Logic puzzle

From “Over 80% of people get this question wrong, can you solve it?“:

Jack is looking at Anne, Anne is looking at George. Jack is married, George is not. Is a married person looking at an unmarried person?

I’ll refer the interested reader to the above link for the answer; I’m happy to report that I got this one right.

Defining Gravity

I’m in nerd heaven: Sir Isaac Newton and Albert Einstein parodying a showstopper from Wicked.

Exponents and the decathlon

During the Olympics, I stumbled across an application of exponents that I had not known before: scoring points in the decathlon or the heptathlon. From FiveThirtyEight.com:

Decathlon, which at the Olympics is a men’s event, is composed of 10 events: the 100 meters, long jump, shot put, high jump, 400 meters, 110-meter hurdles, discus throw, pole vault, javelin throw and 1,500 meters. Heptathlon, a women’s event at the Olympics, has seven events: the 100-meter hurdles, high jump, shot put, 200 meters, long jump, javelin throw and 800 meters…

As it stands, each event’s equation has three unique constants — $latex A$, $latex B$ and $latex C$— to go along with individual performance, $latex P$. For running events, in which competitors are aiming for lower times, this equation is: $latex A⋅(B–P)^C$, where $latex P$ is measured in seconds…

$B$ is effectively a baseline threshold at which an athlete begins scoring positive points. For performances worse than that threshold, an athlete receives zero points.

Specifically from the official rules and regulations (see pages 24 and 25), for the decathlon (where $P$ is measured in seconds):

100-meter run: $25.4347(18-P)^{1.81}$ .
400-meter run: $1.53775(82-P)^{1.81}$ .
1,500-meter run: $0.03768(480-P)^{1.85}$ .
110-meter hurdles: $5.74352(28.5-P)^{1.92}$ .

For the heptathlon:

200-meter run: $4.99087(42.5-P)^{1.81}$ .
400-meter run: $1.53775(82-P)^{1.88}$ .
1,500-meter run: $0.03768(480-P)^{1.835}$ .

Continuing from FiveThirtyEight:

For field events, in which competitors are aiming for greater distances or heights, the formula is flipped in the middle: $latex A⋅(P–B)^C$, where $latex P$ is measured in meters for throwing events and centimeters for jumping and pole vault.

Specifically, for the decathlon jumping events ( $P$ is measured in centimeters):

High jump: $0.8465(P-75)^{1.42}$
Pole vault: $0.2797(P-100)^{1.35}$
Long jump: $0.14354(P-220)^{1.4}$

For the decathlon throwing events ( $P$ is measured in meters):

Shot put: $51.39(P-1.5)^{1.05}$ .
Discus: $12.91(P-4)^{1.1}$ .
Javelin: $10.14(P-7)^{1.08}$ .

Specifically, for the heptathlon jumping events ( $P$ is measured in centimeters):

High jump: $1.84523(P-75)^{1.348}$
Long jump: $0.188807(P-210)^{1.41}$

For the heptathlon throwing events ( $P$ is measured in meters):

Shot put: $56.0211(P-1.5)^{1.05}$ .
Javelin: $15.9803(P-3.8)^{1.04}$ .

I’m sure there are good historical reasons for why these particular constants were chosen, but suffice it to say that there’s nothing “magical” about any of these numbers except for human convention. From FiveThirtyEight:

The [decathlon/heptathlon] tables [devised in 1984] used the principle that the world record performances of each event at the time should have roughly equal scores but haven’t been updated since. Because world records for different events progress at different rates, today these targets for WR performances significantly differ between events. For example, Jürgen Schult’s 1986 discus throw of 74.08 meters would today score the most decathlon points, at 1,384, while Usain Bolt’s 100-meter world record of 9.58 seconds would notch “just” 1,203 points. For women, Natalya Lisovskaya’s 22.63 shot put world record in 1987 would tally the most heptathlon points, at 1,379, while Jarmila Kratochvílová’s 1983 WR in the 800 meters still anchors the lowest WR points, at 1,224.

FiveThirtyEight concludes that, since the exponents in the running events are higher than those for the throwing/jumping events, it behooves the elite decathlete/heptathlete to focus more on the running events because the rewards for exceeding the baseline are greater in these events.

Finally, since all of the exponents are not integers, a negative base (when the athlete’s performance wasn’t very good) would actually yield a complex-valued number with a nontrivial imaginary component. Sadly, the rules of track and field don’t permit an athlete’s score to be a non-real number. However, if they did, scores might look something like this…

Deceiving with Statistics

I really enjoyed a recent Math With Bad Drawings post on how descriptive statistics can be used to deceive. For example:

See the rest of the post for similar picture for mean, median, mode, and variance (equivalent to standard deviation); I’ll be using these in my future statistics classes.

Regression

Source: http://www.xkcd.com/1725/

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

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