Reproducing Polynomials with B-Splines

A B-Spline of order $N$ is known to be able to reproduce any polynomial up to order $N$:

$\sum _{n\in \mathbb {Z} }c_{m,n}\beta _{N}(t-n)=t^{m}$

In words, a proper linear combination of shifted versions of a B-Spline can reproduce any polynomial up to order $N$. This is needed for different applications, for example, for the Sampling at Finite Rate of Innovation (FRI) framework. In this case any kernel $\varphi$ reproducing polynomials (that is, satisfying the Strang-Fix conditions) can be used. However, among all possible kernels, the B-Splines have the smallest possible support.

An important question is how to obtain the coefficients Fehler beim Parsen (MathML mit SVG- oder PNG-Rückgriff (empfohlen für moderne Browser und Barrierefreiheitswerkzeuge): Ungültige Antwort („Math extension cannot connect to Restbase.“) von Server „https://wikimedia.org/api/rest_v1/“:): {\displaystyle c_{m,n}} for the reproduction-formula. In this small article, I describe one way.

Starting from

$\sum _{n\in \mathbb {Z} }c_{m,n}\varphi (t-n)=t^{m}$

the coefficients can be obtained using the dual of $\varphi$, ${\tilde {\varphi }}$ (I set $\beta _{N}=\varphi$ for consistency with my notes):

$c_{m,n}=\int _{-\infty }^{\infty }t^{m}{\tilde {\varphi }}(t-n)\,dt$

However, even if the dual would be known, solving the infinite integral is only feasible when the dual has finite support. This is the case with the B-Spline itself but not with its dual!

A closer look at the formula tells that this is nothing more than a convolution (under the assumption that ${\tilde {\varphi }}$ is symmetric which is the case):

Fehler beim Parsen (MathML mit SVG- oder PNG-Rückgriff (empfohlen für moderne Browser und Barrierefreiheitswerkzeuge): Ungültige Antwort („Math extension cannot connect to Restbase.“) von Server „https://wikimedia.org/api/rest_v1/“:): {\displaystyle c_{m,n} = \int t^m \tilde{\varphi}(-(n-t))\,dt = \int t^m \tilde{\varphi}(n-t)\,dt = (t^m * \tilde{\varphi})(n) }

Now, this can be transformed to fourier domain:

$(t^{m}*{\tilde {\varphi }})(n)={\mathcal {F}}^{-1}\left\{{\mathcal {F}}\left\{t^{m}\right\}{\tilde {\Phi }}(\omega )\right\}={\mathcal {F}}^{-1}\left\{j^{m}{\sqrt {2\pi }}\delta ^{(n)}(\omega ){\tilde {\Phi }}(\omega )\right\}=j^{m}{\sqrt {2\pi }}{\mathcal {F}}^{-1}\left\{\delta ^{(n)}(\omega ){\tilde {\Phi }}(\omega )\right\}$

Writing the inverse of this expression yields:

$j^{m}{\sqrt {2\pi }}{\frac {1}{\sqrt {2\pi }}}\int _{-\pi }^{\pi }\delta ^{(n)}(\omega ){\tilde {\Phi }}(\omega )e^{j\omega n}\,d\omega =j^{m}\int _{-\infty }^{\infty }\delta ^{(n)}(\omega )\underbrace {{\tilde {\Phi }}(\omega )e^{j\omega n}} _{f(\omega )}\,d\omega$

It is known that:

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so that

Fehler beim Parsen (MathML mit SVG- oder PNG-Rückgriff (empfohlen für moderne Browser und Barrierefreiheitswerkzeuge): Ungültige Antwort („Math extension cannot connect to Restbase.“) von Server „https://wikimedia.org/api/rest_v1/“:): {\displaystyle j^m \int_{-\infty}^{\infty} \delta^{(n)}(\omega) f(\omega)\,d\omega = j^m (-1)^m \left. \frac{\partial^m}{\partial \omega^m} f(\omega) \right|_{\omega = 0} }

Now the whole procedure has been reduced to calculate the derivative of $f(\omega )$ and set the result to zero.

An open question is how to obtain the dual of Fehler beim Parsen (MathML mit SVG- oder PNG-Rückgriff (empfohlen für moderne Browser und Barrierefreiheitswerkzeuge): Ungültige Antwort („Math extension cannot connect to Restbase.“) von Server „https://wikimedia.org/api/rest_v1/“:): {\displaystyle \varphi} . As the reproduction formula spans a vector space, the $\varphi$ must be at least bi-orthogonal to Fehler beim Parsen (MathML mit SVG- oder PNG-Rückgriff (empfohlen für moderne Browser und Barrierefreiheitswerkzeuge): Ungültige Antwort („Math extension cannot connect to Restbase.“) von Server „https://wikimedia.org/api/rest_v1/“:): {\displaystyle \tilde{\varphi}} . This translates in fourier domain to:

