Re: [math-fun] Fraction of indivisible numbers
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
On 3/30/08, Gareth McCaughan <gareth.mccaughan@pobox.com> wrote:
On Sunday 30 March 2008, Fred Lunnon wrote:
No. A classic result of number theory says that the i-th prime equals i/log(i) (very roughly)
Or, less roughly, i log i :-).
Ouch --- don't just read what I write, read what I mean! Well, the clocks have just gone forward an hour here ... WFL
* Fred lunnon <fred.lunnon@gmail.com> [Apr 01. 2008 09:31]:
[...]
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?
M(n)=2^(n*(n-1)/2) * prod(i=1,n, 2^i - 1) n=338; 1.0*M(n)/2^(n*n) 0.2887880950866024212788997219292307800889119048406857841 http://www.research.att.com/~njas/sequences/A002884 Whenever you count the OEIS is your friend.
product (1 - z^n) for z = 1/2. 1 - 2^-1 - 2^-2 + 2^-5 + 2^-7 - ... The exponents are (negative) pentagonal numbers. The binary expansion of .2887... should be interesting. Rich ________________________________________ From: math-fun-bounces@mailman.xmission.com [math-fun-bounces@mailman.xmission.com] On Behalf Of Joerg Arndt [arndt@jjj.de] Sent: Tuesday, April 01, 2008 1:15 AM To: math-fun Subject: Re: [math-fun] Fraction of indivisible numbers * Fred lunnon <fred.lunnon@gmail.com> [Apr 01. 2008 09:31]:
[...]
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?
M(n)=2^(n*(n-1)/2) * prod(i=1,n, 2^i - 1) n=338; 1.0*M(n)/2^(n*n) 0.2887880950866024212788997219292307800889119048406857841 http://www.research.att.com/~njas/sequences/A002884 Whenever you count the OEIS is your friend. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
participants (4)
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Fred lunnon -
Gareth McCaughan -
Joerg Arndt -
Schroeppel, Richard