Mean of the derivative of a periodic function

[Robin04]

TL;DR

I'm wondering if given that the mean of a periodic fuction is zero than the mean of all of its derivatives is zero too.

We have a periodic function $f: \mathbb{R} \rightarrow \mathbb{R}$ with period $T, f(x+T)=f(x)$
The statement is the following: $$\frac{1}{T}\int_0^T f(x)dx =0 \implies \frac{1}{T}\int_0^T\frac{d}{dx} f(x)dx =0$$
Can you give me a hint on how to prove/disprove it? The examples I tried all confirmed this.

 

[fresh_42]

Let $F(x)$ be the antiderivative of $\left( \dfrac{d}{dx}\,f(x) \right)$. Then the right hand side is?

 

[Robin04]

Let $F(x)$ be the antiderivative of $\left( \dfrac{d}{dx}\,f(x) \right)$. Then the right hand side is?

So then $F(x)=f(x)+c$, where $c$ is the integration constant.
$$\frac{1}{T}\int_0^T \frac{d}{dx}f(x) dx = \frac{1}{T}[F(x)]_0^T=\frac{1}{T}[f(x)+c]_0^T=\frac{1}{T}(f(T)+c-f(0)-c)=0$$
Is this correct?

 

[pasmith]

Summary: I'm wondering if given that the mean of a periodic fuction is zero than the mean of all of its derivatives is zero too.

We have a periodic function $f: \mathbb{R} \rightarrow \mathbb{R}$ with period $T, f(x+T)=f(x)$
The statement is the following: $$\frac{1}{T}\int_0^T f(x)dx =0 \implies \frac{1}{T}\int_0^T\frac{d}{dx} f(x)dx =0$$
Can you give me a hint on how to prove/disprove it? The examples I tried all confirmed this.

\frac{1}{T}\int_0^T\frac{d}{dx} f(x)dx = 0 is a direct result of the fundamental theorem of caclulus and the fact that f(0) = f(T). It holds irrespective of the value of \frac{1}{T}\int_0^T f(x)dx.

 

[jbriggs444]

Summary: I'm wondering if given that the mean of a periodic fuction is zero than the mean of all of its derivatives is zero too.

We have a periodic function $f: \mathbb{R} \rightarrow \mathbb{R}$ with period $T, f(x+T)=f(x)$
The statement is the following: $$\frac{1}{T}\int_0^T f(x)dx =0 \implies \frac{1}{T}\int_0^T\frac{d}{dx} f(x)dx =0$$
Can you give me a hint on how to prove/disprove it? The examples I tried all confirmed this.

Would you count $\frac{\sin x\ |\cos x|}{\cos x}$ as a periodic function with mean zero for purposes of this question?

 

[Robin04]

\frac{1}{T}\int_0^T\frac{d}{dx} f(x)dx = 0 is a direct result of the fundamental theorem of caclulus and the fact that f(0) = f(T). It holds irrespective of the value of \frac{1}{T}\int_0^T f(x)dx.

Oh, you're right. Interesting, haven't thought about that.

Would you count $\frac{\sin x\ |\cos x|}{\cos x}$ as a periodic function with mean zero for purposes of this question?

Well, that's interesting. I cannot plot its derivative for some reason, but I suppose it's continuity the issue here.

 

[jbriggs444]

Well, that's interesting. I cannot plot its derivative for some reason, but I suppose it's continuity the issue here.

It is continously differentiable over its domain. But its domain misses the odd multiples of $\frac{\pi}{2}$.

That means that the anti-derivative of its derivative over almost any interval of length $\pi$ has two disjoint segments. Two c's, instead of just one. *WHAM* There goes that cancellation of the c's you did in post #3.