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10. set $ A and B $ to the same-order semi-Definite Matrix, $0 \ Leq s \ Leq 1 $. proof: $ \ Bex \ Sen {A ^ Sb ^ s} _ \ infty \ Leq \ Sen {AB} _ \ infty ^ s. \ EEx $ Proof: (1 ). first, it is proved that the spectral norm of $ A $ is the maximum singular value of $ A $. in fact, $ \ beex \ Bea \ Sen {A} _ \ infty ^ 2 =\ Max _ {\ Sen {x} _ 2 = 1} \ Sen {ax} _ 2 ^ 2 \ =\ Max _ {\ Sen {x} _ 2 = 1} x ^ * a ^ * ax \ =\ Max _ {\ Sen {x }_ 2 = 1} x ^ * VV ^ * a ^ * u ^ * uavv ^ * x \ =\ Max

Label: Style Color SP on BS amp ad EF size 8. it is proved that each semi-positive definite matrix has a unique square root of the semi-positive definite number. That is, if $ A \ geq 0 $, there is a unique $ B \ geq 0 $ that satisfies $ B ^ 2 = A $. Proof: $ A \ geq 0 $ indicates the existence of U $, so that $ \ Bex U ^ * Au = \ diag (\ lm_1, \ cdots, \ lm_n ), \ quad \ lm_ I \ geq 0. \ EEx $ \ Bex B = U \ diag (\ SQRT {\ lm_1}, \ cdots, \ SQRT {\ lm_n}) U ^ *, \ EEx $ B \ geq 0 $, an

Label: Style Color SP on BS ad size as nbsp 5. Without the Weierstrass theorem, we can directly prove that the function operation (3.6) of the Hermite matrix has nothing to do with the specific spectral decomposition. Proof: $ h $ also has a spectral decomposition $ \ Bex H = V \ diag (\ lm_1, \ cdots, \ lm_n) V ^ *, \ EEx $ \ Bex w \ diag (\ lm_1, \ cdots, \ lm_n) = \ diag (\ lm_1, \ cdots, \ lm_n) W, \ quad W = V ^ * U. \ EEx $ therefore, $ \ Bex W = \ diag (W_1, \ cdots, w_s ), \ EEx $

Label: Style Color SP on bs ad Size nbsp C 1. Set $ A \ In M_n $. Prove that if $ AA ^ * = a ^ 2 $, $ A ^ * = A $. Proof: By Schur ry triangle theorem, $ U $ exists, making $ \ Bex a = u ^ * Bu, \ EEx $ where $ B = (B _ {IJ}) $ is the upper triangle array. so $ \ Bex U ^ * BB ^ * u = AA ^ * = a ^ 2 = u ^ * B ^ 2u \ rA BB ^ * = B ^ 2. \ EEx $ compare the diagonal yuan between two ends $ \ Bex | B _ {II} | ^ 2 + \ cdots + | B _ {In} | ^ 2 = B _{ II} ^ 2, \ Quad 1 \ Leq I \ Leq n. \ EEx $ whi

":Return ' death 'elif action = = "Slowly Place the bomb":Return ' Escape_pod 'ElseReturn ' The_bridge 'def escape_pod (self):Good_pod = Randint (1,5)Guess = Raw_input ("[Pod #]>>>")if int (guess)! = good_pod:Return ' death 'ElseExit (0)Class Engine (MAP):def __init__ (Self,start):Self.start = Startdef play (self):Next = Self.startWhile True:print "\ n-----------"GetAttr (Self,next)Next = Guest ()A_game = Engine ("Central_corridor")A_game.play () Remark: Beginner young novice, Welcome to the gre

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Exercise 4-7 OriginalExercise 4.7. Let* is similar to let, except that the bindings of the A let variables be performed sequentially from left to right, and E Ach binding is made in a environment in which all of the preceding bindings are visible. For example (let* ( (x 3 ) (y (+ x 2 ) ) Span class= "Hljs-list" > (z ( x y 5 ) ) ) (* x z) ) Returns 39. Explain How a let* expression can is rewritten as a set of nested let expressions,

Exercise 4-8 OriginalExercise 4.8. "Named let" are a variant of let that have the formvar>bindings>body>)The and is just as in ordinary let, except that is bound within to a procedure whose body is and whose parameters are the variables in the . Thus, one can repeatedly execute the by invoking the procedure named . For example, the iterative Fibonacci procedure (section 1.2.2) can be rewritten using named let as follows: (Define (fib N) ( LetFib-it

Analysis of C Programming Exercise questions, c Programming Exercise questionsC program design exercise question parsing has a string containing n characters. Write a function to copy all the characters starting from the m character to another string. Code block: # Include Void cut (char * s, int x); // defines the shear function int main () {char str [20

