Irrationality of pi
We’re getting close! Last time, we defined a new function 
and showed that 
and 
are both integers, and that 
. So, consider the following:
![$ \begin{align*} &\frac{d}{dx} [ F'(x) \sin x - F(x) \cos x ] \\ &= F^{\prime\prime}(x)\sin x + F'(x) \cos x \\ & \qquad - F'(x) \cos x + F(x) \sin x \\ &= F^{\prime\prime}(x) \sin x + F(x) \sin x \\ &= [F^{\prime\prime}(x) + F(x)]\sin x \\ &= f(x) \sin x. \end{align*} $](http://wso.williams.edu/~byorgey/wordpress/wp-content/plugins/latexrender/pictures/e2aa0056c82331c22cdc81cddc6d8680.gif "$ \begin{align*} &\frac{d}{dx} [ F'(x) \sin x - F(x) \cos x ] \\ &= F^{\prime\prime}(x)\sin x + F'(x) \cos x \\ & \qquad - F'(x) \cos x + F(x) \sin x \\ &= F^{\prime\prime}(x) \sin x + F(x) \sin x \\ &= [F^{\prime\prime}(x) + F(x)]\sin x \\ &= f(x) \sin x. \end{align*} $")
The first step uses the product rule for differentiation (recalling that 
and 
); the last step is what we showed last time. Now we see the point of defining : it’s just so that we have a convenient way to talk about the antiderivative of 
. We could just do everything directly in terms of alternating sums of derivatives of 
… but it’s much clearer this way, don’t you agree?
Now that we know the antiderivative of 
, we can use the Fundamental Theorem of Calculus to compute the following integral:
![$ \begin{align*} &\int_0^\pi f(x)\sin x dx \\ &= \left[ F'(x) \sin x - F(x) \cos x \right]_0^\pi \\ &= F'(\pi) \sin \pi - F(\pi) \cos \pi \\ & \qquad - F'(0) \sin 0 + F(0) \cos 0 \\ &= F(\pi) + F(0). \end{align*} $](http://wso.williams.edu/~byorgey/wordpress/wp-content/plugins/latexrender/pictures/dc4b4ef4ced9e046e140679279400b9b.gif "$ \begin{align*} &\int_0^\pi f(x)\sin x dx \\ &= \left[ F'(x) \sin x - F(x) \cos x \right]_0^\pi \\ &= F'(\pi) \sin \pi - F(\pi) \cos \pi \\ & \qquad - F'(0) \sin 0 + F(0) \cos 0 \\ &= F(\pi) + F(0). \end{align*} $")
Note that the value of this integral is an integer, since both and are integers. But next time we’ll show that it is also strictly between 0 and 1 (for a suitable choice of 
), which is clearly nonsense!
is irrational, that is, that it cannot be represented as the ratio of two integers (and hence its decimal expansion goes on forever without repeating). 
and 
are the “numerator” and “denominator” of 
, and in fact that 
. In 
, one for each 
is zero.) Using 
as
, 
turns into 
, and so on. “But wait a minute,” you say. “Shouldn’t the 
at the end of 
in 
is zero, so that term just goes away. Now we note that every term of 
… and indeed, next time we’ll start making some connections with our old trigonometric friends, 
and 
.
, although 
and 
are not necessarily zero, they are always integers. (The notation 
means the 
; that is, take the derivative of 
and 
. 
in the denominator of 
. But let’s take the derivative: 
so 
, which is clearly not an integer. For the derivatives of 
for 
causes every term containing 
. Hence, we must show that the constant term of 
, when expanded out, is a polynomial of the form 
, where the ellipsis contains a bunch of terms with integer coefficients and powers of 
. (In fact, we could use the 
, we see that
is a polynomial with terms of degree 
—but we already knew that.)
is 
, so taking the derivative of a polynomial reduces each of the exponents by one. So the first derivative of 
(and hence a constant term of zero); the second derivative has terms of degree 
through 
(still no constant term); and so on. We can see that none of the first 
(which is certainly an integer) for 
. What about the 
and 
is some integer. When we take the derivative, this term will turn into 
; if we take the derivative again, it will become 
; another derivative gives us 
, and so on. Do you see what is happening? After taking the derivative exactly 
times, we will end up with the constant term
is clearly divisible by 
.
with respect to 
), we get
(the two negatives cancel on the left side), 
, and so on. But the extra negative sign for odd derivatives doesn’t really matter: in either case, 
, which is an integer.
—which (if you recall the punchline) will turn out to be an integer strictly between 
, where 
. It’s not too hard to see that 
as well (remembering that 
, of course, which means that 
):
. That is,
. This also means that 
is an integer, AND
.
