An argument of infinite proportions (pi)

[Masta Squidge]

@unknownuser said:

...and that 1/3 does in fact end, thats why its called a repeating, terminating decimal."

I believe that is what we call a typo I believe it should read "does in fact NOT end, thats why its called a repeating, NON-terminating decimal.

Yeah, I slipped, it was past my bedtime.

[SchreiberBike]

@masta squidge said:

Please by all means message this joke to me! Unless its the one about the Special Olympics of course lol, because that is all too familiar.

Yep, that's the one.

[SchreiberBike]

Not that it will help, but here is a mathematical proof that pi is irrational:

@unknownuser said:

Niven's proof

The proof uses the characterization of π as the the smallest positive zero of the sine function. Like all proofs of irrationality, the argument proceeds by reductio ad absurdum.

Preparation: Suppose that π is rational, i.e. π = a / b for some integers a and b ≠ 0, which may be taken without loss of generality to be positive. Given any positive integer n, we define the polynomial function

f(x) = {x^n(a - bx)^n \over n!},\quad x\in\mathbb{R},

and denote by

F(x) = f(x) + \cdots + (-1)^j f^{(2j)}(x) + \cdots + (-1)^n f^{(2n)}(x). \, 

the alternating sum of f and its first n even derivatives.

Claim 1: F(0) = F(π)

Proof: Since

f(x)=b^n{x^n(\pi - x)^n \over n!}=f(\pi-x),\quad x\in\mathbb{R},

the chain rule and mathematical induction imply

f^{(j)}(x) = (-1)^j f^{(j)}(\pi - x),\quad x\in\mathbb{R},

for all the derivatives, in particular

f^{(2j)}(0)=f^{(2j)}(\pi)\,

for j = 1, 2, ...,n and Claim 1 follows from the definition of F.

Claim 2: F(0) is an integer.

Proof: Using the binomial formula to expand (a – bx)n and the index transformation j = k + n, we get the representation

f(x)={1\over n!}\sum_{j=n}^{2n}{n \choose j-n}a^{2n-j}(-b)^{j-n}x^{j}.\, 

Since the coefficients of x0, x1, ..., xn − 1 are zero and the degree of the polynomial f is at most 2n, we have f (j)(0) = 0 for j < n and j > 2n. Furthermore,

f^{(j)}(0)={j!\over n!}{n \choose j-n}a^{2n-j}(-b)^{j-n}\quad\mbox{for } n\le j\le 2n.

Since j ≥ n, the fraction of the two factorials is an integer. The same holds for the binomial coefficient, as can be seen from its combinatorical interpretation or Pascal's triangle. Hence f and every derivative of f at 0 is an integer and so is F(0).

Claim 3:

\frac12 \int_0^\pi f(x)\sin(x)\,dx=F(0) 

Proof: Since f (2n + 2) is the zero polynomial, we have

F'' + F = f.\, 

The derivatives of the sine and cosine function are given by (sin x)' = cos x and (cos x)' = −sin x, hence the product rule implies

(F'\cdot\sin - F\cdot\cos)' = f\cdot\sin. 

By the fundamental theorem of calculus

\frac12 \int_0^\pi f(x)\sin(x)\,dx= \frac12 \bigl(F'(x)\sin x - F(x)\cos x\bigr)\Big|_{x=0}^{x=\pi}.

Since sin 0 = sin π = 0 and cos 0 = –cos π = 1 (here we use the abovementioned characterization of π as a zero of the sine function), Claim 3 follows from Claim 1.

Conclusion: Since f(x) > 0 and sin x > 0 for 0 < x < π (because π is the smallest positive zero of the sine function), Claims 1 and 3 show that F(0) is a positive integer. Since

x(\pi -x) = \Bigl(\frac\pi2\Bigr)^2-\Bigl(x-\frac\pi2\Bigr)^2\le\Bigl(\frac\pi2\Bigr)^2,\quad x\in\mathbb{R},

and 0 ≤ sin x ≤ 1 for 0 ≤ x ≤ π, we have

\frac12 \int_0^\pi f(x)\sin(x)\,dx\le \frac{b^n}{n!}\Bigl(\frac\pi2\Bigr)^{2n+1},

which is smaller than 1 for large n, hence F(0) < 1 by Claim 3 for these n. This is impossible for the positive integer F(0).

Not that I understand it, but there it is.

[Gaieus]

Aha...
Well, thanks - at least for that very last line (now I can hope I'm not alone in the universe who does not understand it ).