# The Wallis Product

In November, it was announced that a team of physicists had stumbled upon a beautiful formula from mathematics called the Wallis Product while investigating quantum mechanics. The Wallis Product gives a way of expressing the mathematical constant $\pi$ as a fraction with infinite length: $$\frac{\pi}{2}=\frac{2\cdot 2\cdot 4\cdot 4\cdot 6\cdot 6\cdots}{1\cdot 3\cdot 3\cdot 5\cdot 5\cdot 7\cdots}$$ The formula itself takes a while to wrap your head around. After all, what does it even mean to have an infinitely long fraction?

To get a sense of what it means, let’s look at just the initial chunks of the product on the right side. The first chunk, with just the leftmost factor in both the numerator and denominator, is $\dfrac{2}{1}$, which is not terribly far from $\dfrac{\pi}{2}=1.57079633$. If we include the first two factors in both top and bottom, we get closer — the fraction is $\dfrac{2\cdot 2}{1\cdot 3}=\dfrac{4}{3}=1.3333\overline{3}$, which is nearer to $\dfrac{\pi}{2}$ (although the fraction is now an underestimate whereas our first chunk gave an overestimate). We can look at longer and longer chunks to get closer and closer to $\dfrac{\pi}{2}$.

This table is not enough to establish that the formula in the Wallis Product is true. After all, the exact same evidence could be used to guess that the values of the fractions are getting close to, say, $1.5$. If we are to believe the Wallis Product, we will need an airtight mathematical argument.

One of the common arguments given for the Wallis Product being true involve calculus. In fact, I often give this assignment to my advanced calculus students in which they deduce and then apply the formula.

For those not as fluent with calculus (or quantum mechanics, for that matter), there is still hope. A Swedish mathematician named Johan Wastlund gave a simple proof of the Wallis Product. His idea starts by looking at fractions whose numerator is the product of the first $n$ positive odd integers and whose denominator is the product of the first $n-1$ positive even integers. For brevity, we will call these fractions $A_n$. So, $A_1=1$, $A_2=\dfrac{1\cdot 3}{2}=\dfrac{3}{2}$, $A_3=\dfrac{1\cdot 3\cdot 5}{2\cdot 4}=\dfrac{15}{8}$, and in general $A_n=\dfrac{1\cdot 3\cdot 5 \cdots (2n-1)}{2\cdot 4\cdots (2n-2)}$. We then use these fractions to build some boxes in different layers.

The first thing to notice here is that the total area of the rectangles of any color is $1$: the blue square has area $1$, there are two green rectangles with area $1/2$ for a total area of $1$, et cetera. Once we believe this, we can superimpose some quarter-circles over our boxes.

If we consider just the blue area, it contains the smallest quarter-circle and is itself contained in the second-smallest one, so the blue area is between the areas of the smallest two quarter-circles. Since we know the blue area and a formula for the area of a quarter-circle, we can say that $$\dfrac{\pi}{4} A_1^2<1<\dfrac{\pi}{4}A_2^2.$$ Next, we can look at the blue and green regions together. Their areas are not quite between those of the second and third quarter-circles, but we can say that they lie between the first and third quarter-circles, so $$\dfrac{\pi}{4} A_1^2<2<\dfrac{\pi}{4}A_3^2.$$ Similarly, the blue, green, and yellow areas are between the second and fourth of the quarter-circles, so $$\dfrac{\pi}{4}A_2^2<3<\dfrac{\pi}{4}A_4^4.$$ We can continue this line of reasoning as far as we like, but what it really tells us is that, for any positive integer $n$, a comparison of areas shows that $$\dfrac{\pi}{4} A_{n-1}^2<n<\dfrac{\pi}{4}A_{n+1}^2.$$

Here is where we need some algebra. $A_n$ and $A_{n+1}$ are closely related to each other. For example, $$A_3=\dfrac{1\cdot 3\cdot 5}{2\cdot 4}=\dfrac{1\cdot 3}{2}\cdot\dfrac{5}{4}=A_2\cdot\dfrac{5}{4},$$ so $A_3$ had $A_2$ sitting inside of it. This pattern is always the case. More precisely, $$A_{n+1}=A_n\cdot\dfrac{2n+1}{2n}.$$ The same idea shows that $A_n=A_{n-1}\cdot\dfrac{2n-1}{2n-2}$, which becomes $A_{n-1}=A_n\cdot\dfrac{2n-2}{2n-1}$ after rearranging. Let’s substitute these formulas for $A_{n-1}$ and $A_{n+1}$ into our inequality. We get $$\dfrac{\pi}{4}\left(A_{n}\cdot\dfrac{2n-2}{2n-1}\right)^2<n<\dfrac{\pi}{4}\left(A_{n}\cdot\dfrac{2n+1}{2n}\right)^2,$$ or $$\dfrac{\pi}{4} A_n^2\cdot\dfrac{(2n-2)^2}{(2n-1)^2}<n<\dfrac{\pi}{4}\cdot A_{n}^2\cdot\dfrac{(2n+1)^2}{(2n)^2}.$$ Now, since the Wallis Product deals with $\dfrac{\pi}{2}$, we will try to get $\dfrac{\pi}{2}$ in our inequality. We can do this by multiplying through by $\dfrac{2}{A_n^2}$, leaving us with $$\dfrac{\pi}{2}\cdot\dfrac{(2n-2)^2}{(2n-1)^2} <\dfrac{2n}{A_n^2}<\dfrac{\pi}{2}\cdot \dfrac{(2n+1)^2}{(2n)^2}.$$

The fraction $\dfrac{(2n+1)^2}{(2n)^2}$ gets very close to $1$ for large values of $n$, since the $+1$ in the numerator pales in comparison to the $2n$, and if the $+1$ weren’t there at all, the fraction would simplify to exactly $1$. Similarly, $\dfrac{(2n-2)^2}{(2n-1)^2}$ is very close to $1$. The leftmost and rightmost portions of our inequality, then, are close to $\dfrac{\pi}{2}\cdot 1$, or just $\dfrac{\pi}{2}$. If $n$ could somehow be infinitely large, we would have $$\dfrac{\pi}{2}=\dfrac{2n}{A_n^2}.$$ The good news is that when we use the definition of $A_n$ and simplify the fractions, we end up (remembering that $n$ is “infinitely large”) with $$\frac{\pi}{2}=\frac{2\cdot 2\cdot 4\cdot 4\cdot 6\cdot 6\cdots}{1\cdot 3\cdot 3\cdot 5\cdot 5\cdot 7\cdots},$$ thereby establishing the Wallis Product.

Now, there are some steps that I left out here, so anyone wanting an airtight proof will need to work them out separately. In particular, it needs to be verified that

1. Each layer (or color) of rectangles has total area $1$.
2. That each layer of rectangles does, in fact, lie between its two quarter-circles. The proof would break down if some rectangles reached beyond the “outer” one or did not quite reach the “inner” one.
3. The fraction $\dfrac{2n}{A_n^2}$ does actually simplify to what I claimed it did.

All of these facts can be verified with some fraction arithmetic and possibly a tiny bit of algebra. The reader should enjoy some of the discovery, so I will leave these as exercises.