# Pi R Squared

[Another guest blog entry by Dr. Gene Chase.]

You’ve heard the old joke.

Teacher: Pi R Squared.
Student: No, teacher, pie are round. Cornbread are square.

The purpose of this Pi Day note two days early is to explain why $\pi$ is indeed a square.

The customary definition of $\pi$ is the ratio of a circle’s circumference to its diameter. But mathematicians are accustomed to defining things in two different ways, and then showing that the two ways are in fact equivalent. Here’s a first example appropriate for my story.

How do we define the function $\exp(z) = e^z$ for complex numbers z? First we define $a^b$ for integers $a > 0$ and b. Then we extend it to rationals, and finally, by requiring that the resulting function be continuous, to reals. As it happens, the resulting function is infinitely differentiable. In fact, if we choose a to be e, the $\lim_{n\to\infty} (1 + \frac{1}{n})^n \,$ not only is $e^x$ infinitely differentiable, but it is its own derivative. Can we extend the definition of $\exp(z) \,$ to complex numbers z? Yes, in an infinite number of ways, but if we want the reasonable assumption that it too is infinitely differentiable, then there is only one way to extend $\exp(z)$.

That’s amazing!

The resulting function $\exp(z)$ obeys all the expected laws of exponents. And we can prove that the function when restricted to reals has an inverse for the entire real number line. So define a new function $\ln(x)$ which is the inverse of $\exp(x)$. Then we can prove that $\ln(x)$ obeys all of the laws of logarithms.

Or we could proceed in the reverse order instead. Define $\ln(x) = \int_1^x \frac{1}{t} dt$. It has an inverse, which we can call $\exp(x)$, and then we can define $a^b$ as $\exp ( b \ln (a))$. We can prove that $\exp(1)$ is the above-mentioned limit, and when this new definition of $a^b\,$ is restricted to the appropriate rationals or reals or integers, we have the same function of two variables a and b as above. $\ln(x)$ can also be extended to the complex domain, except the result is no longer a function, or rather it is a function from complex numbers to sets of complex numbers. All the numbers in a given set differ by some integer multiple of

[1] $2 \pi i$.

With either definition of $\exp(z)$, Euler’s famous formula can be proven:

[2] $\exp(\pi i) + 1 = 0$.

But where’s the circle that gives rise to the $\pi$ in [1] and [2]? The answer is easy to see if we establish another formula to which Euler’s name is also attached:

[3] $\exp(i z) = \sin (z) + i \cos(z)$.

Thus complex numbers unify two of the most frequent natural phenomena: exponential growth and periodic motion. In the complex plane, the exponential is a circular function.

That’s amazing!

Here’s a second example appropriate for my story. Define the function on integers $\text{factorial (n)} = n!$ in the usual way. Now ask whether there is a way to extend it to (some of) the complex plane, so that we can take the factorial of a complex number. There is, and as with $\exp(z)$, there is only one way if we require that the resulting function be infinitely differentiable. The resulting function is (almost) called Gamma, written $\Gamma$. I say almost, because the function that we want has the following property:

[4] $\Gamma (z - 1) = z!$

Obviously, we’d like to stay away from negative values on the real line, where the meaning of (–5)! is not at all clear. In fact, if we stay in the half-plane where complex numbers have a positive real part, we can define $\Gamma$ by an integral which agrees with the factorial function for positive integer values of z:

[5] $\Gamma (z) = \int_0^\infty \exp(-t) t^{z - 1} dt$.

If we evaluate $\Gamma (\frac{1}{2})$ we discover that the result is $\sqrt{\pi}$.

In other words,

[6] $\pi = \Gamma(\frac{1}{2})^2$.

Pi are indeed square.

That’s amazing!

I suspect that the $\pi$ arises because there is an exponential function in the definition of $\Gamma$, but in other problems involving $\pi$ it’s harder to find where the $\pi$ comes from. Euler’s Basel problem is a good case in point. There are many good proofs that

$1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + ... = \frac{\pi^2}{6}$

One proof uses trigonometric series, so you shouldn’t be surprised that $\pi$ shows up there too.

$\pi$ comes up in probability in Buffon’s needle problem because the needle is free to land with any angle from north.

Can you think of a place where $\pi$ occurs, but you cannot find the circle?

George Lakoff and Rafael Núñez have written a controversial book that bolsters the argument that you won’t find any such examples: Where Mathematics Comes From. But Platonist that I am, I maintain that there might be such places.

# Rationalization Rant

Every high school math student has been taught how to rationalize the denominator. We tell students not to give an answer like

$\frac{1}{\sqrt{2}}$

because it isn’t fully “simplified.” Rather, they should report it as

$\frac{\sqrt{2}}{2}.$

This is fair, even though the second answer isn’t much simpler than the first. What does it really mean to simplify an expression? It’s a pretty nebulous instruction.

