$\|A\| = \max_{\|\underline{x}\|=1}\, \|A\underline{x}\| = \sqrt{\lambda_\max(A^TA)}$
$\|A\underline{x}\| \leq \|A\|\|\underline{x}\|$ $\|AB\| \leq \|A\|\|B\|$ $\|A + B\| \leq \|A\| + \|B\|$
Consider the following series:
$\displaystyle \sum_{j=0}^{\infty} f_j(t)$ where $f_j(t)$ is defined in $t \in [t_0, t_1]$
The infinite series of functions is convergent in $[t_0, t_1]$ iff:
IV Problem (Cauchy Problem)
$\begin{cases} \dot{\underline{x}} = A \underline{x} \\ \underline{x}(0) = \underline{x}_0 \rightarrow IC \end{cases}$
Formally we can interpret the equation directly:
$\displaystyle \underline{x}(t) = \underline{x}_0 + \int_0^t A\underline{x}(\tau) \partial \tau = \underline{x}_0 + A \int_0^t \underline{x}(\tau) \partial \tau$
Formally, $\underline{x}(t) = \underline{x}_0 + A \int_0^t \underline{x}(\tau)\partial\tau$ is an integral equation.
Start from $\underline{x}_0$ and compute $\underline{x}_1(t)$ (approximation):
$\displaystyle \underline{x}_1(t) = \underline{x}_0 + A \int_0^t \underline{x}_0(\tau) \partial\tau + A\underline{x}_0(t)$
then move forward
$\displaystyle \underline{x}_2(t) = \underline{x}_0 + A \int_0^t \underline{x}_1(\tau)\partial\tau = \underline{x}_0 + t A \underline{x}_0 + \frac{t^2}{2} A^2 \underline{x}_0$
$\displaystyle \underline{x}_3(t) = \underline{x}_0 + A \int_0^t \underline{x}_2(\tau)\partial\tau = \underline{x}_0 + t A \underline{x}_0 + \frac{t^2}{2} A^2 \underline{x}_0 + \frac{t^3}{6} A^3 \underline{x}_0$
Eventually we get the following:
$\displaystyle \underline{x}_k(t) = \sum_{j=0}^k \frac{t^j}{j!}A^j\underline{x}_0, \quad \text{where } A^0 = I \text{ (identity)}$
Let's apply the theorem of convergence $\rightarrow$ Look at the $\underline{\text{bound}}$ for the coefficient of the series.
$\displaystyle\left\|\frac{t^j}{j!} A^j \underline{x}_0 \right \| \leq \frac{|t|^j}{j!} \|A^j\| \|\underline{x}_0\| \leq \frac{\big(T\,\|A\|\big)^j}{j!} \|\underline{x}_0\| \text{ if } t\in[-T,T] \leftarrow |t|^j \leq T^j$
Call $\alpha_j = \frac{\big(T\|A\|\big)^j}{j!}\|\underline{x}_0\|$ and look at $\sum_{j=1}^{\infty} \alpha_j$
$\displaystyle \sum_{j=0}^{\infty} \alpha_j = \left(\sum_{j=0}^{\infty} \frac{\big(T\|A\|\big)^j}{j!}\right) \|\underline{x}_0\| = e^{T\|A\|}\|\underline{x}_0\|$
Remember: Taylor series for the exponential is $e^x = \sum_{j=0}{\infty} \frac{x^j}{j!}$ thus we have that the series $\sum_{j=0}^{\infty} \frac{t^j}{j!} A^j \underline{x_0}$ is convergent.
