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".

$f(t)=\sum _{i=0}^{N-1}\underbrace {f(iT)} _{coefficients}\cdot \underbrace {p(t-iT)} _{basisfunctions}$

• 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, $\delta \,\!$

• $\delta (x)={\begin{cases}+\infty ,&x=0\\0,&x\neq 0\end{cases}}$
• $\int _{-\infty }^{\infty }\delta (x)\,dx=1.$

By using the Dirac Delta function the summation becomes an integral

$f(t)=\int _{-\infty }^{\infty }f(u)\cdot \delta (t-u)\,du$

Changing from one orthogonal basis set to another

We have a vector ${\hat {v}}=\sum _{j=1}^{3}a_{j}{\hat {a}}_{j}$ and wish to change it to ${\hat {v}}=\sum _{j=1}^{3}b_{j}{\hat {b}}_{j}$. We know each basis set, and their relationship to each other. We are trying to find the coefficients, (the $b_{j}\,\!$) that go with the new basis set.

• Working from the ${\hat {a}}$ basis set:

${\hat {v}}\cdot {\hat {b}}_{m}=\sum _{j=1}^{3}v_{j}{\hat {a}}_{j}\cdot {\hat {b}}_{m}=\sum _{j=1}^{3}v_{j}\underbrace {\left({\hat {a}}_{j}\cdot {\hat {b}}_{m}\right)} _{projof{\hat {a}}_{j}on{\hat {b}}_{m}}$

• Working from the ${\hat {b}}$ basis set:

${\hat {v}}\cdot {\hat {b}}_{m}=\sum _{j=1}^{3}b_{j}{\hat {b}}_{j}\cdot {\hat {b}}_{m}=\sum _{j=1}^{3}b_{j}\underbrace {\left({\hat {b}}_{j}\cdot {\hat {b}}_{m}\right)} _{projof{\hat {b}}_{j}on{\hat {b}}_{m}}=\sum _{j=1}^{3}b_{j}k_{m}\delta mj=k_{m}\sum _{j=1}^{3}b_{j}\delta mj=b_{m}k_{m}\sum _{j=1}^{3}=b_{m}k_{m}$

• Now taking the ${\hat {v}}\cdot {\hat {b}}_{m}$ that was derived from both basis sets and equating them:

$b_{m}k_{m}=\sum _{j=1}^{3}v_{j}{\hat {a}}_{j}\cdot {\hat {b}}_{m}\Longrightarrow b_{m}={\frac {1}{k_{m}}}\sum _{j=1}^{3}v_{j}\left({\hat {a}}_{j}\cdot {\hat {b}}_{m}\right)$

Defining $k_{m}\,\!$

Taking $k_{m}\,\!$ from the previous section:

${\hat {b}}_{j}\cdot {\hat {b}}_{m}=k_{m}\delta mj\Longrightarrow {\hat {b}}_{m}\cdot {\hat {b}}_{m}=k_{m}\Longrightarrow \left|{\hat {b}}_{m}\right|^{2}=k_{m}$

Thus $k_{m}\,\!$ is the length of ${\hat {b}}_{m}$ squared

Questions

• What does the ${\hat {b}}_{m}$ represent, say compared to ${\hat {b}}_{j}$?
  • ${\hat {b}}_{m}$ is a unit vector in the direction we're interested in finding the new coefficient for the new basis set -- rewrite this
  • ${\hat {b}}_{j}$ is a unit vector for one direction in our new basis set
• When you do the dot product of say ${\vec {A}}\cdot {\vec {B}}$, is it always the projection of ${\vec {A}}$ onto ${\vec {B}}$ and not the opposite way around?
  • ${\vec {A}}\cdot {\vec {B}}=\left|{\vec {A}}\right|\left|{\vec {B}}\right|\cos \theta$
• 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 $k_{m}\,\!$ instead of $k_{j}\,\!$?
  • Completely arbitrary, the end result is the same either way
• ${\hat {x}}$ is a unit vector and ${\vec {x}}$ is a vector