Positivity of the Fourier Transform of the Shortest Maximal Order Convolution Mask for Cardinal B-splines

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1 Positivity of the Fourier Transform of the Shortest Maximal Order Convolution Mask for Cardinal B-splines Markus Sprecher December 8, 08 In [] approximations of functions into manifolds were studied. For the transformation of function values to B-spline coefficients convolution masks were considered. Some of the proofs required that the convolution mask had a positive Fourier transform. This property was used to show that the inverse of the convolution exists and that the spatial depency decays exponentially. For splines of degree m the existence of such convolution masks of length m / were constructively proven. It was posed as an open question if families with shorter sequences could satisfy this property. For m it was computationally verified that the shortest possible sequence that satisfies the polynomial reproduction property also has a positive Fourier transform. This sequence has length m. It was conjectured that this holds true for all odd m > 0. In Section of this work we will prove this fact. In Section we describe how the convolution masks can be computed. Theory We start by defining cardinal B-splines Definition. Cardinal B-splines can recursively be defined by B 0 = [, ] and B m = B m B 0 for all m where [, ] denotes the indicator function on the interval [, ] and denotes the convolution. Next we consider the polynomial reproduction property i Z B m x i)pi j)λ j = px) for all x R )

2 and for all polynomials p of degree m. As in [] we will use the functions N m, Λ: C C defined by N m z) = m )/ j= m )/ B m j)z j, Λz) = λ j z j. ) We will now proof an equivalent formulation of the polynomial reproduction property. Lemma. The polynomial reproduction property ) is equivalent to Proof. We have d l d l dz l N mz)λz)) z= = { l = 0 0 l {,..., m}. 3) dz l N mz)λz)) z= 4) = dl m )/ dz l B m k)z k λ j z j z= k= m )/ 5) = dl S+m )/ dz l B m k j)λ j z k z= k= S m )/ 6) = S+m )/ B m k j)λ j k... k l ))z k l z= 7) = k= S m )/ S+m )/ k= S m )/ B m k j)λ j p l k) 8) where p 0 =, p l x) = x) x )... x l )). If the polynomial reproduction property ) holds Expression 8) is equal to { l = 0 p l 0) = 0 l {,..., m}. On the other hand if 3) holds the polynomial reproduction property ) holds for x = 0 and p = p l for all l {0,..., m}. As the polynomials p 0,..., p m build a basis for the space of polynomials of degree m the polynomial reproduction property ) holds for all polynomials p of degree m at x = 0. By replacing x resp. i by x + resp. i + it follows that the polynomial reproduction property holds at all integer points. Now consider the function fx) = i Z B m x i)pi j)λ j.

3 Since B mx) = B m x + ) B m x ) = δb m x ) where δ is the discrete difference operator δux) = ux + ) ux) it follows inductively that B m) m x) = δ m B 0 x m ). Since δ commutes with the convolution it follows that f m) x) = i Z = i Z = i Z B m) m x i)pi j)λ j 9) δ m B 0 x i m ) pi j)λ j 0) B 0 x i m ) δ m pi j)λ j ) Note that every time δ is applied the polynomial degree decreases by. Hence δ m pi j) and therefore f m) is constant. It follows that f is a polynomial and since p coincides with f on all integers that f = p. Hence the polynomial reproduction property holds for all polynomials of degree m and all x R. We can now formulate and proof our theorem. Theorem. Let k 0 be an nonnegative integer. Then there exist a unique symmetric i.e. λ i = λ i ) sequence λ j ) k j= k R that satisfies the polynomial reproduction property ) for m = k +. Furthermore we have k j= k λ j e πijω > 0 for all ω R. Proof. Since λ and B m are symmetric, i.e. λ i = λ i and B m x) = B m x), both N m z) and Λz) can be written as polynomials of degree k in ) ) x = z+z z ) z ) z ) =, i.e. N m z) = p, Λz) = q z z z for polynomials p, q of degree k. Condition 3) is by Lemma equivalent to px)qx) = + x k+... ), ) i.e. that the constant coefficient of px)qx) is one and coefficients of order up to order k are zero. We first prove uniqueness of q and therefore of λ j ) k j= k R. Let q, q be two polynomials of degree k satisfying ). Then it follows that px)q x) q x)) = x k+... ) and since p0) = 0 that q x) q x) = x k+... ) and since q and q are polynomials of degree k that q = q. The polynomial N m z)z m )/ m! is known as the Eulerian polynomial. By [] the Eulerian polynomial and therefore N m has only negative and simple 3

