From: dlrenfro@gateway.net (Dave L. Renfro) Subject: Re: log 2 Date: 21 Dec 1999 15:31:44 -0500 Newsgroups: sci.math Keywords: algebraic expressions which are nearly integral Zdislav V. Kovarik [sci.math 20 Dec 1999 22:01:50 -0500] wrote (in part): >So x is not equal to 2, just as pi is not >equal to 22/7 or even 355/113, or sqrt(2) >is not equal to 99/70 , and one can go >on and on, with plenty of examples where the >coincidences are much more striking. I remember once reading a Scientific American column by Martin Gardner that wound up being an April's Fool joke in which (it was later revealed) a certain arithmetic combination of well-known constants was (I think) within 10^-24 or so from an integer. I did some net surfing to see if I could locate the example. I didn't find it (I believe it may have been in the April 1975 issue of SI), but what I did find might be of interest to others. I've seen the e, pi, and sqrt(163) example before, but I'm not sure about the others. At the web page one finds the following: (2143/22)^(1/4) = 3.14159265258264... (Reichenbacher, 1900) (9/5)*{7/5 + 2/[1 + (3/7)*5^(3/2)]} ~= Pi (Ramanujan, ~1914) (pi^4 + pi^5) / (e^6) = 0.999999956... e^[pi*163^(1/2)] = 262537412640768743.99999999999925... Fermat's last theorem "counterexample" (JSH, are you there?) [1782^12 + 1841^12]^(1/12) = 1921.999999955... An explanation for the following is given in Angel Garcia's *second* sci.math post on April 25, 1997 at 62^(1/2) * (127/1000) = 0.99999 8 999999 4 99999 4 99999 374... Dave L. Renfro ============================================================================== [The last is just the observation that 999998 is divisible by the comparatively large perfect square 127^2 . The "163" one is explained on other pages at math-atlas.org -- djr] ============================================================================== From: gerry@mpce.mq.edu.au (Gerry Myerson) Subject: Re: Almost Integers Date: Thu, 25 Nov 1999 09:39:34 +1100 Newsgroups: sci.math Keywords: real numbers "unexpectedly close" to integers In article <81gcal\$sr3\$1@mark.ucdavis.edu>, Dean Hickerson wrote: => One example that I like, although the approximations aren't quite as good => as you asked for, is the numbers => => n! => ----------- => n+1 => 2 log(2) Or for those without monotype readers, n! / 2 (log 2)^(n + 1). One reason this is close to an integer is that it is one-half the integral of x^n 2^(-x) from 0 to infinity, which integral is approximately the sum of k^n 2^(-k) for k from 0 to infinity, which sum is an even integer. The Euler-Maclaurin summation formula says that the sum is the integral plus a bunch of terms involving Bernoulli numbers. The first few Bernoulli numbers are small (or zero!), which means the approximation is good; but out past 10 or so the Bernoullis start growing, which explains why the formula starts to depart from integer values. => for 1 <= n <= 15: => => n n!/(2 log(2)^(n+1)) => -- ------------------------ => 0 0.72135 => 1 1.04068 => 2 3.00278 => 3 12.99629 => 4 74.99874 => 5 541.00152 => 6 4683.00125 => 7 47292.99873 => 8 545834.99791 => 9 7087261.00162 => 10 102247563.00527 => 11 1622632572.99755 => 12 28091567594.98157 => 13 526858348381.00125 => 14 10641342970443.08453 => 15 230283190977853.03744 => 16 5315654681981354.51308 => 17 130370767029135900.45799 Gerry Myerson (gerry@mpce.mq.edu.au) ============================================================================== From: Dean Hickerson Subject: Re: Almost Integers Date: 26 Nov 1999 00:33:12 GMT Newsgroups: sci.math I wrote: > One example that I like, although the approximations aren't quite as good > as you asked for, is the numbers > > n! > ----------- > n+1 > 2 log(2) > > for 1 <= n <= 15: Gerry Myerson replied: > One reason this is close to an integer is that it is one-half the integral > of x^n 2^(-x) from 0 to infinity, which integral is approximately the sum > of k^n 2^(-k) for k from 0 to infinity, which sum is an even integer. I hadn't noticed that. The explanation that I had in mind is that the quotient is the dominant term in an infinite series for the number of possible outcomes of a race between n people, if ties are allowed. Calling this number f(n), it's easy to see that n f(n) = SUM C(n,k) f(n-k) k=1 for n >= 1, where C(n,k) is the binomial coefficient "n choose k". From this we obtain the exponential generating function for f: f(n) n 1 SUM ---- z = ------. n >= 0 n! z 2 - e And then by contour integration it can be shown that n! 1 f(n) = -- SUM ---------------------- 2 k n+1 (log(2) + 2 pi i k) for n >= 1, where i is the square root of -1 and the sum is over all integers k. (The imaginary parts of the terms for k and -k cancel each other, so this sum is real.) The k=0 term dominates, so f(n) is asymptotic to n!/(2 log(2)^(n+1)). In fact the other terms are quite small for n from 1 to 15, so f(n) is the nearest integer to n!/(2 log(2)^(n+1)) for these values. The sum that you mentioned is equal to 2 f(n) for n >= 0, as can be seen by showing that it satisfies the same recurrence as f(n). Dean Hickerson dean@math.ucdavis.edu