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

These properties lead to

\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.