I want to write and evaluate an expression something like
Sum[x[i] Product[y[j], {j(!=i), 1, n}], {i, 1, n}] but with correct syntax, where n is a any number (say 4 as a concrete example).
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4 
4 Answers
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A simple, if not elegant way to do it, is
With[{n = 4}, Sum[x[i] Product[If[j == i, 1, y[j]], {j, 1, n}], {i, 1, n}]] x[4] y[1] y[2] y[3] + x[3] y[1] y[2] y[4] + x[2] y[1] y[3] y[4] + x[1] y[2] y[3] y[4]
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Also
Tr[Times @@@ SparseArray[{{i_, i_} -> x@i, {i_, j_} -> y@j}, {4, 4}]] x[4] y[1] y[2] y[3] + x[3] y[1] y[2] y[4] + x[2] y[1] y[3] y[4] + x[1] y[2] y[3] y[4]
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1 Here's a fairly direct implementation of what you wanted:
n = 4; Sum[x[i] Product[y[j], {j, Delete[Range[n], i]}], {i, 1, n}] and here's a way that uses Mathematica's pattern matching facilities:
n = 4; Plus@@ReplaceList[Product[y[j], {j, 1, n}], y[i_] rest_ :> x[i] rest] Both return
x[4] y[1] y[2] y[3] + x[3] y[1] y[2] y[4] + x[2] y[1] y[3] y[4] + x[1] y[2] y[3] y[4]
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ReplaceListversion is the fastest submitted so far: gist.github.com/simonjtyler/6759252 $\endgroup$Simon– Simon2013-09-30 04:04:49 +00:00Commented Sep 30, 2013 at 4:04
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Another approach defines f to be all the x's and g to be all the y's. The sum of the products can then be written concisely as
n = 4; f = Array[x, n]; g = Array[y, n]; Total[f[[#]] (Times @@ g)/g[[#]] & /@ Range[n]] which gives the desired sum
x[4] y[1] y[2] y[3] + x[3] y[1] y[2] y[4] + x[2] y[1] y[3] y[4] + x[1] y[2] y[3] y[4] 
Sum[x[i] Product[y[j], {j, 1, n}]/y[i], {i, 1, n}] /. n-> 4? $\endgroup$j(!=i)(sic) I believe it is. If the answers there do not address your question please edit yours to make clear the difference reply here starting your comment with "@Mr.Wizard" $\endgroup$