Suppose that $u_1 = \left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0,0,\ldots,0\right)\quad\text{and}\quad v_1 = \left(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},0,0,\ldots,0\right)$ and $u_l(n) = \sum_{k=0}^{2^{l-1}-1} u_1\left(n+ \frac{kN}{2^{l-1}}\right)\quad\text{and}\quad v_l(n) = \sum_{k=0}^{2^{l-1}-1} v_1\left(n+ \frac{kN}{2^{l-1}}\right).$ How to show that \begin{align*} u_l(0)&= \frac{1}{\sqrt{2}},\\ u_l(1)&= \frac{1}{\sqrt{2}},\\ u_l(n)&= 0 &\quad&\text{for $2 \leq n \leq \frac{N}{2^{l-1}}-1$,}\\ v_l(0)&= \frac{1}{\sqrt{2}},\\ v_l(1)&= -\frac{1}{\sqrt{2}},\\ v_l(n)&= 0&&\text{for $2 \leq n \leq \frac{N}{2^{l-1}}-1$.} \end{align*}
How to show that $u_l(0)= \frac{1}{\sqrt{2}}$, $u_l(1)= \frac{1}{\sqrt{2}}$ and $u_l(n)= 0$ for $2 \leq n \leq (\frac{N}{2^{l-1}}-1$ $\dots$
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linear-algebra
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0$N$ means that βitβ is $\frac{N}{2^{l-1}}$-periodic( one must assume that $N$ is divisible by $2^p$. $u_1$ and $v_1$ form first stage Haar basis. Yes, $u_1(k) $ mean the $k$:th coordinate of $u_1$. Thus this is my clarification. β 2010-12-30