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. 
Post a Comment

[Python] Answers to: Why error messages are shown a line-by-line way? Why not shown as a block of codes?

-

It would be nice if errors in a program are shown at once.  Unfortunately, (I think) Python's error outputs are structured so that a line by line error messages are shown. 

 

I think the "one-by-one" error message is shown because we are using Python as the programing language. Because 1) Python is an 'interpreter' language, 2) an interpreter language translates a code line by line into a machine language, 3) a machine executes a line by line, 4) when a line contains an error, then the machine stops executing and presents an error message. 

 

I also think the youtube video below helps how an interpreter language (like Python) presents an error message (https://www.youtube.com/watch?v=3iLUls6Z_tw- title: compiler vs interpreter) 

 

Comments

Post a Comment

Popular posts from this blog

[Python] NOT operator examples

-
Image
  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: expression boolean value (true, false) of expression  NOT (boolean value of expression) 1 > 2 false true 3 == 3 true false "rain" == "Rain" false true "snow" == "snow" true false In Python codes:
Read more

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. 
Read more

[Math] A proof of Triangle Inequality (1)

-
Prove: $|x+y| \leq|x|+|y|$ (triangle inequality)  $x y \leq|x y| $  $\Leftrightarrow  x y \leq|x||y| $  $\Leftrightarrow  2 x y \leq 2|x||y| $ $\Leftrightarrow  x^2+y^2+2 x y \leq x^2+y^2+2|x||y| $  $\Leftrightarrow x^2+y^2+2 x y \leq\left.| x\right|^2+|y|^2+2|x||y| $  $\Leftrightarrow  (x+y)^2 \leq(|x|+|y|)^2$ [equation 1] case 1) if $x+y>0$ then $ |x+y| = x+y$  [equation1] $ \Rightarrow (|x+y|)^2 \leq(|x|+|y|)^2 $ [equation 2] $ \Leftrightarrow |x+y| \leq|x|+|y| $  case 2) if $x+y<0$ then $ |x+y| = -(x+y)$  note that $(-(x+y))^2 = (x+y)^2 = |x+y|^2$ $\Rightarrow$ [equation 2] $\Rightarrow$ $|x+y| \leq|x|+|y|$ $\blacksquare$ Published on: January 10, 2023  By: ComputeFinance
Read more