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First question: If $f\colon \mathbb R\to \mathbb R$ is differentiable, how would I prove the implication: $f$ is an odd function $\Rightarrow$ $f \mathrm '$ is an even function?

Also (aka. second question), is the implication "$f \mathrm '$ is an even function $\Rightarrow f$ is an odd function" true?

I tried to solve it by picking a random odd function and find a derivative out of it, but I figured it's not really the most efficient or correct way of solving it. Any ideas?

Rob Arthan
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MathBear
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    Possible duplicate of Prove that the derivative of an even differentiable function is odd, and the derivative of an odd is even. – Chris Culter Oct 06 '17 at 18:29
  • @ChrisCulter Aren't the rules of implication a bit different though? For example True $\to$ False is False, whereas False $\to$ True is True, True $\to$ True is True and False $\to$ False is True. – MathBear Oct 06 '17 at 18:37
  • Ah, I see there are two questions. The second question, i.e. the reverse implication, is https://math.stackexchange.com/questions/98003/derivative-of-a-function-is-odd-prove-the-function-is-even – Chris Culter Oct 06 '17 at 18:42
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    @ChrisCulter I don't think that last link helps here. – zhw. Oct 06 '17 at 18:59
  • @zhw. Thanks, I was too hasty and didn't notice that it's different. It may be possible to salvage the conclusion that the even part of $f$ is a constant. – Chris Culter Oct 06 '17 at 20:45
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    Answer to first question: Suppose ff is odd, then f(x)=f(−x)f(x)=f(−x) for all x∈Rx∈R. f′(−x)=−1f′(−x)=−1 and −f′(x)=−1−f′(x)=−1. So −f′(x)=f′(−x)−f′(x)=f′(−x), and if f′(x)=1f′(x)=1, then f′(x)≠f′(−x)f′(x)≠f′(−x). So that method would not prove the implication: ff is an odd function ⇒⇒ f′f′ is an even function? Can you point out where I'm going wrong?     – MathBear Oct 06 '17 at 20:46

2 Answers2

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To see the second implication fails, let $f(x)= x + 1.$

zhw.
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    Since the OP mentioned that random guessing was not working, the intuition behind this is that differentiating kills constants, so that even if $f$ were odd/even then $f+c$ would have the same derivative but would not be odd/even. – Zach Boyd Oct 06 '17 at 19:15
  • @ZachBoyd Does the answer lie in the fact that since one function is not true, then the whole implication cannot be true? Let $f(x) = x^3$. Therefore $f'(-x) = 3(-x^2) = 3x^2$ and $ff(x) = 3x^2$. Also $f(-x) = -x^3$ and $-f(x) = -x^3$ , which does not fail the implicaton. – MathBear Oct 06 '17 at 21:11
  • Yes. Some functions have the property, but not all. – Zach Boyd Oct 06 '17 at 21:17
  • Sorry, but I fail the see the connection, if $f(x) = x + 1$.

    Asked: "$f \mathrm '$ is an even function $\Rightarrow f$ is an odd function". If $f'(-x) = (-x+1)' = -1$, $-f'(x) = -(x+1)' = -1$ and $f'(x) = (x+1)' = 1$, then it is an odd function, not an even one, which was asked in the problem.

     – MathBear Oct 06 '17 at 21:41
  • Interestingly, "if $f'$ is an odd function then $f$ is an even function" is true (if I'm not mistaken). – David K Oct 06 '17 at 21:45
  • @DavidK But is the statement "if $f(x) = x+1$, then the implication fails" correct? – MathBear Oct 06 '17 at 21:47
  • That's correct. If $f(x)=x+1,$ then the implication "if $f'$ is even then $f$ is odd" is false. – David K Oct 06 '17 at 22:36
  • @DavidK How can $f(x) = x+1$ be correct if it is not true on the first part of the implication? "if $f'$ is even". $f'(x) = 1$ and $f'(-x) = -1$, which shows that $f'$ is not even. If it is not "even" in the first place how does it fail the implication? Shouldn't we find a function where $f'$ is even and then see whether $f$is odd or not? And therefore prove whether the implication fails or not? – MathBear Oct 07 '17 at 08:21
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    You seem to be confused about the role of $x$ in these equations. Let $f$ be the function defined by $f(x)=x+1.$ Then $f'(x)=1$ for all $x.$ For example, $f'(1)=1,$ $f'(2)=1,$ $f'(-1)=1,$ and $f'(-2)=1.$ The expression $f'(-x)$ simply means you define the function $f'$ somehow and then evaluate it at $-x.$ So if $x=2,$ then $f'(-x)=f'(-2)=1,$ not $-1.$ In fact, $f'(-x)=1$ no matter what value of $x$ you try. – David K Oct 07 '17 at 13:41
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Suppose $f$ is odd. Then $f(-x)=-f(x)$ for all $x \in \mathbb{R}$. Now differentiate both sides to express $f'(-x)$ in terms of $f'(x)$. As zhw's answer suggests, the second implication you mention is false.

Clive Newstead
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  • If zhw just proved that the second implication fails, doesn't it also mean that the first implication is not true as well? – MathBear Oct 06 '17 at 19:22
  • No, why would it? The second implication is the converse of the first. – Clive Newstead Oct 06 '17 at 19:22
  • Answer to first question: Suppose $f$ is odd, then $f(x) = f(-x)$ for all $x \in \mathbb R$. $f ' (-x) = -1$ and $-f' (x) = -1$. So $-f ' (x) = f' (-x)$, and if $f'(x) = 1$, then $f'(x) \not = f'(-x)$. So that method would not prove the implication: $f$ is an odd function $\Rightarrow$ $f \mathrm '$ is an even function? – MathBear Oct 06 '17 at 19:35
  • Or did I misunderstand something? – MathBear Oct 06 '17 at 20:40
  • @MathBear Your reasoning is invalid, because not every odd function $f$ satisfies $f'(-x)=-1,$ and no odd function satisfies both $f'(-x)=-1$ and $-f'(x)=-1.$ A link to a proof of the implication has already been given. – David K Oct 06 '17 at 20:47
  • @DavidK True. I suppose the only way to solve it is by using the derivative definition formula. That leads to the second part of the question: "$f \mathrm '$ is an even function $\Rightarrow f$ is an odd function". Zhw answered, that the implication fails, by using $f(x) = x+1$. Can it be true, since not every function $f$ satisfies $f(x)=x+1$? Or does the answer lie in the fact that since one function is not true, then the whole implication cannot be true? – MathBear Oct 06 '17 at 20:57
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    The problem statement is (implicitly) an implication that's supposed to be true for every differentiable function $f.$ To disprove a "for all" statement, one counterexample is enough. – David K Oct 06 '17 at 21:40