Astrum
- 269
- 5
Dyadic Cube C_{k,N} = X \in\ \mathbb{R}^{n} \frac{k_{i}}{2^{N}} \leq x_{i} < \frac{k_{i}+1}{2^{N}} for 1 \leq i \leq n
Where k = \pmatrix {<br /> k_{1} \cr<br /> k_{2} \cr<br /> \vdots \cr<br /> k_{i} \cr<br /> }
I understand that N is the level of the cubes, but what does k equal?
I'm having trouble visualizing this in my head.
A \subset \mathbb{R}^{n}
M_{A}(f)= supp_{x \in A}f(x); m_{A}(f) = inf_{x \in A}f(x) U_{N}(f) = \sum M_{c}(f) vol_{n}C L_{N}(f) = \sum m_{c}(f) vol_{n}C
I get the general idea, but I can't really see this in my head.
If you take the supp value of a function on a given cube, and multiply it by the volume of you cube, you get volume again??
I can't really get the image straight.
I get that this is just an extension of single variable, so it means that U must equal L which must equal I (integral).
Where k = \pmatrix {<br /> k_{1} \cr<br /> k_{2} \cr<br /> \vdots \cr<br /> k_{i} \cr<br /> }
I understand that N is the level of the cubes, but what does k equal?
I'm having trouble visualizing this in my head.
A \subset \mathbb{R}^{n}
M_{A}(f)= supp_{x \in A}f(x); m_{A}(f) = inf_{x \in A}f(x) U_{N}(f) = \sum M_{c}(f) vol_{n}C L_{N}(f) = \sum m_{c}(f) vol_{n}C
I get the general idea, but I can't really see this in my head.
If you take the supp value of a function on a given cube, and multiply it by the volume of you cube, you get volume again??
I can't really get the image straight.
I get that this is just an extension of single variable, so it means that U must equal L which must equal I (integral).