kb autocommit

content/posts/KBhconvolution.md

@@ -0,0 +1,133 @@
++++
+title = "convolution"
+author = ["Houjun Liu"]
+draft = false
++++
+
+For \\(f,g : \mathbb{R} \to \mathbb{C}\\), we have:
+
+\begin{equation}
+(f \* g)(x) = \int\_{\mathbb{R}} f(x-y) g(y) \dd{y} = \int\_{\mathbb{R}} f(y) g(x-y) \dd{y}
+\end{equation}
+
+
+## properties of convolution {#properties-of-convolution}
+
+-   \\((g \* f) (x) = (f \* g) (x)\\)
+-   \\(\mathcal{F}(f \* g) = \mathcal{F}(f)\mathcal{F}(g)\\)
+-   \\(\mathcal{F}^{-1}(\hat{f} \hat{g}) = f \* g\\)
+-   \\((f \* g)' = f \* g' = f' \* g\\)
+-   \\(\lambda ( f \* g ) = (\lambda f) \* g = f \* (\lambda g)\\)
+
+=&gt; "in a convolution, if **ANY ONE** of the two functions are [Differentiable]({{< relref "KBhuniqueness_and_existance.md#differentiable" >}}), both are [Differentiable]({{< relref "KBhuniqueness_and_existance.md#differentiable" >}})."; think about smoothing a jagged function using a [Gaussian]({{< relref "KBhgaussian.md" >}}).
+
+
+## examples {#examples}
+
+
+### rolling average {#rolling-average}
+
+\begin{align}
+U\_{L}(x) = \begin{cases}
+L, |x| \leq \frac{1}{2L} \\\\
+0, |x| > \frac{1}{2L}
+\end{cases}
+\end{align}
+
+The width of the area for which the expression is positive is \\(2L\\), and the height is \\(L\\), so the area (integral) is \\(1\\).
+
+So now let's consider:
+
+\begin{equation}
+(f \* U\_{L})(x)
+\end{equation}
+
+which is:
+
+\begin{equation}
+\int\_{\mathbb{R}} f(x-y) U\_{L}(y) \dd{y}
+\end{equation}
+
+meaning:
+
+\begin{equation}
+L \int\_{-\frac{1}{2}L}^{\frac{1}{2}L} f(x-y) \dd{y}
+\end{equation}
+
+You will note that we are sweeping something of window width \\(\frac{1}{L}\\) over the function, which averages the function \\(f\\) over the window \\(L\\).
+
+So convolving with this function essentially smoothes function over a window \\(\frac{1}{L}\\); as \\(L\\) decreases, we are averaging over a greater interval; vise versa.
+
+
+### signal compression {#signal-compression}
+
+Write your signal in terms of its Fourier transform:
+
+\begin{equation}
+f(t) = \frac{1}{2\pi} \int\_{-\infty}^{\infty} e^{it\lambda} \hat{f}(\lambda) \dd{\lambda}
+\end{equation}
+
+We can write:
+
+\begin{equation}
+\hat{f}(\lambda) \cdot 1\_{J}(\lambda)
+\end{equation}
+
+whose inverse Fourier transform would be:
+
+\begin{equation}
+f(x) \* \mathcal{F}\qty(1\_{J}(\lambda))
+\end{equation}
+
+
+## motivation {#motivation}
+
+What if we want the [Fourier Transform]({{< relref "KBhfourier_transform.md" >}}) of \\(\hat{f}(\lambda)\hat{g}(\lambda)\\) in terms of one expression?
+
+Consider:
+
+\begin{equation}
+\hat{f}(\lambda) \hat{g}(\lambda) = \qty(\int\_{\mathbb{R}} f(x) e^{-i\lambda x} \dd{x}) \qty(\int\_{\mathbb{R}} g(y) e^{-i\lambda y} \dd{y})
+\end{equation}
+
+Notice that because neither integral have dependence on the other, we can actually:
+
+\begin{equation}
+\hat{f}(\lambda) \hat{g}(\lambda) = \int\_{\mathbb{R}} \int\_{\mathbb{R}} f(x) g(y) e^{-i\lambda (x+y)} \dd{x}\dd{y}
+\end{equation}
+
+writing this as a change of variable:
+
+\begin{equation}
+\begin{cases}
+u = x+y \\\\
+x = u-y \\\\
+\dd{x} = \dd{u}
+\end{cases}
+\end{equation}
+
+we can write:
+
+\begin{equation}
+\hat{f}(\lambda) \hat{g}(\lambda) = \int\_{\mathbb{R}} \qty(\int\_{\mathbb{R}} f(u-y) g(y) e^{-i\lambda (u)} \dd{u})\dd{y}
+\end{equation}
+
+Considering they the integrands are isolated and decaying, we can swap them, pulling out also \\(e^{-i\lambda(u)}\\) because it has no \\(y\\) dependence:
+
+\begin{equation}
+\hat{f}(\lambda) \hat{g}(\lambda) = \int\_{\mathbb{R}} \qty(\int\_{\mathbb{R}} f(u-y) g(y) \dd{y})e^{-i\lambda (u)} \dd{u}
+\end{equation}
+
+Notice! The inner part is a function, and the outer part is a Fourier transform! This is similar to a [convolution (probability)]({{< relref "KBhrandom_variables.md#adding-random-variables" >}})!
+
+Meaning:
+
+\begin{equation}
+\hat{f}(\lambda) \hat{g}(\lambda) = \mathcal{F}(f \* g) = \mathcal{F}(f) \mathcal{F}(g)
+\end{equation}
+
+Operating on the inverse, we can obtain a similar result:
+
+\begin{equation}
+\mathcal{F}^{-1}(\hat{f} \hat{g}) = f \* g
+\end{equation}

