On 7/6/14, Warren D Smith <warren.wds@gmail.com> wrote:
h(n) = (-1)^(parity of bit-sum of binary representation of n)
f(x) S=SUM[ f(n) * h(n), for n=0..2^k-1 with k large] ln(x+1) S=-log(2)/2 x^j 0 for any fixed integer j>=0
--Actually, it occurs to me that a slightly better way to prove "Flanelle's theorem" is to prove it using as your basis for polynomials, not the powers x^j, but rather the pseudo-powers x^^j = (x+j-1)! / (x-1)! which will be my ASCII notation for the "rising factorial" aka "Pochammer symbol" which if I had typesetting capability, I would write "x with superscript j, where the j has an overline" as opposed to the falling factorial x! / (x-j)! which I would write "x with superscript j, where the j has an underline" because those are far superior notations to the stupid ones you find in way too many putrid books written by people who like to confuse you with bad notations from hell. Also x^^1 = x, x^^0 = 1. The forward difference of x^^j is (x+1)^^j - x^^j = j * x^^(j-1) if j>0 which is the point -- powers are friendly with differentiation, but rising factorials are friendly with forward differencing. Indeed (Mx+1)^^j - (Mx)^^j = j * (Mx)^^(j-1). Right-o. So proceeding, we evaluate the Wilsonian sum Sum[ n^^j * h(n), for n=0..2^k-1 ] via the layer-by-layer inductive tree approach. At leaf layer we have n^^j. At the father-of leaves layer: -j * (2*n)^^(j-1). At grandfather of leaves layer: (-j) * (1-j) * (4*n)^^(j-2). And so on, at the layer k-1 above leaves we have (-j) * (1-j) * (2-j) * ... * (k-2-j) * (2^(k-1)*n)^^(j-k+1) which provides us with a closed form for the Wilsonian sum of a rising factorial (if power-of-two summands) which is a more powerful statement than Flanelle's theorem (which only told us the answer was 0 is you went too far, but did not tell us what happened if you went less far). -- Furthermore, in my previous evaluation of the Wilsonian sum of C^n, we can differentiate the resulting formula z times with respect to c, in this way evaluating in closed form, the Wilsonian sum of n^z*C^n. -- Now, returning to WIlson's original problem about log(2). Can we assault this problem with our new tools? Specifically, can the log function be written as a series involving rising factorials, or something? -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)