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