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.

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 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.

Thoughts on Infinity: Index

I’m doing something that I should have done a long time ago: collect past series of posts into a single, easy-to-reference post. The following posts formed my series on various lessons I’ve learned while trying to answer the questions posed by gifted elementary school students.

Part 1: Different types of countable sets

Part 2a: Divergence of the harmonic series.

Part 2b: Convergence of the Kempner series.

Part 3a: Conditional convergent series or products shouldn’t be rearranged.

Part 3b: Definition of the Euler-Mascheroni constant $\gamma$ .

Part 3c: Evaluation of the conditionally convergent series $\displaystyle 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} \dots$

Part 3d: Confirmation of this evaluation using technology.

Part 3e: Evaluation of a rearrangement of this conditionally convergent series.

Part 3f: Confirmation of this different evaluation using technology.

Part 3g: Closing thoughts.

Useless Numerology for 2016: Index

I’m doing something that I should have done a long time ago: collecting a series of posts into one single post.

Part 1: Introduction.

Part 2: $2016 = 2^{10}+2^9+2^8+2^7+2^6+2^5 = 2^{11} - 2^5$ .

Part 3: $2016 = 1 + 2 + \dots + 63$ .

Part 4: $2016 = (1 + 2 + \dots + 9)^2 - (1+2)^2 = 3^3 + 4^3 + 5^3 + 6^3 + 7^3 + 8^3 + 9^3$

Part 5: $2016 = \displaystyle \sum_{n=0}^{63} (-1)^{n+1} n^2$

Useless Numerology for 2016: Part 1

The following entertaining (but useless) facts about the number 2,016 appeared in a recent Facebook post (and subsequent comments) by the American Mathematical Monthly.

$2016 = 3^3 + 4^3 + 5^3 + 6^3 + 7^3 + 8^3 + 9^3$

$2016 = 2^{10} + 2^9 + 2^8 + 2^7 + 2^6 + 2^5$

$2016 = 1+2+3 + \dots + 62 + 63$

$2016 = \displaystyle \sum_{n=0}^{63} (-1)^{n+1} n^2$

$2016 = (1+2+...+8+9)^2 - (1+2)^2$

$(2 + 0 + 1)! = 6$

$2016 = 2^{11} - 2^5$

$2016 = 2016 \times 1$

The last tongue-in-check equation is my favorite.

In this series, I’ll explain why these different expressions for $2016$ have to be equal to each other. I’ll begin with tomorrow’s post.

Thoughts on Infinity (Part 3g)

We have seen in recent posts that

$latex \displaystyle 1 – \frac{1}{2} + \frac{1}{3} – \frac{1}{4} + \frac{1}{5} – … = \ln 2$

One way of remembering this fact is by using the Taylor series expansion for $\ln(1+x)$ :

$\ln(1+x) = x - \displaystyle \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} \dots$

“Therefore,” the first series can be obtained from the second series by substituting $x=1$ .

I placed “therefore” in quotation marks because this reasoning is completely invalid, even though it happens to stumble across the correct answer in this instance. The radius of convergence for the above Taylor series is 1, which can be verified by using the Ratio Test. So the series converges absolutely for $|x| < 1$ and diverges for $|x| > 1$ . The boundary of $|x| = 1$ , on the other hand, has to be checked separately for convergence.

In other words, plugging in $x=1$ might be a useful way to remember the formula, but it’s not a proof of the formula and certainly not a technique that I want to encourage students to use!

It’s easy to find examples where just plugging in the boundary point happens to give the correct answer (see above). It’s also easy to find examples where plugging in the boundary point gives an incorrect answer because the series actually diverges: for example, substituting $x = -1$ into the geometric series

$\displaystyle \frac{1}{1-x} = 1 + x + x^2 + x^3 + x^4 + \dots$

However, I’ve been scratching my head to think of an example where plugging in the boundary point gives an incorrect answer because the series converges but converges to a different number. I could’ve sworn that I saw an example like this when I was a calculus student, but I can’t see to find an example in reading Apostol’s calculus text.