It's equivalent to showing that the limit of the series: Sum[(-1)^h(n) (log(2n + 1) - log(2n + 2))] (where n ranges from 0 to infinity) is -log(2)/2, where h(n) is the Hamming weight of n. Interestingly, there is a similar result on polynomials (known as Flanelle's theorem, and rather easy to prove): Sum[(-1)^h(n) p(n)] (where n ranges from 0 to 2^k - 1) is 0, for any polynomial p(n) of degree at most k-1. Can we use Flanelle's theorem together with Weierstrass approximation to prove anything? Sincerely, Adam P. Goucher
Sent: Saturday, July 05, 2014 at 6:42 PM From: "David Wilson" <davidwwilson@comcast.net> To: "'Wouter Meeussen'" <wouter.meeussen@telenet.be>, 'math-fun' <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Problem I never solved
Well, yes, I know the answer is sqrt(2)/2.
What I don't know is how to prove it.
-----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Wouter Meeussen Sent: Saturday, July 05, 2014 12:48 PM To: math-fun Subject: Re: [math-fun] Problem I never solved
nice!
prod( k=1..m, A026147(k) ) / prod( k=1..m, A181155(k) ) approaches sqrt( 2 ) as m-> infinity
goes like 1, 1/2, 2/3, 7/10, 286/405, 144305/204102, 276620298878/391202754597, ...
(* those who abhor Mathematice lines, please look away : *)
k=1/2;NestList[#/(#/. i->i+(k=2k))&,x[i],6] %/. x[i_]->i %/. i->1 %//N
Wouter.
-----Original Message----- From: David Wilson Sent: Saturday, July 05, 2014 5:34 PM To: math-fun Subject: [math-fun] Problem I never solved
I know the answer to this, but not how to derive it:
What is the value of
(((1/2)/(3/4))/((5/6)/(7/8)))/...
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