And, as Neil empiricizes, Sum[2^k/(1 + z^2^k), {k, 0, Infinity}] == 1/(z-1)
1. Why did Julian change his name? 2. do I understand your "it telescopes" correctly as f[z_,n_]:=Sum[2^k/(1 + z^2^k), {k, 0, n}] implies f[z, n+1] == 1/(1 + z) + 2*f[z^2, n] (1) 3. it is easy to show that 1/(z-1) satisfies (1) for n->inf and thus dropped, but how do you generaly solve a recursion like that? Look at it until the solution occurs to you? Wouter. -----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Bill Gosper Sent: vrijdag 1 maart 2013 7:29 To: math-fun@mailman.xmission.com Subject: Re: [math-fun] this HAS to be known Even Mma knows, e.g., In[502]:= Product[1 + a^2^k, {k, -3, Infinity}] Out[502]= 1/(1 - a^(1/8)) (Binary notation principle. Strange that it didn't demand |a|<1.) Easy from here is Product[(1 + b^2^k)/(1 + a^2^k), {k, -oo, oo}] == Subscript[Log,b][a] |a|,|b| <1, a handy but mediocre (1 bit/term) way to compute logs to any base. (Log[1/b,a]==-Log[a]/Log[b].) I originally had the incorrect form Product[(1 + (z^2^k)^b)/(1 + (z^2^k)^a), {k, -oo, oo}] =?= a/b , |z|<1, which is wrong, and even varies with z, unless 0<=z<1, or a=b, or a and b are both integers. (At least I couldn't find any noninteger, unequal a and b that worked. Nor any integer a and b that failed.) --rwg On Wed, Feb 27, 2013 at 3:39 PM, Bill Gosper <billgosper@gmail.com> wrote:
ARGH! Neil proved it telescopes. Then I remembered gosper.org/stanfordn2.pdf page "10" (=4) Never mind. --rwg
On Wed, Feb 27, 2013 at 3:05 PM, Bill Gosper <billgosper@gmail.com> wrote:
Sum[2^k/(1 + z^2^k), {k, -Infinity, Infinity}] == 1/Log[z] Can somebody tell me where? --rwg And, as Neil empiricizes, Sum[2^k/(1 + z^2^k), {k, 0, Infinity}] == 1/(z-1)
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