Daubechies wavelet

Example

For each integer $N \ge 1$, Ingrid Daubechies constructed an orthonormal wavelet $\psi$ from a multiresolution analysis. Its scaling function $\varphi$ satisfies a refinement equation $\varphi(x) = \sqrt2 \sum_{k=0}^{2N-1} h_k \varphi(2x - k)$ with $2N$ real coefficients. Both $\varphi$ and $\psi$ have compact support of length $2N - 1$, and $\psi$ has $N$ vanishing moments. For $N = 1$ this is the Haar wavelet.

The case $N = 2$ has the filter $$h_0 = \frac{1 + \sqrt3}{4\sqrt2}, \quad h_1 = \frac{3 + \sqrt3}{4\sqrt2}, \quad h_2 = \frac{3 - \sqrt3}{4\sqrt2}, \quad h_3 = \frac{1 - \sqrt3}{4\sqrt2}.$$ Check of the filter conditions. The sum is $\frac{8}{4\sqrt2} = \sqrt2$, as needed for $\int \varphi = 1$. The sum of squares is $\frac{(4 + 2\sqrt3) + (12 + 6\sqrt3) + (12 - 6\sqrt3) + (4 - 2\sqrt3)}{32} = 1$. And $h_0 h_2 + h_1 h_3 = \frac{(1 + \sqrt3)(3 - \sqrt3) + (3 + \sqrt3)(1 - \sqrt3)}{32} = \frac{2\sqrt3 - 2\sqrt3}{32} = 0$. So the filter is orthogonal to its shift by $2$, which reflects the orthogonality of the translates $\varphi(\cdot - k)$.

The wavelet is $\psi(x) = \sqrt2 \sum_k g_k \varphi(2x - k)$ with $g_k = (-1)^k h_{1-k}$, so its coefficients are $g_{-2} = h_3$, $g_{-1} = -h_2$, $g_0 = h_1$, $g_1 = -h_0$. Check of two vanishing moments: $\sum_k g_k = h_3 - h_2 + h_1 - h_0 = 0$ and $\sum_k k g_k = -2h_3 + h_2 - h_0 = \frac{-2 + 2\sqrt3 + 3 - \sqrt3 - 1 - \sqrt3}{4\sqrt2} = 0$. These two discrete conditions give $\int \psi = 0$ and $\int x \psi(x) \, dx = 0$. The scaling function of this case is continuous but not smooth, and it has no closed formula; it is computed from the refinement equation.