TITLE: Why is Newton's method faster than gradient descent?
QUESTION [6 upvotes]: Can you provide some intuition as to why Newton's method is faster than gradient descent?
Often we are in a scenario where we want to minimize a function f(x) where x is a vector of parameters. To do that the main algorithms are gradient descent and Newton's method. For gradient descent we need just the gradient, and for Newton's method we also need the hessian. Each iteration of Newton's method needs to  do a linear solve on the hessian:

$$x \leftarrow x - \textrm{hess}(f,x) \backslash \textrm{grad}(f,x)$$

Where \ indicates doing a linear solve (like in matlab). In many cases the hessian is sparse, and in that case we would use an iterative method like conjugate gradient descent to do the linear solve. So what we're actually doing is this:

Compute a quadratic approximation of f around x.
Minimize that quadratic approximation with conjugate gradient descent.
Make a step in the direction of that minimum.

Compare this to normal gradient descent, where we just run (conjugate) gradient descent directly on the unapproximated original function f.
My question is: why is it faster to repeatedly do a quadratic approximation, then minimize that approximation with gradient descent, and then do a step in that direction, than to simply run gradient descent on the original function?

REPLY [2 votes]: I intend to give some glimpses, like one I did here.
Let us consider the minimization problem
$$0=g(\textbf{a})=\min_{\textbf{x}\in A}{g(\textbf{x})},\qquad {g(\textbf{x})}=\frac{1}{2}\|{\bf f}({\bf x})\|^2,$$ to some continuously differentiable function $\textbf{f}:A\to \mathbb{R}^p$, where $A$ is an open set of $\mathbb{R}^m$ containing $\textbf{a}$. Now, if you have some differentiable curve $\textbf{u}:(a,b)\to A$, you can apply the chain rule to obtain
$$\frac{d\, g({\bf u}(t))}{dt}= \left\langle {\bf u}'(t), \nabla g({\bf u}(t))\right\rangle= \left\langle {\bf u}'(t),[J{\bf f}({\bf u}(t))]^*{\bf f}({\bf u}(t))\right\rangle=\left\langle J{\bf f}({\bf u}(t)){\bf u}'(t),{\bf f}({\bf u}(t))\right\rangle,$$ in which $\langle \cdot,\cdot\rangle$ denotes the inner product.
A natural choice to $u(t)$ is given by the
the initial value problem (IVP) $$\left\{\begin{array}{rrrrl}{\bf u}'(t)&=&-\alpha \nabla g({\bf u}(t))\\ {\bf u}(0)&=&{\bf u}_0\end{array}\right.,$$
where $[J{\bf f}({\bf u}(t))]^* {\bf f}({\bf u}(t))=\nabla g({\bf u}(t))$, and $\alpha>0$.
If you use Euler method to solve this PVI numerically, you find the gradient descent method. This method, with step size $h_j$, takes the form
$${\bf u}_{j+1}=\phi({\bf u}_j),$$ to
$$\phi({\bf u})={\bf u}-h_j\alpha\left[J{\bf f}({\bf u})\right]^*{\bf f}({\bf u}),$$ as a fixed point iteration to solve $${\bf f}({\bf a})={\bf 0},\qquad \phi({\bf a})={\bf a}.$$ It converges when $$\|\phi'({\bf a})\|=\max_{1\leq i\leq m}|1-h_j\alpha s_i^2|<1,$$ if you have a good choice to ${\bf u}_0$, in which $s_i$ is a singular value of $J{\bf f}({\bf a})$.
