*Summary* *Problem:* * *(**0:00**)* Fitting non-linear functions to data (e.g., camera projection models). * *(**1:40**)* Minimize the distance between predicted values from the function and actual data points (residuals). *Approach:* 1. *(**2:16**)* *Define Residual Function:* * `r(x) = g(x) - y` * `g(x)`: Non-linear function with parameters `x` * `y`: Observed data 2. *(**2:25**)* *Minimize Sum of Squared Residuals:* * Minimize `F(x) = ||r(x)||^2` (sum of squared residuals) 3. *(**6:09**)* *Iterative Optimization (Gauss-Newton Method):* * Start with an initial guess for parameters `x`. * Approximate `r(x)` using first-order Taylor series (linearization). * Calculate the Jacobian matrix `J` (derivatives of `r(x)` w.r.t `x`). * Update parameters iteratively: * `x_(n+1) = x_n - (J^T * J)^(-1) * J^T * r(x_n)` * This step finds the minimum of the approximated quadratic function. 4. *(**8:20**)* *Repeat* until convergence (minimum is reached). *Key Points:* * Gauss-Newton is an iterative method for finding the minimum of a sum of squared residuals. * It uses a linear approximation of the non-linear function at each iteration. * The Hessian matrix in the update equation is approximated by `J^T * J`, neglecting a smaller term. * The algorithm iteratively refines the parameter estimates until a minimum is reached. i used gemini 1.5 pro to summarize the transcript
You can think of approximating a curve on a particular point by a line. On that particular point, over a small vicinity, the curve and the line are very close and almost equal, you can include higher-order terms (second order, third, ...) to make better approximation, but their effect is small, as the line is close enough to our curve.
Personally I cannot stand this video explanation. You are not really explaining, you are saying what you have written down. Maybe I'm just stupid but I don't follow at all.
Thank you for the clear explanation too!
Thanks for the clear explanation!
love how simple and clear the explanation!
Thank you so much for the neat explanation
Excellent!
*Summary*
*Problem:*
* *(**0:00**)* Fitting non-linear functions to data (e.g., camera projection models).
* *(**1:40**)* Minimize the distance between predicted values from the function and actual data points (residuals).
*Approach:*
1. *(**2:16**)* *Define Residual Function:*
* `r(x) = g(x) - y`
* `g(x)`: Non-linear function with parameters `x`
* `y`: Observed data
2. *(**2:25**)* *Minimize Sum of Squared Residuals:*
* Minimize `F(x) = ||r(x)||^2` (sum of squared residuals)
3. *(**6:09**)* *Iterative Optimization (Gauss-Newton Method):*
* Start with an initial guess for parameters `x`.
* Approximate `r(x)` using first-order Taylor series (linearization).
* Calculate the Jacobian matrix `J` (derivatives of `r(x)` w.r.t `x`).
* Update parameters iteratively:
* `x_(n+1) = x_n - (J^T * J)^(-1) * J^T * r(x_n)`
* This step finds the minimum of the approximated quadratic function.
4. *(**8:20**)* *Repeat* until convergence (minimum is reached).
*Key Points:*
* Gauss-Newton is an iterative method for finding the minimum of a sum of squared residuals.
* It uses a linear approximation of the non-linear function at each iteration.
* The Hessian matrix in the update equation is approximated by `J^T * J`, neglecting a smaller term.
* The algorithm iteratively refines the parameter estimates until a minimum is reached.
i used gemini 1.5 pro to summarize the transcript
Thanks Behnam it was very lucid explanation
I'm glad that you found it useful.
excellent
Does defining the objective function as r=g(x) -y make a difference than r=y-g(x)
no because the residuals get squared. (g(x) - y)^2 = (y - g(x))^2
7:02 Why is the second term so small that it's negligible,thanks
You can think of approximating a curve on a particular point by a line. On that particular point, over a small vicinity, the curve and the line are very close and almost equal, you can include higher-order terms (second order, third, ...) to make better approximation, but their effect is small, as the line is close enough to our curve.
where is 1/2 gone?
A constant plays no roles in finding min/ max.
Personally I cannot stand this video explanation. You are not really explaining, you are saying what you have written down. Maybe I'm just stupid but I don't follow at all.
do not be harsh on yourself...he explained it very well, but you need to have good background in linear algebra and calculus