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Thread: even and odd

  1. #1
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    even and odd

    Recall that $\displaystyle f: \mathbb{R} \to \mathbb{R} $ is an even function if $\displaystyle f(-x)=f(x) , \ \forall x \in \mathbb{R} $
    and is an odd function if $\displaystyle f(-x)=-f(x) , \ \forall x \in \mathbb{R} $
    prove that if $\displaystyle f: \mathbb{R} \to \mathbb{R} $ is an even differentiable function on $\displaystyle \mathbb{R} \ $
    then $\displaystyle f' $ is an odd function
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  2. #2
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    Quote Originally Posted by flower3 View Post
    Recall that $\displaystyle f: \mathbb{R} \to \mathbb{R} $ is an even function if $\displaystyle f(-x)=f(x) , \ \forall x \in \mathbb{R} $
    and is an odd function if $\displaystyle f(-x)=-f(x) , \ \forall x \in \mathbb{R} $
    prove that if $\displaystyle f: \mathbb{R} \to \mathbb{R} $ is an even differentiable function on $\displaystyle \mathbb{R} \ $
    then $\displaystyle f' $ is an odd function

    $\displaystyle f'(x_0):=\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}$ ...and the quotient of even by odd functions is an odd function.

    Tonio
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  3. #3
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    Use the chain rule: the derivative of f(-x) is f'(-x)(-x)'= -f'(-x). The derivative of -f(x) is -f'(x). Put those together.
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    Quote Originally Posted by tonio View Post
    $\displaystyle f'(x_0):=\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}$ ...and the quotient of even by odd functions is an odd function.

    Tonio
    I don't understand your point, tonio. If $\displaystyle g(x)= x- x_0$, then $\displaystyle g(-x)= -x-x_0$ is not equal to either $\displaystyle g(x)= x-x_0$ nor $\displaystyle -g(x)= -x+ x_0$ so g is neither even nor odd. What even and odd functions make up your quotient?
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    Quote Originally Posted by HallsofIvy View Post
    I don't understand your point, tonio. If $\displaystyle g(x)= x- x_0$, then $\displaystyle g(-x)= -x-x_0$ is not equal to either $\displaystyle g(x)= x-x_0$ nor $\displaystyle -g(x)= -x+ x_0$ so g is neither even nor odd. What even and odd functions make up your quotient?

    I wrote that too quick: I meant to write the definition of the derivative as

    $\displaystyle f'(x_0)=\lim_{h\to 0}\frac{f(x_0+h)-f(x_0)}{h}$ , and then

    $\displaystyle f'(-x_0)=\lim_{h\to 0}\frac{f(-x_0+h)-f(-x_0)}{h}$ . Now, here substitute k:=-h, and of course $\displaystyle h\rightarrow 0 \Longleftrightarrow k\rightarrow 0$ , so then (and here enters the numerator even and the denominator odd thing)

    $\displaystyle f'(-x_0)=\lim_{h\to 0}\frac{f(-x_0+h)-f(-x_0)}{h}=\lim_{k\to 0}\frac{f(-x_0-k)-f(-x_0)}{-k}=$ $\displaystyle \lim_{k\to 0}\frac{f(x_0+k)-f(x_0)}{-k}=-\lim_{k\to 0}\frac{f(x_0+k)-f(x_0)}{k}=-f'(x_0)$

    Tonio
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by flower3 View Post
    Recall that $\displaystyle f: \mathbb{R} \to \mathbb{R} $ is an even function if $\displaystyle f(-x)=f(x) , \ \forall x \in \mathbb{R} $
    and is an odd function if $\displaystyle f(-x)=-f(x) , \ \forall x \in \mathbb{R} $
    prove that if $\displaystyle f: \mathbb{R} \to \mathbb{R} $ is an even differentiable function on $\displaystyle \mathbb{R} \ $
    then $\displaystyle f' $ is an odd function
    Essentially the same thing as HallsOfIvy is to define $\displaystyle h(x)=f(x)-f(-x)=0$ and then note that $\displaystyle h'(x)=f'(x)+f'(-x)=0\implies f'(x)=-f'(-x)$
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