Definition: For a matrix $M$, the exponential matrix $e^M = \exp(M)$ is defined via the convergent infinite series:
$$e^M = \sum_{j=0}^{\infty} \frac{1}{j!}M^j$$General Fact: For a given function $f(x)$ and a matrix $A(n \times x)$, $f(A)$ is defined via the infinite series:
$\displaystyle f(x) = \sum_{j=0}^\infty \frac{f^j(0)}{j!}x^j \Rightarrow \sum_{j=0}^\infty \frac{f^j(0)}{j!} A^j$
Thus:
$\displaystyle \begin{cases} \dot{\underline{x}} = A \underline{x} \\ \underline{x}(0) = \underline{x}_0 \end{cases} \Rightarrow \underline{x}(t) = e^{At}\underline{x} \text{ where, } e^{At} = \sum_{j=0}^\infty \frac{t^j}{j!} A^j$
Properties of $e^{At}$:
Claim: $\displaystyle \underline{x}(t) = e^{At}\underline{x}_0 + \int_0^t d^{A(t - \tau}B \underline{u}(\tau) \partial \tau$
Let's see how we prove this is the solution:
If satisfies $IC$:
$\displaystyle\underline{x}(0) = e^{A0}\underline{x}_0 + \int_0^0 e^{A(t-\tau)} B\underline{u}(\tau)\partial\tau = \underline{x}_0$
Take the derivative:
$\displaystyle\dot{\underline{x}} = \frac{\partial}{\partial t} \big(e^{At}\underline{x}_0 + \int_0^t e^{A(t-\tau)}B\underline{u}(\tau) \partial\tau\big) = Ae^{At}\underline{x}_0 + \frac{\partial}{\partial t} \big(e^{A(t-\tau)}B\underline{u}(\tau)\partial\tau\big)$
Leibitz (?????) formula: $\displaystyle\frac{\partial}{\partial t} \int_{\alpha(t)}^{\beta(t)} f(t, \tau)\partial \tau = f(t, \tau) \big |_\beta \beta'(t) - f(t, \tau)\big |_\alpha \alpha'(t) + \int_\alpha^\beta \frac{\partial f}{\partial t} \partial \tau$
Apply:
$\displaystyle\dot{\underline{x}} = A e^{At}\underline{x}_0 + e^A(t-\tau)B\underline{u}(\tau) \big|_{\tau = t} + \int_0^t A e^{A(t - \tau)} B \underline{u}(\tau)\partial\tau = A \big(\underbrace{e^{At}\underline{x}_0 + \int_0^t e^{A(t-\tau)} B\underline{u}(\tau)\partial\tau}_{\underline{x}(t)}\big) + B \underline{u}(t)$
$\Rightarrow \dot{\underline{x}} = A\underline{x} + B\underline{u} \Rightarrow$ it satisfies the state equations.
Find $e^{At}$ for $A = \begin{bmatrix}0 && 1 \\ 0 && 0 \end{bmatrix}$
$\displaystyle e^{At} = \sum_{j=0}^\infty \frac{t^j}{j!}A^j \Rightarrow A^2 = A \cdot A = \begin{bmatrix}0 && 1 \\ 0 && 0\end{bmatrix}\begin{bmatrix}0 && 1 \\ 0 && 0\end{bmatrix} = \begin{bmatrix}0 && 0 \\ 0 && 0\end{bmatrix}$
Thus $A^k = 0$ for $k \geq 2$.
$e^{At} = I + tA = \begin{bmatrix}1 && t \\ 0 && 1\end{bmatrix}$
Find $a^{At}$ for $A = \begin{bmatrix}0 && 1 \\ -1 && 0 \end{bmatrix}$
$A^2 = \begin{bmatrix}0 && 1 \\ -1 && 0\end{bmatrix} \begin{bmatrix}0 && 1 \\ -1 && 0\end{bmatrix} = \begin{bmatrix}-1 && 0 \\ 0 && -1\end{bmatrix} = -I$
$A^3 = A^2 A = -A \Rightarrow A^4 = A^3A = (-A)(A) = -A^2 = I \Rightarrow A^5 = A^4A = A$
So we see a pattern here:
$\displaystyle A^k = \begin{cases} (-1)^{\frac{k-1}{2}}A, && k = 1, 3, 5, \dots \, odd \\ (-1)^{\frac{k}{2}}I, && k = 0, 2, 4, \dots \, even \end{cases}$
The $e^At$ can be split into odd and even:
$\displaystyle e^{At} = \sum_{j=0}^\infty \frac{t^j}{j!}A^j = \big( \underbrace{1 - \frac{t^2}{2!} + \frac{t^4}{4!} \dots}_{\cos t} \big) I + \big ( \underbrace{t - \frac{t^3}{3!} + \frac{t^5}{5!} - \dots}_{\sin t} \big) A$
$e^{At} = \cos(t)I + \sin(t)A = \begin{bmatrix}\cos{t} && 0 \\ 0 && \cos(t)\end{bmatrix} + \begin{bmatrix}0 && \sin(t) \\ -sin(t) && 0\end{bmatrix} = \begin{bmatrix}\cos(t) && \sin(t) \\ -\sin(t) && \cos(t)\end{bmatrix}$
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