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I am confused why people say that the gradient points into the direction of the steepest ascent(Why is gradient the direction of steepest ascent?) because I think that I found a counter example for that.

Take $f(x,y) = x^2 - y^2$ at $(0,0)$: $\nabla f(0,0) = (0,0)$

However, intuitively, from (0,0) we can move in the x direction and increase the function so that would mean that the direction $d=(1,0)$ is the one that increases the function the most.

I am not familiar with the definition of differentiable for multivariate functions. Maybe $f$ is not differentiable at $(0,0)$ and that's why there is this paradox.

There must be a mistake in my reasoning, but I can't spot it. What is the flaw in my logic?

Thank you

Filat Nicolae
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  • Instantaneously, the rate of change of $f(x)=x^2$ at $x=0$ is $0$. The increase is second-order, not first-order. – Ted Shifrin Oct 16 '24 at 19:53
  • @TedShifrin Can you explain better what do you mean by "second-order"? (I assume you mean the second order taylor expansion but I do not see how it relates to the gradient here) The gradient provides the direction of the steepest ascent in the function $f$. However, this is clearly not true in my case. Why is that? Am I interpreting "steepest ascent" incorrectly? – Filat Nicolae Oct 16 '24 at 20:01
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    The statement you quote is meaningful only when the gradient is nonzero. – Ted Shifrin Oct 16 '24 at 20:28
  • Oh. I see. I was suspecting that. – Filat Nicolae Oct 16 '24 at 20:47
  • @TedShifrin In that case I have two more questions:
    1. From an optimization perspective, what mathematical tool allows me know in what direction should I change to input to achieve the highest increases in the function? (Even with saddle points)
    2. You say that the quote is not meaningful but I would even argue is wrong when the gradient is 0.
     – Filat Nicolae Oct 17 '24 at 06:52

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