tea.mathoverflow.net - Discussion Feed (Why my question was closed?)

Anixx comments on "Why my question was closed?" (11708)

2010-12-12T01:11:18-08:00

@ Robin Chapman I did not notice your suggestion with Fourier transform for the first time. My thoughts also went in that direction. I think it would be a good answer if the question not closed. But what about a power function?

Anixx comments on "Why my question was closed?" (11579)

2010-12-09T09:50:50-08:00

Anixx comments on "Why my question was closed?" (11578)

2010-12-09T09:45:40-08:00

@ Robin Chapman

I do not think e^{iax} should evaluate to e^{iax}/(ia). What is definite is that the exponent should evaluate to exponent. I.e from the series perspective, $\sum_{k=1}^{\infty}c(k-1)/k!=1, where c(k) is the natural value in 0 of integral of x^k function. Similar equations can be made from Sine and Cosine series. I am currenctly investigating whether they can lead to finding the c(k) function and hence, the C[x^k] operator. Once the operator for monomials is found, in can be extended to polynomials and analytic functions.

Robin Chapman comments on "Why my question was closed?" (11575)

2010-12-09T08:57:07-08:00

There is no question without some criterion on what "best" means.
Simply insisting that it make cos integrate to sin and sin to -cos
is far too weak and arbitrary a condition. Asking the same for tan is
impossible.

If you want to integrate say a complex exponential
$e^{iax}$ to $e^{iax}/(ia)$ how about this. Write your $f$ as
a Fourier transform: $f(x)=\int F(t)e^{ixt}\,dt$ and then your
notional antiderivative as $f(x)=\int F(t)e^{ixt}/(it)\,dt$
if of course one can make some sense of these integrals?

Anixx comments on "Why my question was closed?" (11574)

2010-12-09T08:47:18-08:00

And my question was if there a paper where some measure is proposed to play role of the "natural integration constant". If doe not know such paper, simply do not answer.

Anixx comments on "Why my question was closed?" (11573)

2010-12-09T08:44:06-08:00

> "The site works best for well-defined questions: math questions that actually have a specific answer. ... [W]e suggest you stick to asking precise math questions ...."

I asked a well-defined question. I asked for an analog of Ramanjuan sum for the case of conventional integration.

jbl comments on "Why my question was closed?" (11572)

2010-12-09T08:29:31-08:00

As hilarious as these threads invariably are, wouldn't it be simpler to stop arguing with someone who has displayed no intention of adapting to community norms and just to point to the handy FAQ?

"The site works best for well-defined questions: math questions that actually have a specific answer. ... [W]e suggest you stick to asking precise math questions ...."

"MathOverflow is not a discussion forum. As a side-effect of being very good for to-the-point questions and answers, the Stack Exchange software is bad for disscusions and designed to minimize them. There's a place for discussion about mathematics, but it isn't MathOverflow. Blogs and threaded discussion forums are a more appropriate place for discussions."

"MathOverflow is not an encyclopedia. MO is a site for questions that have answers. ... When you're stuck, you can come to MathOverflow and say "I'm trying to do X. How can I do that? Does this work? Does anybody have a reference?" The idea being that for an expert, it should take very little effort to understand your confusion and set you on the right path. Or maybe a non-expert has come across the same sticking point and can explain how she resolved it. MathOverflow is not the appropriate place to ask somebody to write an expository article for you."

Anixx comments on "Why my question was closed?" (11571)

2010-12-09T08:28:05-08:00

> Then Anixx made the remark about the condition of Ramanujan on the discrete integral, whose naive generalization to the continuous case is incompatible with the first condition, if you just observe that the average of -cos over the interval (0,1) is not 0.

Well at least it is evident that in the case of conventional integration the formula should be symmetric against zero, which is not the case of discrete integration which operation is not symmetric. That's why there should not be integration over (0,1) interval, definitely.

Anixx comments on "Why my question was closed?" (11568)

2010-12-09T08:23:02-08:00

> Whoops! I should have read Willie's comment more carefully before posting here. I take back my previous post.

Willie just applied the formula from discrete case which I cited as an example, to the case of conventional integration. It is evident that this formula is not suitable here. Otherwise I would not ask. Why Willie decided that antiderivative should be averaged in the case of conventional integration? From the fact that in case of discrete integration the formula uses averaged antiderivative does not follow that the averaged antiderivative should be used in case of conventional integration.

