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Talk:Formal power series: Difference between revisions
No functional equations then
(No functional equations then) |
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:::::So the article originally had kinda poor math. <math>\int \ldots dx</math> is not what we want because it defines an infinite family of functions, all differing by a constant. I changed the article to use what we really meant, which was the definite integral <math>\int_0^x \ldots dx</math>. <math>\cos x = -\int_0^x \int_0^x \cos x\, dx\, dx</math> is not true, but <math>\cos x = 1 - \int_0^x \int_0^x \cos x\, dx\, dx</math> (which is equivalent to the definition in the article) is.
:::::The article doesn't claim that we can use the method to solve things of the form <math>F (x, f_1, f_2, ...) = 0</math>; I don't know where you got that from. It should say more clearly that the power series represent the Taylor series of functions. And it should explain that, since the integration operation can yield the first term in the power series (the constant term) before evaluating its argument, we can define power series recursively in terms of integrals of themselves, and be able to obtain the entire power series, and that is what the example task tests. --[[Special:Contributions/76.167.241.45|76.167.241.45]] 19:43, 18 February 2009 (UTC)
::::::It is not sloppy. <math>\int f</math> is an [http://en.wikipedia.org/wiki/Indefinite_integral indefinite integral] of ''f''. <math>\int\int f</math> is defined up to a linear function, of course. So <math>cos = -\int\int cos</math> is perfectly OK.
::::::My problem is that I don't see what integration of a constant has to do with formal manipulations of Taylor series. Well, constant is a kind of Taylor series. But that looks rather thin to me. It would be better to have a test task with more substance.
::::::Anyway, we seem to agree that this is not about functional equations, but merely about repetitive integration of a constant. The net effect is that laziness and recursion can be thrown over the board and replaced by a plain loop. --[[User:Dmitry-kazakov|Dmitry-kazakov]] 22:21, 18 February 2009 (UTC)
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