It is well-known that using vanilla gradient descent on $f(x) = x^2$ can lead to ping-ponging and non-convergence. I would like to show that convergence can occur for momentum gradient descent.

We have function $f(x) = x^2$ and learning rate $\eta > 1$. Then,

$$\begin{aligned} x_{n+1} &= x_n - \eta g_n \\ g_n &= (1-\gamma)g_{n-1} + \gamma \nabla f(x_n) \end{aligned}$$

with initial condition $g_{-1} = 0$. I would like to show that for any $\eta > 1$, there is a $\gamma > 0$ such that $f(x_n) \to 0$. What I did was find that

$$x_{t+1} = x_t - \eta\big[(1-\gamma)g_{t-1} + \gamma\nabla f(x_t)\big] = x_t(2-\gamma - 2\eta\gamma) - x_{t-1}(1-\gamma)$$

using the facts that $\nabla f(x) = 2x$ and $x_t - x_{t-1} = -2\eta g_{t-1}$. Then I picked $\gamma = \frac12\eta$, which led me to

$$x_{t+1} = \left(1 - \frac{1}{2\eta} \right) (x_t - x_{t-1})$$

and to the recurrence relation

$$x_{t+1} = \alpha(x_t - x_{t-1})$$

where $\alpha < 1$ and $\alpha = 1 - \frac12\eta$. Using this recurrence relation, how can one show that $f(x_t) \to 0$?

I plotted this in Python and it does seem to converge well enough, but I can't show it analytically. Should I be choosing a different value of $\gamma$ instead?


1 Answer 1


$$\begin{bmatrix} x_{k+1} \\ x_k \end{bmatrix} = \begin{bmatrix} \alpha & - \alpha \\ 1 & 0 \end{bmatrix} \begin{bmatrix} x_k \\ x_{k-1} \end{bmatrix}$$

Now find for which values of $\alpha$ the matrix's eigenvalues are strictly inside the unit circle.


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