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MathGroup Archive 2001

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Solving for Sum element

  • To: mathgroup at smc.vnet.net
  • Subject: [mg27260] Solving for Sum element
  • From: Ben Jacobson <bjacobson at illumitech.com>
  • Date: Wed, 14 Feb 2001 03:41:27 -0500 (EST)
  • Sender: owner-wri-mathgroup at wolfram.com

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.




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