Imgae alignment

Newton’s method:

Find the f(x) = 0.
For an initialized point, calculate the tangent line. And set the point that is intersection of x-axis and tangent line as next x.
As the animated picture from wiki:

So write in math:
( x_{n+1} = x_n - \frac{f(x_n)}{f^\prime(x_n)} )

Lucas-Kanade:

When in image alignment that only translation:
( E(u, v) = \sum{x,y}{(I(x+u, y+v) - T(x, y))}^2 )
Do Taylor’s expand in I(x+y, y+v):
( E(u, v) = \sum{x,y}{(I(x, y) - T(x, y) + uI_x + + vIy)}^2 )
And differentiate to find the extremal point:
( 0 = \frac{\partial E}{\partial u} = \sum{x,y}{2I_x(I(x,y)-T(x,y)+uI_x+vIy)} )
( 0 = \frac{\partial E}{\partial v} = \sum{x,y}{2I_y(I(x,y)-T(x,y)+uI_x+vIy)} )
Reformat as matrix:
( \begin{bmatrix} \sum{x,y}{Ix^2} & \sum{x,y}{I_xIy} \ \sum{x,y}{I_xIy} & \sum{x,y}{Iy^2} \end{bmatrix} \begin{bmatrix} u \ v \end{bmatrix} = \begin{bmatrix} \sum{x,y}{Ix(T(x,y)-I(x,y))} \ \sum{x,y}{I_y(T(x,y)-I(x,y))} \end{bmatrix} )
One thing to note:
Although in the math is plus u and v.
But if we write in transformation matrix, it is minus u and v.
Because I(x+u, y+v) means original x, y becomes the x-y, y-v in a transformed image.

circular convolution
1. why don’t just use translation matrix to fit
2. d/dx d/dy use which kernel is better