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

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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. 
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[Math] A proof of Triangle Inequality (1)

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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

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