# 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.

# Proving the “obvious”

from graphjam.com

As Eric Temple Bell said, “‘Obvious’ is the most dangerous word in mathematics.” That being the case, it is often true that we have trouble proving statements that seem self-evident. Many times we are indeed tempted to say “clearly” or “obviously” or “it is trivial” or “the details have been left to the reader” or “this easily follows from Theorems 4.8, 5.1, and Definition 5.8”. For a full list of invalid proof techniques, visit this hilarious site. Here are a few samples (it’s a LONG list!), quoted from  full list on their site:

• Proof by intimidation (“Trivial.”)
• Proof by example (The author gives only the case n = 2 and suggests that it contains most of the ideas of the general proof.)
• Proof by vigorous hand waving (Works well in a classroom or seminar setting. )
• Proof by exhaustion (An issue or two of a journal devoted to your proof is useful. )
• Proof by importance (A large body of useful consequences all follow from the proposition in question.)
• Proof by accelerated course (We don’t have time to prove this… )

Choosing the level of rigor for a proof is often difficult–depending on the mathematical context, and the audience. I’m taking a graduate class in Analysis right now, so I definitely think about this a lot! In fact, I might add one more to the list:

• Proof by beautiful typesetting (Because the proof looks good and is typed in $\LaTeX$, it must be right.)

At least,  I hope my professor feels that’s a valid technique :-).

# Pythagorean Theorem

Just when I said the other day that we care less about the ancient Greek mathematicians than we did 200 years ago, I turn the attention of my readership to the Pythagorean Theorem. It’s a simple repost: I liked the following video that Denise @ Let’s Play Math posted yesterday and wanted to share it. It also connects nicely to the post my dad made the other day about beautiful proofs.

# Irrationals of the form a+b√c

I made the claim in a post last week that the set of irrationals of the form $a+b\sqrt{c}$ is countable. I said that pretty quickly, without justification. I’ve never proved statements like this before, but here I’m going to try.

Theorem. The set of irrationals of the form $a+b\sqrt{c}$, with $a,b \neq 0,c>0\in\mathbb{Q}$, is countable.

Proof. Consider the set of irrationals of the form $a+b\sqrt{c}$, with $a,b \neq 0,c >0\in\mathbb{Q}$. More formally, define

$\mathbb{I}=\left\{a+b\sqrt{c}\in\mathbb{R}: a,b \neq 0,c>0\in\mathbb{Q}\right\}$

And also require that $c$ is ‘square free’–that is, we require that neither the numerator or denominator of $c$ contain factors that are perfect squares. So $a+b\sqrt{c}$ is in ‘simplest’ form. We aim to show that $\mathbb{I}$ is countable.

Now, consider the function

$\mathbb{I} \overset{f}{\rightarrow} \mathbb{Q}^3$

defined

$f(a+b\sqrt{c})=(a,b,c)$

This function is one-to-one since we require $c$ to be in simplest form–that is, the image of any number $a+b \sqrt{c}$ under $f$ is unique. So $f$ is an injection from $\mathbb{I}$ into$\mathbb{Q}^3$.

We know that $\mathbb{Q}$ is countable. Since a finite Cartesian product of countable sets is countable, $\mathbb{Q}^3$ must also be countable. And we have constructed a function $f$ which is an injection from $\mathbb{I}$ into $\mathbb{Q}^3$. So the cardinality of $\mathbb{I}$ must be no greater than the cardinality of $\mathbb{Q}^3$. Thus $\mathbb{I}$ must also be countable, as desired.

I think I did that right. Any suggestions, math readers?