update

pages/CSE559A/CSE559A_L6.md

@@ -118,7 +118,7 @@ BRDF dataset:
 
 #### Digital Camera block diagram
 
-![Digital Camera block diagram](https://static.notenextra.trance-0.com/images/CSE559A/DigitalCameraBlockDiagram.png)
+![Digital Camera block diagram](https://static.notenextra.trance-0.com/CSE559A/DigitalCameraBlockDiagram.png)
 
 Scanning protocols:
 
@@ -126,3 +126,88 @@ Scanning protocols:
 - Interlaced: odd and even lines are exposed at different times
 - Rolling shutter: each line is exposed as it is read out
 
+#### Eye
+
+- Pupil
+- Iris
+- Retina
+- Rods and cones
+- ...
+
+#### Eye Movements
+
+- Saccade
+  - Can be consciously controlled. Related to perceptual attention.
+  - 200ms to initiation, 20 to 200ms to carry out. Large amplitude.
+- Smooth pursuit
+  - Tracking an object
+  - Difficult w/o an object to track!
+- Microsaccade and Ocular microtremor (OMT)
+  - Involuntary. Smaller amplitude. Especially evident during prolonged
+    fixation.
+
+#### Contrast Sensitivity
+
+- Uniform contrast image content, with increasing frequency
+- Why not uniform across the top?
+- Low frequencies: harder to see because of slower intensity changes
+- Higher frequencies: harder to see because of ability of our visual system to resolve fine features
+
+### Color Perception
+
+Visible light spectrum: 380 to 780 nm
+
+- 400 to 500 nm: blue
+- 500 to 600 nm: green
+- 600 to 700 nm: red
+
+#### HSV model
+
+We use Gaussian functions to model the sensitivity of the human eye to different wavelengths.
+
+- Hue: color (the wavelength of the highest peak of the sensitivity curve)
+- Saturation: color purity (the variance of the sensitivity curve)
+- Value: color brightness (the highest peak of the sensitivity curve)
+
+#### Color Sensing in Camera (RGB)
+
+- 3-chip vs. 1-chip: quality vs. cost
+
+Bayer filter:
+
+- Why more green?
+  - Human eye is more sensitive to green light.
+
+#### Color spaces
+
+Images in python:
+
+As matrix.
+
+```python
+import matplotlib.pyplot as plt
+
+from mpl_toolkits.mplot3d import Axes3D
+from skimage import io
+
+def plot_rgb_3d(image_path):
+    image = io.imread(image_path)
+    r, g, b = image[:,:,0], image[:,:,1], image[:,:,2]
+    fig = plt.figure()
+    ax = fig.add_subplot(111, projection='3d')
+    ax.scatter(r.flatten(), g.flatten(), b.flatten(), c=image.reshape(-1, 3)/255.0, marker='.')
+    ax.set_xlabel('Red')
+    ax.set_ylabel('Green')
+    ax.set_zlabel('Blue')
+    plt.show()
+
+plot_rgb_3d('image.jpg')
+```
+
+Other color spaces:
+
+- YCbCr (fast to compute, usually used in TV)
+- HSV
+- L\*a\*b\* (CIELAB, perceptually uniform color space)
+
+Most information is in the intensity channel.

