Talk:Secondary polynomials
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- To see that the functions q_n(x) are indeed polynomials,
... seems like belaboring the obvious: antidifferntiate a polynomial function, get a polynomial function. Michael Hardy 20:12, 31 May 2007 (UTC)
- Is not quite that. First, we have a ratio, so it is not clear if the result is a polynomial. Second, this is not an antiderivative, but rather a definite integral, with the variable of integration not the same as the indeterminate of the polynomial (y and x, respectively).
- And third, that text I added is a filler in a sense to an otherwise barebone stubby stub. If you have more to add, like uses, references, etc., that derivation may be reduntant and may be either removed or moved down the article. Oleg Alexandrov (talk) 01:33, 1 June 2007 (UTC)
- Well, the quotient seems rather obviously a polynomial since the numerator vanishes at the only point where the denominator vanishes, and the fundamental theorem of calculus tells you that if the antiderivative is a polynomial then so is the integral as a function of one of the bounds of integration. Michael Hardy 22:14, 1 June 2007 (UTC)
- It took you an entire paragraph to explain why that's true. Perhaps it is not that obvious to people without a PhD in math. :)
- Well, the quotient seems rather obviously a polynomial since the numerator vanishes at the only point where the denominator vanishes, and the fundamental theorem of calculus tells you that if the antiderivative is a polynomial then so is the integral as a function of one of the bounds of integration. Michael Hardy 22:14, 1 June 2007 (UTC)
- More to the point, this article needs expansion and references. Let's worry later about what existing material is obvious. :) Oleg Alexandrov (talk) 01:55, 2 June 2007 (UTC)