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I do not understand a step in the derive of Newton method in my lecture notes:

enter image description here

When it applies derivative on both sides of $q(x) = f(\bar{x}) + \bigtriangledown f(\bar{x})^{T}(x-\bar{x}) + \frac{1}{2}(x-\bar{x})H(\bar{x})(x-\bar{x})$

Should not it become:

$\bigtriangledown q(x) = \bigtriangledown f(\bar{x}) + \bigtriangledown^{2} f(\bar{x})^{T}(x-\bar{x}) + \{\frac{1}{2}(x-\bar{x})H(\bar{x})(x-\bar{x})\}^{'}$ ?

I do not know what the derivative of the red-box term and I also do not know why it disappears after taking derivatives.

I hope my question is clear enough. I found few books in the library and all versions of proof came like this. I appreciate your help so much.

nam
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  • $\bar x$ is fixed, while $x$ is the variable for the purpose of differentiating $q$. – daw Sep 17 '14 at 09:46
  • @daw Thanks for replying but I still do not know why the term disappears. – nam Sep 17 '14 at 09:49
  • you have to SOLVE the equation $\nabla q(x)=0$ for $x$, the derivative $\nabla q(x)$ is not zero in general. – daw Sep 17 '14 at 09:50
  • yes, I know that I should set $\bigtriangledown q$ to be zero but my question where the derivative of $\frac{1}{2} (x-\bar{x})H(\bar{x})(x-\bar{x})$ is? – nam Sep 17 '14 at 10:01
  • this is a quadratic function in $x$ - it should not be too hard to differentiate... – daw Sep 17 '14 at 10:08
  • @daw I believe I understand what you mean by $\bar{x}$ is fixed now. Thank you very much. – nam Sep 17 '14 at 10:09

1 Answers1

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please note that $\tilde{x}$ is here a fixed point, all derivatives are taken with respect to $x$. So you get for $\nabla q(x)$: $$ \nabla f(\tilde{x}) + H(\tilde{x}) (x-\tilde{x})$$ The first two terms on the RHS of your last equation are wrong, differentiating the first term on RHS of the equation before gives zero and the second gives $\nabla f(\tilde{x})$ and differentiating the third gives $H(\tilde{x}) (x-\tilde{x})$. If you do not believe it, try it out with a simple two-dimensional quadratic function. Ciao Karl

Karl
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