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Re: Extracting terms of a polynomial into a list and then multiply each
*To*: mathgroup at smc.vnet.net
*Subject*: [mg90374] Re: [mg90354] Extracting terms of a polynomial into a list and then multiply each
*From*: "Thomas Dowling" <thomasgdowling at gmail.com>
*Date*: Tue, 8 Jul 2008 02:22:50 -0400 (EDT)
*References*: <200807070905.FAA00034@smc.vnet.net>
Hello,
As regards the first part,
poly = Integrate[
Normal[Series[(1 - t^2)^(-1/2), {t, 0, 50}]], {t, 0, x}]
Apply[List, poly]
or (infix syntax)
List @@ poly
Tom Dowling
On Mon, Jul 7, 2008 at 10:05 AM, Bob F <deepyogurt at gmail.com> wrote:
> Can anyone suggest a way to extract the terms of a polynomial into a
> list. For example the integral of the series expansion of
>
> 1
> --------------------
> (1 - t^2) ^(1/2)
>
> could be expressed in Mathematica (the first 50 terms) as
>
> Integrate[Normal[Series[(1 - t^2)^(-1/2), {t, 0, 50}]], {t, 0,
> x}]
>
> and gives the polynomial
>
> x + x^3/6 + (3 x^5)/40 + (5 x^7)/112 + (35 x^9)/1152 + (63 x^11)/
> 2816 + (231 x^13)/13312 + (143 x^15)/10240 +
> (6435 x^17)/557056 + (12155 x^19)/1245184 + (46189 x^21)/
> 5505024 + . . .
>
> And I would like to extract each term of this polynomial into a list
> like
>
> { x, x^3/6, (3 x^5)/40, (5 x^7)/112, (35 x^9)/1152, (63 x^11)/
> 2816, (231 x^13)/13312, (143 x^15)/10240,
> (6435 x^17)/557056, (12155 x^19)/1245184, (46189 x^21)/
> 5505024, . . . }
>
> Then I would like to take this list and multiply each element in the
> list by the integrated polynomial in order to get a list of
> polynomials that shows all of the components of the fully multiplied
> polynomial in an expanded form. In other words I would like to show
> the term by term expansion of the integral multiplied by itself, ie
>
> Expand[ Integrate[Normal[Series[(1 - t^2)^(-1/2), {t, 0, 50}]],
> {t, 0, x}] *
> Integrate[Normal[Series[(1 - t^2)^(-1/2), {t, 0,
> 50}]], {t, 0, x}]]
>
> Was working thru an example of what Euler did to compute Zeta[2] and
> was looking for patterns in the polynomial coefficients.
>
> Thanks very much ...
>
> -Bob
>
>
>
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