Even more math testing

I'm matching up this: 
$$\int_{\Gamma} dD \tag 1$$
With what I found in this section of the Wikipedia article on line integrals:
$$\int\limits_{C} f\, ds = \int_{a}^{b} f(\mathbf{r}(t))\,|\mathbf{r}' (t)|\,dt \tag 2$$
It seems $\Gamma$ corresponds to the curve (or path) $C,$ the constant $1$ corresponds to the integrand $f,$ and $dD$ corresponds to what they call an "elementary arc length" $ds.$ That all seems clear enough. Apparently $\mathbf{r}$ get's you from one end of the curve $C$ to the other via its parameter $t$ when $t$ varies from $a$ to $b$ with $a < b.$ The result should be independent of the parameterization $\mathbf{r}$ of $C.$ That all makes sense. However, I'm not familiar with this notation:
$$\int_{\partial \Gamma} D \tag 3$$
What is the meaning of that exactly?
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fun with arrays:

$$
\left[
\begin{array}{c|c}
A_d & B_d \\
\hline
C_d & D_d
\end{array}
\right]_{SIM_d} =
\left[
\begin{array}{c|c}
1-\frac{\alpha_2 \theta}{1 - \alpha_1 (1 - \theta)} & 1-\frac{\theta}{1 - \alpha_1 (1 - \theta)} \\[1.5ex]
\hline
\frac{\alpha_2}{1 - \alpha_1 (1 - \theta)} & \frac{1}{1 - \alpha_1 (1 - \theta)} \\[1.5ex]
\frac{\alpha_2 \theta}{1 - \alpha_1 (1 - \theta)} & \frac{\theta}{1 - \alpha_1 (1 - \theta)} \\[1.5ex]
\frac{\alpha_2 (1-\theta)}{1 - \alpha_1 (1 - \theta)} & \frac{1-\theta}{1 - \alpha_1 (1 - \theta)} \\[1.5ex]
\frac{\alpha_2}{1 - \alpha_1 (1 - \theta)} & \frac{\alpha_1 (1-\theta)}{1 - \alpha_1 (1 - \theta)}
\end{array}
\right]
$$

$$
\left[
\begin{array}{c|c}
A_c & B_c \\
\hline
C_c & D_c
\end{array}
\right]_{SIM_c} =
\left[
\begin{array}{c|c}
\frac{1}{T_s} \log\left(1-\frac{\alpha_2 \theta}{1 - \alpha_1 (1 - \theta)}\right) & \frac{\theta + \alpha_1 (1-\theta) - 1}{T_s \alpha_2 \theta} \log\left(1-\frac{\alpha_2 \theta}{1 - \alpha_1 (1 - \theta)}\right)   \\[1.5ex]
\hline
\frac{\alpha_2}{1 - \alpha_1 (1 - \theta)} & \frac{1}{1 - \alpha_1 (1 - \theta)} \\[1.5ex]
\frac{\alpha_2 \theta}{1 - \alpha_1 (1 - \theta)} & \frac{\theta}{1 - \alpha_1 (1 - \theta)} \\[1.5ex]
\frac{\alpha_2 (1-\theta)}{1 - \alpha_1 (1 - \theta)} & \frac{1-\theta}{1 - \alpha_1 (1 - \theta)} \\[1.5ex]
\frac{\alpha_2}{1 - \alpha_1 (1 - \theta)} & \frac{\alpha_1 (1-\theta)}{1 - \alpha_1 (1 - \theta)}
\end{array}
\right]
$$

Where $T_s$ is the sample period of the discrete time model.

---------------------------------

$$
\left[
\begin{array}{c}
x_{n+1} \\
\hline
y_n
\end{array}
\right] =
\left[
\begin{array}{c|c}
A_d & B_d \\
\hline
C_d & D_d
\end{array}
\right]
\left[
\begin{array}{c}
x_{n} \\
\hline
u_n
\end{array}
\right]
$$
$$
\left[
\begin{array}{c}
\dot{x}(t) \\
\hline
y(t)
\end{array}
\right] =
\left[
\begin{array}{c|c}
A_c & B_c \\
\hline
C_c & D_c
\end{array}
\right]
\left[
\begin{array}{c}
x(t) \\
\hline
u(t)
\end{array}
\right]
$$