pages/Math4121/Math4121_L8.md

@@ -1 +1,83 @@
-# Lecture 8
+# Lecture 8
+
+## Continue on Riemann-Stieltjes Integral
+
+### Integrable Functions
+
+#### Theorem 6.9
+
+If $f$ is monotonic (increasing) on $[a, b]$ and $\alpha$ is continuous on $[a, b]$, then $f\in \mathscr{R}(\alpha)$ on $[a, b]$.
+
+Proof:
+
+Given a partition $P = \{a = x_0, x_1, \cdots, x_n = b\}$, we have
+
+$$
+M_i = \sup_{x\in [x_{i-i}, x_i]} f(x)\leq f(x_{i})
+$$
+
+$$
+m_i = \inf_{x\in [x_{i-1}, x_i]} f(x)\geq f(x_{i-1})
+$$
+
+So,
+
+$$
+\begin{aligned}
+U(P,f,\alpha) - L(P,f,\alpha) &= \sum_{i=1}^{n} (M_i - m_i)\Delta \alpha_i \\
+&\leq \sum_{i=1}^{n} \left[ f(x_i) - f(x_{i-1}) \right] \left[ \alpha(x_i) - \alpha(x_{i-1}) \right] \\
+
+&\leq \sum_{i=1}^{n} \left[ f(x_i) - f(x_{i-1}) \right](\sup_{x\in [x_{i-1}, x_i]} \alpha(x) - \inf_{x\in [x_{i-1}, x_i]} \alpha(x)) \\
+&=U(P,\alpha,f) - L(P,\alpha,f)
+\end{aligned}
+$$
+
+By Theorem 6.8, $\alpha\in \mathscr{R}(f)$, so for any $\epsilon > 0$, there exists a partition $P$ such that
+
+$$
+U(P,\alpha,f) - L(P,\alpha,f) < \epsilon
+$$
+
+Therefore, $U(P,f,\alpha) - L(P,f,\alpha)<U(P,\alpha,f) - L(P,\alpha,f) < \epsilon$, so $f\in \mathscr{R}(\alpha)$ on $[a, b]$.
+
+EOP
+
+#### Theorem 6.10
+
+Suppose $f$ is bounded on $[a, b]$ and has finitely many points $\{y_1, y_2, \cdots, y_J\}$ of discontinuity, and $\alpha$ is continuous on $[a, b]$. Then $f\in \mathscr{R}(\alpha)$ on $[a, b]$.
+
+Proof:
+
+Since $f$ is bounded, there exists a $M>0$ such that $|f(x)|\leq M$ for all $x\in [a, b]$.
+
+Let $\epsilon > 0$. Since $\alpha$ is continuous on $[a, b]$, we can find some intervals $[u_j,v_j]\subset (a,b)$ and $y_j\in [u_j,v_j]$ and $|\alpha(u_j) - \alpha(v_j)| < \epsilon$ for all $j=1,2,\cdots,J$.
+
+Set $K=[a,b]\setminus \bigcup_{j=1}^{J}(u_j,v_j)$. Since $K$ is compact, $f$ is uniformly continuous on $K$. Hence, there exists a $\delta > 0$ such that for any $s,t\in K$ and $|s-t|<\delta$, we have $|f(s)-f(t)|<\epsilon$.
+
+Let $P=\{x_0,x_1,\cdots,x_n=b\}$ containing all the points $u_j,v_j,\forall j=1,2,\cdots,J$ and $\Delta x_i<\delta,\forall x_i\notin \{u_j,v_j,\forall j=1,2,\cdots,J\}$.
+
+Then,
+
+If $x_i=u_j$ for some $j=1,2,\cdots,J$, then $M_i-m_i\leq M:=2\sup|f_x|$. But $\Delta \alpha_i\leq \epsilon$ for all $i=1,2,\cdots,n$.
+
+If $x_i\neq u_j$ for all $j=1,2,\cdots,J$, then by uniform continuity of $f$ on $K$, we have $M_i-m_i\leq \epsilon$.
+
+In either case, we have
+
+$$
+\begin{aligned}
+U(P,f,\alpha) - L(P,f,\alpha) &= \sum_{i=1}^{n} (M_i - m_i)\Delta \alpha_i \\
+&\leq J M\epsilon + \sum_{i=1}^{n} \epsilon \Delta \alpha_i \\
+&= \epsilon(J M + \sum_{i=1}^{n} \Delta \alpha_i)
+\end{aligned}
+$$
+
+Since $\epsilon$ is arbitrary, we have $U(P,f,\alpha) - L(P,f,\alpha) < \epsilon$.
+
+Therefore, $f\in \mathscr{R}(\alpha)$ on $[a, b]$.
+
+EOP
+
+#### Theorem 6.11
+
+Suppose $f\in \mathscr{R}(\alpha)$ on $[a, b]$, $m\leq f(x)\leq M$ for all $x\in [a, b]$, and $\phi$ is continuous on $[m, M]$, and let $h(x)=\phi(f(x))$ on $[a, b]$. Then $h\in \mathscr{R}(\alpha)$ on $[a, b]$.