Source: https://www.facebook.com/Mathhappiness/photos/a.170688796611683.1073741829.101136273566936/321994341481127/?type=3&theater

Combinatorics and Jason’s Deli (Part 2)

Jason’s Deli is one of my family’s favorite places for an inexpensive meal. Recently, I saw the following placard at our table advertising their salad bar:

The small print says “Math performed by actual rocket scientist”; let’s see how the rocket scientist actually did this calculation.

The advertisement says that there are 50+ possible ingredients; however, to actually get a single number of combinations, let’s say there are exactly 50 ingredients. Lettuce will serve as the base, and so the 5 ingredients that go on top of the lettuce will need to be chosen from the other 49 ingredients.

Also, order is not important for this problem… for example, it doesn’t matter if the tomatoes go on first or last if tomatoes are selected for the salad.

Therefore, the number of possible ingredients is

$\displaystyle {49 \choose 5}$ ,

or the number in the 5th column of the 49th row of Pascal’s triangle. Rather than actually finding the 49th row of Pascal’s triangle by direct addition, it’s simpler to use factorials:

$\displaystyle {49 \choose 5} = \displaystyle \frac{49!}{5! \times 44!} = \displaystyle \frac{49 \times 48 \times 47 \times 46 \times 45 \times 44!}{5 \times 4 \times 3 \times 2 \times 1 \times 44!}$

$= \displaystyle \frac{49 \times 48 \times 47 \times 46 \times 45}{5 \times 4 \times 3 \times 2 \times 1}$

$= 49 \times 12 \times 47 \times 23 \times 3$

$= 1,906,884$ .

Under the assumption that there are exactly 50 ingredients, the rocket scientist actually got this right.

Combinatorics and Jason’s Deli (Part 1)

Jason’s Deli is one of my family’s favorite places for an inexpensive meal. Recently, I saw the following placard at our table advertising their salad bar:

I share this in the hopes that this might be reasonably engaging for students learning about different methods of counting.

Lessons from teaching gifted elementary students (Part 8g)

Every so often, I’ll informally teach a class of gifted elementary-school students. I greatly enjoy interacting with them, and I especially enjoy the questions they pose. Often these children pose questions that no one else will think about, and answering these questions requires a surprisingly depth of mathematical knowledge.

Here’s a question I once received, in the students’ original handwriting. They wanted me to add adjacent numbers on the bottom row to produce the number on the next row, building upward until I reached the apex of the triangle. Then, after I reached the top number, they wanted me to take the square root of that number. (Originally, they wanted me to first multiply by 80 before taking the square root, but evidently they decided to take it easy on me.)

And, just to see if I could do it, they wanted me to do all of this without using a calculator. But they were nice and allowed me to use pencil and paper.

So far, I’ve used Pascal’s triangle to obtain

$y = \displaystyle \sum_{k=0}^{11} (k+1)^2 {11 \choose k}$

$= \displaystyle \sum_{k=2}^{11} k(k-1) {11 \choose k} + \sum_{k=1}^{11} 3k {11 \choose k} + \sum_{k=0}^{11} {11 \choose k}$ .

$= \displaystyle \sum_{k=2}^{11} k(k-1) \left( \frac{11!}{k!(11-k)!} \right) + \sum_{k=1}^{11} 3k \left( \frac{11!}{k!(11-k)!} \right) + \sum_{k=0}^{11} \left( \frac{11!}{k!(11-k)!} \right)$ .

$= \displaystyle \sum_{k=2}^{11} \frac{11!}{(k-2)!(11-k)!} + 3 \sum_{k=1}^{11} \frac{11!}{(k-1)!(11-k)!} + \sum_{k=0}^{11} \frac{11!}{k!(11-k)!}$ .

$= \displaystyle \sum_{i=0}^{9} \frac{11!}{i!(9-i)!} + 3 \sum_{j=0}^{10} \frac{11!}{j!(10-j)!} + \sum_{k=0}^{11} \frac{11!}{k!(11-k)!}$ .

