10/01 - Vectors & Functions: Difference between revisions

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:<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>  
:<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 \c\! </math>==
==Defining <math> k_m \,\! </math>==
Define <math> k_m = \left | \hat a_m \right |^2 </math>
Define <math> k_m = \left | \hat a_m \right |^2 </math>
*How did you get the last two lines of the last page?
*How did you get the last two lines of the last page?

Revision as of 15:32, 10 November 2008

Vectors & Functions

  • How to related the vector v to the sampling?

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

f(t)=∑i=0N−1f(iT)⏟coefficients⋅p(t−iT)⏟basisfunctions

  • Where p(t) 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, δ

  • δ(x)={+∞,x=00,x≠0
  • ∫−∞∞δ(x)dx=1.

By using the Dirac Delta function the summation becomes an integral

f(t)=∫−∞∞f(u)⋅δ(t−u)du

Changing from one orthogonal basis set to another

We have a vector v^=∑j=13aja^j and wish to change it to v^=∑j=13bjb^j. We know each basis set, and their relationship to each other. We are trying to find the coefficients, (the bj) that go with the new basis set.

  • Working from the a^ basis set:
v^⋅b^m=∑j=13vja^j⋅b^m=∑j=13vj(a^j⋅b^m)⏟projofa^jonb^m
  • Working from the b^ basis set:
v^⋅b^m=∑j=13bjb^j⋅b^m=∑j=13bj(b^j⋅b^m)⏟projofb^jonb^m=∑j=13bjkmδmj=km∑j=13bjδmj=bmkm∑j=13=bmkm
  • Now taking the v^⋅b^m that was derived from both basis sets and equating them:
bmkm=∑j=13vja^j⋅b^m⟹bm=1km∑j=13vj(a^j⋅b^m)

Defining km

Define km=|a^m|2

  • How did you get the last two lines of the last page?
  • What does the b^m represent, say compared to b^j?
  • When you do the dot product of say A \cdot B, is it always the projection of A onto B and not the opposite way around?
  • Why did you decide to make it k_m instead of k_j?