0
$\begingroup$

I am learning Newton's method for second-order optimization in ML. I encountered this formula, but I do not understand how we get it. I guess it is from the Taylor series, but I still cannot fully explain this formula.

$$f(x + \Delta x) \approx f(x) + \langle \nabla f(x), \Delta x \rangle + { 1 \over 2} \langle \Delta x, B(x) \Delta x \rangle$$

$$B(x) = \nabla^2 f(x)$$

$\endgroup$
2
  • $\begingroup$ What level do you not understand? The symbols, the terms, or how the terms fit together? One could also write this as $f(x+v)=f(x)+f'(x)[v]+\frac12f''(x)[v,v]+O(|v|^3)$, where $()$ contain the point of evaluation and $[]$ the vector input for the derivatives as multi-linear maps. $\endgroup$ Commented Nov 7, 2022 at 8:43
  • $\begingroup$ @LutzLehmann I don't understand how we get this formula. I think it's from Taylor series but I don't understand how to get from Taylor series formula to this $\endgroup$ Commented Nov 7, 2022 at 19:21

1 Answer 1

1
$\begingroup$

Yes, this is exactly the start of the multi-dimensional Taylor formula. You can reduce that to the scalar formula by considering the value evolution along a line $x+tv$, setting $\phi(t)=f(x+tv)$. Then the start of the scalar Taylor formula gives $$ \phi(t)=\phi(0)+\phi'(0)t+\frac12\phi''(0)t^2. $$ By the chain rule you then get $$ \phi'(0)=\sum_i\frac{\partial f(x)}{\partial x_i}v_i $$ and $$ \phi''(0)=\sum_{i,j}\frac{\partial^2 f(x)}{\partial x_i\partial x_j}v_iv_j $$ Now one can arrange the quadratic expression as a vector-matrix-vector product $v^TH_fv$ or as a multi-linear form $f''(x)[v,v]$. The first is compact for calculations that stop with the second-order term, the other can easily be extended for higher-order derivatives.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.