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 ln(x+2) S = -0.1379330125 ln(x+3) -0.07070756527 x^j 0 for any fixed integer j>=0 1/(x+1) 0.398761088108 1/(x+2) 0.1049709156499 sqrt(x) -0.63407426 1/sqrt(x+1) 0.1983140804979
--reminiscent of "Flanelle's theorem" Sum[ n^j * h(n), for n=0..2^k-1 ] = 0 if 0<=j<k with j=integer (and perhaps Adam P. Goucher will inform us... who the helle was Flanelle? I think it was actually Adam P. Goucher in disguise?) I find now a new closed form sum Sum[ C^n * h(n), for n=0..2^k-1 ] = (1-C)*(1-C^2)*(1-C^4)*...*(1-C^(2^(k-1))) To prove it, you work inductively layer by layer in the "Wilson binary tree" starting from leaves. At the leaf layer, we have C^n. At the next layer, fathers of leaves, we have by differencing (1-C) * C^(2*n). At the next layer, grandfathers of leaves, we have (1-C)*(1-C^2) * C^(4*n). And so on... at the the layer k+1 above the leaves we have (1-C) * (1-C^2) * (1-C^4) * ... * (1-C^(2^k)) * C^(2^(k+1) * n) ...and now evaluate this when n=0 at the layer k-1 above leaves, QED. You can also prove Flanelle by the same kind of layer-by-layer inductive approach, noting that differencing a polynomial of degree d, yields 0 after you do it more than d times. Note that our new sum also can be viewed as a "generating function(C)" for h(n). Well, isn't that nifty. Looking around for another nail to hit with my shiny new hammer, consider Sum[ h(n)*C^n/n!, for n=0..2^k-1 ] = ?? work inductively layer by layer in the "Wilson binary tree" starting from leaves. At the leaf layer, we have C^n / n!. At the next layer, fathers of leaves, we have by differencing (1-C/(2*n+1)) * C^(2*n) / (2*n)! Oh dear, this is getting yukky. To try to keep it simple, let's do C=1. Sum[ h(n)/n!, for n=0..2^k-1 ] = ?? At the leaf layer, we have 1 / n!. At the next layer, fathers of leaves, we have by differencing (1-1/(2*n+1)) / (2*n)! = 1/(2*n)! - 1/(2*n+1)! At the next layer, grandfathers of leaves, we have ...well... this one can be done. Actually the general C one can be done. You just have to stick it out, which I haven't. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)