It holds the inequality
$$\frac{d\, g({\bf u}(t))}{dt}= -\alpha\|\nabla g({\bf u}(t))\|^2\leq -2\alpha \sigma_{min}(t)^2g({\bf u}(t))\leq 0,$$ using the inequality
$$\|\nabla g({\bf u}(t))\|^2=\|[J{\bf f}({\bf u}(t))]^*{\bf f}({\bf u}(t))\|^2\geq \sigma_{min}(t)^2\|{\bf f}({\bf u}(t))\|^2,$$ in which $\sigma_{min}(t)$ is the smallest singular value of $J{\bf f}({\bf u}(t))$. You can prove that this produces the inequality $$g({\bf u}(t))\leq g({\bf u}(0))e^{-2\alpha \lambda(t)},\qquad \lambda(t)=\int_0^t\sigma_{min}(s)^2\,ds.$$
Another choice is the curve satisfing the initial value problem (IVP) $$\left\{\begin{array}{rrl}J{\bf f}({\bf u}(t)){\bf u}'(t)&=&-\alpha {\bf f}({\bf u}(t))\\ {\bf u}(0)&=&{\bf u}_0\end{array}\right.,$$ to some $\alpha>0$. You find that $$\frac{d\, g({\bf u}(t))}{dt}= -2\alpha g({\bf u}(t))$$ or $$g({\bf u}(t))=g({\bf u}(0))e^{-2\alpha t}.$$
In both cases, it follows that, ${g(\textbf{u}}(t))$ is a non increasing function. This also means that, if $g(\textbf{u}(t))> 0$, then $g(\textbf{u}(t+h))<g(\textbf{u}(t))$, when $0<h<h_t$, to some  $h_t>0$ close enough to $0$. Please see Picard Theorem and Lyapunov stabylity theory.
If $m=p$ and $J{\bf f}({\bf x})$ has bounded inverse matrix, to all $\textbf{x}\in A$, the previous IVP becomes
$$\left\{\begin{array}{lll}{\bf u}'(t)&=&-\alpha \left[J{\bf f}({\bf u}(t))\right]^{-1}{\bf f}({\bf u}(t))\\ {\bf u}(0)&=&{\bf u}_0\end{array}\right..$$
We can use the Euler method $$\left\{\begin{array}{rll}J{\bf f}({\bf u}_j) {\bf w}_j&=&-\alpha_j {\bf f}({\bf u}_j)\\ {\bf u}_{j+1}&=&{\bf u}_j+{\bf w}_j\end{array}\right.,$$ to solve the previous IVP numerically, where $\textbf{u}_0=u(0)$, $t_{j+1}=t_j+h_j$, $0<h_j$, ${\bf u}(t_{j+1}) \approx {\bf u}_{j+1}$  and $\alpha_j=\alpha h_j$.
This method, with step size $h_j$, takes the form
$${\bf u}_{j+1}=\psi({\bf u}_j),$$ to
$$\psi({\bf u})={\bf u}-h_j\alpha\left[J{\bf f}({\bf u})\right]^{-1}{\bf f}({\bf u}),$$ as a fixed point iteration to solve $${\bf f}({\bf a})={\bf 0},\qquad \psi({\bf a})={\bf a}.$$ It converges when $$\|\psi_j'({\bf a})\|=|1-h_j\alpha|<1,$$ if you have a good choice to ${\bf u}_0$.
We call $\alpha_j$ as the tuning parameter, as we call it in gradient descent method, and you should be carefully to choose it to have $g({\bf u}_{j+1})<g({\bf u}_j)$. Otherwise you can has a "bad" approximation  ${\bf u}(t_{j+1}) \approx {\bf u}_{j+1}$ in which $g({\bf u}_{j+1})>g({\bf u}_j)$.
But, when the things works well, $\alpha_j$ can be 1, to $j$ big enough.
Remark: You can rewrite this text using $$g(\textbf{x})=\frac{1}{2}\|\nabla \textbf{f}(\textbf{x})\|^2$$ instead, when you are working with the implication
$$ \min_{\textbf{x}\in A}{f(\textbf{x})}=f(\textbf{a})\Longrightarrow \nabla \textbf{f}(\textbf{a})=\textbf{0}.$$
The hessian matrix arises in this case.