--------------------------- 2016.04.23: Below is the last bit of what used to be post SIM8: I removed it until I have time to fix it ----------------

Where the variables with a dot above them indicate rates (time derivatives) and $\dot{t}$ indicates the tax rate, and in general, for sample period $n$, we have:
$$
\begin{align}
H_n &=  \int_\limits{-\infty}^{n T_s} \dot{h}(\tau) d\tau &= h(n T_s) &&\text{cash money in existence by end of period n}\tag {11}\\[1.9ex]
G_n & = \int_\limits{(n-1)T_s}^{n T_s} \dot{g}(\tau) d\tau & = g(n T_s) - g((n-1)T_s) &&\text{government spending in period n}\tag {12}\\[1.9ex]
Y_n & = \int_\limits{(n-1)T_s}^{n T_s} \dot{y}(\tau) d\tau & = y(n T_s) - y((n-1)T_s) &&\text{national income in period n}\tag {13}\\[1.9ex]
T_n & = \int_\limits{(n-1)T_s}^{n T_s} \dot{t}(\tau) d\tau & = t(n T_s) - t((n-1)T_s) &&\text{taxes collected in period n}\tag {14}\\[1.9ex]
Y_{Dn} & = \int_\limits{(n-1)T_s}^{n T_s} \dot{y}_D(\tau) d\tau & = y_D(n T_s) - y_D((n-1)T_s) &&\text{household disposable income in period n}\tag {15}\\[1.9ex]
C_n & = \int_\limits{(n-1)T_s}^{n T_s} \dot{c}(\tau) d\tau & = c(n T_s) - c((n-1)T_s) &&\text{consumption in period n}\tag {16}
\end{align}
$$
For a more detailed descriptions of what the discrete time symbols at the left mean see the section entitled Notations Used in the Book near the beginning of Godley & Lavoie's (G&L's) Monetary Economics, starting at page 10 (page 10 of the .pdf). Or see the extracts from that book in this post by Jason Smith.

Note that $H_n$ are samples of $h$ at the sample times provided $\dot{g}$ is constant between sample times (only changing at the sample times themselves, if at all)

--------------------------------------- End 2016.04.23 SIM8 extract ---------------------------------------
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Table 1: SIM Parameters
Symbol	Value
$T_{s_0}$	1
α1	0.6
α2	0.4
θ	0.2

Table 2: Useful Expressions
Name	Expression	Value
P	1 - α1(1-θ)	0.52
$Q$	$A_c/(A_d-1)$	1.085852
$V$	$(B_c-B_d)/(A_d-1)$	-0.34341

Table 3: Discrete Time State Space Matrices
Matrix	Elements	Name	Expression	Value
$A_d$	$A_{d_{1,1}}$	$A_d$	1 - α2θ/P	0.84615385
$B_d$	$B_{d_{1,1}}$	$B_d$	1 - θ/P	0.61538462
$C_d$	$C_{d_{1,1}}$	$C_Y$	α2/P	0.76923077
	$C_{d_{2,1}}$	$C_T$	α2θ/P	0.15384615
	$C_{d_{3,1}}$	$C_{Y_D}$	α2(1-θ)/P	0.61538462
	$C_{d_{4,1}}$	$C_C$	α2/P	0.76923077
$D_d$	$D_{d_{1,1}}$	$D_Y$	1/P	1.92307692
	$D_{d_{2,1}}$	$D_T$	θ/P	0.38461538
	$D_{d_{3,1}}$	$D_{Y_D}$	(1-θ)/P	1.53846154
	$D_{d_{4,1}}$	$D_C$	α1(1-θ)/P	0.92307692

