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*Organization*: Universidad del Valle*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[1]:= D[(1 - x^2) y''[x], {x, k}][[2]] Out[1]= -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 about the following version: 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[2]:= D[(1 - x^2)y''[x], {x, -1}] $IterationLimit::"itlim":"Iteration limit of \!\(4096\) exceeded." Out[2]= Hold[D[(1 - x^2)*Derivative[2][y][x], {x, -1}]] doesn't try to solve a problem Mathematica solves by it's own, and even allows things like In[3]:= D[(1 - x^2) y''[x] z[x], {x, k}] Out[3]= -((-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 > 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