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>|| | 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 ?
A The minimization of the square root of the norm is difficult, but it's doable by just taking the derivative of the function; though one will have the same result by minimizing just , where with this notation we mean .
Note that in the literature, is generally already defined as .
--Elisa
