Let x be a sequence in R. Then there exists a subsequence y of x such that y is monotone.

Proof. If x is monotone, then construct any increasing sequence j of integers and then y_n = x_(j_n) is also a monotone sequence. If x is not monotone, then there exists a least integer k_1 such that x_(k_1) is neither an upper or a lower bound for x_M where M > k_1. Now consider the sequence of x beyond x_(k_1); if that sequence attains its minimum, let

Let f:R->R be differentiable with the (n+1)th derivative of f being identically zero, then f is a polynomial of degree at most n.

For notational clarity in ascii, the nth derivative of f will be called "f^(n)", and the value of that function at a point x will be called "f^(n)(x)".

Proof. Since the (n+1)th derivative is identically zero, we know that f', f'', f''', ..., f^(n) all exist on R. Thus, the Taylor Expansion of f can be computed as the sum of a polynomial of degree n and a remainder term. That is, given some u in R,

f(x) = f(u) + f'(u)(x - u) + (1/2!)f''(u)(x - u)^2 + ... + (1/n!)f^(n)(u)(x - u)^n + (1/(n+1)!)f^(n+1)(e)(x - u)^(n+1)

for some e between u and x. However, since f^(n+1) is identically zero, f^(n+1)(e) = 0, so the remainder term in the taylor expasion is also zero. Thus,

f(x) = f(u) + f'(u)(x - u) + (1/2!)f''(u)(x - u)^2 + ... + (1/n!)f^(n)(u)(x - u)^n

Which is to say that f behaves like a polynomial in the variable (x - u) on the entire interval between x and u. Since u was chosen arbitrarily, let it be zero for simplicity's sake. Then,

f(x) = f(0) + f'(0)x + (1/2!)f''(0)x^2 + ... + (1/n!)f^(n)(0)x^n

so that f is a polynomial of degree at most n. QED, bitches.