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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
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
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.
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?
Sep
29
comment Comb Function — First Derivative at a Point
As far as I can tell, the function at hand is defined on the domain of all real integer numbers. Also, there is no limiting process characteristic for a derivative of a continuous function.
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?