## happy pi day!

Is it really Pi Day again already? Happy 3.1415926535897932384626433832795028841971. Happy 3.1415926535897932384626433832795028841971, everyone. I considered watching the movie Pi (which I love, even though it doesn’t have much to do with the number), but I couldn’t find my DVD. So instead, I watched Vanishing Point, which has fast cars and a naked hippie chick on a motorcycle.

Anyway, here is a proof of the irrationality of π. (Proof of the transcendence of is π left as an exercise for the reader. Hint: first prove e is transcendental, then use Euler’s formula!)

In other news, the last few years of grad school have led me to the conclusion that advanced math is essentially a black hole into which time and self-esteem are sucked, and from which nothing good ever escapes.

Assume π = a/b with positive integers a and b.

Now, for some natural number n define the functions f and F as follows. Strictly speaking, f and F should each have n as an index as they depend on n but this would render things unreadable; remember that n is always the same constant throughout this proof.

Let

• f(x) = xn(abx)n/n!

and let

• F(x) = f(x) + … + (-1)jf(2j)(x) + … + (-1)nf(2n)(x)

where f(2j) denotes the 2j-th derivative of f.

Then f and F have the following properties:

1. f is a polynomial with coefficients that are integer, except for a factor of 1/n!

2. f(x) = f(π-x)

3. 0 <= f(x) <= πnan/n! for 0 <= x <= π

4. For 0 <= j < n, the j-th derivative of f is zero at 0 and π.

5. For n <= j, the j-th derivative of f is integer at 0 and π
(inferred from (1.)).

6. F(0) and F(π) are integer (inferred from (4.) and (5.)).

7. F + F ” = f

8. (F ‘·sin – F·cos)’ = f·sin   (inferred from (7.))

Hence, the integral over f·sin, taken from 0 to π, is integer.

For sufficiently large n, however, inequality (3.) tells us that this integral must be between 0 an 1. Hence, we could have chosen n such that the assumption is led ad absurdum.

Blatantly stolen from here.