Exercise 3-67 OriginalExercise 3.67. Modify The pairs procedure so (pairs integers integers) would produce the stream of all pairs of integers (i,j) (Witho UT the condition i Code(Define (all-pairs s T) (Cons-stream (list (stream-car s) (stream-car T)) (Interleave (interleave(stream-map (Lambda (x) ( list ( stream-car s) x ) ( stream-cdr t))( All-pairs (stream-cdr s) (stream-cdr t )))

Exercise 3-68 OriginalExercise 3.68. Louis Reasoner thinks that building a stream of the pairs from three parts is unnecessarily complicated. Instead of separating the pair (s0,t0) from the rest of the pairs on the first row, he proposes to work with the whole FIR St row, as follows: (define (pairs s t ) (interleave ( Stream-map (lambda (x ) (list (stream-car s) x) ) t ) (pairs (stream-cdr s) ( stream-cdr t ) ) ) ) Does

Exercise 3-69 OriginalCode(Define (triples s T u) (Cons-stream (list(stream-car s)(stream-car t)( Stream-car u)) (Interleave (stream-map (Lambda (x) ( cons (stream-car s) x))(stream-cdr (pairs t u) )) (triples (stream-cdr s)(stream-cdr t) (stream-cdr u)) ))); Value:triples(Define (phythagorean-numbers) (define (square x) (* x x)) (define numbers (triles integers integers integers)) (stream-filter

Exercise 3-70 OriginalCode(Define (merge-weighted S1 s2 weight) (Cond ((stream-null s1) S2) ((stream-null S2) s1) (Else ( Let ((cars1 (stream-car s1)) (cars2 (stream-car S2) )) (Cond ( ( weight cars1 ) ( weight cars2 ) ( cons-stream cars1 (merge-weighted (stream-cdr s1) s2 weight)) ) ((= (weight cars1) (weight cars2)) (cons-stream cars1 (merge-weighted (stream-cdr s1) s2 weight ))) (else (cons-stream cars2 (merge-weighted s1 (stream-c

Exercise 2.66This problem is equivalent to two fork tree in the actual project of one application, we still need to use the previous three processes learned: entry, Left-branch, Right-branch. The three functions are taking out the node, the left branch and the right branch. And according to the topic request, here also need a key that gets the value. Of course, as shown in the book above, key does not need to be written out, and here is a desire to th

Python exercise Questions 4-digit three-digit non-repetition, python exercise questions I heard that doing exercises is the best way to master a programming language, so I will try to have over 100 questions first. ---------------------------------------------------------------------- [Python exercise 001]There are 1, 2, 3, and 4 numbers. How many three number

Oracle Database exercise example and answer sharing, oracle exercise questions I. Create a table 1. create table empgj (empno number () primary key, ename varchar2 (10), 2.job varchar2 (9), mgr number (), hiredate date, sal number ), comm 3. number (7, 2), deptno number (2, 0); 4. -- 1 5. insert into empgj (empno, ename, job, mgr, hiredate, sal, comm, deptno) values 6. (7369, 'Smith ', 'cler', 7902, to_dat

Exercise 3-1 Score StatisticsEnter some student's score, which score appears the most times? If there are multiple parallel, from small to large output.Task 1: Non-negative integers with fractions not exceeding 100Task 2: The score does not exceed 100 of the nonnegative real number, but retains a maximum of two decimal places.Task 1#include Task 2#include Summary: Task 2 degree set to double, then to int, with floor (m+0.5), previously encountered, re

{If (I! = K I! = J J! = K)/* Ensure that I, j, and K are different */Printf ("% d, % d, % d \ n", I, j, k );}} Aside from many other issues in this Code, the question "How many three numbers can be composed of different and no repeated numbers" is obviously ignored by programmers. Such code has no value. Therefore, we strongly recommend that you copy the question before writing the code.Some people may think that the question cannot be compiled. It doesn't matter. This can be an excellent com

Ansible (10) first exercise and ansible first exerciseSummary We have learned apt, yum, zypper, copy, and other modules. You must be bored.The following is a practical example.Actual case: Batch tomcat configuration Only the knowledge of the above modules is enough for us to orchestrate a Playbook to configure a large number of physical/virtual servers.Actual scenarios A company has applied for 100 cloud computing virtual machines and needs

, to Vertex y, up to the n-1 edge (to consider connectivity, each vertex up to 1 times), so the shortest possible X-to-y is up to n-1 relaxation operation (that is, the update length) should appear, if the first N times relaxation can also be optimal, then this diagram is definitely the existence of the ring (directly with Dijkstra can not be optimal, the existence of the ring will affect the existence of the optimal solution). Here gives the vertex when the English name of the currency, in ord