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

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.


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

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