On 3/30/08, Steve Witham <sw@tiac.net> wrote:
oo ------ | | 1 | | 1 - ----- | | f(i) i = 1
f(i) = i + 1 (1/2) (2/3) (3/4) (4/5)... = 0
f(i) = (i+1)^2 (1/4) (8/9) (15/16) (24/25) ... = 0
Is the result always zero when f(i) is a positive polynomial of i?
The last equation should apparently read (3/4) (8/9) (15/16) (24/25) ... = 1/2 So no, the product is not always zero --- in fact, taking logarithms and approximating the log by its Taylor expansion log(1-x) = -x + ... , the product is actually nonzero for degree > 1 --- unless one of the factors vanishes "accidentally" because f(i) - 1 happens to have a positive integer root.
f(i) = i'th prime (1/2) (2/3) (4/5) (6/7) (10/11) (12/13) ... = 0
(The fraction of counting numbers that aren't divisible by any prime.) Does that equation have a name?
f(i) = 2^i (1/2) (3/4) (7/8) (15/16) ... ~= 0.288788095087
This number turns up as the probability that a large random binary matrix is nonsingular. I don't know of a closed form for it --- does anybody else?
Is the result always > 0 when f(i) = k^i and k > 1?
Yes, by the same reasoning as before.
If so, does this put the primes between polynomials and exponentials?
No. A classic result of number theory says that the i-th prime equals i/log(i) (very roughly) --- this is the "Prime Number Theorem", see Hardy & Wright "Introduction to the Theory of Numbers" for a detailed account [and no doubt Wikipedia for a quick 'n' dirty summary]. Fred Lunnon