# 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: $latex A(BP)^C$, 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: $latex A(PB)^C$, 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…

# Lessons from teaching gifted elementary students (Part 6b)

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 surprising depth of mathematical knowledge.

Here’s a question I once received:

255/256 to what power is equal to 1/2? And please don’t use a calculator.

Here’s how I answered this question without using a calculator… in fact, I answered it without writing anything down at all. I thought of the question as

$\displaystyle \left( 1 - \epsilon \right)^x = \displaystyle \frac{1}{2}$.

$\displaystyle x \ln (1 - \epsilon) = \ln \displaystyle \frac{1}{2}$

$\displaystyle x \ln (1 - \epsilon) = -\ln 2$

I was fortunate that my class chose 1/2, as I had memorized (from reading and re-reading Surely You’re Joking, Mr. Feynman! when I was young) that $\ln 2 \approx 0.693$. Therefore, we have

$x \ln (1 - \epsilon) \approx -0.693$.

Next, I used the Taylor series expansion

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

to reduce this to

$-x \epsilon \approx -0.693$,

or

$x \approx \displaystyle \frac{0.693}{\epsilon}$.

For my students’ problem, I had $\epsilon = \frac{1}{256}$, and so

$x \approx 256(0.693)$.

So all I had left was the small matter of multiplying these two numbers. I thought of this as

$x \approx 256(0.7 - 0.007)$.

Multiplying $256$ and $7$ in my head took a minute or two:

$256 \times 7 = 250 \times 7 + 6 \times 7$

$= 250 \times (8-1) + 42$

$= 250 \times 8 - 250 + 42$

$= 2000 - 250 + 42$

$= 1750 + 42$

$= 1792$.

Therefore, $256 \times 0.7 = 179.2$ and $256 \times 0.007 = 1.792 \approx 1.8$. Therefore, I had the answer of

$x \approx 179.2 - 1.8 = 177.4 \approx 177$.

So, after a couple minutes’ thought, I gave the answer of 177. I knew this would be close, but I had no idea it would be so close to the right answer, as

$x = \displaystyle \frac{\displaystyle \ln \frac{1}{2} }{\displaystyle \ln \frac{255}{256}} \approx 177.0988786\dots$

# Lessons from teaching gifted elementary school students (Part 6a)

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 surprising depth of mathematical knowledge.

Here’s a question I once received:

255/256 to what power is equal to 1/2? And please don’t use a calculator.

Answering this question is pretty straightforward using algebra:

$\displaystyle \left( \frac{255}{256} \right)^x = \displaystyle \frac{1}{2}$.

$\displaystyle x \ln \frac{255}{256} = \ln \displaystyle \frac{1}{2}$

$x \displaystyle \frac{ \displaystyle \ln \frac{1}{2} }{\ln \displaystyle \frac{255}{256}}$

However, doing this without a calculator — and thus maintaining my image in front of these elementary school students — is a little formidable.

I’ll reveal how I did this — getting the answer correct to the nearest integer — in tomorrow’s post. In the meantime, I’ll leave a thought bubble if you’d like to think about it on your own.

# Engaging students: Negative and zero exponents

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission again comes from my former student Jennifer Elliott. Her topic, from Algebra: negative and zero exponents.

• Technology Engage
• I found the website, https://www.mangahigh.com/en-us/math_games/number/exponents/negative_exponents. It is an interactive game that gives a brief explanation of what negative and zero exponents are. Then you can select the difficulty level and the number or questions you wish the children to try. If this a new topic introduced, then the student may miss several. That is ok. As a teacher, you are setting a ground level for the direction of your teach. At the end of the lesson, you can utilize the same game to check the students’ new level of understanding for the topic.

• Activity Engage
• The students will engage in prior knowledge that might be needed to understand the idea behind negative and zero exponents. First I will make different notecards, some with definitions such as negative number, fractions, number line, and reciprocals and others. Then I will have some index cards with different exponents including positive, negative, and zero. The cards will have different values such as one might say 10^-1 and one might say 1/10. Every student will have a note card. I will have different sections set up in the room. Example would be definitions, 1, <1, and >1 and have students find which section they belong in. I could also have them find their card partner (different way of writing the same number) and the word matching the definition. Then maybe from there, that group find their counter-partner (I would maybe not use definitions for this part) such that the group with 10^-2 would find the group with 10^2. This would set up groups for them to explore the idea of negative and zero exponents.
• This activity came from myself but I had some ideas from different pictures on Pinterest, but nothing in particular to source.)