$= \displaystyle \sum_{i=0}^{9} 11 \times 10 \frac{9!}{i!(9-i)!} + 3 \sum_{j=0}^{10} 11 \frac{10!}{j!(10-j)!} + \sum_{k=0}^{11} \frac{11!}{k!(11-k)!}$

$= \displaystyle 110 \sum_{i=0}^{9} {9 \choose i} + 33 \sum_{j=0}^{10} {10 \choose j} + \sum_{k=0}^{11} {11 \choose k}$

$= 110 \times 2^9 + 33 \times 2^{10} + 2^{11}$

$= 92,160$ .

I’m almost done… except my students wanted me to find the square root of this number without using a calculator.

There are a couple ways to do this; the method I chose was directly extracting the square root by hand… a skill that was taught to children in previous generations but has fallen out of pedagogical disfavor with the advent of handheld calculators. I lost my original work, but it would have looked something like this (see the above website for details on why this works):

And so I gave my students their answer: $x \approx 303.578\dots$

Lessons from teaching gifted elementary students (Part 8f)

And, just to see if I could do it, they wanted me to do all of this without using a calculator. But they were nice and allowed me to use pencil and paper.

So far, I’ve used Pascal’s triangle to obtain

$y = \displaystyle \sum_{k=0}^{11} (k+1)^2 {11 \choose k}$

$= \displaystyle \sum_{k=2}^{11} k(k-1) {11 \choose k} + \sum_{k=1}^{11} 3k {11 \choose k} + \sum_{k=0}^{11} {11 \choose k}$ .

$= \displaystyle \sum_{k=2}^{11} k(k-1) \left( \frac{11!}{k!(11-k)!} \right) + \sum_{k=1}^{11} 3k \left( \frac{11!}{k!(11-k)!} \right) + \sum_{k=0}^{11} \left( \frac{11!}{k!(11-k)!} \right)$ .

$= \displaystyle \sum_{k=2}^{11} \frac{11!}{(k-2)!(11-k)!} + 3 \sum_{k=1}^{11} \frac{11!}{(k-1)!(11-k)!} + \sum_{k=0}^{11} \frac{11!}{k!(11-k)!}$ .

$= \displaystyle \sum_{i=0}^{9} \frac{11!}{i!(9-i)!} + 3 \sum_{j=0}^{10} \frac{11!}{j!(10-j)!} + \sum_{k=0}^{11} \frac{11!}{k!(11-k)!}$ .

$= \displaystyle \sum_{i=0}^{9} 11 \times 10 \frac{9!}{i!(9-i)!} + 3 \sum_{j=0}^{10} 11 \frac{10!}{j!(10-j)!} + \sum_{k=0}^{11} \frac{11!}{k!(11-k)!}$

$= \displaystyle 110 \sum_{i=0}^{9} {9 \choose i} + 33 \sum_{j=0}^{10} {10 \choose j} + \sum_{k=0}^{11} {11 \choose k}$

To numerically evaluate $y$ , I use the identity

$\sum_{r=0}^n {n \choose r} = 2^n$ ;

this identity can be proven by using the binomial theorem

$\sum_{r=0}^n {n \choose r} x^r y^{n-r} = (x+y)^n$

and then plugging in $x = 1$ and $y = 1$ . Using this identity, I conclude that

$y = 110 \times 2^9 + 33 \times 2^{10} + 2^{11}$

$= 55 \times 2 \times 2^9 + 33 \times 2^{10} + 2 \times 2^{10}$

$= 55 \times 2^{10} + 33 \times 2^{10} + 2 \times 2^{10}$

$= (55+33+2) \times 2^{10}$

$= 90 \times 2^{10}$ .

Since I know that $2^{10} = 1024$ , it’s now a simple matter of multiplication:

$y = 90 \times 1024 = 92,160$ .

(Trust me; after I showed my students this answer about five minutes after it was posed, I was ecstatic when I confirmed this answer with Mathematica.)

Lessons from teaching gifted elementary students (Part 8e)

And, just to see if I could do it, they wanted me to do all of this without using a calculator. But they were nice and allowed me to use pencil and paper.

