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I'm thinking about gradient descent, but I don't get it.
I understand that it can overshoot the minimum when the learning rate is too large. But I can't understand why it would diverge.
Let's say we have

$$J(\theta_0, \theta_1) = \frac{1}{2m}\sum_{i=1}^m (h_\theta(x^i)-y^i)^2$$

$$\theta_1 := \theta_1-\alpha\frac{\partial}{\partial\theta_1}J(\theta_1)$$

When the slope is negative the cost will converge to the minimum from the left of the graph, as $\theta_1$ will increase.
When the slope is positive it will converge from the right of the graph, as $\theta_1$ will decrease.
Now it might overshoot when the learning rate $\alpha$ is too large.
In that case it should overshoot again.
But not by much, should it not circle around the minimum? Why would it diverge?

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    $\begingroup$ If $\alpha$ is too large it could continually overshoot and not converge. You can dynamically adjust the learning rate by using a simple line search algorithm that satisfies the Wolfe condition for each step. $\endgroup$ – Nicholas Mancuso Mar 16 '16 at 17:46
  • $\begingroup$ I hope that this might be of some help: medium.com/@prash24goel/… $\endgroup$ – Prash Goel Jul 6 '19 at 17:44
  • $\begingroup$ Take a look at this. $\endgroup$ – Rodrigo de Azevedo Aug 8 '19 at 8:52
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If you overshoot by 10% you are fine, for example if the solution is 0 and the starting value 1000, you go to -100, +10, -1, +0.1 etc. If you overshoot by 80%, you converge but much smaller. If you overshoot by 110% you go from 1,000 to -1,100, +1,210, -1,331 etc. - divergence.

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I think it's useful to understand first why gradient descent converges. We assume that our function has $L$-Lipshitz gradient:

$$\|\nabla f(x) - \nabla f(y)\| \le L \|x - y\|,$$

or, in 1-dimensional case:

$$|f'(x) - f'(y)| \le L |x-y|$$

The main tool in non-convex optimization is the Descent Lemma: $$f(y) \le f(x) + \langle \nabla f(x), y - x \rangle + \frac L2 \|y - x\|^2$$ or, in 1-dimensional case: $$f(y) \le f(x) + f'(x) (y - x) + \frac L2 (y - x)^2$$ (Descent Lemma makes an intuitive sense because of Taylor expantion and since $L$ is an upper bound on $|f''(x)|$ by Mean-Value theorem applied to $f'$)

Now, let's make a gradient descent step: $y \gets x - \gamma \nabla f(x)$. Substituting this into descent Lemma, we have:

\begin{align} f(y) &\le f(x) + \langle \nabla f(x), -\gamma \nabla f(x) \rangle + \frac L2 \|\gamma \nabla f(x)\|^2 \\ &= f(x) - \gamma \|\nabla f(x)\|^2 + \frac {L \gamma^2}2 \|\nabla f(x)\|^2 \\ &= f(x) - \gamma(1 - \frac {L \gamma} 2) \|\nabla f(x)\|^2 \end{align}

Therefore, when $\gamma \le \frac 2L$, we have $f(y) \le f(x)$. Something like $\gamma = \frac 1L$ suffices for convergence of canonical gradient descent.

But what if $\gamma > \frac 2L$? The thing is, there are cases when Descent Lemma is tight: for example, when $f$ is quadratic. Let $f(x) = x^2$. It has $f'(x) = 2x$ and $L=2$ (since $|2x - 2y| \le 2 |x-y|$). Then, by selecting $\gamma > \frac 2L = 1$, we have:

$$(x - \gamma 2 x)^2 = (2 \gamma - 1)^2 x^2 > (2 - 1)^2 x^2 > x^2,$$

I.e. gradient descent diverges.

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