### 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