TITLE: Is gradient descent nothing other than discretized gradient flow?
QUESTION [7 upvotes]: I can find definitions of the gradient flow of a scalar field $f$ as
$$\frac{d \xi}{dt} = - \nabla_\xi f$$
in here and here.
Gradient descent can be used to find a minimum in $f$ and can be written as
$$\xi_{i+1} = \xi_{i} - \lambda \nabla_\xi f,$$
where $\lambda$ is a constant scalar.
As far as I understand, gradient descent is nothing other than discretized gradient flow. So I set $\lambda = 1$ to obtain
$$\frac{d \xi}{dt} = \lim_{\Delta t \rightarrow 0} \frac{\xi_{i+1} - \xi_i}{\Delta t} = - \lim_{\Delta t \rightarrow 0} \frac{\nabla_\xi f}{\Delta t} = - \frac{d}{dt} \nabla_\xi f \stackrel{???}{=} - \nabla_\xi f = \frac{d \xi}{dt}.$$
This does not make much sense to me... Shouldn’t both $\frac{d \xi}{dt}$ give the same? What mistake am I making? Could anyone help me, please?

REPLY [4 votes]: What you are missing is that $\lambda$ "is" the time scale $\Delta t$. So you should rather interpret the first formula as
$$\xi_{t+\Delta t}:=\xi_t - (\Delta t)\nabla_{\xi_t}f,$$
and with this you get what you want.

REPLY [3 votes]: I think the confusion might stem from the ambiguity in notation. The definition of gradient flow is $\frac{d \xi}{dt}(t_0) = - \nabla f(\xi(t_0))$, where the right-hand side is the gradient of $f$ evaluated at $\xi(t_0)$. (This is closer to the notation used in the second link.)
Further, the discrete step is $\xi_{i+1} = \xi_i - \lambda \nabla f(\xi_i)$. This is simply a rearrangement of $$-\nabla f(\xi(t_i)) = \frac{d\xi}{dt}(t_i) \approx \frac{\xi(t_i + \lambda) - \xi(t_i)}{\lambda},$$
where $t_i$ is such that $\xi_i = \xi(t_i)$.
In particular, $\lambda$ is the parameter that should be driven to zero to see the asymptotic correspondence with gradient flow; you cannot just set it at $\lambda=1$.