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Mathematical interpretation of why Gradient descent works | Linlin

# Mathematical interpretation of why Gradient descent works

Mathematical interpretation of why Gradient descent works | Linlin

For minimizing a differentiable multivariable function $$f(\mathbf x)$$ with initial value of $$\mathbf x_0$$, the fastest decreasing would be the direction of negative gradient of $$f(x)$$ since its gradient points to fastest ascending direction. The updating rule

$$$\mathbf x_{t+1} = \mathbf x_t - \eta\nabla f(\mathbf x_t), t=1,2~\cdots$$$

with sufficiently small $$\eta$$ leads to $$f(\mathbf x_{t+1})<f(\mathbf x_t)$$. The scalar constant $$\eta$$ is the step size determining how far the updating moves in the negative gradient direction, which is usually called learning rate in machine learning model training.

### But, why $$\eta$$ has to be small?

To gain the mathematical interpretation, assume the general context of

$$$\min_{\mathbf x}f(\mathbf x), \mathbf x\in \mathcal R^m$$$

with initial condition $$\mathbf x=\mathbf x_0$$.

The goal of first updating is to find $$\mathbf x_1=\mathbf x_0+\Delta\mathbf x$$ such that

$$$f(\mathbf x_1) \le f(\mathbf x_0),$$$

where $$\Delta\mathbf x$$ is the updating vector.

First we approximate $$f(\mathbf x_1)$$ with the first-order Taylor expansion as

$$$f(\mathbf x_1) = f(\mathbf x_0 + \Delta\mathbf x)=f(\mathbf x_0)+\Delta\mathbf x^T\nabla f(\mathbf x_0),$$$

$$$f(\mathbf x_1) - f(\mathbf x_0)=\Delta\mathbf x^T\nabla f(\mathbf x_0).$$$

Now to ensure the semi-negative definite of $$f(\mathbf x_1) - f(\mathbf x_0)$$, we can simply choose

$$$\Delta\mathbf x = -\eta\nabla f(\mathbf x_0)$$$

and then we have

$$$f(\mathbf x_1) - f(\mathbf x_0)=-\eta\|\nabla f(\mathbf x_0)\|^2 \le 0,$$$

for some scalar constant $$\eta$$, and $$\|\nabla f(\mathbf x_0)\|^2=0$$ if and only if $$f(\mathbf x_0)$$ is already the minimum.

To sum up and generalize to step $$t$$, the updating rule

$$$\mathbf x_{t+1} = \mathbf x_t - \eta \nabla f(\mathbf x_t)$$$

drives the searching towards the minimum given that the proper value of $$\eta$$ can ensure that $$x_{t+1}$$ is close enough to $$x_t$$ such that first-order approximate is a valid approximate with tolerable error.