So far, I’ve used Pascal’s triangle to obtain

$y = \displaystyle \sum_{k=0}^{11} (k+1)^2 {11 \choose k}$

$= \displaystyle \sum_{k=2}^{11} k(k-1) {11 \choose k} + \sum_{k=1}^{11} 3k {11 \choose k} + \sum_{k=0}^{11} {11 \choose k}$ .

$= \displaystyle \sum_{k=2}^{11} k(k-1) \left( \frac{11!}{k!(11-k)!} \right) + \sum_{k=1}^{11} 3k \left( \frac{11!}{k!(11-k)!} \right) + \sum_{k=0}^{11} \left( \frac{11!}{k!(11-k)!} \right)$ .

$= \displaystyle \sum_{k=2}^{11} \frac{11!}{(k-2)!(11-k)!} + 3 \sum_{k=1}^{11} \frac{11!}{(k-1)!(11-k)!} + \sum_{k=0}^{11} \frac{11!}{k!(11-k)!}$ .

$= \displaystyle \sum_{i=0}^{9} \frac{11!}{i!(9-i)!} + 3 \sum_{j=0}^{10} \frac{11!}{j!(10-j)!} + \sum_{k=0}^{11} \frac{11!}{k!(11-k)!}$ .

In the first series, I’ll rewrite $11!$ as $11 \times 10 \times 9!$ . Also, in the second series, I’ll rewrite $11!$ as $11 \times 10!$ . Therefore,

$y = \displaystyle \sum_{i=0}^{9} 11 \times 10 \frac{9!}{i!(9-i)!} + 3 \sum_{j=0}^{10} 11 \frac{10!}{j!(10-j)!} + \sum_{k=0}^{11} \frac{11!}{k!(11-k)!}$

$y = \displaystyle 110 \sum_{i=0}^{9} \frac{9!}{i!(9-i)!} + 33 \sum_{j=0}^{10} \frac{10!}{j!(10-j)!} + \sum_{k=0}^{11} \frac{11!}{k!(11-k)!}$

We now see that binomial coefficients appear in each of these series:

$y = \displaystyle 110 \sum_{i=0}^{9} {9 \choose i} + 33 \sum_{j=0}^{10} {10 \choose j} + \sum_{k=0}^{11} {11 \choose k}$

I’ll conclude the evaluation of $y$ in tomorrow’s post.

Lessons from teaching gifted elementary students (Part 8d)

And, just to see if I could do it, they wanted me to do all of this without using a calculator. But they were nice and allowed me to use pencil and paper.

So far, I’ve used Pascal’s triangle to obtain

$y = \displaystyle \sum_{k=0}^{11} (k+1)^2 {11 \choose k}$

$= \displaystyle \sum_{k=2}^{11} k(k-1) {11 \choose k} + \sum_{k=1}^{11} 3k {11 \choose k} + \sum_{k=0}^{11} {11 \choose k}$ .

I now use the definition of the binomial coefficient:

$= \displaystyle \sum_{k=2}^{11} k(k-1) \left( \frac{11!}{k!(11-k)!} \right) + \sum_{k=1}^{11} 3k \left( \frac{11!}{k!(11-k)!} \right) + \sum_{k=0}^{11} \left( \frac{11!}{k!(11-k)!} \right)$ .

Since $k! = k(k-1) \times (k-2)!$ and $k! = k \times (k-1)!$ , this simplifies as

$y = \displaystyle \sum_{k=2}^{11} \frac{11!}{(k-2)!(11-k)!}+ 3 \sum_{k=1}^{11} \frac{11!}{(k-1)!(11-k)!} + \sum_{k=0}^{11} \frac{11!}{k!(11-k)!}$ .

In the first series, I’ll use the change of index $i = k-2$ , so that $k = i+2$ and $11-k = 11-(i+2) = 9-i$ . Also, in the first series, the index will change from $k = 2$ to $k = 11$ to $i = 0$ to $i = 9$ .

In the second series, I’ll use the change of index $j = k-1$ , so that $k = j+1$ and $11-k = 11-(j+1) = 10-j$ . Also, in the first series, the index will change from $k = 1$ to $k = 11$ to $j = 0$ to $j = 10$ .

