RE: Solving for Sum element

*To*: mathgroup at smc.vnet.net*Subject*: [mg27270] RE: [mg27260] Solving for Sum element*From*: "David Park" <djmp at earthlink.net>*Date*: Fri, 16 Feb 2001 03:58:20 -0500 (EST)*Sender*: owner-wri-mathgroup at wolfram.com

Ben, Here is one approach: Attributes[SumTermSolve] = {HoldFirst}; SumTermSolve[sumequation_, index_] := Module[{seqn = Hold[sumequation], i, t, start, end}, seqn = ReleaseHold[seqn /. Sum -> sum]; seqn = seqn /. sum[t_, {i_, start_, end_}] -> sum[t, {i, start, index - 1}] + term[index] + sum[t, {i, index + 1, end}]; Solve[seqn, term[index]] /. sum -> Sum] f[i_] := i SumTermSolve[Sum[f[i], {i, 1, n}] == 0, 3] {{term[3] -> -3 - 1/2*(-3 + n)*(4 + n)}} SumTermSolve[Sum[a[i], {i, 0, 6}] == x, 2] {{term[2] -> x - a[0] - a[1] - a[3] - a[4] - a[5] - a[6]}} David Park djmp at earthlink.net http://home.earthlink.net/~djmp/ > From: Ben Jacobson [mailto:bjacobson at illumitech.com] To: mathgroup at smc.vnet.net > > I want to solve for one element in an equation involving long sums. The > actual equation is of course complicated, but a very simplified example > would be > > Sum[a[i],{i,0,n}]==x > > where I want to solve for a[0]. The solution should be something like > a[0]->x-Sum[a[i],{i,1,n}. > > Solve[Sum[a[i], {i, 0, n}] == x, a[0]] > > works fine if, for example, n=3, but it returns an empty result for > symbolic n. Is there a way to tell Solve to assume that n>1 and to solve > accordingly? The best I've been able to do so far is to explicitly remove > the element I'm interested in from the sum: > > In: Solve[a[0] + Sum[a[i], {i, 1, n}] == x, a[0]] > Out: {{a[0] -> x - Sum[a[i], {i, 1, n}]}} > > This works nicely, but becomes awkward when I try to solve for more > elements and longer sums. Thanks. > > Ben Jacobson > Illumitech Inc. >