content/posts/KBhos_index.md

@@ -128,6 +128,11 @@ how do we [trust]({{< relref "KBhprivacy.md#trust" >}}) software?
 
 -   [modern OS]({{< relref "KBhmodern_os.md" >}})
 
+
+#### trust and OS {#trust-and-os}
+
+-   [trust]({{< relref "KBhprivacy.md#trust" >}})
+
 ---
 
 An example of a good time:

content/posts/KBhprivacy.md

@@ -21,7 +21,58 @@ draft = false
     -   "fiduciary": proxy between you and a company
     -   "should anyone who has access to personal info have a fiduciary responsibility?"
 
----
+
+## key trust questions {#key-trust-questions}
+
+-   who/what do we trust?
+-   what do we do if trust isn't upheald?
+-   how to approach building trust
+
+
+## trust {#trust}
+
+**trust**: to stop questioning the responsibility of something
+
+-   intentions
+-   dependence
+-   extensions of agency
+
+We mostly don't trust software; instead, we trust the people that developed the software.
+
+
+## accountability {#accountability}
+
+a lot of people who are accountable in this chain:
+
+-   hardware designer (intel)
+-   OS developer (iOS, ec.)
+-   app developer
+-   users
+
+
+## stakeholder {#stakeholder}
+
+1.  **direct stakeholders** (people who are operating, technicians, etc.)
+2.  **indirect stakeholders**: patients
+
+purchase = long-term support ---- what do you do to get it fixed/repaired.
+
+
+## time {#time}
+
+-   support duration
+-   obsolescence (how long is the end of support)
+    -   products/services may not be garanteed forever
+    -   problems with halting use---requires deleting entire pentagram account
+
+
+## meltdown vulnerability {#meltdown-vulnerability}
+
+**meltdown**: hardware vulnerability that allows an user program to access kernel level pages of system memory.
+
+**potential ways of fixing a vulnerability/violation of trust**:
+
+<https://www.cs.cmu.edu/~rdriley/487/papers/Thompson_1984_ReflectionsonTrustingTrust.pdf>
 
 
 ## loss of privacy {#loss-of-privacy}

content/posts/KBhsu_math53_mar112024.md

@@ -4,56 +4,145 @@ author = ["Houjun Liu"]
 draft = false
 +++
 
-## convolution {#convolution}
+## heat equation on the entire line {#heat-equation-on-the-entire-line}
 
-For \\(f,g : \mathbb{R} \to \mathbb{C}\\), we have:
+\begin{equation}
+\pdv{u}{t} = \frac{1}{2} \pdv[2]{u}{x}
+\end{equation}
+
+We can try to find a:
 
 \begin{equation}
-(f \* g)(x) = \int\_{\mathbb{R}} f(x-y) g(y) \dd{y}
+U(0,x) = f(x)
 \end{equation}
 
+if we write:
+
+\begin{equation}
+\hat{U}(t,\lambda) = \int e^{-i x \lambda} U(t,x) \dd{x}
+\end{equation}
 
-### properties of convolution {#properties-of-convolution}
+which means we can write, with initial condtions:
 
--   \\((g \* f) (x) = (f \* g) (x)\\)
--   \\(\mathcal{F}(f \* g) = \mathcal{F}(f)\mathcal{F}(g)\\)
+\begin{equation}
+\hat{U} (t, \lambda) = \hat{f}(\lambda) e^{- t \frac{\lambda^{2}}{2}}
+\end{equation}
 
+We want to reach a close form:
+
+\begin{equation}
+U (t, x) =  \frac{1}{\sqrt{2\pi} t} \int\_{-\infty}^{\infty} f(y) e^{-\frac{(x-y)^{2}}{2t}} \dd{y}
+\end{equation}
+
+---
+
+Steps: recall we ended up at
+
+\begin{equation}
+\hat{U} (t, \lambda) = \hat{f}(\lambda) e^{- t \frac{\lambda^{2}}{2}}
+\end{equation}
+
+Let's call:
+
+\begin{equation}
+\hat{g}(\lambda) = e^{- t \frac{\lambda^{2}}{2}}
+\end{equation}
+
+so we have:
+
+\begin{equation}
+\hat{U} (t, \lambda) = \hat{f}(\lambda) \hat{g}(\lambda)
+\end{equation}
+
+we can use [convolution]({{< relref "KBhconvolution.md#convolution" >}}) to figure \\(U(t,x)\\).
+
+Recall that the 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}
 
