Computer Science, Finance, Mathematics

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. 

Question: Suppose f(x) and g(x) are even functions. Prove that (f+g)(x) is also an even function.  Answer: 1. Strategy Definition: a function h(x) is an even function if h(x) = h(-x) To prove (f+g)(x) is an even function, we need to show (f+g)(x) = (f+g)(-x)  2. Explanation  Recall $(f+g)(x)=f(x)+g(x)$ (by definition of sum of functions) Since $f(x)$ and $g(x)$ are even functions $\Rightarrow f(x)=f(-x)$ and $g(x) = g(-x)$ $\Rightarrow(f+g)(x)=f(x)+g(x)=f(-x)+g(-x)=(f+g)(-x)$ $\Rightarrow(f+g)(x)=(f+g)(-x)$ $\Rightarrow(f+g)$ is an even function. Q.E.D. 

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...

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