📂Fourier Analysis

B-spline Scaling Equation

Formulas¹

(a) B-spline scaling equation:

For a B-spline of order $m\in N$, the following equation holds:

$$ \widehat{N_{m}}(2\gamma)=H_{0}(\gamma)\widehat{N_{m}}(\gamma),\quad \forall \gamma \in \mathbb{R} $$

Where $H_{0}$ is a function with period $1$, which is described as follows:

$$ H_{0}(\gamma)=\left( \frac{1+e^{-2\pi i \gamma}}{2} \right)^{m} $$

Also, the definition of the Fourier transform of $f$, $\widehat{f}$, is as follows:

$$ \widehat{f}(\gamma):=\int _{-\infty} ^{\infty} f(x)e^{-2\pi i x \gamma}dx $$

(b) Central B-spline scaling equation:

Similarly, for a central B-spline of order $m$, the following equation holds:

$$ \widehat{B_{m}}(2\gamma) = H_{0}(\gamma)\widehat{B_{m}}(\gamma),\quad \forall \gamma \in \mathbb{R} $$

Where, again, $H_{0}$ is a function with period $2$, which is described as follows:

$$ H_{0}(\gamma)=\left( \frac{e^{\pi i \gamma}+e^{-\pi i \gamma}}{2} \right)^{m}=\cos^{m}(\pi \gamma) $$

Proof

(a)

The Fourier transform of the B-spline is as follows:

$$ \widehat{N_{m}}(\gamma)=\left( \frac{1-e^{-2\pi i\gamma}}{2\pi i \gamma} \right)^{m} $$

Therefore,

$$ \begin{align*} \widehat{N_{m}}(2\gamma) =&\ \left( \frac{1-e^{-2\pi i2\gamma}}{2\pi i 2\gamma} \right)^{m} \\ =&\ \frac{(1+e^{-2\pi i \gamma})^{m}(1-e^{-2\pi i \gamma})^{m}}{2^{m}(2\pi i \gamma)^{m}} \\ =&\ \left( \frac{1+e^{-2\pi i \gamma}}{2} \right)^{m} \left( \frac{1-e^{-2\pi i \gamma}}{2\pi i \gamma} \right)^{m} \\ =&\ H_{0}(\gamma) \widehat{N_{m}}(\gamma) \end{align*} $$

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(b)

The Fourier transform of the central B-spline is as follows:

$$ \widehat{B_{m}}(\gamma)=\left( \frac{e^{\pi i \gamma}-e^{-\pi i \gamma}}{2\pi i \gamma} \right)^{m}=\left( \frac{\sin (\pi\gamma)}{\pi \gamma} \right)^{m} $$

Therefore,

$$ \begin{align*} \widehat{B_{m}}(2\gamma) =&\ \left( \frac{e^{\pi i 2\gamma} - e^{-\pi i 2\gamma}}{2\pi i 2\gamma} \right)^{m} \\ =&\ \frac{(e^{\pi i \gamma}+e^{-\pi i \gamma})^{m}(e^{\pi i \gamma}-e^{-\pi i \gamma})^{m} }{2^{m}(2\pi i \gamma)^{m}} \\ =&\ \left( \frac{e^{\pi i \gamma}+e^{-\pi i \gamma}}{2} \right)^{m} \left( \frac{e^{\pi i \gamma}-e^{-\pi i \gamma} }{2\pi i \gamma} \right)^{m} \\ =&\ H_{0}(\gamma)\widehat{B_{m}}(\gamma) \end{align*} $$

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