-### motivation {#motivation}
+Let's first set:
 
-What if we want the [Fourier Transform]({{< relref "KBhfourier_transform.md" >}}) of \\(\hat{f}(\lambda)\hat{g}(\lambda)\\) in terms of one expression?
+\begin{equation}
+a = \frac{1}{t}
+\end{equation}
 
-Consider:
+Which will give us that:
 
 \begin{equation}
-\hat{f}(\lambda) \hat{g}(\lambda) = \qty(\int\_{\mathbb{R}} f(x) e^{-i\lambda x} \dd{x}) \qty(\int\_{\mathbb{R}} g(y) e^{-i\lambda y} \dd{y})
+g(x) = \frac{1}{\sqrt{2\pi t} } e^{-\frac{x^{2}}{2t}}
 \end{equation}
 
-Notice that because neither integral have dependence on the other, we can actually:
+Meaning, with convolution:
 
 \begin{equation}
-\hat{f}(\lambda) \hat{g}(\lambda) = \int\_{\mathbb{R}} \int\_{\mathbb{R}} f(x) g(y) e^{-i\lambda (x+y)} \dd{x}\dd{y}
+\mathcal{F}^{-1}(\hat{f} \hat{g}) = f \* g
 \end{equation}
 
-writing this as a change of variable:
+
+### why does this make sense {#why-does-this-make-sense}
+
+We are convolving a Gaussian against \\(f(x)\\). Meaning, at very small \\(t\\) , we are taking a very small window of size \\(1\\) against.
+
+
+### Heavyside function {#heavyside-function}
 
 \begin{equation}
-\begin{cases}
-u = x+y \\\\
-\dd{x} = \dd{u}
+f(x) = \begin{cases}
+1, x\geq  0 \\\\
+0, x<0
 \end{cases}
 \end{equation}
 
-we can write:
+This gives: if we split the room by \\(x\\). Recall:
 
 \begin{equation}
-\hat{f}(\lambda) \hat{g}(\lambda) = \int\_{\mathbb{R}} \qty(\int\_{\mathbb{R}} f(u-y) g(y) e^{-i\lambda (u)} \dd{u})\dd{y}
+U (t, x) =  \frac{1}{\sqrt{2\pi} t} \int\_{-\infty}^{\infty} f(y) e^{-\frac{(x-y)^{2}}{2t}} \dd{y}
 \end{equation}
 
-Considering they the integrands are isolated and decaying, we can swap them, pulling out also \\(e^{-i\lambda(u)}\\) because it has no \\(y\\) dependence:
+Given our \\(f\\), this becomes:
+
+\begin{equation}
+U (t, x) =  \frac{1}{\sqrt{2\pi} t} \int\_{0}^{\infty}  e^{-\frac{(x-y)^{2}}{2t}} \dd{y}
+\end{equation}
+
+If we change variables:
+
+\begin{align}
+\frac{(x-y)^{2}}{2t} - \qty( \frac{x}{\sqrt{2t}} - \frac{y}{\sqrt{2t}})^{2}
+\end{align}
+
+which means:
+
+\begin{equation}
+z = \frac{y}{2\sqrt{t}}
+\end{equation}
 
 \begin{equation}
-\hat{f}(\lambda) \hat{g}(\lambda) = \int\_{\mathbb{R}} \qty(\int\_{\mathbb{R}} f(u-y) g(y) \dd{y})e^{-i\lambda (u)} \dd{u}
+\frac{1}{\sqrt{\pi}} \int\_{0}^{\infty} e^{^{-\qty(\frac{x}{\sqrt{2t}} - z)^{2}}} \dd{z}
 \end{equation}
 
-Notice! The inner part is a function, and the outer part is a Fourier transform! This is similar to a [convolution (probability)]({{< relref "KBhrandom_variables.md#adding-random-variables" >}})!
+and we will also apply:
+
+\begin{equation}
+w = z - \frac{x}{\sqrt{2t}}
+\end{equation}
+
+which will give:
+
+\begin{equation}
+\frac{1}{\sqrt{\pi}} \int\_{-\frac{x}{\sqrt{2t}}}^{\infty}  e^{-w^{2}} \dd{w}
+\end{equation}
+
+notice, as \\(x\\) increases, we are integrating more of a Gaussian, which will be exceedingly close to \\(1\\); as \\(x\\) decreases, we'll get closer to \\(0\\). And also, \\(t\\) smoothed \\(x\\) out, which means as \\(t\\) increases the interface between \\(0\\) and \\(1\\) becomes smoother.
+
+
+## erf {#erf}
+
+[erf](#erf)
+
+
+## [convolution]({{< relref "KBhconvolution.md#convolution" >}}) {#convolution--kbhconvolution-dot-md}
+
+see [convolution]({{< relref "KBhconvolution.md#convolution" >}})