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. 
Post a Comment

[Python] NOT operator examples

-

 NOT operator takes 1 argument and gives you output based on the validity of the argument.

  • If the argument is true, then the result of NOT operator(argument) is false.  
  • If the argument is false, then the result of NOT operator(argument) is true. 

I will provide a few examples:

expressionboolean value (true, false) of expression NOT (boolean value of expression)
1 > 2falsetrue
3 == 3truefalse
"rain" == "Rain"falsetrue
"snow" == "snow"truefalse

In Python codes:



Comments

Post a Comment

Popular posts from this blog

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. 
Read more

[Math] Taking a square root of a number: finding an upper limit for factors of a non-prime number (Eratosthenes)

-
Eratosthenes has discovered that "a non-prime number has its factor (other than 1) that is less than the square root of the number."  1) Task: finding an upper limit for factors of a non-prime number.    2) Motivation: suppose number = 100 square root (100) = 10 factors of 100 = {1, 2, 4,5, 10, 20, 25, 50, 100}  Note that factors of 100 exist  before the square root of 100 (= 10). Namely, 1, 2, 4, 5 are those. With exception of 1, we have found prime factors of 10 that are less than 2,4, and 5. That's good enough information to determine that 100 is not a prime number.   3) An idea for mathematical proof would be: 1. suppose a number is a non-prime number. That is, this number = M * N where M>N > 0  2. multiplying N to both sides of the inequality  M>N results M*N > N^2  3. by design, this number = M*N > N^2 4. taking the square root, we have (M*N)^(1/2) > N  conclusion: a factor of a non-prime number is l...
Read more