inf ==== \ N! N! 4 2 pi > ------ = - + --------- . / (2 N)! 3 9 SQRT(3) ==== N=0PROBLEM: Evaluate in closed form
inf ==== \ N! N! N! > -------- , which = / (3 N)! ==== N=0 1 / [ I (P + Q arccos (R)) dT , where ] / 0 2 3 2 (8 + 7 T - 7 T ) P = ------------------- and 2 3 2 (4 - T + T ) 2 2 3 4 (T - T ) (5 + T - T ) Q = ------------------------------------------ and 2 3 2 2 3 (4 - T + T ) SQRT((4 - T + T ) (1 - T)) 2 3 T - T R = 1 - ------- . 2
2 3 4 X X X gamma = - ln X + X - ---- + ---- - ---- ... + ERROR 2 2! 3 3! 4 4!Where ERROR is of the order of (e^-X)/X.
-Y Y E = X to get Y + Y X Y' - X Y' = 0.Substitute for Y a power series in X with coefficients to be determined. One observes the curious identity:
N ==== \ J - 1 N - J N 0 > J (N - J) BINOMIAL(N, J) = N (0 = 1) / ==== J=1and thus
==== N - 1 N \ N X Y(X) = > --------- / N! ==== N=1
2 ==== pi 1 1 \ 1 -- = -- + -- + ... = > -- ? 6 2 2 / 2 1 2 ==== N
(Abramowitz & Stegun, se. 3.6.27)
==== K K \ (-1) (delta A0) A0 - A1 + A2 - ... = > ----------------- / K+1 ==== 2 K ==== K \ K-m (where (DELTA A0) = > BINOMIAL(K, m) (-1) Am = Kth forward difference on A0) / ==== m=0when applied to
pi 1 1 -- = 1 - - + - - ... 4 3 5gives
==== N-1 2 pi \ 2 N! -- = > ---------- 4 / (2 N + 1)! ==== N=0Applied to the formula for gamma in Amer. Math. Monthly (vol. 76, #3, Mar69 p273) =
==== T \ (- 1) [LOG2(T)] > ---------------- ([ ] means integer part of) / T ==== T=0we get
inf K-1 ==== ==== \ -(K+1) \ 1 > 2 > ------------------ / / I ==== ==== BINOMIAL(2 + J, J) K=1 J=0(Gosper) which converges fast enough for a few hundred digits.
The array of reciprocals of the terms follows, with powers of 2
factored out to the left from all member of each row.
4 1
8 1 3
16 1 5 6
32 1 9 15 10
64 1 17 45 35 15
128 1 33 153 165 70 21
256 1 65 561 969 495 126 28
The next to left diagonal is 2^(N+1); the perpendicular one 3rd from
the right is 1, *9/1= 9, *10/2= 45, *11/3= 165, *12/4= 495.
1 1 1 1 4 1 1 11 11 1 1 26 66 26 1 1 57 302 302 57 1This bears an interesting relationship to Pascal's triangle. The 302 in the 4th southeast diagonal and the 3rd southwest one = 4*26 + 3*66. Note that rows then sum to factorials rather than powers of 2. If the nth row of the triangle is dotted with any n consecutive elements of (either) n+1st diagonal of Pascal's triangle, we get the nth Bernoulli polynomial: for n = 5, 1(6,i) + 26(6,i+1) + 66(6,i+2) + 26(6,i+3) + 1(6,i+4) = sum of 5th powers of 1 thru i+5, where (j,i) = BINOMIAL (j+i j).
