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It is now necessary to evaluate explicitly the largest eigenvalue of V. Using (15.117), we obtain 1 2= lim -log A n-"oo n (15.118) y(v) = -(2k - 1) (15.119) ~ 00, (15.120) 2= - 1 12~ dvy(v) = - 1 1~ dvy(v) 4'TT 0 2'TT 0 where the last step results from (15.105), which states that y(v) = y(2'TT - v). To express 2 in a more convenient form, we recall that y( v) is the positive solution FIGURE 7.36 This is typical price action following a trendline break, as prices push up against the former supporting trendline. (15.121) .net code 39 reader Code 39 Reader In VB. NET - OnBarcode
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