I'm trying to learn the conjugate gradient method, and I'm reading this article. The author presents the line search algorithm as the procedure where we choose a step size $\alpha$ that minimizes our function $f$ along a line. I understand the mathematics behind it: that is, finding the directional derivative and setting it to zero and solving for the optimal $\alpha$. I have issues with visualizing this process. Note the two figures.

enter image description here enter image description here

Figure on the left is the plot of the function and a plane. The intersection is the line along which we seek to minimize $f$. We start from $\vec{x_0}=\{-2,-2\}$, which is roughly where I have drawn the red dot. Also, the green arrow represents the direction of $-\nabla{f}$ at that point. The second figure shows the intersection of the two surfaces as a function of $\alpha$.

My issue is here. If we start at $\vec{x_0}$ and move $\alpha$ units along the direction of $-\nabla{f}$ evaluated at $\vec{x_0}$, aren't we just going to move along a straight line, away from $f$? How do we get the curved line to be a function of $\alpha$? Am I interpreting the meaning of $\alpha$ correctly here?


$\alpha$ causes you to move along the domain of $f$, which is a subset of $\mathbb{R}^2$.

The green arrow (which I'm assuming is $\nabla f(\mathbf{x}_i)$) should technically lie in $\mathbb{R}^2$, and not as a vector in $\mathbb{R}^3$. So, when we move along the negative gradient, we're just moving along the domain, and not away from the surface defined by $z = f(\mathbf{x})$.

  • $\begingroup$ Yes, $\nabla{f}$ should lie in $\mathbb{R}^2$. So is the negative gradient in the direction along the plane (from left to right)? $\endgroup$ – Ptheguy May 24 '18 at 0:46
  • $\begingroup$ Yes, if you translate that green vector down to $\mathbb{R}^2$, then that gives you the direction that $\mathbf{x}_i + \alpha\nabla f$ moves along. $\endgroup$ – youngsmasher May 24 '18 at 0:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.