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

• To: mathgroup at smc.vnet.net
• Subject: [mg14996] 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 03:59:15 -0500
• References: <9097585452@wiskunde_1.win.tue.nl>
• Sender: owner-wri-mathgroup at wolfram.com

```Fred,
I apologize for my stupid question. Your solution is nice. The thing is
I used your formula with n changed for k, and it didn't work:

In:= D[(1 - x^2) y''[x], {x, k}][]

Out= -2 y''[x] - 2x y''[x] + (1 - x^2) y''[x]

I should have realized that this was just a minor bug. What do you think

D[p_ f_, {x_Symbol, n_Symbol}] /; PolynomialQ[p, x] :=
Module[{k}, Sum[Binomial[n, k] D[p, {x, k}]
D[f, {x, n - k}], {k, 0, Exponent[p, x]}]]

It avoids the error above, avoids strange answers like

In:= D[(1 - x^2)y''[x], {x, -1}]

\$IterationLimit::"itlim":"Iteration limit of \!\(4096\) exceeded."

Out= Hold[D[(1 - x^2)*Derivative[y][x], {x, -1}]]

doesn't try to solve a problem Mathematica solves by it's own, and even
allows things like

In:= D[(1 - x^2) y''[x] z[x], {x, k}]

Out= -((-1 + k) k D[z[x] y''[x], {x, -2 + k}]) -
2 k x D[z[x] y''[x], {x, -1 + k}] +
(1 - x^2) D[z[x] y''[x], {x, k}]

The formula for n < Exponent[p] doesn't bother me too much, the way
Mathematica works (and many mathematicians think), Binomial takes care
of that.

Jurgen

Fred Simons wrote:
>
> Jurgen,
>
> I do not completely understand your question, so if this reply is not
>
> 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} ] []
>
> 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} ] ]
>