We also don’t consider

$\frac{12}{1+\sqrt{5}}$

to be rationalized because of the square root in the denominator, so we multiply by the conjugate to obtain

$2-2\sqrt{5}.$

In this particular example, multiplying by the conjugate was really fruitful and the resulting expression does indeed seem much more desirable than the original expression.

But here’s where it gets a little ridiculous. Our Algebra 2 book also calls for students to rationalize the denominator when (1) a higher root is present and (2) roots containing variables are present. Let me show you an example of each situation, and explain why this is going a little too far.

## Rationalizing higher roots

First, when a higher root is present like

$\sqrt[5]{\frac{15}{2}},$

the book would have students multiply the top and bottom of the fraction inside the radical by $2^4$ so as to make a perfect fifth root in the denominator. The final answer would be

$\frac{\sqrt[5]{240}}{2}.$

Simpler? You decide.

This becomes especially problematic when we encounter sums involving higher roots. It’s certainly possible, using various tricks, to rationalize the denominator in expressions like this:

$\frac{1}{2-\sqrt[3]{5}}.$

But is that really desirable? The result here is

$\frac{1}{2-\sqrt[3]{5}}\cdot\frac{4+2\sqrt[3]{5}+\sqrt[3]{25}}{4+2\sqrt[3]{5}+\sqrt[3]{25}}=\frac{4+2\sqrt[3]{5}+\sqrt[3]{25}}{3},$

which is, arguably, more complex than the original expression. Can anyone think of a good reason to do this, except just for fun?

## Rationalizing variable expressions

Now, let’s think about variable expressions. Here is a problem, directly from our Algebra 2 book (note the directions as well):

Write the expression in simplest form. Assume all variables are positive.

$\sqrt[3]{\frac{x}{y^7}}$

The method that leads to the “correct” solution is to multiply the fraction under the radical by $\frac{y^2}{y^2}$, and to finally write

$\frac{\sqrt[3]{xy^2}}{y^3}.$

This is problematic for two reasons. (1) This isn’t really simpler than the original expression and (2) this expression isn’t even guaranteed to have a denominator that’s rational! (Suppose $y=\sqrt{2}$ or even $y=\pi$.) Once again I ask, can anyone think of a good reason to do this, except just for fun??

## So how far do we take this?

Is it reasonable to ask someone to rationalize this denominator?

$\frac{1}{2\sqrt{2}-\sqrt{2}\sqrt[3]{5}+2\sqrt{5}-5^{5/6}}$

You can rationalize the denominator, but I’ll leave that as an exercise for the reader. So how far do we take this? I had to craft the above expression very carefully so that it works out well, but in general, most expressions have denominators that can’t be rationalized (and I do mean “most expressions” in the technical, mathematical way–there are are an uncountable number of denominators of the unrationalizable type). All that being said, I think this would make a great t-shirt:

And I rest my case.

# Leap Day Birthday Math

## Happy leap day!!!

Here are some leap-day birthday thoughts I discussed with my colleagues and students today:

### What’s the probability of a leap year birthday?

The probability that someone is born on a leap day is $\frac{1}{365\cdot 4+1}=\frac{1}{1461}\approx 0.000684$. Oh wait, that’s not completely true. Leap years don’t really occur every four years. Years divisible by 100 are not leap years, unless also divisible by 400. So, the actual probability is

$\frac{100-4+1}{365\cdot 400+100-4+1}= \frac{97}{146097}\approx 0.000639$.

### What’s the probability of having triplets on a leap day?

One of our RM students is a triplet, born today. What are the chances of that occurring? Well, the statistics on triplets are pretty hard to get right. But let’s say the occurrence of a triplet birth is 1 in 8000. (That’s my informal estimate based on this site and this site.) I think we can say that the probability of being a triplet is 3 times that (right?). Then, the probability of being a triplet born on a leap day is

$\left(\frac{100-4+1}{365\cdot 400+100-4+1}\right)\left(\frac{3}{8000}\right)= \frac{291}{1168776000}\approx\frac{1}{4016412} \approx 0.249 \times 10^{-7}$.

The current US population is 311,591,917, so that means there are roughly 77 triplets in the US with leap day birthdays. Happy birthday to all of you!

Bonus thought question: Iif you have quadruplets born on a leap day, you get to celebrate 4 birthdays every four years, so doesn’t that average out to one birthday a year?

### Half-birthday for those born on August 29

One of my other colleagues has a birthday on August 29th. So today is her half birthday! But it only comes around every four years (roughly). Hooray!

But then that got us thinking about half birthdays: Some people, like those born on August 30th or 31st NEVER have a half birthday. How sad!! This happens to anyone born on August 30th, August 31st, March 31st, October 31st, May 31st, or December 31st. That’s a lot of people without half birthdays.

