Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $c=0$.

share|cite|improve this question
Without knowing the relation between $\;f,g\;$ this doesn't look doable in all generality. – DonAntonio Sep 6 '13 at 18:08
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 6 '13 at 18:38

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

share|cite|improve this answer
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 6 '13 at 18:25
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 6 '13 at 18:30
You are right Micah. My example is wrong. – Pocho la pantera Sep 6 '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 6 '13 at 21:01

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?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.