A multi-resolution analysis is defined by scaling function and the corresponding wavelet. From the scaling relations

$\displaystyle \phi_{j,k}(x) = \frac{1}{\sqrt{2}} \sum_{l} g_0(l) \phi_{j+1,2k+l}(x)$

$\displaystyle \psi_{j,k}(x) = \frac{1}{\sqrt{2}} \sum_{l} g_1(l) \phi_{j+1,2k+l}(x)$

the scaling function and wavelet can be defined from the scaling coefficient $g_0, g_1$

The coefficients are constrained due to the properties of wavelet and scaling function.

$\displaystyle \int \phi(x) dx = 1$

$\displaystyle \int \phi_{j,k}(x) \phi_{j,k'}(x) dx = \delta_{kk'}$

$\displaystyle \int \psi(x) dx = 0$

$\displaystyle \int \psi_{j,k}(x) \psi_{j,k'}(x) dx = \delta_{kk'}$

$\displaystyle \int \psi_{j,k}(x) \phi_{j,k'}(x) dx = 0$

$\displaystyle \sum g_0(l) = 2$

$\displaystyle \sum g_1(l) = 0$

$\displaystyle \sum_{l,n} g_0(l) g_0(l+2n) = \begin{matrix} 2, & n=0 \\ 0, & else \end{matrix}$

$\displaystyle \sum_{l,n} g_1(l) g_1(l+2n) = \begin{matrix} 2, & n=0 \\ 0, & else \end{matrix}$

$\displaystyle \sum_{l,n} g_0(l) g_1(l+2n) = 0$

The 3rd and 4th constrains requires the numbers of non-zero element in $g_0, g_1$ are even.

One of the solution is setting

$g_1(k) = (-1)^k g_0 (1-k)$

so that we don’t need to worry $g_1$ and the 4th constrain becomes the 3rd constrain, and the 5th constrain is always satisfied. Now, only the 1st, 2nd, and 3rd constrains are needed. This is equivalent to $1+m/2$ equations with number of non-zero elements in $g_0$ is $m$.

$m$ $1 + \frac{m}{2}$ Degree of Freedom
2 2 0
4 3 1
6 4 2
8 5 3

For size of 4, the solution is

$\displaystyle g_0 = \left(a, \frac{1-\sqrt{1+4a-4a^2}}{2}, 1-a, \frac{1+\sqrt{1+4a-4a^2}}{2}\right)$

In fact, the coefficient for $g_0$ can be grouped as even and odd, so that

$\displaystyle \sum g_0(2k) = \sum g_0(2k+1) = 1$

and the constrain 3rd can lead to,

$\displaystyle (\sum g_0(2k))^2 + (\sum g_0(2k+1)^2 = 2$,

which is automatically fulfill.