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A139146
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Interpolation one-half polynomials based on Chebyshev T(x.n) polynomial coefficients(A053120 ): even-> 2*T(x,n); odd->T(x,n) + T(x,n+1).
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0
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2, 1, 1, 0, 2, -1, 1, 2, -2, 0, 4, -1, -3, 2, 4, 0, -6, 0, 8, 1, -3, -8, 4, 8, 2, 0, -16, 0, 16, 1, 5, -8, -20, 8, 16, 0, 10, 0, -40, 0, 32
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OFFSET
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1,1
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COMMENTS
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Row sums are all 2.
The rationale behind this interpolation is that Bessel functions have 1/2 values, so what about other orthogonal polynomials?
The integration shows that they are "mostly" orthogonal when three away from the diagonal.
TableForm[Table[Integrate[p[x, m]*p[x, n]/Sqrt[1 - x^2], {x, -1, 1}], {n, 0, 10}, {m, 0, 10}]]
These polynomials would also be related to two-dimensional Chladni-Chebyshev type standing waves as:
Chladni[x,y,n,m]=ChebyshevT[n, x] + ChebyshevT[m, y].
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LINKS
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FORMULA
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even->p(x.m)= 2*T(x,n); odd->p(x,m)=T(x,n)+T(x,n+1); out_n,m=Coefficients(p(x,m).
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EXAMPLE
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{2},
{1, 1},
{0, 2},
{-1, 1, 2},
{-2, 0, 4},
{-1, -3, 2, 4},
{0, -6, 0, 8},
{1, -3, -8, 4,8},
{2, 0, -16, 0, 16},
{1, 5, -8, -20, 8, 16},
{0, 10, 0, -40, 0, 32}
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MATHEMATICA
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Clear[p, x] p[x, 0] = 2*ChebyshevT[0, x]; p[x, 1] = ChebyshevT[0, x] + ChebyshevT[1, x]; p[x, 2] = 2*ChebyshevT[1, x]; p[x_, m_] := p[x, m] = If[Mod[m, 2] == 0, 2*ChebyshevT[Floor[m/2], x], ChebyshevT[Floor[m/2], x] + ChebyshevT[Floor[m/2 + 1], x]]; Table[ExpandAll[p[x, n]], {n, 0, 10}]; a = Table[CoefficientList[p[x, n], x], {n, 0, 10}]; Flatten[a] Table[Apply[Plus, CoefficientList[p[x, n], x]], {n, 0, 10}]
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CROSSREFS
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KEYWORD
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uned,tabf,sign
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AUTHOR
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STATUS
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approved
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