WillieWong comments on "Why my question was closed?" (11567)

2010-12-09T08:21:09-08:00

@Henry Wilton: My main problem with this question is this. Given a list of functions F, and a list of their "preferred" anti-derivatives F', you can certainly make a function h: F -> C such that \int_0^x f(t) dt + h(f) is the preferred antiderivative. Anixx provided one condition: that the integral of sin is -cos, and the integral of cos is sin. Just with that the question is under-constrained. Then Anixx made the remark about the condition of Ramanujan on the discrete integral, whose naive generalization to the continuous case is incompatible with the first condition, if you just observe that the average of -cos over the interval (0,1) is not 0. So maybe this condition is one that must be thrown out. Then what other good properties of the Ramanujan formula is there? We are not told.

Basically, the point is that there should be some motivation from which we can infer what the appropriate function h should be, or even whether such a function should exist. Without it, an arbitrary rule that says "if you are integrating the function cosine, the lower limit should start from 0; but for sine, the lower limit should start with \pi / 2" is technically a solution to Anixx's question, but is not very enlightening at all.

HJRW comments on "Why my question was closed?" (11565)

2010-12-09T08:14:15-08:00

Like I said in the comment to your question, which you promptly ignored, the "natural" definition of making integration of trigonometric functions "circular" is incompatible with the "best" condition of making the average over (0,1) of the antiderivative vanish.

Whoops! I should have read Willie's comment more carefully before posting here. I take back my previous post.

Anixx comments on "Why my question was closed?" (11564)

2010-12-09T08:06:30-08:00

> the "natural" definition of making integration of trigonometric functions "circular" is incompatible with the "best" condition of making the average over (0,1) of the antiderivative vanish.

Why do you think that in the case of the conventional integration the formula should be the same as in case of discrete integration? If the one formula was suitable for the both cases I would not ask this question. I only mentioned the discrete case as an example where a similar question has been already resolved.

Transition from discrete integration to conventional integration is not trivial in any sense.

WillieWong comments on "Why my question was closed?" (11563)

2010-12-09T08:00:18-08:00

Consistency, my friend. Consistency.

Like I said in the comment to your question, which you promptly ignored, the "natural" definition of making integration of trigonometric functions "circular" is incompatible with the "best" condition of making the average over (0,1) of the antiderivative vanish.

Make up your mind.

Anixx comments on "Why my question was closed?" (11562)

2010-12-09T07:55:47-08:00

@ WillieWong For example, Euler Gamma constant appears in mathematics as a natural discrete integration constant of the function f(x)=1/(x-1). (hence eulergamma=psi(1))

HJRW comments on "Why my question was closed?" (11561)

2010-12-09T07:55:32-08:00

But the main reason it is not a real question is that you never gave a consistent definition/requirement of what is natural or nice.

I think this is a little harsh. Although precise questions with a definite answer are preferred on MO, there are many situations where this requirement is relaxed. Anixx did give examples of what (s)he meant by 'natural'. I haven't checked if they're consistent - they apply in slightly different contexts, so one has to think about what that would mean. No one has explained why they're inconsistent, if they are.

I'm not sure whether the question is `MO-level' - it's well out of my area of expertise - but in my opinion it's comfortably a real question.

Anixx comments on "Why my question was closed?" (11560)

2010-12-09T07:49:44-08:00

> Because it was not a real question.

The question is exactly formulated. What is "real question" then?

> if you are referring to something well-known in the literature, go ahead and say so in the text of the question

I said it in the text. Any explanation of Ramanujan's theory of diverging series includes this.

> But the main reason it is not a real question is that you never gave a consistent definition/requirement of what is natural or nice.

I would know if there are such proposals. I already said, that natural definition would make integration of trigonometric functions circular.

It should be in accord with the following expressions for differintegral, for example:

$$\mathbb{D}^{q}(\sin(t))=\sin \left( t+\frac{q\pi}{2} \right)$$
$$\mathbb{D}^{q}(e^{at})=a^{q}e^{at}$$

WillieWong comments on "Why my question was closed?" (11558)

2010-12-09T06:45:52-08:00

FWIW: if you are referring to something well-known in the literature, go ahead and say so in the text of the question, and possibly provide a reference.

But the main reason it is not a real question is that you never gave a consistent definition/requirement of what is natural or nice.

Robin Chapman comments on "Why my question was closed?" (11557)

2010-12-09T06:39:33-08:00

Because it was not a real question.

Anixx comments on "Why my question was closed?" (11556)

2010-12-09T06:19:20-08:00

I wonder why people closed my question here: http://mathoverflow.net/questions/48742/is-there-natural-integration-constant

It seems any non-trivial non-standard question is going to be closed on this site.

In short, for the case of indefinite summation (discrete integration) there is a known way to determine the most natural integration constant. This was found by Ramanjuan. But for conventional integration there is no such way known (at least for me) so I asked a question.

Also Willie Wong said in the comment that I "propose" some solution which is completely incorrect: for the discrete case the solution is known long before me.