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Unformatted text preview: + 1 = j h 2 sin kj n + 1 cos j n + 1 + cos kj n + 1 sin j n + 1 + 2 sin kj n + 1 sin kj n + 1 cos j n + 1 cos kj n + 1 sin j n + 1 = j h 2 2 2 cos j n + 1 sin kj n + 1 = 1 h 2 2 2 cos j n + 1 v k,j = j v k,j . Stacking these elements we obtain B x v j = j v j . (b) This follows by writing the k th component of Vy . CHALLENGE 30.5. See the website for the programs. The results are shown in Figures 30.1 and 30.2. All of the algorithms give accurate results, but as n gets large, the eciency of the fast algorithm of Challenge 4 becomes more apparent....
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This note was uploaded on 01/21/2012 for the course MAP 3302 taught by Professor Dr.robin during the Fall '11 term at University of Florida.
- Fall '11