0
$\begingroup$

This question already has an answer here:

I am trying to derive ELM going through the basics , please help me out here :

$$f = x^Tx$$ $$g = Ax-b $$

The constraint is $Ax-b = 0$

I calculated $J' = f'+\lambda^T g'$ which is $2x+(\lambda^T A)^T = 0 $ and $Ax-b=0$ . I dont know what to do next please help me out .

$\endgroup$

marked as duplicate by whuber Feb 25 '14 at 21:55

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Please tell us what ELM means. It looks like you are asking for help solving a Lagrange multiplier problem, right? $\endgroup$ – whuber Feb 25 '14 at 21:46
  • $\begingroup$ Extreme Learning Machine $\endgroup$ – abkds Feb 25 '14 at 21:47
  • $\begingroup$ That term is irrelevant , its just an optimization problem $\endgroup$ – abkds Feb 25 '14 at 21:47
  • 1
    $\begingroup$ This is solved in many, many places on this site: the key is to recognize that your are trying to solve the least squares equations and the solutions are given by normal equations. Follow the links in this search for your choice of demonstrations. I'll close this thread and provide a link to the first hit in the search (ordered by relevance). $\endgroup$ – whuber Feb 25 '14 at 21:55
  • 1
    $\begingroup$ TrafalgarLaw - it may help to check out the link that whuber suggests, and if it's still not clear, to ask a much more specific question. $\endgroup$ – Glen_b Feb 25 '14 at 22:01

Browse other questions tagged or ask your own question.