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I'd like to run past you also this problem, connected with the discrete function problems I posted earlier:

It would be interesting to look for conditions whereby the product of the non-zero part of a periodic function and its first derivative, integrated over a subinterval within the period might be less than zero. Thus, observe the skewed Gaussian function between $t = 0.5$ and $t = 5$ \begin{equation} f(t)=-ae^{-\frac{(t - b)^2}{2c^2}} -ae^{-\frac{(0.5t - b)^2}{2c^2}} \end{equation} which for $a=1$, $b=1$ and $c=1$ is \begin{equation} f(t) = -e^{-\frac{(0.5t-1)^2}{2}} -e^{-\frac{(t-1)^2}{2}} \end{equation} The first derivative of the simplified $f(t)$ over $t$ is \begin{equation} \frac{df(t)}{dt} = 0.5e^{-\frac{(0.5t-1)^2}{2}} (0.5t-1) + e^{-\frac{(t-1)^2}{2}} (t-1) \end{equation} We now want to integrate the product $f(t)f'(t)$ for which we will get \begin{equation} \int\limits_{0.5}^{5} f(t)f'(t)dt = \frac{1}{2} \int\limits_{0.5}^{5} d (f(t))^2 = (f(t))^2|_{0.5}^{5} = \end{equation} \begin{equation} \frac{1}{2}(-e^{-\frac{(0.5t-1)^2}{2}} -e^{-\frac{(t-1)^2}{2}})^2|_{0.5}^{5} = -1.28763 < 0 \end{equation} Thus, we have obtained a non-zero value of $\int\limits_{0.5}^5 f(t)f'(t)dt$ for a part of the period $[0,10]$ of the periodic function $f(t)$ whose values are zero for $t>5$ and for $0<t<0.5$. Thus, for the average value over the whole period we get \begin{equation} \frac{1}{T} \int\limits_{0.5}^{5} f(t)f'(t)dt = \frac{1}{10} \frac{1}{2} \int\limits_{0.5}^{5} d (f(t))^2 = \frac{1}{10} (f(t))^2|_{0.5}^{5} = \end{equation} \begin{equation} \frac{1}{10} \frac{1}{2}(-e^{-\frac{(0.5t-1)^2}{2}} -e^{-\frac{(t-1)^2}{2}})^2|_{0.5}^{5} = \frac{1}{10} (-1.28763) < 0 \end{equation} I have started the function at $t = 0.5$ to avoid questions regarding the value of $f(10)f'(10)$ and how that value should be included in the obtained average value.

I'd like to hear objections to the above approach.

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Why the tag discrete-math? – enzotib Oct 20 '12 at 8:03
    
$f(t)$ is periodic? – Gerry Myerson Oct 20 '12 at 8:49
    
$f(t)$ is, indeed, periodic. It is with a discrete-math tag because it resembles the earlier discussed discrete periodic functions. Notice, the integral over the entire $[0,T]$ period is zero and yet there is a section within $[0,T]$ where the integral isn't zero. Why should one ignore that fact when carryin out integration over the entire $[0,T]$? – ganzewoort Oct 20 '12 at 12:08
    
You have $f(t)=e^{-((1/2)t-1)^2/2}-e^{-(t-1)^2/2}$, and you claim that it is periodic with period 10, and you claim it is zero for $t\gt5$, and you claim it is zero for $0\lt t\lt1/2$. Aren't all three of these claims false? – Gerry Myerson Oct 21 '12 at 23:28
    
How else can you express the fact that there will be a non-zero burst of that particular form, lasting for $\Delta t = 4.5$, starting from $t = 0$? The first burst will be, as said, from $t=0,5$ to $t = 5$. The next non-zero burst will occur from $t= 10.5$ to $t = 15$, the third from $t = 20.5$ to $t = 25$ and so on. – ganzewoort Oct 22 '12 at 4:05

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