• Curriculum Engage
• To show how this might be used later in class, I will work on the idea of decay. The idea of decay can be introduced in science and history off the top of my head. Although the students might be years away from the idea of physics and decay value, this will be a fun way to engage students and hopefully recall the information when a lesson on decay comes in the future. The idea is found on several different websites and has to do with the idea of exponential decay using M&M’s. The idea is to create (or use one of the several choices) of a table to record the data from the trials. The group(s) count the total number of M&M’s. The total is the starting number for trial 0. Trial number would be the first column. The second column would be the number of M&M’s. For trial one, you would dump the bag/cup of candy and the student would remove all the M&M’s that do not have the M showing. Shake the candy up again, and dumb out. Continue with trials until you do not have any M&M’s left. Then the third column will be what percentage of the bag they have left (example maybe ½ of the M&M’s remain.) This activity will lead to the discovery of decay and how it uses zero and negative exponents. The starting point of trial 0 has us with “1” bag/cup of candy and then it will decrease from there. Just like x^0=1 which is great than x^-2=1/2 and so on. At the end, of the complete lesson the idea of using negative exponents in sports, sound, radioactive waste, and scientific notation will be a start of what that students will learn in other subjects in the future.

# 1729

The following little anecdote probably deserves to be known by every secondary mathematics teacher. From Wikipedia (see references therein for more information):

1729 is known as the Hardy–Ramanujan number after a famous anecdote of the British mathematician G. H. Hardy regarding a visit to the hospital to see the Indian mathematician Srinivasa Ramanujan. In Hardy’s words:

I remember once going to see him when he was ill at Putney. I had ridden in taxi cab number 1729 and remarked that the number seemed to me rather a dull one, and that I hoped it was not an unfavorable omen. “No,” he replied, “it is a very interesting number; it is the smallest number expressible as the sum of two cubes in two different ways.”

The two different ways are these:

1729 = 13 + 123 = 93 + 103

The quotation is sometimes expressed using the term “positive cubes”, since allowing negative perfect cubes (the cube of a negative integer) gives the smallest solution as 91 (which is a divisor of 1729):

91 = 63 + (−5)3 = 43 + 33

Numbers that are the smallest number that can be expressed as the sum of two cubes in n distinct ways have been dubbed “taxicab numbers”. The number was also found in one of Ramanujan’s notebooks dated years before the incident, and was noted by Frénicle de Bessy in 1657.

# Lessons from teaching gifted elementary school students (Part 5d)

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.

A bright young student of mine noticed that multiplication is repeated addition:

$x \cdot y = x + x + x \dots + x$,

$x \uparrow y = x^y = x \cdot x \cdot x \dots \cdot x$,

The notation $x \uparrow y$ is unorthodox, but it leads to the natural extensions

$x \upuparrows y = x \uparrow x \uparrow x \dots \uparrow x$,

$x \uparrow^3 y = x \upuparrows x \upuparrows x \dots \upuparrows x$,

and so. I’ll refer the interested reader to Wikipedia and Mathworld (and references therein) for more information about Knuth’s up-arrow notation. As we saw in yesterday’s post, these numbers get very, very large… and very, very quickly.

When I was in elementary school myself, I remember reading in the 1980 Guiness Book of World Records about Graham’s number, which was reported to be the largest number ever used in a serious mathematical proof. Obviously, it’s not the largest number — there is no such thing — but the largest number that actually had some known usefulness. And this number is only expressible using Knuth’s up-arrow notation. (Again, see Wikipedia and Mathworld for details.)

From Mathworld, here’s a description of the problem that Graham’s number solves:

Stated colloquially, [consider] every possible committee from some number of people $n$ and enumerating every pair of committees. Now assign each pair of committees to one of two groups, and find $N$, the smallest $n$ that will guarantee that there are four committees in which all pairs fall in the same group and all the people belong to an even number of committees.

In 1971, Graham and Fairchild proved that there is a solution $N$, and that $N \le F(F(F(F(F(F(F(12)))))))$, where

$F(n) = 2 \uparrow^n 3$.

For context, $2 \uparrow^4 3$ is absolutely enormous. In yesterday’s post, I showed that $2 \uparrow^3 = 65,536$. Therefore,

$2 \uparrow^4 3 = 2 \uparrow^3 (2 \uparrow^3 2)$

$= 2 \uparrow^3 65,536$

$= 2 \upuparrows 2 \upuparrows 2 \upuparrows \dots \upuparrows 2$,

repeated 65,536 times.

