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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
4
revised Comb Function — First Derivative at a Point
added 501 characters in body
Oct
3
comment Comb Function — First Derivative at a Point
@leonbloy, I'm not comfortable calling it a train of Dirac deltas because its values are finite. I think it's all right to call it discrete function and speak of finite differences but am I right that no matter how you form these finite differences (or difference quotients which I denote by f'(t); f'(t) are not derivatives), there will always be conditions whereby the product f(t)f'(t) is negative?
Oct
3
awarded  Commentator
Oct
2
comment Comb Function — First Derivative at a Point
@Emmad Kareem, I'm denoting by f'(t) conditionally any comb function that can be arrived at on the basis of f(t). Since in this case f(t) is of only 9 values so must f'(t) be. Then I'm summing up the 9 f(t)f'(t) products and I think that there will always be a condition whereby this sum will be negative. Am I right?
Oct
2
comment Comb Function — First Derivative at a Point
@alfC, I edited the question with which I think the problem is formulated more precisely.
Oct
2
awarded  Editor
Oct
2
revised Comb Function — First Derivative at a Point
improved formulation of the problem
Sep
30
comment Comb Function — First Derivative at a Point
Correction -- of course, I don't mean any continuous periodic function but a sine function resembling the discussed comb function.
Sep
30
comment Comb Function — First Derivative at a Point
What puzzles me is that when I sum up the 9 central difference quotients I mentioned above (to find the value of the integral within the period [0,T]), I get a non-zero sum. The integral over a period of any periodic function, however, is zero. What do you think about this difference in the outcome when integrating comb function compared to continuous function?