Two Variables

Let’s begin with the equations:

\begin{equation} \begin{cases} 2x+y = 3 \\ x - y = 0 \end{cases} \end{equation}

We will first change this into a matrix equation:

\begin{equation} \begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 3 \\ 0 \end{pmatrix} \end{equation}

We need to find, then, the inverse of:

\begin{equation} \begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix} \end{equation}

Suppose \(\mathbb{F} = \mathbb{R}\), and \(V \neq \{0\}\). Replace the positivity condition with the condition that \(\langle v,v \rangle > 0\) for some \(v \in V\). Show that this change in definition does not change the set of functions from \(V \times V\) to \(\mathbb{R}\) that are inner products on \(V\).

We hope to show that \(\langle v,v \rangle >0\) for some \(v \in V\) implies that \(\langle v,v \rangle \geq 0\) for all \(v \in V\) in real vector spaces.

Proof: identity of a group is unique

Assume for contradiction that there exists two identities \(e_1\) and \(e_2\) which are identities of the group \(A\). Take also an \(a \in A\).

Given both \(e_1\) and \(e_2\) are identities, we have that:

\begin{equation} a * e_1 = a \end{equation}

as well as

\begin{equation} a * e_2 = a \end{equation}

Therefore, we have by the transitive property that:

\begin{equation} a * e_1 = a*e_2 \end{equation}

We declare known battery voltage \(E(t)\).

Here are the \(y\) values.

\begin{equation} \begin{cases} \dv{x_1}{t} = y_{4}\\ \dv{x_2}{t} = y_{3}\\ \dv{x_3}{t} = y_{1}\\ \dv{x_4}{t} = y_{2}\\ \end{cases} \end{equation}

And here are some of the \(x\) values.

\begin{equation} \begin{cases} \dv{x_4}{t}=-\frac{2}{RC}x_2-\frac{1}{RC}x_{3}-\frac{2E(t)}{R} \\ \dv{y_1}{t}=-\frac{1}{LC}x_2-\frac{E(t)}{C} \\ \dv{y_4}{t} = -\frac{R}{L}y_2-\frac{2E(t)}{L} \end{cases} \end{equation}

Right off the bat, we can see that we can make one substitution. That, given:

\begin{equation} \begin{cases} \dv{x_4}{t}=-\frac{2}{RC}x_2-\frac{1}{RC}x_{3}-\frac{2E(t)}{R} \\ \dv{x_4}{t} = y_{2} \end{cases} \end{equation}

We need to solve this system:

\begin{equation} \begin{cases} \dv{I}{t} = -0.73U + 0.0438 + 0.4 \dv{M}{t} \\ \dv{U}{t} = 0.4I - 0.012 \\ \dv{G}{t} = \dv{M}{t}- I \\ M(t)=0.02\sin (1.15t + \phi) \end{cases} \end{equation}

To be able to work on this, let us create some functions:

# variable
t, dm = var("t dm")

# functions
I = function("_I")(t) # _I because i is imaginary
U = function("U")(t)
G = function("G")(t)

# parameter
phi = var("phi", latex_name="\phi")

# our equations
eqns = [
    diff(I,t) == -0.73*U + 0.0438 + 0.4*dm,
    diff(U,t) == 0.4*I - 0.012,
    diff(G,t) == dm - I
]

eqns
desolve(eqns, U, ivar=t, algorithm="fricas").expand()

Great, now, we will run the laplace transform upon these equations: