Revision as of 13:33, 27 January 2015 by Li163 (Talk | contribs)


QE2013_AC-3_ECE580-1

Part 1,2,3,4,5

(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.

Back to QE2013 AC-3 ECE580

Alumni Liaison

Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett