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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. 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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    CORRECTION: edit time expired so I couldn't remove the $\frac{1}{4.5}$'s and the $\frac{1}{10}$ typo.2012-10-24

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