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 Oct 20 comment Periodic Function with Discrete Values The way I understand it is as follows: if there is one non-zero point of the function and its non-zero first derivative, it is enough to carry out the summation (which is in fact integration) of $f(t)f'(t)$ over the entire interval. All the other, infinite in number products within the period, are zero. In the case at hand, it is undeniable that $f(t)f'(t) < 0$. How is that fact accounted for if the integral over the period is taken to be zero? Oct 20 comment $f(t)f'(t)$ where $f$ is part of a Gaussian $f(t)$ 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 $[0,T]$? Oct 20 asked $f(t)f'(t)$ where $f$ is part of a Gaussian Oct 20 asked Periodic Function with Discrete Values Oct 11 accepted second-order divided differenc Oct 10 awarded Scholar Oct 10 awarded Supporter Oct 10 comment second-order divided differenc 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? Oct 8 comment second-order divided differenc 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. Oct 8 comment second-order divided differenc 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? Oct 8 asked second-order divided differenc Oct 3 awarded Commentator Oct 2 awarded Editor Jan 4 awarded Student Sep 23 comment Is the following substitution legitimate? Emmad, I quess your reply also answers my question about the seeming similarity with boundary value problems in differential equations. In the boundary value problems the equalities given as boundary conditions are absolute, unlike the case at hand, right? Sep 23 comment Is the following substitution legitimate? Thanks, Emmad to you too. Sep 23 comment Is the following substitution legitimate? Ted, but the first equation shows that $b_1$ depends on $x$, doesn't it? Therefore, Deven's reply holds. Sep 23 comment Is the following substitution legitimate? Thanks a bunch, Deven. This appeared to be something akin to the boundary value problems in differential equations. How would you comment the seeming similarity of the above incorrect substitution to the boundary value problems? Sep 23 asked Is the following substitution legitimate?