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FIGURE F.13 LMF.
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12.7. EARLY COMPENSATORS This section on null compensators is adapted from the original chapter in earlier versions of this book, written by late Abbe Offner. The corresponding test of a
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3.1 1 PROBLEMS 3.1 Show for discrete stochastic variables that the expectation of the Fisher score vector is equal to the null vector.
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print_r( $authors ); echo $nextCell; print_r( array_splice( $authors, 2, 0, $arrayToAdd ) ); echo $nextCell; print_r( $arrayToAdd ); echo $nextCell; print_r( $authors ); echo $rowEnd; echo {$headingStart}2. Replacing two elements with a new element{$headingEnd} ; $authors = array( Steinbeck , Kafka , Tolkien ); $arrayToAdd = array( Bronte ); echo $rowStart; print_r( $authors ); echo $nextCell; print_r( array_splice( $authors, 0, 2, $arrayToAdd ) ); echo $nextCell; print_r( $arrayToAdd ); echo $nextCell; print_r( $authors ); echo $rowEnd; echo {$headingStart}3. Removing the last two elements{$headingEnd} ; $authors = array( Steinbeck , Kafka , Tolkien ); echo $rowStart; print_r( $authors ); echo $nextCell; print_r( array_splice( $authors, 1 ) ); echo $nextCell; echo Nothing ; echo $nextCell; print_r( $authors ); echo $rowEnd; echo {$headingStart}4. Inserting a string instead of an array{$headingEnd} ; $authors = array( Steinbeck , Kafka , Tolkien ); echo $rowStart; print_r( $authors ); echo $nextCell; print_r( array_splice( $authors, 1, 0, Orwell ) ); echo $nextCell; echo Orwell ; echo $nextCell; print_r( $authors ); echo $rowEnd; echo </table> ; > </body> </html>
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Appendix: Using the CD-ROM . . . . . . . . . . . . . . . . . . . . . . . 893
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and with xn = b for n [b 2h, b], we have [via (9.105b)] 1 1 [3f (xn ) 4f (xn 1 ) + f (xn 2 )] + h2 f (3) ( n ). 2h 3 (9.106b) Thus, (9.104) becomes (approximate corrected trapezoidal rule) f (1) (xn ) = f (1) (b) = TC (n) = T (n) h [3f (xn ) 4f (xn 1 ) + f (xn 2 ) + 3f (x0 ) 4f (x1 ) + f (x2 )] 24 (9.107)
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But x = 0 as it is an eigenvector of A, and also a = 0. Immediately, = , contradicting our assumption that these eigenvalues are distinct. Thus, y = ax is impossible; that is, we have proved that y is independent of x. Theorem 11.2 leads us to the next theorem. Theorem 11.3: If A Cn n has n distinct eigenvalues, then A has n linearly independent eigenvectors. Proof Uses mathematical induction (e.g., Stewart [6]). We have already seen the following theorem. Theorem 11.4: If A Rn n and A = AT , then all eigenvalues of A are realvalued. Proof See Hill [1], or see the appropriate footnote in 4. In addition to this theorem, we also have the following one. Theorem 11.5: If A Rn n , and if A = AT , then eigenvectors corresponding to distinct eigenvalues of A are orthogonal. Proof Suppose that ( , x) and ( , y) are eigenpairs of A with = . We wish to show that x T y = y T x = 0 (recall De nition 1.6). Now x = Ax = AT x so that y T x = y T AT x = (Ay)T x = y T x, implying that ( )y T x = 0. But = so that y T x = 0, that is, x y. Theorem 11.5 states that eigenspaces corresponding to distinct eigenvalues of a symmetric, real-valued matrix form mutually orthogonal vector subspaces of Rn . Any vector from one eigenspace must therefore be orthogonal to any eigenvector from another eigenspace. If we recall De nition 11.2, it is apparent that all symmetric, real-valued matrices are nondefective, if their eigenvalues are all distinct.1 In fact, even if A Cn n and is not symmetric, then, as long as the eigenvalues are distinct, A will be nondefective (Theorem 11.3).
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Figure 2-6 The IEEE 802.5 Token Ring frame structure
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