Re: [math-fun] pi cf record
Bill Gosper:
Has anyone listed the largest terms in Hans's data?
Hans Havermann:
I have. :)
For his 13th birthday, funster Neil Bickford has just computed and verified 485M partial quotients (http://www.neilbickford.com/picf.htm). The longest previous CF we can find remains Hans Havermann's 180M, wherein Neil found two glitches--a "q" instead of a "1" at position 4,311,037 (a picked 2^6 bit) and a semicolon following term 100,000,000. The probably biggest news is the survival of 878783625 as the largest known partial quotient, seeming to signal the end of a pattern of "premature" high water marks, assuming the lg 1+t distribution for the tails of almost all reals. If pi is to maintain its reputation, the next high should be a doozy. (I dimly recall deriving that lg 1+t implies that, in an n term burst, the largest should be about e*n. Can anyone help here?) Neil also reports "Taking 9,701 seconds (161 minutes), the longest string of second difference zeroes is of length 18, starting at 173,533,907 and ending at 173,533,924." I.e., he has significantly increased the numerators and denominators necessary for any rational linear relationship of pi to e. Expect details shortly at http://nbickford.wordpress.com/ . Is there reason to expect the longest burst of vanishing second differences should have a difference interval of 1? Or did Neil neglect to mention that it was >1? --rwg
Bill Gosper: "The longest previous CF we can find remains Hans Havermann's 180M, wherein Neil found two glitches--a 'q' instead of a '1' at position 4,311,037 (a picked 2^6 bit) and a semicolon following term 100,000,000." I will take blame for the semicolon which (like the 100 terms per line) was fashioned into the Mathematica output.
Actually, it's only 458M terms. Sorry for the confusion... --Neil On Sun, Oct 24, 2010 at 5:17 AM, Hans Havermann <pxp@rogers.com> wrote:
Bill Gosper: "The longest previous CF we can find remains Hans Havermann's 180M, wherein Neil found two glitches--a 'q' instead of a '1' at position 4,311,037 (a picked 2^6 bit) and a semicolon following term 100,000,000."
I will take blame for the semicolon which (like the 100 terms per line) was fashioned into the Mathematica output.
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On Sun, Oct 24, 2010 at 1:24 AM, Bill Gosper <billgosper@gmail.com> wrote:
Bill Gosper:
Has anyone listed the largest terms in Hans's data?
Hans Havermann:
I have. :)
For his 13th birthday, funster Neil Bickford has just computed and verified 485M partial quotients (http://www.neilbickford.com/picf.htm). The longest previous CF we can find remains Hans Havermann's 180M, wherein Neil found two glitches--a "q" instead of a "1" at position 4,311,037 (a picked 2^6 bit) and a semicolon following term 100,000,000.
The probably biggest news is the survival of 878783625 as the largest known partial quotient, seeming to signal the end of a pattern of "premature" high water marks, assuming the lg 1+t distribution for the tails of almost all reals. If pi is to maintain its reputation, the next high should be a doozy. (I dimly recall deriving that lg 1+t implies that, in an n term burst, the largest should be about e*n. Can anyone help here?)
The Lord heps they that heps theyselves. The probability that a generic tail of a CF is > k is
(c43) p(k) = logb(2,1+1/k) 1 log(- + 1) k (d43) p(k) = ---------- log(2) Therefore, the probability that the max of n independent terms exceeds k is (c44) 1-(1-p(k))^n n (d44) 1 - (1 - p(k)) E.g., for three terms, (c45) makelist(float(1-(1-log(1/k+1)/log(2))^3),k,1,9) (d45) [1.0, 0.92851, 0.79984, 0.68824, 0.59974, 0.5298, 0.47375, 0.42806, 0.39021] The probability that the max of three *consecutive* terms exceeds k is (c46) p(k)+sum(p(j)-p(1/k+j),j,1,k-1)+sum(sum(p(1/(1/k+j)+i)-p(1/j+i),i,1,k-1),j,1,k-1) k - 1 ==== \ 1 (d46) p(k) + > (p(j) - p(- + j)) / k ==== j = 1 k - 1 k - 1 ==== ==== \ \ 1 1 + > > (p(----- + i) - p(- + i)) / / 1 j ==== ==== - + j j = 1 i = 1 k E.g., for 1<=k<=9: (c47) makelist(float(apply_nouns(apply_nouns(log(1/k+1)/log(2)+sum(log(1/j+1)/log(2)-log(1/(1/k+j)+1)/log(2),j,1,k-1)+sum(sum(log(1/(1/(1/k+j)+i)+1)/log(2)-log(1/(1/j+i)+1)/log(2),i,1,k-1),j,1,k-1)))),k,1,9) (d47) [1.0, 0.94111, 0.81825, 0.70548, 0.6144, 0.54199, 0.48389, 0.43657, 0.39741] So the (false) independence assumption seems to be within about 1%. This says that the prob. that the max of n terms exceeds k*n is about (c48) 1-%e^-(1/(log(2)*k)) 1 - -------- log(2) k (d48) 1 - %e For various landmarks in pi, (c49) makelist(sfloat(1-%e^-(1/(log(2)*k))),k,[3/1,7/2,15/3,292/5,20776/432,878783625/11505030,878783625/458585379]) (d49) [0.38177, 0.33781, 0.25064, 0.0244, 0.02955, 0.0187, 0.52898] so Neil had only a 53% chance to beat my record. (Which I had only a 1.9% chance of setting.) --rwg
participants (3)
-
Bill Gosper -
Hans Havermann -
Neil Bickford