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A289358
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The sequence a(n,m) of the m polynomial coefficients of the n-th order B-spline scaled by n!, read by rows, with n in {0,1,2,...} and m in {1,2,3,...,(n+1)^2}.
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2
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1, 1, 0, -1, 2, 1, 0, 0, -2, 6, -3, 1, -6, 9, 1, 0, 0, 0, -3, 12, -12, 4, 3, -24, 60, -44, -1, 12, -48, 64, 1, 0, 0, 0, 0, -4, 20, -30, 20, -5, 6, -60, 210, -300, 155, -4, 60, -330, 780, -655, 1, -20, 150, -500, 625, 1, 0, 0, 0, 0, 0, -5, 30, -60, 60, -30, 6, 10, -120, 540, -1140, 1170, -474, -10, 180, -1260, 4260
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OFFSET
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0,5
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COMMENTS
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The n-th order B-spline N_n(x) may be calculated with the expression
N_n(x) = (1/n!) Sum_{k=0..n+1} (-1)^k binomial(n+1,k) (x-k)^n step(x-k),
where
* n! is n factorial, which is defined as n! = n(n-1)(n-2)...(1),
* binomial(n,k) is the binomial coefficient, which can be defined as
binomial(n,k) = n!/((n-k)!k!),
* step(x) is the step function defined as step(x) = {1 for x >= 0
{0 otherwise.
From these definitions, it is apparent that the coefficients of the polynomials induced by n!*N_n(x) are integers and can be "flattened" (as in the Pascal triangle A007318) to form an integer sequence, part of which is listed above.
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REFERENCES
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Ole Christensen, Frames and bases: An Introductory Course, 2008, isbn13:9780817646776, page 142, Theorem 6.1.3.
Charles K. Chui, An Introduction to Wavelets, 1992, isbn13: 9780121745844, page 84, equation (4.1.12).
Daniel J. Greenhoe, Wavelet Structure and Design, 2013, isbn13: 9780983801139, page 318, Theorem H.1.
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LINKS
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Daniel J. Greenhoe, Wavelet Structure and Design, [version 1.20], January 2017, "Mathematical Structure and Design" series, volume 3, Theorem H.1, pages 267--268.
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FORMULA
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The n-th order B-spline N_n(x) may be calculated with the expression
N_n(x) = (1/n!) Sum_{k=0..n+1} (-1)^k binomial(n+1,k) (x-k)^n step(x-k).
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EXAMPLE
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The m=(n+1)^2 coefficients for the n-th order B-spline N_n(x) begin at the sequence index value p=Sum_{k=0..n}k^2=(1/6)n(n+1)(2n+1) and end at index value p+(n+1)^2-1.
Each set of m=(n+1)^2 coefficients for n=0,1,2,... can be written in the form of an (n+1)X(n+1) matrix A_n as...
for n=0 (index values 0 to 0):
A_0 = [1]
for n=1 (index values 1 to 4):
A_1 = [ 1 0]
[-1 2]
for n=2 (index values 5 to 13):
[ 1 0 0 ]
A_2 = [-2 6 -3 ]
[ 1 -6 9 ]
for n=3 (index values 14 to 29):
[ 1 0 0 0]
A_3 = [ -3 12 -12 4]
[ 3 -24 60 -44]
[ -1 12 -48 64]
That is, the sequence of integers induces a sequence of (n+1)X(n+1) square matrices (A_0, A_1, A_2, ...).
Taking the specific case of n=3, for example, the coefficients for N_3(x) begin at index value p=0+1+4+9=14 and end at index value p+4^2-1=29.
Using the coefficients from this range of indices yields the following expression for N_3(x):
[ 1 0 0 0 : for 0 <= x < 1] [x^3]
3!N_3(x) = [-3 12 -12 4 : for 1 <= x < 2] [x^2]
[ 3 -24 60 -44 : for 2 <= x < 3] [ x ]
[-1 12 -48 64 : for 3 <= x < 4] [ 1 ]
[ 0 0 0 0 : otherwise ]
{ x^3 :for 0 <= x < 1
{-3x^3 +12x^2 -12x + 4 :for 1 <= x < 2
= { 3x^3 -24x^2 +60x -44 :for 2 <= x < 3
{- x^3 +12x^2 -48x +64 :for 3 <= x < 4
{ 0 :otherwise
Note: Sum_{k=1..n}k^2 is called a "power sum".
For proof that p=Sum_{k=0..n}k^2=(1/6)n(n+1)(2n+1) (as stated above), see Appendix B of the Technical Report link.
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PROG
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(Maxima)
n:2;
Nnx:(1/n!)*sum((-1)^k*binomial(n+1, k)*(x-k)^n*unit_step(x-k), k, 0, n+1);
assume(x<=0); print(n!, "N(x)= ", expand(n!*Nnx), " for x<=0"); forget(x<=0);
for i:0 thru n step 1 do(
assume(x>i, x<(i+1)),
print(n!, "N(x)= ", expand(n!*Nnx), " for ", i, "<x<", i+1), forget(x>i, x<(i+1))
);
assume(x>(n+1)); print(n!, "N(x)= ", expand(n!*Nnx), " for x>", n+1); forget(x>(n+1));
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CROSSREFS
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KEYWORD
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sign,tabf
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AUTHOR
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STATUS
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approved
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