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I've been working on the following example:

Is the following even, odd or neither: $f_{0}(x^2)$, where $f_{0}(x)$ can be any unknown function

I've tried the following:

1) for example I assume $$f_{0}(x^2)=x^3$$ Then: $$f_{0}(x^2)=x^2 \cdot x$$ $$f_{0}(x^2)=x^2 \sqrt{x^2}$$ $$f_{0}(x)=x \sqrt{x}$$

Now I take $f_{0}(-x)$ which is: $$f_{0}(-x)=-x^{1.5}$$

This is neither $f_{0}(x)$(would be even) nor $-f_{0}(x)$(would be odd) so it is neither even nor odd. Is this true?

2) My second attempt is:

$f_{0}((-x)^2)=f_{0}(x^2)$ which shows that it is even.

I got 2 opposite results. Which of my attempts is true?

user50224
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  • $g(x):=f_0(x^2)$ is even by attempt 2). – lemon Aug 03 '14 at 17:13
  • Now, I got 2 opposite comments. Which one is true? @ozo: Why is my first attempt wrong? – user50224 Aug 03 '14 at 17:14
  • @Hakim Of course you can tell. The composition $f\circ g$ is always even when $g$ is even, with no condition on $f$. – Amitai Yuval Aug 03 '14 at 17:16
  • We are not asked to show that $f_0$ is even. We are asked about the composition $f_0(x^2)$. – Amitai Yuval Aug 03 '14 at 17:17
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    You are mixed two things. $g(x)=f(x^2)$ is even because $g(-x)=f((-x)^2)=f(x^2)=g(x)$, but in your argue you calculated $f(-x)=(-x)^{1.5}$ not $g(-x)$ – Mojtaba Golshani Aug 03 '14 at 17:18
  • Your first attempt is wrong because it violates the rules of logic. You are not allowed to "assume" that $f_0(x^2)=x^3$ unless that equation is a hypothesis, which it is not. – Lee Mosher Aug 03 '14 at 17:19
  • @AmitaiYuval: Why is the following wrong: If I try the test for even functions (as in my second attempt) I put the minus sign somewhere else: $f_{0}(-(x^2)). Now it is not even any longer. Why is this incorrect? – user50224 Aug 03 '14 at 17:20
  • Because, as you say, you put the minus sign somewhere else, thus you are no longer loyal to the definition of an even function. – Amitai Yuval Aug 03 '14 at 18:43

4 Answers4

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I think you are getting confused by $f(x^2)$. $f$ is a function $f:x\longmapsto f(x)$.

Consider $g:x\longmapsto g(x)=f(x^2)$.

$\forall x\in\mathbb{R},\ g(-x)=g(x)$. Therefore $g$ is even. So is the function $x\longmapsto f(x^2)$ of course.

In your first attempt you did not compare $f((-x)^2)$ and $f(x^2)$.

anderstood
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  • Thanks, but I have got some more questions: 1)In my first attempt the function $f(x)=x\sqrt{x}$ is only defined for positive numbers so I can't make $f(-x)$.
    1. Isn't my second atttempt wrong because: In the function $f(x^2)$ I have to treat $x^2$ as $x$ in $m(x)$. If I want to test $m(x)$ for being even I have to take $m(-x)$. Now, if I do the same with $f(x)$ and then I get: $f(-(x^2))$ which is not $f((-x)^2)$. Why is this reasoning incorrect?
     – user50224 Aug 03 '14 at 17:34
  • @user50224 1) $f$ is defined on $\mathbb{R}^+$ so if $x<0$ you cannot apply $f$ to $x$. 2) Your second attempt is correct, and my answer is only a more detailed explanation of what you said. – anderstood Aug 03 '14 at 17:37
  • Why is the following wrong: $$g(x)=f(x^2)$$ Then: $$g(-x)=f(-(x^2))$$ – user50224 Aug 03 '14 at 17:40
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    @user50224: because $g(-x)=f((-x)^2)\neq f(-(x^2))$! – anderstood Aug 03 '14 at 17:41
  • Thanks! Now I know that my reasoning was wrong but I still don't know why it is wrong. 1. Could you maybe prove your equation: $$g(-x)=f((-x)^2)\neq f(-(x^2))$$?

    2.Why was my reasoning in the comment above ( $g(x)=f(x^2)$ then $g(-x)=f(-(x^2))$ wrong?

    Up to now I understand that it is wrong but I have no idea why it is wrong.

    Why is the following wrong?: I have $g(x)=f(x^2)$ Then I can see this also as $g(a)=f(b)$ where $a=x$ and $b=x^2$. Then I want to test for being even: $g(-a)=g(a)$ if it is even. Then f(-b)=f(-(x^2))=f(b) which is not true. Where is the mistake?

     – user50224 Aug 04 '14 at 08:15
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    It should be obvious, take e.g. $f$ such that $f(x)=f(x^3)$. Then $f(-(x^2))=(-(x^2))^3$ while $f((-x)^2)=f(x^2)=(x^2)^3$. If you don't get it I can just advise your to go back to the definition of a function... – anderstood Aug 04 '14 at 09:16
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Let's just prove a more general proposition:

Given two functions $f,g$ with $g$ even, the composition $f\circ g$ is even, with no conditions on $f$.

Proof: $\forall x\quad f\circ g(-x)=f(g(-x))=f(g(x))=f\circ g(x)$ QED

Amitai Yuval
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Let $g(x)=f(x^2)$ then $$g(-x)=f((-x)^2)=f(x^2)=g(x)$$ so it is even

Adi Dani
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In your first attempt, you attempt to "solve" for a function $f_0$ such that $f_0(x^2) = x^3$. You come to the conclusion that if we define $f_0(x) = x^{1.5}$, then we'd have $f_0(x^2) = x^3$.

This is true if you only consider $x \geq 0$, but it is not true for $f_0$ over all of $\Bbb R$. In your third line, you make the substitution $x = \sqrt{x^2}$, which only applies when $x \geq 0$ (otherwise, we'd have $x = -\sqrt {x^2}$). As a result, the conclusion you reached was invalid.

Ben Grossmann
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