Relationship Between Cut vectors and gradients

Previously, we mistakenly thought that the tangent vector is a gradient. Due to the recent look at the method of the Laplace, we encountered this problem and found out the relationship between the tangent vector and the gradient.

In simple terms, the tangent vector and the gradient are the derivatives, but it depends on who is the derivative. For example:

Y = x ^ 2, simple.

Returns the tangent vector at the vertex (A, B.

In this case, you need to write the parameter formula X = x and y = x ^ 2 to obtain the (1, 2x) derivative of x and bring (a, B) into it.

Calculates the gradient at the point (A, B.

In this case, we need to write F (x, y) = x ^ 2-y and evaluate the partial direction of X and Y to get (2x,-1) respectively ).

At this point, we understand that the gradient is not the same as the cut vector, and it is still vertical. Because only the method vector is perpendicular to the cut vector, the gradient is the method vector .. Maybe the biggest difference with the normal vector is the direction ..

Let's look at the effect of the figure below, respectively indicating that the cut vector and the gradient are at the point (1, 1 ).

There is a problem that needs to be understood:

In many gradient descent methods, what we see is just a curve that is used to obtain the partial direction. We always think that this is the tangent direction. In fact, we just make a component of the gradient, at this time, this component is indeed in the tangent direction of X, but the gradient is finally obtained as a vector jointly determined by N components. The final jointly determined vector is perpendicular to the tangent.