8/3/2019 Zak New Errata

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ECE 680 Original Textbook Errata April 21, 2008 1

Additional Textbook Errata

p. 6 line 9. The quintuple should be enclosed in parentheses not curly brackets. Curlybrackets indicate a set, i.e. that the elements are unordered.

p. 7 line 16. The triple should be enclosed in parentheses not curly brackets.

pp. 21-22, Figure 1.11 would be clearer ifL sin were used rather than sin. It is easyenough to let L = 1 later, but the picture really should show the forces and distances.Also (1.34) is dimensionally inconsistant because while L is present on the left hand side,it has been omitted from the first term on the right hand side. This makes the derivationvery hard to follow. It would be much clearer if the L were included in the denominatorof the second and third terms on the rhs of the equation between (1.35) and (1.36) andin the equation with m on the lhs after (1.37) and in (1.38).

pp. 76-77, Example 2.14. Same problem as in Section 1.7.1. Need to change all equationsfor x2 (first one only last two terms, second one only half of first and all of second entries,third and fourth ones all three entries) Also equation for N2x2

3eand same in text on p.77.

p. 99 line -11. Contrapositive is for some not for all t... Argument appears to assumethat if the n rows are linearly dependent for some t they are linearly dependent for all t.This is not true. Consider M = [1 1 1;cos(t) cos(2*t) cos(3*t);sin(t) sin(2*t)sin(3*t)]. It has rank 1 for t = 0, rank 3 for t = 1, and rank 2 for t = . Now,unfortunately, once it can no longer be assumed that the value of the vector qTeAtB isconstant, the argument that the derivatives are also zero no longer holds.

p. 107, line -3. side should be sides.

pp. 107-108. notation keeps switching between x(t0) and x0.

p. 117. (1,3) element of the L1 matrix should be negative.

p. 138. I should be L in line 1, line 13, line 20, and line 23 (twice). I should be L in line14, line 16, and line 27 (twice).

p. 167, line 5 r = 0 not k = 0.

p. 190, proof of Theorem 4.11, define S to be the set ofx within a radius r of the origin(so that g(x) is bounded in norm by x). Then in line -3 of the proof, add in S afterV(x) is negative definite.

p. 193, (4.70) change x2 to x1 in denominator of (2,1) element of rhs.

p. 240, 5th line up from bottom, exits should be exists.p. 264, the sign of the second element of v2 is wrong.

p. 278, 8th line up from bottom, xN1 should be x(N 1).

p. 316, the lhs of (6.2) should be u not a.