[mg90392] Re: Extracting terms of a polynomial into a list and then multiply

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Parent Message unknown [mg90392] Re: Extracting terms of a polynomial into a list and then multiply

by dh-8 :: Rate this Message:

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Hi Bob,

the magic word is: "CoefficientList"

Daniel



Bob F 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

>

>





--



Daniel Huber

Metrohm Ltd.

Oberdorfstr. 68

CH-9100 Herisau

Tel. +41 71 353 8585, Fax +41 71 353 8907

E-Mail:<mailto:dh@...>

Internet:<http://www.metrohm.com>



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