[math-fun] this HAS to be known
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)
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)
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)
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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Both of these are incredibly beautiful identities, especially if true. Which, given their source, seems very likely. If anyone has proved this already, my guess is it would be Euler. Have his complete works been collected yet, and even better, are they freely available in English translation on the Web? --Dan On 2013-02-27, at 3:05 PM, Bill Gosper 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) _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
On 2013-02-27, at 3:05 PM, Bill Gosper 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)
This reminds me of something Neil Sloane mentions in his OEIS talks, namely that the function f(n) = floor(2n/log(2)) has the same value as the function g(n) = ceiling(2/(2^(1/n)-1)) (Sloane's sequence A78608) for all integer n from 1 to 77451915729367, but differs at n=77451915729368 (and at some other larger values, see oeis.org/A129935). -- Robert Munafo -- mrob.com Follow me at: gplus.to/mrob - fb.com/mrob27 - twitter.com/mrob_27 - mrob27.wordpress.com - youtube.com/user/mrob143 - rilybot.blogspot.com
On 3/1/2013 2:17 PM, Dan Asimov wrote:
Both of these are incredibly beautiful identities, especially if true. Which, given their source, seems very likely.
If anyone has proved this already, my guess is it would be Euler.
Have his complete works been collected yet, and even better, are they freely available in English translation on the Web?
In the year of Euler's bicentennial, 1907, the Swiss Academy of Sciences formed the Euler Commission (or "Committee", their own translation varies) to publish his complete works ("Opera Omnia" in Latin). The first volume appeared in 1911, and in the next century they managed to gather nearly all of Euler's published books and articles into 72 large volumes. Two final volumes--on perturbation theory--have been announced for this year. All of these are available under Springer's Birkhauser imprint at prices intended for well-endowed research libraries. They are still only halfway through publishing a planned nine volumes of Euler's correspondence. (The Euler-Goldbach volume has been announced for 2010, 2011 and, currently, 2013.) Then there will be four or five volumes worth of previously unpublished manuscripts, notes and diaries. The most extensive online resource is the MAA's Euler Archive: http://eulerarchive.maa.org/. It currently contains 866 books, articles and selected letters. Most are available only as PDF scans of varying quality, but 164 have been translated into English. Volunteers are solicited for additional translations; although most of the items are in Latin, some are in French, German and even English. Some Euler translations are in the arXiv; I was amused to see an article (http://arxiv.org/abs/math/0412062) in which Euler shows that, contrary to a published table, 1000009 is composite. He does so by showing that it is a sum of two squares in two different ways: 1000^2 + 3^2 = 972^2 + 235^2, so its primality would contradict the uniqueness statement in Fermat's two-squares theorem. When I was in high school, I discovered the same idea and showed that 11041 was composite: 105^2 + 4^2 = 104^2 + 15^2. It then took me quite a few years to realize that the two expressions can be used to produce the actual factorization. Euler, however, sees the trick immediately. -- Fred W. Helenius fredh@ix.netcom.com
participants (5)
-
Bill Gosper -
Dan Asimov -
Fred W. Helenius -
Meeussen Wouter (bkarnd) -
Robert Munafo