[seqfan] Re: israel at math.ubc.ca
israel at math.ubc.ca
israel at math.ubc.ca
Tue Jun 26 08:43:17 CEST 2012
The roots of U_n(z) are cos(k pi/(n+1)), k=1..n.
Robert Israel
University of British Columbia
On Jun 25 2012, David Wilson wrote:
>On 6/24/2012 9:33 PM, israel at math.ubc.ca wrote:
>> The formula seems to be correct. It arises as follows. Let A be the 64
>> x 64 adjacency matrix of a chessboard, with edges corresponding to
>> king-moves. Then A025597(n) is the matrix element of A^n corresponding
>> to two opposite corners of the board, and thus a certain linear
>> combination of the n'th powers of the eigenvalues of A. Now A + I = (B
>> + I) tensor (B + I) where B is the adjacency matrix of [1,2,...,8]
>> with nearest-neighbour edges. The characteristic polynomial of B is
>> t^8-7*t^6+15*t^4-10*t^2+1, which is U_8(t/2) where U_8 is the 8'th
>> Chebyshev polynomial of the second kind.
>>
>> Robert Israel
>> University of British Columbia
>
>Clearly I'm not at your level of mathematical understanding.
>
>The Chebyshev form of this polynomial implies something nice about the
>eigenvalues?
>
>
>
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