kb autocommit

content/posts/KBhfourier_transform.md

@@ -44,7 +44,7 @@ Where,
 \end{equation}
 
 \begin{equation}
-f(x) = \frac{1}{2\pi} \int\_{\infty}^{\infty} e^{ix\lambda} \hat{f}(\lambda) \dd{\lambda}
+f(x) = \frac{1}{2\pi} \int\_{\infty}^{-\infty} e^{ix\lambda} \hat{f}(\lambda) \dd{\lambda}
 \end{equation}
 
 We sometimes write:
@@ -91,6 +91,66 @@ Consider also:
 you can show this in a similar way, by attempting to distribute a \\(\dv \lambda\\) into the Fourier transform and showing that they are equal.
 
 
+### Fourier Transform of a Gaussian {#fourier-transform-of-a-gaussian}
+
+\begin{equation}
+\mathcal{F}\qty(e^{-\frac{ax^{2}}{2}}) = \sqrt{\frac{2\pi}{a}}e^{-\frac{\lambda^{2}}{2a}}
+\end{equation}
+
+and:
+
+\begin{equation}
+\mathcal{F}^{-1}\qty(e^{-a\frac{\lambda^{2}}{2}}) = \frac{e^{-\frac{x^{2}}{2a}}}{\sqrt{2\pi a}}
+\end{equation}
+
+---
+
+we obtain this:
+
+\begin{equation}
+u = e^{-\frac{x^{2}}{2}}
+\end{equation}
+
+and:
+
+\begin{equation}
+\dv{u}{x} = -xe^{-\frac{x^{2}}{2}} = -xu
+\end{equation}
+
+and if we took a Fourier transform on both sides, we obtain:
+
+\begin{equation}
+\mathcal{F}\qty(\dv{u}{x} + xu) = 0 = i \lambda \hat{u} + i \pdv{\hat{u}} = 0
+\end{equation}
+
+and note that this is the same equation. Meaning:
+
+\begin{equation}
+\mathcal{F}\qty(\dv{u}{x} + xu) = \dv{\lambda}{x} + \lambda u
+\end{equation}
+
+this gives:
+
+\begin{equation}
+\mathcal{F}(u) = Cu
+\end{equation}
+
+which is what we see.
+
+
+### Look! A table {#look-a-table}
+
+{{< figure src="/ox-hugo/2024-03-06_21-36-14_screenshot.png" >}}
+
+where:
+
+\begin{equation}
+\Lambda\_{a}
+\end{equation}
+
+is the triangle between \\([-a, a]\\), that goes up to \\(1\\).
+
+
 ### interpreting \\(\lambda\\) {#interpreting-lambda}
 
 

