### Archive

Archive for December, 2012

## Compendium of Claims and Proofs, Including New Ones, Part I

December 3rd, 2012 No comments

I've condensed this exceptional mathematical wisdom here, which is still transforming as I organize and jot down ideas.

Part I v16Â (latest, but very unorganized afterÂ @Dynamics)

Part I v15

Part I v14

Part I v13

Part I v12

Part I v11

Part I v10

Part I v9

Part I v8

Part I v7

Part I v6

Part I v5

Part I v4

Part I v3

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## On Patch Stationariness II (RWLA,MCT,GT,AM Part X)

December 2nd, 2012 No comments

Claim: Â Suppose $p(x,y) = f_1(x) \cdot g_1(y) + f_2(x) \cdot g_2(y)$ and $\int p(x,y) dx = 1$. Â This last restriction causes freedom of choice of three functions, say the first three, and then $g_2(y) = \frac{1 - g_1 (y) \int_0^1 f_1(x) dx}{\int_0^1 f_2(x) dx} = \frac{1 - g_1(y) F_1}{F_2}$. Â This basic construction of $p(x,y)$ is the premise of all claims regarding patches I've proven here on my blog. Â Take an $a(x)$ with $\int_0^1 a(x) dx = 1$, with form that will be defined in the course of the proof. Â  Then $B = \int a(1-y) g_1(y) dy = \sum_{i=0}^{\infty} a^i(1-y) G_1^{i+1} \vert_0^1$ provided the sum converges. Â If it does, this gives rise to the "stationary" $p_\infty (x,y) = \frac{f_2(x)}{F_2} - \left( \frac{f_2(x) F_1}{F_2} - f_1(x) \right) B$ (and it is also a patch).

Proof: Since we are looking for stationary $p_\infty (x,y)$, we are looking for $a(x) \star p(x,y) = \lambda a(x)$ with $\lambda = 1$ in direct analogy with discrete Markov chains. Â The star operator is explicitly as I've defined it previously implies $\int_0^1 a(1 - y) p(x,y) dy = a(x)$. Â By the definition of $p(x,y)$, we have that $\int_0^1 a(1 - y) \left( f_1(x) g_1(y) + f_2(x) \left( \frac{1 - g_1(y) F_1}{F_2} \right) \right) dy = a(x)$. Â Expansion results in $a(x) = f_1(x) \int_0^1 a(1-y) g_1(y) dy + \frac{f_2(x)}{F_2} \cdot 1 - \frac{F_1}{F_2} f_2(x) \int_0^1 a(1-y) g_1(y) dy$, where we have simplified $\int_0^1 a(1-y) dy$ to $1$ because the transformation to the y-axis does not change the integral result.

Next we have the expression

$a(x) = \frac{f_2(x)}{F_2} - \left( \frac{F_1 f_2(x)}{F_2} - f_1(x) \right) \int_0^1 a(1-y) g_1(y) dy$

or

$a(x) = \frac{f_2(x)}{F_2} - \left( \frac{F_1 f_2(x)}{F_2} - f_1(x) \right) B$ and derivatives

$a^i(x) = \frac{f_2^i(x)}{F_2} - \left( \frac{F_1 f_2^i(x)}{F_2} - f_1^i(x) \right) B$

We want to obtain $B$, so that the expression $\int_0^1 a(1-y) g_1(y) dy$ can be clearly defined. Â We use the tabular method to simplify the integration by parts.

and obtain

$B = a(1-y) G_1^1(y) + a^{'}(1-y) G_1^2(y) + \ldots \vert_0^1 = \sum_{i=0}^\infty a^i (1-y) G_1^{i+1} \vert_0^1$

Lastly the steady-state "stationary" patch is exactly this $a(x)$, so that $p_\infty(x,y) = a(x)$ as we desired to show.

Notice that $\int_0^1 p_\infty(x,y) dx = 1$ by hypothesis, since it is equal toÂ $a(x)$.

Existence. Â Lastly, notice that $p_\infty (x,y)$ exists provided the sum converges. Â In previous exercises, I have shown that such converges in two specific examples: when derivatives of $a(x)$ vanish (when $a(x)$ is a finite polynomial) and when derivatives of $a(x)$ are periodic but vanish for the periodic terms.