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Re: Re: Way to evaluate D[(1-x^2)y''[x],{x,n}] ?
*To*: mathgroup at smc.vnet.net
*Subject*: [mg14975] Re: [mg14961] Re: [mg14914] Way to evaluate D[(1-x^2)y''[x],{x,n}] ?
*From*: Jurgen Tischer <jtischer at col2.telecom.com.co>
*Date*: Wed, 2 Dec 1998 03:58:58 -0500
*Organization*: Universidad del Valle
*References*: <199811280856.DAA07470@smc.vnet.net.>
*Sender*: owner-wri-mathgroup at wolfram.com
Fred,
in what new behavior consists your extension of D?
Jurgen
Fred Simons wrote:
>
> Phillip Kent raised the question:
>
> > I'm wondering if and how to make Mathematica evaluate derivatives like
> >
> > D[(1-x^2)y''[x],{x,n}]
> >
> > y[x] is an unspecified function, n is a +ve integer.
> >
> > It seems as though the system ought to "know" that this reduces to three
> > terms only, provided that n is constrained?
> >
> The following extension of D does the job for any polynomial p times
> a derivative of a function. The condition Floor[n]==n tests whether n
> is an integer; if we would have used IntegerQ instead an infinite
> loop would result since in that case the condition evaluates to
> False.
>
> D[p_ Derivative[m_][y_][x_], {x_Symbol, n_} ] /;PolynomialQ[p, x]:=
> If[ Floor[n]==n&& n>=Exponent[p,x],
> Sum[ Binomial[n,k]D[p, {x,k}] Derivative[m+n-k][y][x], {k, 0,
> Exponent[p, x]}] // Evaluate, D[p Derivative[m][y][x], {x,
> n} ] ]
>
> Fred Simons
> Eindhoven University of Technology
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**nice Gotcha!**
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