inf ==== 1 \ PARITY(N) - > --------- 2 / N ==== 2 N=0where the parity of N is the sum the bits of N mod 2. The parity number's value is .4124540336401075977..., or, for hexadecimal freaks, .6996966996696996... . It can be written (base 2) in stages by taking the previous stage, complementing, and appending to the previous stage:
.0 .01 .0110 .01101001 .0110100110010110 .01101001100101101001... radix 2i.e.,
stage 0 = 0 N -2 1 - 2 - stage N stage N+1 = stage N + ----------------- . N 2 2If
NUM 0 = 0, DEN 0 = 2 NUM N+1 = ((NUM N)+1) * ((DEN N)-1) N+1 2 2 DEN N+1 = (DEN N) = 2then
N N NUM N+1 -2 -2 ------- = stage N+1 = (stage N + 2 ) * (1 - 2 ) . DEN N+1Or, faster, by substituting in the string at any stage:
Its regular continued fraction begins: 0 2 2 2 1 4 3 5 2 1 4 2 1 5 44 1 4 1 2 4 1 1 1 5 14 1 50 15 5 1 1 1 4 2 1 4 1 43 1 4 1 2 1 3 16 1 2 1 2 1 50 1 2 424 1 2 5 2 1 1 1 5 5 2 22 5 1 1 1 1274 3 5 2 1 1 1 4 1 1 15 154 7 2 1 2 2 1 2 1 1 50 1 4 1 2 867374 1 1 1 5 5 1 1 6 1 2 7 2 1650 23 3 1 1 1 2 5 3 84 1 1 1 1284 ... and seems to continue with sporadic large terms in suspicious patterns. A non-regular fraction is
1/(3 -1/(2 -1/(4 -3/(16 -15/(256 -255/(65536 -65535/ N N 2 2 (...2 -(2 -1)/(... .This fraction converges much more rapidly than the regular one, its Nth approximant being
1 + NUM N --------- , 1 + DEN Nwhich is, in fact, an approximant of the regular fraction, roughly the 2^N'th.
In addition, 4*(parity number) =
1 3 15 255 65535 2 - - * - * -- * --- * ----- * ... 2 4 16 256 65536This gives still another non-regular fraction per the product conversion item in the CONTINUED FRACTION section.
For another property of the parity number, see the spacefilling curve item in the TOPOLOGY section.
n === 2 \ z f(z) = > --- . / n === 2This is physically analogous to a series of clock hands placed end to end. The first hand rotates around the center (0,0) at some rate. the next hand is half as long and rotates around the end of the first hand at twice this rate. The third hand rotates around the end of the second at four times this rate; etc. It would seem that the end of the "last" hand (really there are infinitely many) would sweep through space very fast, tracing out an (infinitely) long curve in the time the first hand rotates once. The hands shrink, however, because of the 2^n in the denominator. Thus it is unclear whether the curve's arc length is really infinite.
Also, it is a visually interesting curve, as are
=== n! === FIB(n) \ z \ z f(z) = > --- and f(z) = > ------- . / n! / FIB(n) === ===Gosper has programmed the one mentioned first, which makes an intriguing display pattern. see following illustrations. If you write a program to display this, be sure to allow easy changing of:
inf === n! \ z f(z) = > --- . / n! === n=1Figure 6(b). Image of circles |z| = 1/8, 2/8, ..., 8/8 under the function
inf n === 2 \ z f(z) = > --- . / n === 2 n=0Both [original] plots by Salamin on the RLE PDP-1.
==== ==== ==== \ 1 \ 1 1 \ 1 1 > -- = > (-- - ------) + > (----- - -----) = / 2 / 2 2 1 / 1 1 ==== N ==== N N - - ==== N - - N + - 4 2 2 ==== \ 1 2 - > ------------- . / 2 2 ==== N (4 N - 1)Take the last sum and re-apply this transformation. This may be a winner for computing the original sum. For example, the next iteration gives
==== 31 \ 1 -- - 9 > -------------------------------- 18 / 2 2 4 2 ==== N (4 N - 1) (25 N + 5 N + 9)where the denominator also =
2 2 2 N (2 N - 1) (2 N + 1) (5 N - 5 N + 3) (5 N + 5 N + 3)
Reference: Polya, Mathematics and Plausible Reasoning, volume 2, page 46.