Table 4: Continuous Time State Space Matrices					& Resistors
Matrix	Elements	Name	Expression	Value	Name	Expression	Value
$A_c$	$A_{c_{1,1}}$	$A_c$	$\ln(A_d)/T_{s_0}$	-0.16705	$R_a$	$-{A_c}^{-1}$	5.986085
$B_c$	$B_{c_{1,1}}$	$B_c$	$Q B_d$	0.668216	$R_b$	${B_c}^{-1}$	1.496521
$C_c$	$C_{c_{1,1}}$	$C_y$	$Q C_d$	0.83527	$R_{C_y}$	${C_y}^{-1}$	1.197217
	$C_{c_{2,1}}$	$C_{\tau}$		0.167054	$R_{C_{\tau}}$	${C_{\tau}}^{-1}$	5.986085
	$C_{c_{3,1}}$	$C_{y_D}$		0.668216	$R_{C_{y_D}}$	${C_{y_D}}^{-1}$	1.496521
	$C_{c_{4,1}}$	$C_c$		0.83527	$R_{C_c}$	${C_c}^{-1}$	1.197217
$D_c$	$D_{c_{1,1}}$	$D_y$	$D_d + V C_d$	1.658918	$R_{D_y}$	${D_y}^{-1}$	0.602802
	$D_{c_{2,1}}$	$D_{\tau}$		0.331784	$R_{D_{\tau}}$	${D_{\tau}}^{-1}$	3.014012
	$D_{c_{3,1}}$	$D_{y_D}$		1.327135	$R_{D_{y_D}}$	${D_{y_D}}^{-1}$	0.753503
	$D_{c_{4,1}}$	$D_c$		0.658918	$R_{D_c}$	${D_c}^{-1}$	1.517639

------------------------------------------------------------------------------------------------------------------

Table 2: Useful Expressions
Name	Expression	Value
P	1 - α1(1-θ)	0.52
$Q$	$A_c/(A_d-1)$	1.085852
$V$	$(B_c-B_d)/(A_d-1)$	-0.34341

------------------------------------------------------------------------------------------------------------------

1. When $T_s = T_{s_0}$ it reduces to the original $SIM_d$ model.
2. At $T_s = 0$ the flows of $SIM_{T_s}$ will also be $0$ since the integration time will be $0$, and thus $C_{T_s} = 0$ and $D_{T_s} = 0$
3. $d C_{T_s} / d T_s$ evaluated at $T_s = 0$ will be equal to $C_c$ and $d D_{T_s} / d T_s$ evaluated at $T_s = 0$ will be equal to $D_c$. This is because $C_c$ and $D_c$ form the measurement equation for instantaneous flows, which is what happens in the limit for flows over a sample period $T_s$ as $T_s \rightarrow 0$ normalized by $T_s$.
4. If as $t \rightarrow \infty,\; g(t) \rightarrow \overline{g} > 0$ (where $\overline{g} \equiv$ the steady state value of g(t)) then as $T_s \rightarrow \infty$ we should have $SIM_{T_s}$ flows also going to $\infty$ because we're integrating over an infinite window of the non-zero steady sate of the model.
5. If as $t \rightarrow \infty,\; g(t) \rightarrow \overline{g} > 0$ then as $T_s \rightarrow \infty$ we should have $SIM_{T_s}$ flows averaged over an interval of length $T_s$ going to the steady state value of the flows divided by $T_{s_0}$.

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1. When $T_s = T_{s_0}$ it reduces to the original $SIM_d$ model.
2. At $T_s = 0$ the flows of $SIM_{T_s}$ will also be $0$ since the integration time will be $0$, and thus $C_{T_s} = 0$ and $D_{T_s} = 0$
3.  $d C_{T_s} / d T_s$ evaluated at $T_s = 0$ will be equal to $C_c$ and $d D_{T_s} / d T_s$ evaluated at $T_s = 0$ will be equal to $D_c$. This is because $C_c$ and $D_c$ form the measurement equation for instantaneous flows, which is what happens in the limit for flows over a sample period $T_s$ as $T_s \rightarrow 0$ normalized by $T_s$.
4. If as $t \rightarrow \infty,\; g(t) \rightarrow \overline{g} > 0$ (where $\overline{g} \equiv$ the steady state value of g(t)) then as $T_s \rightarrow \infty$ we should have $SIM_{T_s}$ flows also going to $\infty$ because we're integrating over an infinite window of the non-zero steady sate of the model.
5. If as $t \rightarrow \infty,\; g(t) \rightarrow \overline{g} > 0$ then as $T_s \rightarrow \infty$ we should have $SIM_{T_s}$ flows averaged over an interval of length $T_s$ going to the steady state value of the flows divided by $T_{s_0}$.

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