I Mean of the derivative of a periodic function

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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

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

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

jbriggs444

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

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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.
 

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