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Let's have the periodic function




where $t_0$, $t_1$ and $t_2$ are equally spaced. All I need to know is the sign of the sum

$\sum\limits_{i=0}^{i=2} {f(t_i)f′(t_i)}$

Obviously, however, the above function is not differentiable and therefore $f′(t_i)$ cannot be determined. Can I substitute the above sum by the product of the average function value for the period and its second-order divided difference and thus make a conclusion about the sign of that sum -- say, in the case at hand claim that it's negative?

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You must interpolate the periodic function $f(t_n)$ first. – doraemonpaul Oct 9 '12 at 22:52
up vote 0 down vote accepted

I think we can assume that $f'(t_0)<0$, $f'(t_1)=0$ and $f'(t_2) = -f'(t_0)$.

Then $\sum\limits_{i=0}^{i=2} {f(t_i)f′(t_i)} = 0$

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Under that assumprion, said sum (with $i$ running from 0 to 4) will be negative for a function having the values -5, 0, -5, -10, -5, right? Is the assumption acceptable, though? – ganzewoort Oct 8 '12 at 10:47
I thought in this way: $i_1$ is max (where f(i_1)=0) => $f'(i_1)=0$, in point $i_0$ function increase so $f'(i_0)>0$ and so on. – Nikita Evseev Oct 8 '12 at 15:54
This seems to me to be the case as well but I'd like to hear objections, if a all, from other mathematicians. If there are none, I'll accept the answer. – ganzewoort Oct 8 '12 at 16:03
@nikiat2 You certainly can not make that assumption unless you have further information about $f$. Samples of $f$ at finitely many points say nothing about the possible values of $f'$. – fgp Oct 20 '12 at 12:26
@fgp Ok, in this case we can not say anything about the sum. – Nikita Evseev Oct 20 '12 at 16:58

In fact $f(t_n)$ is not difficult to interpolate, just $f(t_n)=\dfrac{5\cos\pi n-15}{2}$

Then $\sum\limits_{i=0}^{2}f(t_i)f′(t_i)=\sum\limits_{i=0}^{2}\dfrac{5\cos\pi i-15}{2}\left(-\dfrac{5\pi\sin\pi i}{2}\right)=0$

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Thanks a lot, doraemonpaul. This is similar to @nikita2's suggestion, isn't it? By the same token $\sum\limits_{i = 0}^{i = 4} {\left(5 sin \left( i \frac{\pi}{2} \right) - 10 \right) \left(5 cos \left( i \frac{\pi}{2} \right) \right)} < 0$ which shows that treating $f'(t)$ in terms of $\delta$-functions is incorrect since it yields 0. Do you agree with that? – ganzewoort Oct 10 '12 at 10:52
What reason is there to pick this particular interpolation, and not any other of the gazillion possible interpolations? You might just as well pick a desired value for the sum in question, and then construct an interpolation which produces it... – fgp Oct 20 '12 at 12:29

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