That’s just $2 \uparrow^4 3$. Now try to imagine $F(12) = 2 \uparrow^{12} 3$. That’s a lot of arrows.

Now try to imagine $F(F(12)) = 2 \uparrow^{F(12)} 3$, which is even more arrows.

Now try to imagine $F(F(F(F(F(F(F(12)))))))$. I bet you can’t. (I sure can’t.)

Graham and Fairchild also helpfully showed that $N \ge 6$. So somewhere between 6 and Graham’s number lies the true value of $N$.

A postscript: according to Wikipedia, things have improved somewhat since 1971. The best currently known bounds for $N$ are

$13 \le N \le 2 \uparrow^3 6$.

# Lessons from teaching gifted elementary school students (Part 5c)

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.

A bright young student of mine noticed that multiplication is repeated addition:

$x \cdot y = x + x + x \dots + x$,

$x^y = x \cdot x \cdot x \dots \cdot x$,

So, my student asked, why can’t we define an operation that’s repeated exponentiation? Like all good explorers, my student claimed naming rights for this new operation and called it $x \hbox{~playa~} y$:

$x \hbox{~playa~} y = x^{x^{x^{\dots}}}$

For example,

$4 \hbox{~playa~} 3 = 4^{4^4} = 4^{256} \approx 1.34 \times 10^{154}$

Even with small numbers for $x$ and $y$, $x \hbox{~playa~} y$ gets very large.

Unfortunately for my student, someone came up with this notion already, and it’s called Knuth’s up-arrow notation. I’ll give some description here and refer the interested reader to Wikipedia and Mathworld (and references therein) for more information. Surprisingly, this notion has only become commonplace since 1976 — within my own lifetime.

Let’s define $x \uparrow y$ to be ordinary exponentiation:

$x \uparrow y = x \cdot x \cdot x \dots \cdot x$.

Let’s now define $x \upuparrows y$ to be the up-arrow operation repeated $y$ times:

$x \upuparrows y = x \uparrow x \uparrow x \dots \uparrow x$.

In this expression, the order of operations is taken to be right to left.

Numbers constructed by $\upuparrows$ get very, very big and very, very quickly. For example:

$2 \upuparrows 2 = 2 \uparrow 2 = 2^2 = 4$.

Next,

$2 \upuparrows 3 = 2 \uparrow (2 \uparrow 2)$

$= 2 \uparrow (2^2)$

$= 2 \uparrow 4$

$= 2^4$

$= 16$

Next,

$2 \upuparrows 4 = 2 \uparrow (2 \uparrow (2 \uparrow 2))$

$= 2 \uparrow 16$

$= 2^{16}$

$= 65,536$

Next,

$2 \upuparrows 5 = 2 \uparrow (2 \uparrow (2 \uparrow (2 \uparrow 2)))$

$= 2 \uparrow 65,536$

$= 2^{65,536}$

$\approx 2.0035 \times 10^{19,728}$

Next,

$2 \upuparrows 6 = 2 \uparrow (2 \uparrow (2 \uparrow (2 \uparrow (2 \uparrow 2))))$

$= 2 \uparrow 2^{65,536}$

$= 2^{2^{65,536}}$

$\approx 10^{6.031 \times 10^{19,727}}$

We see that $2 \upuparrows 6$ is already far larger than a googolplex (or $10^{10^{100}}$), which is often (and erroneously) held as the gold standard for very large numbers.

I’ll refer the interested reader to a previous post in this series for a description of how logarithms can be used to write something like $2^{65,536}$ in ordinary scientific notation.

Knuth’s up-arrow notation can be further generalized:

$x \uparrow^3 y = x \upuparrows x \upuparrows x \dots \upuparrows x$,

repeated $y$ times. The numbers $x \uparrow^4 y$, $x \uparrow^5 y$, etc., are defined similarly.

These numbers truly become large quickly. For example,

$2 \uparrow^3 2 = 2 \upuparrows 2 = 4$, from above.

Next,

$2 \uparrow^3 3 = 2 \upuparrows (2 \upuparrows 2)$

$= 2 \upuparrows 4$

$= 65,536$, from above

Next,

$2 \uparrow^3 4 = 2 \upuparrows (2 \upuparrows (2 \upuparrows 2))$

$= 2 \upuparrows 65,536$

$= 2 \uparrow 2 \uparrow 2 \uparrow \dots \uparrow 2$,

where there are 65,536 repeated 2’s on this last line. It’d be nearly impossible to write this number in scientific notation, and we’ve only reached $2 \uparrow^3 4$.

