Question About Gradient Descent Gradient descent is numerical optimization method for finding local/global minimum of function. It is given by following formula: $$ x_{n+1} = x_n - \alpha \nabla f(x_n) $$
For sake of simplicity let us take one variable function $f(x)$. In that case, gradient becomes derivative $\frac {df} {dx} $ and formula for gradient descent becomes: $$ x_{n+1} = x_n - \alpha \frac {df} {dx} $$
My question is: How can we get new iterands $x_{n+1}$ from change in value of $f$? Gradient  defines both direction and value of biggest increase of $f$ at certain point, not how much $x$ changes and so it has no sense to me that we use it in formula to compute new values of $x$.

Suppose We want to find $f(x,y,z)=0$
Then we want $df=\nabla f \cdot d\vec{s}=0$.
I terms of a finite difference, $f(\vec{u}+d\vec{u})-f(\vec{u})=\frac{\partial f}{\partial x}(\Delta x)+\frac{\partial f}{\partial y}(\Delta y)+\frac{\partial f}{\partial z}(\Delta z)$.
If we want $f(\vec{u}+d\vec{u})=0$
,for a single variable, this becomes $\Delta x=x_{n+1}-x_n=\frac{-f}{df/dx}$
For ascent/descent we know the gradient is direction of steepest ascent. We can take a controlled step in that direction. In that case, the gradient and our step are parallel.
If we want an approximate change of $\Delta v$ in f, then we want $\Delta f\approx\frac{\nabla f}{|\nabla f|}\cdot \frac{\Delta v\nabla f}{|\nabla f|}$, i.e. the gradient as unit vector dotted with its direction vector scaled by the desired change in $f$. So the vector coordinate change  is $\Delta v \frac{\nabla f}{|\nabla f|}$. If you want to restrict $\Delta f=\Delta r$, you can use $\Delta v=\Delta r/(\nabla f/|\nabla f|)$. Sometimes you want to change your step size as a function fo $|\nabla f|$. Chaging $\Delta r$ heps you dot that.
