Yes, you're definitely right that the partial derivative with respect to w1 depends on both w2 and w3. (And perhaps the slide is a bit misleading in that respect.)
In the example on this slide, the partial derivative of a3 (output of a particular layer) with respect to f2 (input of a particular layer) is a function of w3. Similarly, the partial derivative of a2 with respect to f1 is a function of w2. So as a result, the weights w2 and w3 will appear when computing the gradient for w1.
Why don't we depend on our weights w2, w3 in computing the gradient for w1? Aren't those essentially constants in our function?
If I had something like y = f(5*g(x)), dy/dx would include the 5 in there, right?
There appears to be an error on this slide: in the third equation, $T(x) - I(W(x;p))$ should be $I(W(x;p)) - T(x)$.
Yes! That's all there is to it.
Is lambda * W the partial derivative of just L_reg with respect to W?
Sounds correct to me! Let me add some details.
The camera center (in world coordinates) is a homogeneous 3D coordinate $\mathbf{c}$ that satisfies the equation $\mathbf{0} = \mathbf{P}\mathbf{c}$. That is, it's the only point in 3D space whose projection onto the image plane is undefined.
As a result, there are a couple of ways to solve for the camera center $\mathbf{c}$. As you stated, one approach is to compute the smallest eigenvector of a 4x4 matrix $\mathbf{P}^T \mathbf{P}$. Or compute the SVD $\mathbf{P} = \mathbf{U}\Sigma \mathbf{V}^T$, and grab the smallest singular vector of $\mathbf{V}$.
To find the camera center, do we take the eigenvector v in V in U?V^T that corresponds to an eigenvalue of 0, and that vector by itself is the coordinates of the camera center?
Great question! This is described in a bit more detail on page 743 of the Szeliski textbook. It might be helpful to refer to the matrix cookbook, which contains a number of identities related to matrices and vectors.
For the first step, we should note that $||\mathbf{x}||^2 = \sum_{k=1}^{K} \mathbf{x}_k^2 = \mathbf{x}^T \mathbf{x}$. Using this fact, we can expand out the first term as follows: $||\mathbf{A}\mathbf{x} - \mathbf{b}||^2 = (\mathbf{A}\mathbf{x} - \mathbf{b})^T (\mathbf{A}\mathbf{x} - \mathbf{b}) = \mathbf{x}^T (\mathbf{A}^T \mathbf{A})\mathbf{x} - 2\mathbf{x}^T \mathbf{A}^T\mathbf{b} + \mathbf{b}^T \mathbf{b}$.
After that, the last few lines take the derivative with respect to $\mathbf{x}$, set the derivative to 0, and solve for $\mathbf{x}$. If you would like more details on how to do this (e.g., computing derivatives of matrices and vectors), I would recommend looking at Section 2.4 of the matrix cookbook.
How did we expand out E_LLS = ||Ax-b||^2 to get the line below it?
Instead of convolving an image with a 2D separable filter, we can instead perform two separate 1D convolution operations to our image. The first convolution is between a $M \times M$ image and a $1 \times N$ kernel. The second convolution is between the resulting $M \times M$ image (output from the first step) and a $N \times 1$ kernel. The cost of a convolution operation is $M^2 N$ in both cases (for every pixel in our image, we have to perform $N$ multiplication and addition operations). So the total cost is $2M^2 N$.
How do you get the cost of convolution with separable filter in 2 X N X M^2? How would you multiply the row/column vectors in to get it?
As mentioned in class (someone pointed it out), d is indeed $\rho$. Thanks for clearing that up, @lemonyum!
I think it's the values of rho
what is d here
Hello!
Good catch---this is a typo: I = g.
what is I and g here?
Helloooo : )
Hey!
Thanks! That makes a lot of sense!
The $h=2$ represents the distance between samples. So in this case, we're sampling two values $f(x+1)$ and $f(x-1)$ at points $x+1$ and $x-1$, which have a distance of $(x+1)-(x-1) = 2$.
We could change this value though. For example, we could sample $f(x+1)$ and $f(x)$ (corresponding to $h=1$), or $f(x-2)$ and $f(x+2)$ (corresponding to $h=4$). However, $h > 0$ should be as small as possible for the derivative to be approximately correct.
Hi!
Why do we set h=2 here? Where does 2 come from?
Hello!
Hi!
Hi!
Yup, should be just fine.
Hi, Is the first version of the book okay?
Hi!
Yeah, that's a more apt description of what's going on here (think I'll steal it in the future).
This kind of blurring is like projecting my PC on a low-resolution screen instead of taking pictures from my camera.
Hi everyone!
hi yall
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Hiiii
It appears that the domains over which the integrals are being performed are mixed up. The last equation should be written as follows:
$S = \int_{0}^{2\pi} \int_{0}^{\pi} \sin(\theta) d\theta d\phi = 4\pi$