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<math>\therefore \alpha_k = - \frac {g^{(k)T} d^{(k)}} {d^{(k)T} Q d^{(k)}} </math> | <math>\therefore \alpha_k = - \frac {g^{(k)T} d^{(k)}} {d^{(k)T} Q d^{(k)}} </math> | ||
| + | |||
| + | '''(ii)''' | ||
| + | <br> '''Solution: ''' <br> | ||
| + | <math>\because Q > 0,\ \therefore d^{(k)T} Q d^{(k)} > 0</math> | ||
| + | |||
| + | <math>\therefore \alpha_k > 0 \Leftrightarrow -g^{(k)T} d^{(k)} = g^{(k)T} H_k g^{(k)} > 0</math> | ||
| + | |||
| + | Therefore a sufficient condition is <math>H_k</math> is positive definite. | ||
| + | |||
[[ QE2013 AC-3 ECE580|Back to QE2013 AC-3 ECE580]] | [[ QE2013 AC-3 ECE580|Back to QE2013 AC-3 ECE580]] | ||
Revision as of 12:33, 27 January 2015
QE2013_AC-3_ECE580-1
(i)
Solution:
$ \alpha_k $ is the solution to $ {d \over d\alpha}f(x^{(k)} + \alpha d^{(k)}) = 0 $
$ {d \over d\alpha}f(x^{(k)} + \alpha d^{(k)}) = (x^{(k)T} + \alpha d^{(k)T}) Q d^{(k)} - d^{(k)T} b = 0 $
$ \therefore \alpha d^{(k)T} Q d^{(k)} = -x^{(k)T} Q d^{(k)} + d^{(k)T} b = (b - Qx^{(k)})^T d^{(k)} = - g^{(k)T} d^{(k)} $
$ \therefore \alpha_k = - \frac {g^{(k)T} d^{(k)}} {d^{(k)T} Q d^{(k)}} $
(ii)
Solution:
$ \because Q > 0,\ \therefore d^{(k)T} Q d^{(k)} > 0 $
$ \therefore \alpha_k > 0 \Leftrightarrow -g^{(k)T} d^{(k)} = g^{(k)T} H_k g^{(k)} > 0 $
Therefore a sufficient condition is $ H_k $ is positive definite.
