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Talk:Formal power series: Difference between revisions
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::::These integrals are OK, the implementations of functions using Taylor series are not, '''provided''' they should serve the purpose of solving functional equations like <math>F (x, f_1, f_2, ...) = 0</math>, where x is a variable, <math>f_i</math> are Taylor series and F is a combination of functional operators from the task (+, - etc). This is what <math>cos x + \int\int cos x=0</math> makes me suggest. But that does not looks to me like a use case of the Taylor series representation. If it were a case, then for example: <math>cos^2 x + sin^2 x=1</math> should work as well. Will it?
:::::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)
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