content/posts/KBhgaussian.md

@@ -0,0 +1,172 @@
++++
+title = "Gaussian"
+author = ["Houjun Liu"]
+draft = false
++++
+
+The [Gaussian]({{< relref "KBhgaussian.md" >}}), in general, gives:
+
+\begin{equation}
+e^{-\frac{ax^{2}}{2}}
+\end{equation}
+
+which is a Bell-Shaped curve. It's pretty darn important
+
+
+## [Heat Equation]({{< relref "KBhheat_equation.md" >}}) and [Gaussian]({{< relref "KBhgaussian.md" >}}) {#heat-equation--kbhheat-equation-dot-md--and-gaussian--kbhgaussian-dot-md}
+
+\begin{equation}
+H(t,x) = \frac{1}{\sqrt{2\pi} t}e^{-\frac{x^{2}}{2t}}
+\end{equation}
+
+You will note that \\(H\\) **does** satisfy the heat equation:
+
+\begin{equation}
+\pdv{U}{t} = \pdv[2]{U}{x}
+\end{equation}
+
+
+### solving [Heat Equation]({{< relref "KBhheat_equation.md" >}}) without boundary {#solving-heat-equation--kbhheat-equation-dot-md--without-boundary}
+
+Consider the partial [Fourier Transform]({{< relref "KBhfourier_transform.md" >}}) on the \\(x\\) variable of the heat equation.
+
+\begin{equation}
+U(t,x) = \frac{1}{2\pi} \int\_{\mathbb{R}} e^{ix\lambda} \hat{U} \qty(t,\lambda) \dd{\lambda}
+\end{equation}
+
+Taking derivatives of this:
+
+\begin{equation}
+\pdv{U}{t} (t,x) = \frac{1}{2\pi} \int\_{\mathbb{R}} e^{i\lambda x} \pdv{\hat{U}}{t} (t,\lambda) \dd{\lambda}
+\end{equation}
+
+and:
+
+\begin{equation}
+\pdv[2]{U}{x} = \frac{1}{2\pi} \int\_{\mathbb{R}} \qty(-\lambda^{2}) e^{ix \lambda } \hat{U}(t,\lambda) \dd{\lambda}
+\end{equation}
+
+Because these two are equal, it gives us that:
+
+\begin{equation}
+\hat{U}(t,\lambda) = -\lambda^{2} \hat{U}(t,\lambda)
+\end{equation}
+
+meaning:
+
+\begin{equation}
+\hat{U}(t,\lambda) = a(\lambda)e^{-\lambda^{2}(t)}
+\end{equation}
+
+Finally, at:
+
+\begin{equation}
+\hat{U}(0,\lambda) = a(\lambda) = \hat{f}(\lambda)
+\end{equation}
+
+We see that:
+
+\begin{equation}
+\hat{U}(t,\lambda) = \hat{f}(\lambda)e^{-\lambda^{2}(t)}
+\end{equation}
+
+To get our original function back, we need to inverse Fourier transform it:
+
+\begin{equation}
+U(t,x) = \frac{1}{2\pi} \int\_{\mathbb{R}} e^{ix\lambda - \lambda^{2}t} \hat{f}(\lambda) \dd{\lambda}
+\end{equation}
+
+
+### closed form solution {#closed-form-solution}
+
+\begin{equation}
+U(t,x) = \frac{1}{\sqrt{2\pi} t} \int\_{\mathbb{R}} f(y) e^{-\frac{(x-y)^{2}}{2t}} \dd{y}
+\end{equation}
+
+this is exactly:
+
+\begin{equation}
+\int\_{\mathbb{R}}f(y) H(t,(x-y)) \dd{y} = \int\_{\mathbb{R}}\frac{1}{\sqrt{2\pi} t}e^{-\frac{(x-y)^{2}}{2t}} f(y) \dd{y}
+\end{equation}
+
+We can understand this when \\(t \to 0\\), where there is a single, narrow, area \\(1\\) band which we sweep across all of \\(y\\). Because its thin and \\(1\\), its basically \\(f(x)\\) at each \\(y\\).
+
+
+## Integrating Gaussian, more Generally {#integrating-gaussian-more-generally}
+
+Let's integrate:
+
+\begin{equation}
+\int\_{-\infty}^{\infty} e^{-\frac{{ax}^{2}}{2}}  \dd{x}
+\end{equation}
+
+Let's replace: \\(s = \sqrt{a} x\\)
+
+This gives us that (based on [Integrating Gaussian](#integrating-gaussian)):
+
+\begin{equation}
+x = \sqrt{\frac{2\pi}{a}}
+\end{equation}
+
+If we replace \\(a\\) by \\(\frac{1}{t}\\), we obtain:
+
+\begin{equation}
+\frac{1}{\sqrt{2\pi}t} \int\_{-\infty}^{\infty} e^{-\frac{x^{2}}{2t}} \dd{x} = 1
+\end{equation}
+
+by rescaling \\(x(a)\\) function above.
+
+If \\(t\\) increases, you will see that this function diffuses from a single point at \\(0\\) and spreading out. Notice, that over the whole real line, no matter what the \\(t\\) is, you always end up with integral \\(1\\).
+
+
+## Integrating Gaussian {#integrating-gaussian}
+
+Let's integrate:
+
+\begin{equation}
+\int\_{-\infty}^{\infty} e^{-\frac{x^{2}}{2}}  \dd{x}
+\end{equation}
+
+---
+
+computing this is funny:
+
+\begin{equation}
+A \cdot A = \int\_{-\infty}^{\infty} e^{-\frac{x^{2}}{2}}  \dd{x} \int\_{-\infty}^{\infty} e^{-\frac{y^{2}}{2}}  \dd{y}
+\end{equation}
+
+We can think of this as a double integral:
+
+\begin{equation}
+A \cdot A = \int\_{-\infty}^{\infty} \int\_{-\infty}^{\infty} e^{-\frac{x^{2}}{2}}   e^{-\frac{y^{2}}{2}} \dd{x}  \dd{y}
+\end{equation}
+
+meaning we get:
+
+\begin{equation}
+A \cdot A = \int\_{-\infty}^{\infty} \int\_{-\infty}^{\infty} e^{-\frac{x^{2}+y^{2}}{2}} \dd{x}  \dd{y}
+\end{equation}
+
+Its polar time; recall:
+
+\begin{equation}
+x^{2} + y^{2} = r^{2}
+\end{equation}
+
+we can now go over this whole thing by converting into polar (notice the extra factor \\(r\\)):
+
+\begin{equation}
+A \cdot A = \int\_{0}^{2\pi} \int\_{0}^{\infty} e^{-\frac{r^{2}}{2}} r \dd{r} \dd{\theta}
+\end{equation}
+
+very suddenly we can use u sub on \\(r\\) to obtain:
+
+\begin{equation}
+2\pi \int\_{0}^{\infty} e^{-u} \dd{u} = 2\pi \cdot 1 = 2\pi
+\end{equation}
+
+Meaning:
+
+\begin{equation}
+A = \sqrt{2\pi}
+\end{equation}

content/posts/KBhgaussian_distribution.md

@@ -89,7 +89,7 @@ Z=\mathcal{N}(0,1)
 mean 0, variance 1. You can transform anything into a standard normal via the following linear transform:
 
