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 $\Rightarrow f(-x) =-f(x)$ 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) $\Rightarrow(f+g)(-x) =-(f+g)(x)$ $\Rightarrow f+g$ is an odd function. Q.E.D. 

[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:



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