       Re: Extracting terms of a polynomial into a list and then multiply

• To: mathgroup at smc.vnet.net
• Subject: [mg90384] Re: Extracting terms of a polynomial into a list and then multiply
• From: Bob F <deepyogurt at gmail.com>
• Date: Tue, 8 Jul 2008 02:24:45 -0400 (EDT)
• References: <g4sm99\$p\$1@smc.vnet.net>

On Jul 7, 3:05 am, Bob F <deepyog... 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^2=
1)/
> 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 and
> was looking for patterns in the polynomial coefficients.
>
> Thanks very much ...
>
> -Bob

Found it -- one way is to use the "Apply[List,polynomial]]". So in
this example:

Apply[List, Integrate[Normal[Series[(1 - t^2)^(-1/2), {t, 0, 50}]],
{t, 0, x}]]

would give the desired list, then what I was really after would be

Expand[Apply[List,
Integrate[
Normal[Series[(1 - t^2)^(-1/2), {t, 0, 20}]], {t, 0, x}]]*
Integrate[Normal[Series[(1 - t^2)^(-1/2), {t, 0, 20}]], {t, 0, x}]]

Anyone have a different way? Thanks...

-Bob

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