 
-#### transformation into [standard normal](#standard-normal) {#transformation-into-standard-normal--orgb5839a8}
+#### transformation into [standard normal](#standard-normal) {#transformation-into-standard-normal--org430b977}
 
 \begin{equation}
 X \sim \mathcal{N}(\mu, \sigma^{2})

content/posts/KBhheat_equation.md

@@ -24,6 +24,12 @@ and with [Neumann Conditions]({{< relref "KBhsu_math53_feb232024.md#neumann-cond
 u\_{k}(t,x) =  \sum b\_{k} e^{ - \frac{k^{2} \pi^{2}}{l^{2}} t } \cos \qty( \frac{k \pi x}{l})
 \end{equation}
 
+with infinite boundaries:
+
+\begin{equation}
+U(t,x) =\frac{1}{\sqrt{4 \pi \alpha t}} \int\_{\mathbb{R}} f(y)  e^{-\frac{(x-y)^{2}}{4\alpha t}} \dd{y}
+\end{equation}
+
 general system:
 
 \begin{equation}
@@ -47,7 +53,7 @@ you can remove the constant by finanglisng because the constant drops out when s
 
 ## damping {#damping}
 
-[damped heat equation]({{< relref "KBhdamped_heat_equation.md#damped-heat-equation" >}})
+[damped heat equation]({{< relref "KBhdamped_heat_equation.md" >}})
 
 
 ## Solving Heat Equation {#solving-heat-equation}

content/posts/KBhlanguage_information_index.md

@@ -136,3 +136,9 @@ DP costs \\(O(nm)\\), backtrace costs \\(O(n+m)\\).
 ### Neural Nets {#neural-nets}
 
 -   [Neural Networks]({{< relref "KBhneural_networks.md" >}})
+
+
+### The Web {#the-web}
+
+-   [Web Graph]({{< relref "KBhweb_graph.md" >}})
+-   [Social Network]({{< relref "KBhsocial_network.md" >}})

content/posts/KBhmarkov_chain.md

@@ -0,0 +1,45 @@
++++
+title = "Markov Chain"
+author = ["Houjun Liu"]
+draft = false
++++
+
+A [Markov Chain]({{< relref "KBhmarkov_chain.md" >}}) is a chain of \\(N\\) states, with an \\(N \times N\\) transition matrix.
+
+1.  at each step, we are in exactly one of those states
+2.  the matrix \\(P\_{ij}\\) tells us \\(P(j|i)\\), the probability of going to state \\(j\\) given you are at state \\(i\\)
+
+And therefore:
+
+\begin{equation}
+\sum\_{j=1}^{N} P\_{ij} = 1
+\end{equation}
+
+
+## Ergotic Markov Chain {#ergotic-markov-chain}
+
+a markov chain is [Ergotic](#ergotic-markov-chain) if...
+
+1.  you have a path from any one state to any other
+2.  for any start state, after some time \\(T\_0\\), the probability of being in any state at any \\(T > T\_0\\) is non-zero
+
+Every [Ergotic Markov Chain](#ergotic-markov-chain) has a long-term visit rate:
+
+i.e. a steady state visitation count exists. We usually call it:
+
+\begin{equation}
+\pi = \qty(\pi\_{i}, \dots, \pi\_{n})
+\end{equation}
+
+
+### Computing steady state {#computing-steady-state}
+
+Fact:
+
+let's declare that \\(\pi\\) is the steady state to a transition matrix \\(T\\); recall that the FROM states are the rows, which means that \\(\pi\\) has to be a row vector; \\(\pi\\) being a steady state makes:
+
+\begin{equation}
+\pi T = \pi
+\end{equation}
+
+This is a left e.v. with [eigenvalue]({{< relref "KBheigenvalue.md" >}}) \\(1\\), which is the principle [eigenvector]({{< relref "KBheigenvalue.md" >}}) of \\(T\\) as transition matricies always have [eigenvector]({{< relref "KBheigenvalue.md" >}}) eigenvalue to \\(1\\).