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Suppose that $x=f(t)$ and $y=g(t)$ define $y$ as a function of $x$ and that this is known to be an even function.

If $g$ is an even function is it true that $f$ must be an odd function? (Here I should have said $f$ cannot be an odd function + a constant.)

I believe that it is true and that it is obvious. Am I right? Is it in fact true and if so does it require proof?

I've tried to answer my own question below but I'm not sure it's quite right so I won't post it as an answer. Being new here I'm not sure of the site etiquette, sorry if I get anything wrong.

May I let $o$ be any odd function and $e$ be any even function?

So if $x=o(t)+c$ where $c$ is a constant then $t=o^{-1}(x-c)$ and so $y=e(o^{-1}(x-c))$.

Now if $o$ has an inverse its inverse must also be odd and I think $e(o^{-1}(x-c))$ is only even if $c=0$.

Trebor Nosnibor
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  • Without knowing the relation between $;f,g;$ this doesn't look doable in all generality. – DonAntonio Sep 06 '13 at 18:08
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    Believing things are "true and obvious" based on feelings is OK sometimes, but it sure isn't very convincing. That's what proofs are for. – rschwieb Sep 06 '13 at 18:38

2 Answers2

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It is not true. Example:

$y=t^4$ and $x=t^2$, then $y=x^2$ (edit: doesn't work)

Here is (I believe) a correct counterexample.

$y(x)=1$, $y=g(t)=1$, and $$x(t)=\left\{\begin{array}{ll} x & x>0 \\ x^3 & x\le0\end{array}\right.$$

Pocho la pantera
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  • Thank you. :) I think I tried to overgeneralise. What I was originally thinking was that f cannot be an odd function plus a constant. Sorry to move the goalposts. – Trebor Nosnibor Sep 06 '13 at 18:25
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    Writing $y$ as a function of $x$ for $(x,y)=(t^2, t^4)$, gives you $y=x^2$ on a restricted domain (namely $x \geq 0$). This is not an even function; its domain isn't even symmetric about the origin. – Micah Sep 06 '13 at 18:30
  • You are right Micah. My example is wrong. – Pocho la pantera Sep 06 '13 at 19:58
  • Sorry for any confusion. What I was thinking of before I basically asked the wrong question was things like $x=t^3+1, y=t^2$. Now y(x) is clearly not an even function. – Trebor Nosnibor Sep 06 '13 at 21:01
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Perhaps my understanding is flawed, does this example disprove the theory?

Both of these functions satisfy $f(x) = f(-x)$ making them even
$f(t) = cos(t)$
$g(t) = /sin(t)/$

$x=f(t), y=g(t)$ will define a semicircle from (-1,0) through (0,1) to (1,0), which is also even, correct?

Joe
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