But wait. When is your actual half birthday? Shouldn’t it be 182.5 days before/after your birthday? That’s not necessarily the same date in the month. For instance, my birthday is May 15. So my half birthday should be November 15, right? Wrong. My half birthday is (May 15 + 182.5 days), which is November 13th or November 14th, depending on if you round up or down. Even accounting for a leap year, it’s still not quite right.

Who else is miscalculating their half birthday? Unless your birthday is in June, April, October, or December, you’re half-birthday isn’t what you think it is. To calculate your half birthday, use this amazing half birthday calculator I just discovered!

# Square One

I was raised on this show, and it’s so fun that you can find clips of it on youtube, including these three gems. At the end of every episode of Square One there was a mathematical music video. I just showed my Precalculus class this first video, but I spared them the other two. If you’ve never seen Square One, these will give you a little bit of a feel for the show, or if you know and loved the show, these will help you take a walk down memory lane.

# Inverse functions and the horizontal line test

I have a small problem with the following language in our Algebra 2 textbook. Do you see my problem?

Horizontal Line Test

If no horizontal line intersects the graph of a function f more than once, then the inverse of f is itself a function.

Here’s the issue: The horizontal line test guarantees that a function is one-to-one. But it does not guarantee that the function is onto. Both are required for a function to be invertible (that is, the function must be bijective).

Example. Consider $f:\mathbb{R}\to\mathbb{R}$ defined $f(x)=e^x$. This function passes the horizontal line test. Therefore it must have an inverse, right?

Wrong. The mapping given is not invertible, since there are elements of the codomain that are not in the range of $f$. Instead, consider the function $f:\mathbb{R}\to (0,\infty)$ defined $f(x)=e^x$. This function is both one-to-one and onto (bijective). Therefore it is invertible, with inverse $f^{-1}:(0,\infty)\to\mathbb{R}$ defined $f(x)=\ln{x}$.

This might seem like splitting hairs, but I think it’s appropriate to have these conversations with high school students. It’s a matter of precise language, and correct mathematical thinking. I’ve harped on this before, and I’ll harp on it again.

Love this.

# Why are infinite series so hard to grasp?

I’ve posted on infinite series a few times before. But I was inspired to touch on the topic again because I saw this post, yesterday, over at the Math Less Traveled. Actually, the post isn’t really about infinite series as much as it is about p-adic numbers and zero divisors. I’m excited to read more from Brent on this subject. But I digress.

The point I want to make with this post is that students struggle with wrapping their minds around convergent infinite series, and yet they live with them all the time. Students have inconsistently held beliefs about infinite sums.

The simplest convergent series is a geometric series $\sum_{n=1}^\infty a_n=a_1r^{n-1}$ which converges to $\frac{a_1}{1-r}$. The easy proof of this fact goes like this: we look at the sum formula for a finite geometric series, $s_n=\frac{a_1(1-r^n)}{1-r}$ and we notice that

$\lim_{n\to\infty}\frac{a_1(1-r^n)}{1-r}=\frac{a_1}{1-r}$

for $|r|<1$.

But this proof isn’t very satisfying for the student encountering infinite series for the first time ever. Evaluating the limit feels like ‘magic.’ The idea of adding up an infinite amount of things and getting a finite value is unsettling. I admit, it sounds like quite a lot to swallow. That being said, however, students have no problem declaring the infinite series

$0.3 + 0.03 + 0.003 + 0.0003 + \cdots$

to be $1/3$. It’s not “close to” $1/3$, it’s not “approaching” $1/3$, it IS EQUAL TO $1/3$. And my Precalculus students already accept this as fact. So without even thinking about it, they’ve been living with convergent infinite series all along. Hah!

Once they finally shake their denial, they can more easily accept the convergence of other infinite series like $\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$. At first when students encounter a series like this, they think, “surely we can’t say the sum is EQUAL to $\frac{\pi^2}{6}$. It must be close to $\frac{\pi^2}{6}$ or approach it, but equal to?” But the same students make no such distinction with $0.3+0.03+0.003+\cdots = \frac{1}{3}$.

So there it is. An inconsistently held belief about infinite sums. To the student: You cannot have it both ways. Either you must agree with, or deny, both of the following equations:

$0.3+0.03+0.003+\cdots = \sum_{n=1}^\infty 0.3(0.1)^{n-1}=\frac{1}{3}$

$1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\cdots = \sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$

But to believe one equation is true and the other is only ‘kind of’ true is inconsistent. I rest my case. 🙂

# The Arc Cotangent Controversy

I love this discussion at squareCircleZ. All my readers should check it out. Which is the graph of arccot(x)?

from squarecircleZ

from squarecircleZ

I especially like this controversy because some big players have weighed in on each side. Mathcad and Maple prefer the first interpretation, Mathematica and Matlab prefer the second.

For a more thorough treatment, check out the original post here. Three cheers for great math blogging! 🙂