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<math>\alpha_k</math> is the solution to <math>{d \over d\alpha}f(x^{(k)} + \alpha d^{(k)}) = 0 </math> | <math>\alpha_k</math> is the solution to <math>{d \over d\alpha}f(x^{(k)} + \alpha d^{(k)}) = 0 </math> | ||
| + | <math>{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</math> | ||
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| + | <math>\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)}</math> | ||
| + | |||
| + | <math>\therefore \alpha_k = - \frac {g^{(k)T} d^{(k)}} {d^{(k)T} Q d^{(k)}} </math> | ||
[[ QE2013 AC-3 ECE580|Back to QE2013 AC-3 ECE580]] | [[ QE2013 AC-3 ECE580|Back to QE2013 AC-3 ECE580]] | ||
Revision as of 13:18, 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)}} $ Back to QE2013 AC-3 ECE580
