Unless the closed form solution is extremely expensive to compute, it generally is the way to go when it is available. However,
For most nonlinear regression problems there is no closed form solution.
Even in linear regression (one of the few cases where a closed form solution is available), it may be impractical to use the formula. The following example shows one way in which this can happen.
For linear regression on a model of the form $y=X\beta$, where $X$ is a matrix with full column rank, the least squares solution,
$\hat{\beta} = \arg \min \| X \beta -y \|_{2}$
is given by
$\hat{\beta}=(X^{T}X)^{-1}X^{T}y$
Now, imagine that $X$ is a very large but sparse matrix. e.g. $X$ might have 100,000 columns and 1,000,000 rows, but only 0.001% of the entries in $X$ are nonzero. There are specialized data structures for storing only the nonzero entries of such sparse matrices.
Also imagine that we're unlucky, and $X^{T}X$ is a fairly dense matrix with a much higher percentage of nonzero entries. Storing a dense 100,000 by 100,000 element $X^{T}X$ matrix would then require $1 \times 10^{10}$ floating point numbers (at 8 bytes per number, this comes to 80 gigabytes.) This would be impractical to store on anything but a supercomputer. Furthermore, the inverse of this matrix (or more commonly a Cholesky factor) would also tend to have mostly nonzero entries.
However, there are iterative methods for solving the least squares problem that require no more storage than $X$, $y$, and $\hat{\beta}$ and never explicitly form the matrix product $X^{T}X$.
In this situation, using an iterative method is much more computationally efficient than using the closed form solution to the least squares problem.
This example might seem absurdly large. However, large sparse least squares problems of this size are routinely solved by iterative methods on desktop computers in seismic tomography research.
No.
These two methods both solve the same problem: minimizing the sum of squares error. One method is much faster than the other, but they are both arriving at the same answer.
This would be akin to asking "which gives a better answer to 10/4: long division or a calculator?"
Best Answer
The computational cost of gradient descent depends on the number of iterations it takes to converge. But according to the Machine Learning course by Stanford University, the complexity of gradient descent is $O(kn^2)$, so when $n$ is very large is recommended to use gradient descent instead of the closed form of linear regression.
source: https://www.coursera.org/learn/machine-learning/supplement/bjjZW/normal-equation