Re: [math-fun] (a1 + a2 + ...)^2 = a1^2 + a2^2 + ... = Pi^2/8
(*) (a_1 + a_2 + . . . )^2 = a_1^2 + a_2^2 + . . . is indeed a fascinating equation to ponder. Assume hereon that a_i -> 0 as i -> oo. Note that since sum (a_i) may not be absolutely convergent (or perhaps *must* not?), the LHS may (must?) depend on the order in which its terms are summed. Maybe the intelligentest order is to do (**) sum {n = 2 to oo} of (sum {i+j=n} of a_i*a_j). The truth of (*) <=> the cross terms (sum {i <>j} a_i*a_j) = 0. Of course, if the cross terms are NOT absolutely convergent, then there are always some (inf'ly many) orderings of them that sum to 0. (Or any other chosen sum.) So unless we make some assumption about the ordering of (a_1 + a_2 + . . .)^2, (*) is not that interesting; I propose (**). What would be really cool to see would be a *geometrical* proof of sum (1/n^2) = pi^2 / 6. That is, some object of volume pi^2 / 6 that can be partitioned into pieces of volume 1/n^2. P.S. Also note that 1 + 1/3 - 1/5 - 1/7 + 1/9 + 1/11 - 1/13 . . . . = integral {0 to 1} (1+x^2) / (1+x^4) dx --Dan
----- Original Message ----- From: <dasimov@earthlink.net> To: "math-fun" <math-fun@mailman.xmission.com> Sent: Wednesday, February 22, 2006 19:27 Subject: Re: [math-fun] (a1 + a2 + ...)^2 = a1^2 + a2^2 + ... = Pi^2/8
What would be really cool to see would be a *geometrical* proof of sum (1/n^2) = pi^2 / 6. That is, some object of volume pi^2 / 6 that can be partitioned into pieces of volume 1/n^2.
You might be interested in <http://www.pisquaredoversix.force9.co.uk/Tiling.htm>. BTW, before Clive found that, I had found another conjectured packing, using an algorithm slightly different from his. If interested, see <http://groups.google.com/group/sci.math/msg/d92d4a0ea46b4762> and other messages in that thread and its parent. Unfortunately this cannot (obviously) be considered to "be a *geometrical* proof of sum (1/n^2) = pi^2 / 6". Cheers, David
* David W. Cantrell <DWCantrell@sigmaxi.org> [Feb 22. 2006 22:46]:
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You might be interested in <http://www.pisquaredoversix.force9.co.uk/Tiling.htm>. BTW, before Clive found that, I had found another conjectured packing, using an algorithm slightly different from his. If interested, see <http://groups.google.com/group/sci.math/msg/d92d4a0ea46b4762> and other messages in that thread and its parent.
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Going along the top side: Pi^2/6-(1/1+1/2+1/7) == 0.00207692399108358 How does the series continue? Similarly, half way down: Pi^2/6-(1/1+1/3+1/4+1/17) == 0.00277720410312840 -- p=2^q-1 prime <== q>2, cosh(2^(q-2)*log(2+sqrt(3)))%p=0 Life is hard and then you die.
Joerg Arndt <jj@suse.de> [Feb 25, 2006 12:05]:
* David W. Cantrell <DWCantrell@sigmaxi.org> [Feb 22. 2006 22:46]:
[...]
You might be interested in <http://www.pisquaredoversix.force9.co.uk/Tiling.htm>. BTW, before Clive found that, I had found another conjectured packing, using an algorithm slightly different from his. If interested, see <http://groups.google.com/group/sci.math/msg/d92d4a0ea46b4762> and other messages in that thread and its parent.
[...]
Going along the top side: Pi^2/6-(1/1+1/2+1/7) == 0.00207692399108358 How does the series continue?
You're asking me about Clive's packing. Offhand, I don't know how that series continues. Ask Clive, or implement his algorithm and see for yourself how it continues.
Similarly, half way down: Pi^2/6-(1/1+1/3+1/4+1/17) == 0.00277720410312840
But, if you wish, I can fairly easily answer similar questions about the packing produced by my simple as-far-to-the-left-and-down-as-possible algorithm. Here's a picture of squares of side lengths 1, 1/2, 1/3,..., 1/99 packed in a rectangle 1 by pi^2/6. The bounding rectangle is shown as a dashed line. <http://img518.imageshack.us/img518/1297/990ng.gif> And below is a detail of the lower right corner after squares down to a side length 1/999 have been packed. (BTW, to the immediate left of the square of side length 1/67 is that of side length 1/8, visible in the previous figure.) <http://img119.imageshack.us/img119/2027/999lowerright9up.gif> So, at the bottom of the packing produced by my algorithm, we have, thus far, a gap of width Pi^2/6 - (1/1 + 1/2 + 1/8 + 1/67 + 1/293 + 1/993) = 0.0005886+. David
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dasimov＠earthlink.net -
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