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MathGroup Archive 1998

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Re: Re: Way to evaluate D[(1-x^2)y''[x],{x,n

  • To: mathgroup at smc.vnet.net
  • Subject: [mg14982] Re: [mg14961] Re: [mg14914] Way to evaluate D[(1-x^2)y''[x],{x,n
  • From: "Fred Simons" <wsgbfs at win.tue.nl>
  • Date: Wed, 2 Dec 1998 03:59:03 -0500
  • In-reply-to: <366032AD.1C0B956D@col2.telecom.com.co>
  • Sender: owner-wri-mathgroup at wolfram.com

Jurgen,

I do not completely understand your question, so if this reply is not 
an answer to your question, please ask again.

I think Phillips question was to find the three terms formula for the 
n-th derivative of (1-x^2)y''[x] automatically, and Mathematica can 
not do that. So I extended the Mathematica D-function in such a way 
that the desired result wil be produced. First Unprotect[D], enter  the
line and Protect[D] again.

Now this formula hold only for integer values of n at least equal to 
the degree of the polynomial. So I used an If-statement with this 
condition, to make sure that the extension of D does not have an 
undesired impact on other computations.

The formula Phillips was looking for is now obtained in the following 
way:

D[ (1-x^2) y''[x] , {x, n} ] [[2]]

One of my colleagues remarked that the implementation is not  complete;
in the situation with y''[x] replaced with y[x] it does not  work. So
the following line is better:

D[p_ f_[x_], {x_Symbol, n_} ] /;PolynomialQ[p, x]:= 
   If[ Floor[n]==n&& n>=Exponent[p,x], 
     Sum[ Binomial[n,k]D[p, {x,k}] Derivative[n-k][f][x], {k, 0, 
           Exponent[p, x]}] // Evaluate, D[p f[x], {x,
        n} ] ]

Please ask again if I did not reply your question satisfactory.


Fred Simons
Eindhoven University of Technology


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