Fehler beim Parsen (MathML mit SVG- oder PNG-Rückgriff (empfohlen für moderne Browser und Barrierefreiheitswerkzeuge): Ungültige Antwort („Math extension cannot connect to Restbase.“) von Server „https://wikimedia.org/api/rest_v1/“:): {\displaystyle \tilde{\Phi}(\omega) = \frac{\Phi(\omega)}{\sum_{k \in \mathbb{Z}} |\Phi(\omega + 2\pi k)|^2} }

The fourier transform of a B-Spline of order $N$ is (e.g. ):

$\mathrm {B} _{N}(\omega )=\Phi (\omega )=\left({\frac {\sin(\omega /2)}{\omega /2}}\right)^{N+1}=\mathrm {sinc} ^{N+1}(\omega /2)$

The following derivation of the sum is borrowed from . For this derivation to work, I set $L=N+1$ temprarily:

$\sum _{k\in \mathbb {Z} }|\Phi (\omega +2\pi k)|^{2}=\sum _{k\in \mathbb {Z} }\left|\mathrm {sinc} \left({\frac {1}{2}}(\omega +2\pi k)\right)^{L}\right|^{2}=\sum _{k\in \mathbb {Z} }\left|\mathrm {sinc} \left({\frac {1}{2}}(\omega +2\pi k)\right)\right|^{2L}$

and because $2L$ is always even:

Fehler beim Parsen (MathML mit SVG- oder PNG-Rückgriff (empfohlen für moderne Browser und Barrierefreiheitswerkzeuge): Ungültige Antwort („Math extension cannot connect to Restbase.“) von Server „https://wikimedia.org/api/rest_v1/“:): {\displaystyle = \sum_{k \in \mathbb{Z}}\frac{\sin^{2L}(\frac{1}{2}(\omega + 2\pi k))}{\left(\frac{1}{2}(\omega + 2\pi k)\right)^{2L}} = \sum_{k \in \mathbb{Z}}\frac{\sin^{2L}(\frac{\omega}{2} + \pi k))}{(\frac{\omega}{2} + \pi k)^{2L}} }

Because of the periodicity it is known that

Fehler beim Parsen (MathML mit SVG- oder PNG-Rückgriff (empfohlen für moderne Browser und Barrierefreiheitswerkzeuge): Ungültige Antwort („Math extension cannot connect to Restbase.“) von Server „https://wikimedia.org/api/rest_v1/“:): {\displaystyle \sin^{2L}(x + \pi k) = \sin^{2L}(x) }

such that

$=\sin ^{2L}({\frac {\omega }{2}})\sum _{k\in \mathbb {Z} }{\frac {1}{({\frac {\omega }{2}}+\pi k)^{2L}}}$

And finally the following relation is used:

$\sum _{k}{\frac {1}{(x+\pi k)^{2L}}}=-{\frac {1}{(2L-1)!}}{\frac {d^{2L-1}}{dx^{2L-1}}}\cot {x}$

in order to finally obtain:

$\sum _{k\in \mathbb {Z} }\left|\mathrm {sinc} \left({\frac {1}{2}}(\omega +2\pi k)\right)^{L}\right|^{2}=-\sin ^{2L}\left({\frac {\omega }{2}}\right){\frac {1}{(2L-1)!}}{\frac {d^{2L-1}}{d\left({\frac {\omega }{2}}\right)^{2L-1}}}\cot {\left({\frac {\omega }{2}}\right)}$

and with $L=N+1$:

$\sum _{k\in \mathbb {Z} }|\Phi (\omega +2\pi k)|^{2}=-\sin ^{2(N+1)}\left({\frac {\omega }{2}}\right){\frac {1}{(2N+1)!}}{\frac {d^{2N+1}}{d\left({\frac {\omega }{2}}\right)^{2N+1}}}\cot {\left({\frac {\omega }{2}}\right)}$

Therefore, together with $\Phi (\omega )$ this yields:

${\tilde {\Phi }}(\omega )={\frac {(2N+1)!}{\left({\frac {\omega }{2}}\right)\sin \left({\frac {\omega }{2}}\right)^{N+1}{\frac {d^{2N+1}}{d\left({\frac {\omega }{2}}\right)^{2N+1}}}\cot {\left({\frac {\omega }{2}}\right)}}}$

and finally substituting for $t(\omega )$:

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As this function is not well defined it is better to use the limit:

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For a cubic spline (N=3) the coefficients are:

Fehler beim Parsen (MathML mit SVG- oder PNG-Rückgriff (empfohlen für moderne Browser und Barrierefreiheitswerkzeuge): Ungültige Antwort („Math extension cannot connect to Restbase.“) von Server „https://wikimedia.org/api/rest_v1/“:): {\displaystyle c_{0,n} = 1 }

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$c_{2,n}={\frac {1}{3}}\left(-1+3n^{2}\right)$

$c_{3,n}=-n+n^{3}$