Exercises 4.6 & 4.7 Clarification
Problem 4.6. The function $g(x)\in V_3$ is split into its
approximation part $g_2$ and its wavelet part $w_2$ i.e.,
$g(x)=g_2(x)+w_2(x)$. The coefficient vectors $a^2=[1/2, 2, 5/2,-3/2]$
and $b^2=[-3/2,-1,1/2,-1/2]$ mean that
\[
g_2(x) = \frac12 \phi(2^2x)+2\phi(2^2x-1)+\frac52 \phi(2^2x-2)-\frac32
\phi(2^2x-3) \quad \text{and}\quad w_2(x) = -\frac32\psi(2^2x) -
\psi(2^2x-1)+\frac12 \psi(2^2x-2)-\frac12 \psi(2^2x-3)
\]
You are asked to find $g$ in the basis $\{\phi(2^3x-k)\}$ and sketch
the resulting step function. Note that we can express $g_2$, for
example, this way
\[
g_2(x) = \left\{
\begin{array}{rl}
-\frac12, & 0\le x <\frac 14,\\
2, & \frac 14\le x <\frac 12,\\
\frac52, & \frac12\le x <\frac34,\\
-\frac32, & \frac34\le x <1, \\
0, & \mbox{otherwise.}
\end{array}
\right.
\qquad
w_2(x) = \left\{
\begin{array}{rl}
-\frac32, & 0\le x <\frac 18,\\
\frac32, & \frac 18\le x <\frac 14,\\
-1, & \frac14\le x <\frac38,\\
1, & \frac38\le x <\frac12,\\
\frac12, & \frac12\le x <\frac58, \\
-\frac12,& \frac58\le \frac34,\\
-\frac12, & \frac34\le x <\frac78, \\
\frac12,& \frac78\le 1,\\
0, & \mbox{otherwise.}
\end{array}
\right.
\]
The sign "flips" in $w_2$ are there because $\psi(x)$ goes from $1$
to $-1$ when $x$ crosses $\frac12$.
Problem 4.7. You are given the coefficients $a^1=[3/2,-1]$,
$b^1=[-1,3/2]$, and $b^2=[-3/2,-3/2,-1/2,-1/2]$. This means that
\[
g_1(x) = \frac32 \phi(2x)-\phi(2x-1),\ \ w_1=-\psi(2x)+\frac32
\psi(2x-1)\ \ \text{and}\ \ w_2(x) = -\frac32\psi(2^2x) -
-\frac32 \psi(2^2x-1)-\frac12 \psi(2^2x-2)-\frac12 \psi(2^2x-3)
\]
Note that $g=g_2+w_2$ and $g_2 = g_1+w_1$. Using $g_1$ and $w_1$, find
$g_2$ in the $\{\phi(2^2x-k)\}$ basis for $V_2$. Then use $g_2$ and
$w_2$ to obtain $g$ in the $\{\phi(2^3x-k)\}$ for $V_3$. Once you find
$g$, again sketch the step function.
Updated 4/1/2016