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- Thread starter gsingh2011
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[tex] \int_{a}^{b}dxT_{n}(x)T_{m}(x)u(x) = \delta _{n,m} [/tex]

so you could expand any function f(x) as

[tex] f(x)= \sum_{n=0}^{\infty}c_{n} T_{n} (x) [/tex] (1)

here T(x) are POlynomials so (1) can be regarded also as a power series different from the Taylor one

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A "Taylor series" is a particular way of getting

For example, if you are asked to write [itex](1- x)^{-1}[/itex] as a power series in x, there are two ways to do that:

1) Find the Taylor's series for 1/(1- x) around x= 0 (the MacLaurin series) by taking the derivatives.

2) Recall that the sum of the geometric series [itex]\sum a r^n[/itex] is given by a/(1- r) so that [itex]1/(1- x)= \sum x^n[/itex].

Those are two different ways of forming a power series but they give exactly the same power series for the same function. Even the series of polynomials that zetafunction refers to, once you combine like powers, are the same power series.

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