Here when we talk about $f(\mathbf{x})$, $\mathbf{x}$ is essentially a vector in 2 dimensions.
I understood that the peak is not differentiable, but is the gradient defined on the lines of the pyramid structure(connecting the vertices of the base square to the peak) as well?
Is the gradient undefined on one of the dimensions on those lines?
marshmallow
I always had a hard time trying to understand the relationship between gradients and directional directives, but this clarified my understanding!
keenan
@tarangs Right: the function is not differentiable along the lines $x_1 = x_2$ and $x_1 = -x_2$ either. However, it does have a well-defined directional derivative along those same lines. In constrast, the point $\mathbf{x} = (0,0)$ does not have a well-defined derivative along any line.
Here when we talk about $f(\mathbf{x})$, $\mathbf{x}$ is essentially a vector in 2 dimensions. I understood that the peak is not differentiable, but is the gradient defined on the lines of the pyramid structure(connecting the vertices of the base square to the peak) as well? Is the gradient undefined on one of the dimensions on those lines?
I always had a hard time trying to understand the relationship between gradients and directional directives, but this clarified my understanding!
@tarangs Right: the function is not differentiable along the lines $x_1 = x_2$ and $x_1 = -x_2$ either. However, it does have a well-defined directional derivative along those same lines. In constrast, the point $\mathbf{x} = (0,0)$ does not have a well-defined derivative along any line.