visioning exercise

$ \ Bex R =\frac {1} {\ SQRT {2 }}\ sex {\ BA {CC}-1 1 \ 1 1 \ EA} \ EEx $ make $ \ Bex R ^ t \ sex {\ BA {CC} s _ {2j-1} 0 \ 0 S _ {2j} \ EA} R = \ sex {\ BA {CC} B _j C_j \ C_j B _j \ EA }, \ quad 2b_j = S _ {2j-1} + S _ {2j}, \ quad 2c_j = S _ {2j}-S _ {2j-1 }. \ EEx $ so, $ \ beex \ Bea \ quad \ diag (R, \ cdots, R) ^ t \ diag (S_1, \ cdots, s_n) \ diag (R, \ cdots, R) \\\=\ diag \ sex {\ BA {CC} B _1 C_1 \ C_1 B _1 \ EA }, \ cdots, \ sex {\ BA {CC} B _k c_k \ c_k B _k \ EA }\\\=

Label: style color for SP on BS amp AD Size 3. Prove that the numerical radius $ W (\ cdot) $ is a norm on $ M_n $. Proof: (1 ). $ \ beex \ Bea W (a) \ geq 0; \ W (A) = 0 \ rA x ^ * AX = 0, \ quad \ forall \ x \ \ rA x ^ * Ay = \ frac {1} {4} \ sum _ {k = 0} ^ 3 I ^ k (x + I ^ KY) ^ * a (x + I ^ KY) = 0, \ quad \ forall \ X, Y \ \ rA ay = 0, \ quad \ forall \ Y \ \ rA a = 0. \ EEA \ eeex $(2 ). $ \ beex \ Bea \ quad | x ^ * (\ Al A) x | = | \ Al | \ cdot | x ^ * ax |, \ quad \ forall

5. (Gelfand) set $ A \ In M_n $. Proof: $ \ Bex \ rock () = \ vlm {k} \ Sen {A ^ k} _ \ infty ^ \ frac {1} {k }. \ EEx $ Proof: (1 ). for $ \ forall \ lm \ In \ sigma (a) $, $ \ Bex \ exists \ x \ NEQ 0, \ st AX = \ lm X, \ EEx $ and $ \ Bex a ^ kx = \ lm ^ k x \ Ra \ Sen {A ^ k x} _ 2 = | \ lm | ^ k \ Sen {x} _ 2, \ EEx $ \ Bex \ Sen {A ^ k} _ \ infty = \ Max _ {\ Sen {x} _ 2 = 1} \ Sen {A ^ kx} _ 2 \ geq | \ lm | ^ K. \ EEx $ then $ \ Bex \ Sen {A ^ K }_{\ infty} ^ \ frac {1} {k} \ geq | \ l

Label: Style Color ar SP on BS amp ad ef 7. (Marcus-ree) a non-negative matrix is called a double random matrix. If the sum of each element in each row is equal to $1 $, and the elements in each column are also equal to $1 $. if $ A = (A _ {IJ}) $ is $ N $, $1, 2, \ cdots, an arrangement of N $ \ Sigma $ makes every $ I = 1, \ cdots, N $, $ \ Bex a _ {I \ sigma (I )} \ geq \ sedd {\ BA {ll} \ cfrac {1} {K (k + 1)}, n = 2 K, \\\ cfrac {1} {(k + 1) ^ 2}, n = 2 k + 1. \ EA} \ EEx $ Proof

I don't know why. I may not be able to concentrate when I relax. The two simple questions just took an hour and my start point was too low. Although it is necessary to practice programming more, it is also a matter of white. It would be a little more efficient to be fully engaged in your work. Is this the reason? Exercise 9: output the num item data before the Fibonacci. Of course, int is used here, which will overflow. I marked my mistakes. The detai

. in fact, $ \ beex \ Bea \ sum _ {I = 1} ^ K s_ I (B) =\ sum _ {I = 1} ^ K s_ I (p + (-V )) \\ \ Leq \ sum _ {I = 1} ^ k \ SEZ {s_ I (p) + s_ I (-v )} \ quad \ sex {\ mbox {theorem 4.9 }\\\=\ sum _ {I = 1} ^ K [\ lm_ I (P) + 1] \ quad \ sex {\ lm_ I (P) \ mbox {is} p \ mbox {'s feature value }}\\\=\ sum _ {I = 1} ^ k \ lm_ I (p + I) \\ =\ sum _ {I = 1} ^ K s_ I (p + I) = \ sum _ {I = 1} ^ K s_ I (c ), \ Quad 1 \ Leq k \ Leq n. \ EEA \ eeex $ the first non-equal sign of the card \ eqref {4_16_s

{Theorem 4.3 }\\\=\ sum _ {j = 1} ^ n s_j (X ). \ EEA \ eeex $(3 ). the proof theorem 4.9 is as follows. for $1 \ Leq k \ Leq N $, $ \ beex \ Bea \ sum _ {I = 1} ^ K s_ I (A + B) =\ sum _ {I = 1} ^ k \ MAX \ sed {| \ tr (a + B) G) |; g \ mbox {is a rank} k \ mbox {partial offset matrix ,} g \ In M_n }\\\ \ Leq \ sum _ {I = 1} ^ k \ MAX \ sed {|\ tr (AG) |; g \ mbox {is a rank} k \ mbox {partial offset matrix ,} g \ In M_n }\\\ \ quad + \ sum _ {I = 1} ^ k \ MAX \ sed {|\ tr (BG) |; g \ mbox {i

