View source for 10/01 - Vectors & Functions

==Vectors & Functions==
*How to related the vector v to the sampling?

We could sample a continuous function every T seconds, creating a "bar graph".

<math> f(t)= \sum_{i=0}^{N-1} \underbrace{f(i T)}_{coefficients} \cdot \underbrace{p(t - i T)}_{basis functions}</math>
*Where <math> p(t) \,\! </math> is a rectangle 1 unit high and T units wide

In an effort to make this more exact, will will continue to shrink the rectangle down to the Dirac Delta function, <math> \delta \,\!</math>
*<math>\delta(x) = \begin{cases} +\infty, & x = 0 \\ 0, & x \ne 0 \end{cases}</math>
*<math>\int_{-\infty}^\infty \delta(x) \, dx = 1.</math>

By using the Dirac Delta function the summation becomes an integral

<math> f(t) = \int_{-\infty}^{\infty} f(u) \cdot \delta (t - u)\, du </math>

==Changing from one orthogonal basis set to another==
We have a vector <math> \hat v = \sum_{j=1}^3 a_j \hat a_j </math> and wish to change it to <math> \hat v = \sum_{j=1}^3 b_j \hat b_j </math>. We know each basis set, and their relationship to each other. We are trying to find the coefficients, (the <math> b_j \,\!</math>) that go with the new basis set.
*Working from the <math> \hat a </math> basis set:
:<math>\hat v \cdot \hat b_m= \sum_{j=1}^3 v_j \hat a_j \cdot \hat b_m = \sum_{j=1}^3 v_j \underbrace{\left (\hat a_j \cdot \hat b_m \right )}_{proj of \hat a_j on \hat b_m} </math>
*Working from the <math> \hat b </math> basis set:
:<math> \hat v \cdot \hat b_m= \sum_{j=1}^3 b_j \hat b_j \cdot \hat b_m = \sum_{j=1}^3 b_j \underbrace{\left (\hat b_j \cdot \hat b_m \right )}_{proj of \hat b_j on \hat b_m}= \sum_{j=1}^3 b_j k_m \delta mj = k_m \sum_{j=1}^3 b_j \delta mj= b_m k_m \sum_{j=1}^3 = b_m k_m </math>
*Now taking the <math> \hat v \cdot \hat b_m </math> that was derived from both basis sets and equating them:
:<math> b_m k_m = \sum_{j=1}^3 v_j \hat a_j \cdot \hat b_m \Longrightarrow b_m = \frac{1}{k_m} \sum_{j=1}^3 v_j \left (\hat a_j \cdot \hat b_m \right ) </math> 

==Defining <math> k_m \,\! </math>==
Taking <math> k_m \,\! </math> from the previous section:
:<math> \hat b_j \cdot \hat b_m = k_m \delta mj \Longrightarrow \hat b_m \cdot \hat b_m = k_m \Longrightarrow \left | \hat b_m \right |^2 = k_m </math>

Thus <math> k_m \,\! </math> is the length of <math>\hat b_m</math> squared

==Questions==
*What does the <math> \hat b_m </math> represent, say compared to <math> \hat b_j</math>?
**<math> \hat b_m </math> is a unit vector in the direction we're interested in finding the new coefficient for the new basis set -- rewrite this
**<math> \hat b_j </math> is a unit vector for one direction in our new basis set
*When you do the dot product of say <math>\vec A \cdot \vec B</math>, is it always the projection of <math>\vec A</math> onto <math>\vec B</math> and not the opposite way around?
**<math>\vec A \cdot \vec B = \left | \vec A \right |  \left | \vec B \right | \cos \theta </math>
*Then is the picture assuming something is a unit vector?
**You will have to choose whether you're interested in projecting A onto B or B onto A. The lengths will be the same, but the direction will be different.
*Why did you decide to make it  <math>k_m \,\!</math> instead of <math>k_j\,\!</math>?
**Completely arbitrary, the end result is the same either way
*<math> \hat x </math> is a unit vector and <math> \vec x </math> is a vector