Proving odd and even function statements (1 Viewer)

kpad5991 · Apr 19, 2017

1. Suppose that h(x) = f(x) x g(x)
A) show that if f and g are both even or both odd then h(x) is even
B) show that if one of f(x) and g(x) is even and the other odd, then h(x) is odd.

2. Suppose that h(x) = f(x) + g(x)
A) show that if f(x) and g(x) are both even then h(x) is even
B) show that if f(x) and g(x) are both odd then h(x) is odd

1729 · Apr 19, 2017

kpad5991 said:
1. Suppose that h(x) = f(x) x g(x)
A) show that if f and g are both even or both odd then h(x) is even
B) show that if one of f(x) and g(x) is even and the other odd, then h(x) is odd.

2. Suppose that h(x) = f(x) + g(x)
A) show that if f(x) and g(x) are both even then h(x) is even
B) show that if f(x) and g(x) are both odd then h(x) is odd

$$\noindent \textbf{Q1.} $h(x) = f(x)g(x) \\ \begin{align*}f \text{ and } g &\text{ both even} \\ h(x) &= f(-x)g(-x) \\ &= h(-x) \end{align*} \\ \begin{align*}f \text{ and } g &\text{ both odd} \\ h(x) &= -f(-x) \times -g(-x) \\ &= f(-x)g(-x) \\ &= h(-x) \end{align*} \\ \begin{align*}f \text{ and } g &\text{ not both even nor both odd} \\ h(x) &= -f(-x) \times g(-x) \\ &= f(-x) \times -g(-x) \\ &= -h(-x) \end{align*}$

$$\noindent \textbf{Q2.} $h(x) = f(x) + g(x) \\ \begin{align*}f \text{ and } g &\text{ both even} \\ h(x) &= f(-x) + g(-x) \\ &= h(-x) \end{align*} \\ \begin{align*}f \text{ and } g &\text{ both odd} \\ h(x) &= -f(-x) - g(-x) \\ &= -(f(-x) + g(-x)) \\ &= -h(-x) \end{align*}$

Mahan1 · Apr 23, 2017

kpad5991 said:
1. Suppose that h(x) = f(x) x g(x)
A) show that if f and g are both even or both odd then h(x) is even
B) show that if one of f(x) and g(x) is even and the other odd, then h(x) is odd.

2. Suppose that h(x) = f(x) + g(x)
A) show that if f(x) and g(x) are both even then h(x) is even
B) show that if f(x) and g(x) are both odd then h(x) is odd

$\textrm{This is another way of looking at it}$
$\textrm{Well, it's obvious that if } \ f(x) \textrm{or} \ g(x) \ \textrm{is a zero function then h(x) is also zero and the statement follows}$
$h(x) = h(-x) = 0$
$\textrm{we can make the following observation that if f(x) is odd then }$
$f(x) f(-x) \leq 0$
$\textrm{we can make the following observation that if f(x) is even then }$
$f(x) f(-x) \geq 0$
$\textrm{also note whether f is even or odd }\ f^{2}(x) = f^{2}(-x) \textrm{which implies }$
$h^{2}(x) - h^{2}(-x) = (h(x)+h(-x))(h(x)-h(-x))= 0 \ \textrm{so h(x) is either odd or even}$

$\textrm{i) prove by contradiction: Suppose h(x) is odd then :}$
$h(x)h(-x) \leq 0 \Rightarrow f(x)f(-x)g(x)g(-x) \leq 0$
$\textrm{which implies either} \ f(x)f(-x)\leq 0 \ \textrm{or}\ g(x)g(-x) \leq 0\ \textrm{but not both}$
$\textrm{without loss of generality assume } \ f(x)f(-x) \leq 0 \ \textrm{which implies f(x) is odd}$
$\textrm{ But if f is odd then g is odd which means } g(x)g(-x)\leq 0 \ \textrm{and that is a contradiction}$
$\textrm{therefore h(x) is even. Also note we just proved if f and g don't have the same evenness then h(x) is odd}$

Proving odd and even function statements (1 Viewer)

kpad5991

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Mahan1

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