With these changes, I obtain

$y = \displaystyle \sum_{i=0}^{9}\frac{11!}{(i!(9-i)!} + 3 \sum_{j=0}^{10} \frac{11!}{j!(10-j)!} + \sum_{k=0}^{11} \frac{11!}{k!(11-k)!}$ .

I’ll continue the simplification of these series in tomorrow’s post.

Lessons from teaching gifted elementary students (Part 8c)

And, just to see if I could do it, they wanted me to do all of this without using a calculator. But they were nice and allowed me to use pencil and paper.

In yesterday’s post, I explained how Pascal’s triangle can be used to conclude

$y = \displaystyle \sum_{k=0}^{11} (k+1)^2 {11 \choose k}$ ,

thus allowing me to get the top number without getting all of the intermediate steps.

To compute this sum without a calculator, I’ll start rearranging the terms. The reasons for rearranging the terms in this way will become evident later.

$y = \displaystyle \sum_{k=0}^{11} (k+1)^2 {11 \choose k}$

$= \displaystyle \sum_{k=0}^{11} (k^2 + 2k + 1) {11 \choose k}$

$= \displaystyle \sum_{k=0}^{11} ([k^2 -k] + 3k + 1) {11 \choose k}$

$=\displaystyle \sum_{k=0}^{11} [k(k-1) + 3k + 1] {11 \choose k}$

$= \displaystyle \sum_{k=0}^{11} k(k-1) {11 \choose k} + \sum_{k=0}^{11} 3k {11 \choose k} + \sum_{k=0}^{11} {11 \choose k}$ .

The terms of the first sum are clearly equal to 0 when $k = 0$ and $k =1$ . Also, the $k=0$ term of the second sum is clearly 0. Therefore,

$y = \displaystyle \sum_{k=2}^{11} k(k-1) {11 \choose k} + \sum_{k=1}^{11} 3k {11 \choose k} + \sum_{k=0}^{11} {11 \choose k}$ .

It doesn’t look like I’ve improved matters much with this rearrangement of $y$ ; I’ll continue the solution in tomorrow’s post.

Lessons from teaching gifted elementary students (Part 8b)

And, just to see if I could do it, they wanted me to do all of this without using a calculator. But they were nice and allowed me to use pencil and paper.

Here’s how I started the problem, using a trick that I use in my mathematical magic show. Suppose that there are only six numbers instead of twelve, and let the six numbers be $a$ , $b$ , $c$ , $d$ , $e$ , and $f$ . Then here’s how the triangle unfolds (turning the triangle upside down):

$a \qquad \qquad \qquad \quad b \qquad \qquad \qquad \quad c \qquad \qquad \qquad \quad d \qquad \qquad \qquad \quad e \qquad \qquad \qquad \quad f$

$a+b \qquad \qquad \qquad b+c \qquad \qquad \qquad c+d \qquad \qquad \qquad d+e \qquad \qquad \qquad e+f$

$a+2b+c \qquad \qquad b+2c+d \qquad \qquad c+2d+e \qquad \qquad d+2e+f$

$a+3b+3c+d \qquad \quad b+3c+3d+e \qquad \quad c+3d+3e+f$

$a+4b+6c+4d+e \qquad b+4c+6d+5e+f$

$a+5b+10c+10d+5e+f$

In other words, the top number can be obtained by using the numbers on the fifth row of Pascal’s triangle (recall that the fifth row of Pascal’s triangle has six numbers on it). Specifically, if I multiply the bottom numbers by the corresponding number in a row of Pascal’s triangle and add them up, I’ll get the number on top without having to compute all of the intermediate steps.

For the problem my students gave me, the bottom row has 12 numbers, which means I’ll need to use the 11th row of Pascal’s triangle. Also, as we’ll see, I was fortunate that my students gave me a simple pattern of consecutive squares for the numbers on the bottom row. Since the numbering in Pascal’s triangle starts on zero, the numbers in the bottom row are $(k+1)^2$ as $k$ varies from 0 to 11.

Putting all this together, I can conclude that

$y = \displaystyle \sum_{k=0}^{11} (k+1)^2 {11 \choose k}$ .

Beginning with tomorrow’s post, I’ll discuss how I computed this sum without a calculator.