4
$\begingroup$

I have a question that gives me a 3d function and asks me to calculate the gradient vector of it. This part I understand. It then asks me to indicate the points at which it does not exist. When does a gradient vector not exist? Is it when it equals to zero? Or does it mean when the function DNE.

Thanks.

1 Answers 1

4

Well the gradient is defined as the vector of partial derivatives so that it will exist if and only if all the partials exist. A zero gradient is still a gradient (it's just the zero vector) and we sometimes say that the gradient vanishes in this case (note that vanish and does not exist are different things). A function can still be well defined at a point without the gradient existing, in fact a function can still be continuous without the gradient existing.

  • 0
    One might define $\nabla f(x) $ to be the unique vector $g$ such that $f(x +h) =f(x) +g^T h +o(h) $ as $|h| \to 0$ (assuming such a $g$ exists).2012-11-06
  • 0
    Thanks! From what I understand tho, if a partial derivative of the gradient does not exist, it means that it is defined as a critical point.2012-11-06
  • 0
    @littleO That definition holds only for differentiable functions in which case it just reduces to the normal definition.2012-11-06
  • 0
    Is this critical point the point where the gradient does not exist? I don't believe so since, from what I have learned, this point is either a local/absolute maximum, minimum or saddle point.2012-11-06
  • 0
    A critical point is typically defined as a point in which the gradient vanishes. It'll be a local max/min or saddle depending on the definiteness of the Hessian.2012-11-06
  • 0
    So basically, a critical point is a point where the gradient does not exist?2012-11-06
  • 0
    No, I've emphasized that the gradient not existing and the gradient being zero are different. A critical point is a point in which the gradient is zero (but it still exists!). A point where the gradient does not exist is essentially a cusp where the function experiences some sort of turbulence in some direction. It's a point where the function fails to be partially differentiable in some direction.2012-11-06
  • 0
    With the definition I gave, which I like because it is valid in any finite dimensional inner product space over $\mathbb{R}$, existence of partial derivatives is not enough to guarantee existence of a gradient vector.2012-11-06
  • 1
    Oh. Thanks for the clarification! An example of a gradient being non existing then is f(x,y,z) = ln| 2x - y + z| when all variables are equal to 0? Since you ln of 0 does not exist, that means that it cannot be partially differentiable at this point. Therefore, the gradient doesn't exist at this point.2012-11-06
  • 1
    Yes, that would be one example. The absolute value function itself is another.2012-11-06