10. set $ a, B \ In M_n $ and $ AB $ as the Hermite matrix, then, for any undo norm $ \ Bex \ Sen {AB} \ Leq \ Sen {\ re (BA )}. \ EEx $ Proof: (1 ). first prove $ \ Bex x \ prec Y \ rA | x | \ prec_w | Y |. \ EEx $ in fact, the $ x \ prec y $ knows $ \ beex \ Bea X = Ay \ quad \ sex {: \ mbox {double random matrix }\\\=\ sum \ al_kp ^ Ky \ quad \ sex {\ al_k \ geq 0, \ sum \ al_k = 1, \ P ^ k \ In \ pi_n, \ mbox {By Birkhoff theorem }\\\=\ sum _ {\ Sigma \ In s_n} \ Al _ \ Sigma Y _ \ Sigma

} ^ {-1}-\ tr (B _j ^ {-2} B _ {J-1 }) \ quad \ sex {\ tr (xy) = \ tr (Yx )} \ =\ frac {1} {2} \ tr B _ {J-1} ^ {-1}-\ frac {1} {2} \ tr (B _j ^ {- 2} B _ {J-1 }) \ =\ frac {1} {2} \ tr B _ {J-1} ^ {-1}-\ frac {1} {2} \ tr (B _j ^ {- 2} (B _j-A_j )) \ =\ frac {1} {2} \ tr B _ {J-1} ^ {-1}-\ frac {1} {2} \ tr B _j ^ {-1} + \ frac {1} {2} \ tr (B _j ^ {-2} a_j ). \ EEA \ eeex $ hence $ \ beex \ Bea \ tr B _j ^ {-2} a_j =\ tr B _ {J-1} ^ {-1}-\ tr B _j ^ {-1 }, \ tr \ sum _ {j = 1} ^ k \ sex {\

Simple SQL partially strengthens exercise questions, simple SQL strengthens exercise questions Simple query of some SQL exercises -- Select * from emp where deptno = 30 for all employees in department 30; -- list the names, numbers, and department numbers of all clerks (clers. ename, e. empno, e. deptno from emp e where e. job = 'cler'; -- select * from emp where comm> sal for employees with higher bonus

JSP simple exercise-timed refresh page and jsp exercise refresh page The date method is used in the program. You can use the import command at the beginning of the page to import java. util. date class; the program uses the setHeader () method of the response object to set the value of refresh information in the HTTP header, so that the webpage is continuously refreshed to get the latest time, and the page

[Huawei machine trial exercise questions] 9. coordinate movement, Huawei exercise questions Question Develop A coordinate computing tool. A indicates moving to the left, D indicates moving to the right, W indicates moving up, and S indicates moving down. Start from (0, 0), read some coordinates from the input string, and output the final input result to the output file. Input: Valid coordinate is A (or D,

[Huawei machine trial exercise questions] 57. Object Manager, Huawei exercise questions Question Code /* ------------------------------------- * Date: 2015-07-05 * Author: SJF0115 * Subject: Object Manager * Source: Huawei machine trial exercises */# include Copyright Disclaimer: This article is an original article by the blogger and cannot be reproduced without the permission of the blogger.

Basic algorithm Exercise 1, basic algorithm Exercise 1 Recursive and non-recursive traversal of Binary Trees:1 # include

 (define (Cube x) (* x x x))(define (P x) (-(* 3 x) (* 4 (Cube x)))(Define (sine angle) (if (Not (> (ABS angle) 0.1)) Angle (P (sine (/angle 3.0))))The code in the topic is written to Edwin , and trace can be used to trace P 's call, which is available in Visual Studio , and I've only recently known trace the.(TRACE-ENTRYP); Unspecifiedreturn value(sine12.15)[Entering#[compound-procedure P] Args:4.9999999999999996e-2][Entering#[compound-procedure P] Args:. 1495][Entering#[compound-procedure P]

-structure Branch)) (mobile-balance?) (Branch-structure Branch)) #t))C we have finished the small problem, let's victory finish the last small question. d Small problem changed the original list to cons. So at the beginning of the right-branch and branch-structure to use cadr instead of Cdr, but here with Cdr That's right. (Define (Left-branch mobile) (Car mobile))(Define (Right-branch mobile) (Cdr mobile)) (define (Branch-length branch) (Car branch))(Define (Branch-structure branch) (CDR branch

 Practice 2.41This problem is actually a variant of the prime example in the book, the essence of which is the same. So we're going to do it in the same order. First, let's complete the ternary group that produces 3 distinct integers. In the previous question, however, we have written a two-tuple that can produce 2 different integers. So as long as we produce a more I, so that it and the resulting two-tuple combination, it can produce ternary group. So, let's get started.(Define (unique-triple

) (4 69))If there are any accumulate changes here, you can go back and look at the answer to the practice 2.33 . and Accumulate-n is just a accumulate shell, in the transformation of Accumulate-n will slowly become accumulate.In linear algebra we have learned that the value of the first row of the first column of the product of MN of the two matrix is equal to the dot product of the first column of M and the first row of N , and the value of the second row of the first column ofmn equals the fir

Label: Style Color SP on BS size res nbsp C5. (Friedland) given $ A \ In M_n $, $ \ lm_ I \ In \ BBC $, $ I = 1, \ cdots, N $. proof: There is a diagonal matrix $ d \ In M_n $ to make $ \ sigma (a + d) =\sed {\ lm_1, \ cdots, \ lm_n} $, and only a limited number of diagonal matrices $ d $ meet the preceding conditions.Proof: see [S. Friedland, matrices with prescribed off-diagonal elements, Israel J. Math., 11 (1972), 184--189].[Zhan Xiang matrix theory exer