# Lessons from teaching gifted elementary school students (Part 5b)

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 (though I probably changed the exact wording somewhat):

Exponentiation is to multiplication as multiplication is to addition. In other words,

$x^y = x \cdot x \cdot x \dots \cdot x$,

$x \cdot y = x + x + x \dots + x$,

where the operation is repeated $y$ times.

My kneejerk answer was that there was no answer… while exponents can be thought of as repeated multiplication and multiplication can be thought of as repeated addition, addition can’t be thought of as some other thing being repeated. But it took me a few minutes before I could develop of proof that could be understood by my bright young questioner.

Suppose $y = 1$. Then the expressions above become

$x^1 = x$

and

$x \cdot 1 = x$

However, we know full well that

$x + 1 \ne x$.

Therefore, there can’t be an operation analogous to addition as addition is to multiplication or as multiplication is to exponentiation.

# Lessons from teaching gifted elementary school students (Part 5a)

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 (though I probably changed the exact wording somewhat):

Exponentiation is to multiplication as multiplication is to addition. In other words,

$x^y = x \cdot x \cdot x \dots \cdot x$,

$x \cdot y = x + x + x \dots + x$,

where the operation is repeated $y$ times.

My kneejerk answer was that there was no answer… while exponents can be thought of as repeated multiplication and multiplication can be thought of as repeated addition, addition can’t be thought of as some other thing being repeated.

Which then naturally led to my student’s next question, which I was dreading:

Can you prove that?

This led to another kneejerk reaction, but I kept this one quiet: “Aw, nuts.”

I suggested that $x + y$ can be thought of as starting with $x$ and then adding $1$ repeatedly $y$ times, but my bright student wouldn’t hear of this. After all, in the repeated renderings of $x^y$ and $x \cdot y$, there’s no notion of starting with a number and then doing something with a different number $y$ times.

So I had to put my thinking cap on, and I’m embarrassed to say that it took me a good five minutes before I came up with a logically correct answer that, in my opinion, could be understand by the bright young student who asked the question.

I’ll reveal that answer in tomorrow’s post. In the meantime, I’ll leave a thought bubble if you’d like to think about it on your own.

# Proving theorems and special cases (Part 5): Mathematical induction

Today’s post is a little bit off the main topic of this series of posts… but I wanted to give some pedagogical thoughts on yesterday’s post concerning the following proof by induction.

Theorem: If $n \ge 1$ is a positive integer, then $5^n - 1$ is a multiple of 4.

Proof. By induction on $n$.

$n = 1$: $5^1 - 1 = 4$, which is clearly a multiple of 4.

$n$: Assume that $5^n - 1$ is a multiple of 4, so that $5^n - 1 = 4q$, where $q$ is an integer. We can also write this as $5^n = 4q + 1$.

$n+1$. We wish to show that $5^{n+1} - 1$ is equal to $4Q$ for some (different) integer $Q$. To do this, notice that

$5^{n+1} - 1 = 5^1 5^n - 1$

$= 5 \times 5^n - 1$

$= 5 \times (4q + 1) - 1$ by the induction hypothesis

$= 20q + 5 - 1$

$= 20q + 4$

$= 4(5q + 1)$.

So if we let $Q = 5q +1$, then $5^{n+1} - 1 = 4Q$, where $Q$ is an integer because $q$ is also an integer.

My primary observation is that even very strong math students tend to have a weak spot when it comes to simplifying exponential expressions (as opposed to polynomial expressions). For example, I find that even very good math students can struggle through the logic of this sequence of equalities:

$2^n + 2^n = 2 \times 2^n = 2^1 \times 2^n = 2^{n+1}$.

The first step is using the main stumbling block. Students who are completely comfortable with simplifying $x + x$ as $2x$ can be perplexed by simplifying $2^n + 2^n$ as $2 \times 2^n$. I attribute this to lack of practice with this kind of simplification in lower grade levels.

Here’s another algoebraic stumbling block that I’ve often seen: at the beginning of the $n+1$ case, some students will make the following mistake:

$5^{n+1} - 1 = 5^1 5^n - 1 = 5 (4q) = 4 (5q) = 4Q$.

Because these students end with a multiple of 4, they fail to notice that the second equality is incorrect since

$5^1 5^n - 1 \ne 5^1 (5^n - 1)$.

Again, I attribute this to lack of practice with simplifying exponential expressions in lower grade levels… as well as being a little bit over-excited upon seeing $5^n - 1$ and wishing to use the induction hypothesis as soon as possible.