Re: Way to evaluate D[(1-x^2)y''[x],{x,n}] ?
- To: mathgroup at smc.vnet.net
- Subject: [mg14952] Re: [mg14914] Way to evaluate D[(1-x^2)y''[x],{x,n}] ?
- From: BobHanlon at aol.com
- Date: Fri, 27 Nov 1998 03:49:49 -0500
- Sender: owner-wri-mathgroup at wolfram.com
In a message dated 11/25/98 10:34:21 PM, p.kent at ic.ac.uk writes:
>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?
>
Phillip,
I do not know a general solution; however, for the specific example
nmax = 4;
deriv = TableForm[Table[{n, D[(1 - x^2)*y''[x], {x, n}]},
{n, 0, nmax}], TableHeadings -> {None, {"n", "f[n]"}}]
[Table deleted]
By inspection
f[n_, x_Symbol:x, y_Symbol:y] :=
-n*(n-1)*D[y[x], {x, n}] -
2*n*x*D[y[x], {x, n+1}] +
(1 - x^2)*D[y[x], {x, n+2}]
If you can't determine the exact form by inspection but can "guess" the
rough form (undetermined coefficients)
eqn = (a*n^2 + b*n + c)*D[y[x], {x, n}] -
(d*n + e)*x*D[y[x], {x, n+1}] +
(1 - x^2)*D[y[x], {x, n+2}];
Solve[Table[eqn == D[(1 - x^2)*y''[x], {x, n}],
{n, 0, 4}], {a, b, c, d, e}];
eqn /. % // Simplify
{-(-1 + n)*n*Derivative[n][y][x] - 2*n*x*(y^Derivative[1 + n])[x] -
(-1 + x^2)*(y^Derivative[2 + n])[x]}
Verifying
deriv == TableForm[Table[{n, f[n]}, {n, 0, nmax}],
TableHeadings -> {None, {"n", "f[n]"}}]
True
Proof by induction
D[f[n], x] == f[n+1] // Simplify
True
Bob Hanlon