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Prove that (f+g)(x) is an odd function, if f and g are odd functions (Stewart, Calculus)

Suppose f(x) and g(x) are odd functions. Prove that (f+g)(x) is also an odd function.  Answer:  1. Strategy By definition, f is an odd function if and only if f(-x) = - f(x) To show (f+g) is an odd function, we need to show (f+g)(-x) = - (f+g)(x) 2. Explanation Since f(x) and g(x) are odd functions f(x)=f(x) and g(x)=g(x) By definition of sum of functions. (f+g)(x)=f(x)+g(x) =f(x)g(x) =(f(x)+g(x)) =(f+g)(x) (by definition of sum of functions) (f+g)(x)=(f+g)(x) f+g is an odd function. Q.E.D. 

[Java] Excercise: compare two given arrays and determine their equality

Motivation: In Java, c heck if Two Arrays are Equal or not Equal. a={1,2,3,4,5,6} b={2,3,1,4,5,4,4} Time complexity: O(n2) (because of the bubble sort algorithm) - Version 1: 1 process. double-nested loop statement. - Version 2: 2 processes. 1) bubble sort algorithm: O(n2), 2) single-nested loop statement O(n) - Version 3: 1 process. 1) Arrays . equals ( a , b ): O(n2) (still n2) - Version 4: 2 processes. 1) check lengths of arrays, 2) sort arrays, 3) compare elements of arrays. Still O(n2) because of sorting arrays. Java codes import java.util.Arrays ; public static void main ( String [] args ) { int a [] = { 1 , 2 , 3 , 4 , 5 , 6 }; int b [] = { 2 , 3 , 1 , 4 , 5 , 4 , 4 }; boolean result = true ; // version 1 // determine whether two arrays are equal, by comparing elements one-by-one: version 1 // this is theta(n^2) algorithm for ( int i = 0 ; i < a . length ; i ++) { for ( int j = 0 ; j < b . len...