Difference between revisions of "Hw5 ex1 - norm minimization"

From Murray Wiki
Jump to navigationJump to search
 
Line 3: Line 3:
 
'''A''' The minimization of the square root of the norm is difficult, but it's doable by just taking the derivative of the <math>\sqrt{\cdot}</math> function; though one will have the same result by minimizing just <math>|| \dot{x} || </math>, where with this notation we mean <math> \dot{x}^T \dot{x} </math>.
 
'''A''' The minimization of the square root of the norm is difficult, but it's doable by just taking the derivative of the <math>\sqrt{\cdot}</math> function; though one will have the same result by minimizing just <math>|| \dot{x} || </math>, where with this notation we mean <math> \dot{x}^T \dot{x} </math>.
  
  Note that in the literature, <math>|| \dot{x} || </math> is generally already defined as <math>\sqrt{ \dot{x}^T \dot{x} }</math>.
+
  Note that in the literature, <math>|| x || </math> is generally already defined as <math>\sqrt{ x^T x }</math>.
  
  

Latest revision as of 02:34, 12 February 2007

Q How do we solve the minimization problem for \(\sqrt{ || \dot{x} ||}\)?

A The minimization of the square root of the norm is difficult, but it's doable by just taking the derivative of the \(\sqrt{\cdot}\) function; though one will have the same result by minimizing just \(|| \dot{x} || \), where with this notation we mean \( \dot{x}^T \dot{x} \).

Note that in the literature, \(|| x || \) is generally already defined as \(\sqrt{ x^T x }\).



--Elisa