Re: Pattern matching problem

*To*: mathgroup at smc.vnet.net*Subject*: [mg88548] Re: Pattern matching problem*From*: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>*Date*: Thu, 8 May 2008 04:13:08 -0400 (EDT)*Organization*: Uni Leipzig*References*: <fvs2kf$eo4$1@smc.vnet.net>*Reply-to*: kuska at informatik.uni-leipzig.de

Hi, *when* you implement it, using pattern matching than there is no Sum[] and no Count[] but patterns instead, i.e., myCount[u_[x_], u_, x_] := 0 myCount[Derivative[n_][u_][x_], u_, x_] := n myCount[expr_, _, x_] /; FreeQ[expr, x] := 0 myCount[expr_Plus, u_, x_] := myCount[#, u, x] & /@ expr myCount[expr_Times, u_, x_] := myCount[#, u, x] & /@ (Plus @@ expr) myCount[Power[expr_, i_Integer], u_, x_] := i*myCount[expr, u, x] and myCount[1 + u[x] + D[u[x], {x, 2}]*D[u[x], {x, 3}]^2, u, x] gives 8 as you expect. Regards Jens Charlie Brummitt wrote: > Hi all, > Here is my problem: Given a polynomial in the variables u[x,t] and its > spatial derivatives (for example, the polynomial could be 1 + u + > u_xx*u_xxx^2), count the number of spatial derivatives with multiplicity. > That is, after writing the above polynomial as > > > 1 + u + u_xx * u_xxx * u_xxx > > the output should be 2 + 3 + 3 (basically, you count the number of x's). > > I have tried implementing this using a pattern matching approach. Here is > what I tried: > > f[equation_ ] := Sum[ k * j * Count[ equation , D[ u[x, t], {x, j} ] ^ k , > {0, \infinity} ], {j, 1, 50}, {k, 1, 50}] > > This fails to work on, for example, the input u_xx^2, because it outputs 6 > when it should output 4. This is because the u_xx is counted (+2 to the > sum), and the u_xx^2 is counted (+4 to the sum). This is because the u_xx is > nested inside the Power[ , 2] in its representation in Mathematica and so > it gets counted too many times in my formula. I can't seem to figure out a > way to use the "provided that" operator /; to make this formula work. > > I've also tried doing some replacement methods, but to no success. > > Thanks for any help you may be able to provide. > > -Charlie > >