Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For an assignment I had, I had to prove that the derivative of an odd function is even. In the assignment we also had to prove that $F(x)=\int_0^x f(t)dt$ is odd given that $f$ is even, which I did do. Using that fact I stated the following:

Let us define $F(x)=\int_0^xf(t)dt$ such that $F(x)$ is odd. \begin{equation} F'(x)=\frac{d}{dx}\int_0^xf(t)dt=f(x) \nonumber \end{equation} Using 1.1 (the section where I proved that $F(x)=\int_0^x f(t)dt$ is odd given that $f$ is even) we know that $f(x)$ is even.

However, the teacher felt that the answer was not rigorous enough and that I was simply going backwards. Am I indeed simply moving backwards and not proving anything in which case: could someone point out places where I could make it more succinct and rigorous or alternatively supply better proof altogether.

share|improve this question
    
@AlexBecker We were supposed to prove that for any odd function $f$ then $f'$ is even. I chose to define an odd function $F$ as the integral of an even function $f$, which basically means that $F'(x)=f(x)$. I might as well have named them $g(x)=\int_0^xr(t)dt$, but that really is trivial. My intent all along was to prove that if $F$ is odd, then $f$ is even. –  E.O. Feb 20 '12 at 10:32
    
You are quite right, I missed the word "also". –  Alex Becker Feb 20 '12 at 10:33
    
@AlexBecker No harm done :) –  E.O. Feb 20 '12 at 10:34
1  
Differentiating $f(x)=-f(-x)$ using the chain rule will also give you the result. –  David Mitra Feb 20 '12 at 15:32
    
@E. O. : I admit I haven't read every word of your question, but defining an odd function as the integral (a more precise term is "the primitive") of an even function is a bad idea, because then any odd function would have to be continuous. Also, it may trivialize the problem, and it differs from everyone else's definition of "odd". David Mitra's hint gives you a slick way to solve the problem. –  Stefan Smith Oct 20 '13 at 16:55
add comment

2 Answers

up vote 7 down vote accepted

The proof is quite simple from the definition of the derivative: if $f$ is odd then $$ f'(-x) = \lim\limits_{h\to 0}\frac{f(-x+h)-f(-x)}{h} = -\lim\limits_{h\to 0}\frac{f(x-h)-f(x)}{h} = -f'(x). $$

W.r.t. your proof. You have showed that if $f$ is even, then $F = \int f$ is odd. You proved it - but you didn't prove that any odd function is an anti-derivative of the even function. That would be a reverse statement, as Alex has already told you.

Generally, you have $A\Rightarrow B$ where $A = \{f\text{ is even}\}$ and $B = \{F\text{ is odd}\}$ but to prove that the derivative of the odd function is even you need $B\Rightarrow A$ which you don't know at the moment.

share|improve this answer
1  
That is indeed a very nice proof! Would it be possible to comment on the validity of my proof as well? –  E.O. Feb 20 '12 at 10:37
    
@Emile: sure, I've edited. Please tell me if it is clear –  Ilya Feb 20 '12 at 11:08
add comment

I think your teacher is right that the argument is not quite rigorous. You've proved that the integral of an even function is odd. However, you haven't proved (or don't say specifically that you've proved) that the integral of a function that is not even is not odd. (Every dog has four legs, but it is not true that everything that is not a dog does not have four legs.)

Logically, there could be (a) even functions whose integrals are odd, (b) odd functions whose integrals are odd, and (c) functions that are neither even nor odd, whose integrals are odd. To complete your proof, you would have to show that cases b and c don't exist.

share|improve this answer
add comment

Your Answer

 
discard

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.