Mean of the derivative of a periodic function

In summary: Summary: The function ##\frac{\sin x\ |\cos x|}{\cos x}## may not have a mean of zero due to discontinuities in its derivative.
  • #1
Robin04
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TL;DR 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.
 
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  • #2
Let ##F(x)## be the antiderivative of ##\left( \dfrac{d}{dx}\,f(x) \right)##. Then the right hand side is?
 
  • #3
fresh_42 said:
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?
 
  • #4
Robin04 said:
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].
 
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  • #5
Robin04 said:
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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  • #6
pasmith said:
[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.

jbriggs444 said:
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.
 
  • #7
Robin04 said:
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.
 

Related to Mean of the derivative of a periodic function

1. What is the meaning of the derivative of a periodic function?

The derivative of a periodic function is the rate of change of the function at any given point. It represents the slope of the function's graph at that point.

2. How is the mean of the derivative of a periodic function calculated?

The mean of the derivative of a periodic function is calculated by taking the average of all the derivative values over one period of the function. This can be done by finding the integral of the derivative over one period and dividing it by the length of the period.

3. Why is the mean of the derivative of a periodic function important?

The mean of the derivative of a periodic function is important because it provides information about the overall trend of the function. It can also help in analyzing the behavior of the function over time and identifying any critical points.

4. Can the mean of the derivative of a periodic function be negative?

Yes, the mean of the derivative of a periodic function can be negative. This indicates that the function is decreasing on average over one period.

5. How does the mean of the derivative of a periodic function relate to the function's average rate of change?

The mean of the derivative of a periodic function is equal to the function's average rate of change over one period. This means that it represents the average slope of the function over that period.

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