4 real roots. If z is a root of N m then x i = z i + z i 4 is a root of p. Furthermore all k roots x,..., x k of p can be constructed in this way. Therefore the roots of p are all smaller or equal to 4. Note that for x < 4 we have px) = ) = k i= x x i k x j i= j=0 x j i Define q by the truncating this power series at order k +, i.e. such that Then we have px)qx) = px) px) = qx) + xk+... ) ) px) + xk+... ) = + x k+... ),. which shows that ) is satisfied, i.e. the sequence λ j ) k j= k q satisfies the polynomial reproduction property. The statement for the Fourier transform is equivalent to corresponding to Λz) for all z = which is equivalent to qx) for all 4 x 0. Since all roots x,..., x k of p are negative all terms xj px) px) x j i in the power series of are positive for x 4, 0). Therefore also all terms of the power series of and therefore also all polynomial terms of qx) are positive and since the zero-order term is one we have qx). The special cases x = 4 and x = 0 follow from continuity. Construction In this section, we describe how to construct the coefficients. It is tedious even for small m to compute the roots of the polynomial p and the power series of /p used in the proof in the previous section. To get the coefficients it is easier to determine the polynomial coefficients the polynomial q recursively. To compute the coefficients of the N m, i.e. the values of the B-splines at integers de Boors 3-point recursion can be used which for our case of uniform grids reads m+ + x ) B m x + B m x) = ) + m+ x ) B m x ). 3) m 4

5 . The cases m = 3 and m = 5 For m = 3 we have N m z) = 6 z z = + 6 px) = + x 6 px) z + z ) 4) 5) = x 6 + x 6 ) 6) qx) = 6 x 7) Λz) = z + z ) = 6 6 z z 8) λ j ) k j= k = 6, 4 3, ) 9) 6 For m = 5 we have N m z) = 0 z z z + 0 z 0) = + z + z ) + z + z ) ) 4 0 px) = + x 4 + x 0 ) qx) = x 4 + 3x 40 3) Λz) = z + z ) + 3 z + z ) ) = 3 40 z 7 5 z z z 5) λ j ) k 3 j= k = 40, 7 5, 73 40, 7 5, 3 ) 6) 40. Code to construct convolution mask for arbitrary m Below an octave/matlab code for computing the convolution mask. function lambda = conv_maskm) assertmodm,)==, m must be an odd positive integer ) k=m-)/; %compute B-spline at integer by de Boor three point recursion b=; for j=:m x=j-)/*linspace-,,j); 5

6 b=j+)/+x).*[b 0]+j+)/-x).*[0 b])/j; %coefficient wrt z of powers of x=/z+z- powx=cell,k); powx{}=[ - ]; for j=:k powx{j}=convpowx{},powx{j-}); %determine recursively coeeficients of px) cp=zeros,k+); for j=:k cpj)=b)/powx{k+-j}); b=b-b)*powx{k+-j}; %remove first and last coefficient which is zero b[ ])=[]; cpk+)=b); %determine coefficients of q cq=zeros,k+); cqk+)=; for j=k:-: cqj)=-sumcqj+:).*cp-:-:j)); %determine coefficients of lambdaz) lambda=cq); for j=:k lambda=[0 lambda 0]+cq-j)*powx{j}; References [] G. Frobenius. Über die Bernoullischen Zahlen und die Eulerschen Polynome, 90. [] P. Grohs and M. Sprecher. Projection-based quasiinterpolation in manifolds. Technical Report 03-3, Seminar for Applied Mathematics, ETH Zürich, Switzerland, 03. 6