A finite-dimensional vector space is a vector space where some actual list (which remember, has finite length) of vectors spans the space.

An infinite-demensional vector space is a vector space that’s not a finite-dimensional vector space.

additional information

every finite-dimensional vector space has a basis

Begin with a spanning list in the finite-dimensional vector space you are working with. Apply the fact that all spanning lists contains a basis of which you are spanning. Therefore, some elements of that list form a basis of the finite-dimensional vector space you are working with. \(\blacksquare\)

Fireside Chats are a group of broadcasts by Franklin D. Roosevelt (FDR) which allowed him to speak directly to the people.

First Order ODEs are Differential Equations that only takes one derivative.

Typically, by the nature of how they are modeled, we usually state it in a equation between three things:

\begin{equation} t, y(t), y’(t) \end{equation}

as in—we only take one derivative.

Sometimes the solution may not be analytic, but is well-defined:

\begin{equation} y’ = e^{-x^{2}} \end{equation}

we know that, by the fundamental theorem of calculus, gives us:

\begin{equation} y(x) = \int_{0}^{x} e^{-s{2}} \dd{s} \end{equation}

Consider the case where there are two functions interacting with each other:

\begin{equation} y_1(t) \dots y_{2}(t) \end{equation}

So we have more than one dependent function, with functions \(y_1, y_1’, y_2, y_2’\) and so forth. To deal with this, we simply make it into a matrix system:

\begin{equation} y(t) = \mqty(y_1(t) \\ \dots \\ y_{n}(t)) \end{equation}

For instance, should we have:

\begin{equation} \begin{cases} y_1’ = 3y_1 - 2y_2 \\ y_2’ = -y_1 + 5y_2 \end{cases} \end{equation}