Re: [math-fun] Maple 11
----- Original Message ---- From: Fred lunnon <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Monday, January 7, 2008 11:47:53 AM Subject: Re: [math-fun] xkcd points out dangers of math fun On 1/7/08, Eugene Salamin <gene_salamin@yahoo.com> wrote:
... Maple 11 can do these. I will say that Maple 11 is a winner, while Maple 9 is a loser. ...
That's useful to know. Maple 10 was less appallingly flaky than Maple 9, but their @£$% licensing software kept clobbering it. I might try upgrading again, I suppose (groan!)
The denominators in the diagonal terms, R[p,p] seem to fit the pattern 1*3*5*...*(2p -1).
As I remarked earlier (attachment Jan 3rd), nobody seems previously to have noticed that u^{kk} = ( 1 + 1/3 + 1/5 + ... + 1/(2k-1) ) 2/\pi --- all the remaining values follow from these and u^{01} = 1/2, via \del^2 u^{kl} = 0. WFL _______________________________________________ Check maplesoftsoft.com. See if you can qualify for the student edition. It's identical to the professional edition, but costs only $160 or so. When I bought mine, I had to provide a school and student ID number, which I could do because I'm a distance student at U of Arizona. Do you have a proof for the value of u^{kk} ? Gene ____________________________________________________________________________________ Be a better friend, newshound, and know-it-all with Yahoo! Mobile. Try it now. http://mobile.yahoo.com/;_ylt=Ahu06i62sR8HDtDypao8Wcj9tAcJ
On 1/7/08, Eugene Salamin <gene_salamin@yahoo.com> wrote:
Do you have a proof for the value of u^{kk} ?
Cserti's paper equation (32) quotes a recurrence equivalent to (2*k+1) u^{k+!,k+1} - 2(2*k) u^{k,k} + (2*k-1) u^{k-1,k-1} = 0 from which u^{kk} = ( 1 + 1/3 + 1/5 + ... + 1/(2k-1) ) 2/\pi follows easily, given initial values u^{00} = 0, u^{11} = 2/\pi. He credits this to T. Morita, J. Math. Phys. 12, 1744–1747 (1971), but gives no proof. WFL
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Eugene Salamin -
Fred lunnon