Comment #26 on issue 3128 by someb...@bluewin.ch: Sum and Product
manipulations
http://code.google.com/p/sympy/issues/detail?id=3128
No, the correct answer is:
In [103]: [ Sum(Sum(1,(i,-2+k,2-k)),(k,0,j)).doit() for j in range(5)]
Out[103]: [5, 8, 9, 8, 5]
If we look at the individual summands:
In [106]: [ [ Sum(1,(i,-2+k,2-k)) for k in range(j+1) ] for j in range(5) ]
Out[106]:
[[Sum(1, (i, -2, 2))],
[Sum(1, (i, -2, 2)), Sum(1, (i, -1, 1))],
[Sum(1, (i, -2, 2)), Sum(1, (i, -1, 1)), Sum(1, (i, 0, 0))],
[Sum(1, (i, -2, 2)),
Sum(1, (i, -1, 1)),
Sum(1, (i, 0, 0)),
Sum(1, (i, 1, -1))],
[Sum(1, (i, -2, 2)),
Sum(1, (i, -1, 1)),
Sum(1, (i, 0, 0)),
Sum(1, (i, 1, -1)),
Sum(1, (i, 2, -2))]]
In [107]: [ [ Sum(1,(i,-2+k,2-k)).doit() for k in range(j+1) ] for j in
range(5) ]
Out[107]: [[5], [5, 3], [5, 3, 1], [5, 3, 1, -1], [5, 3, 1, -1, -3]]
If we look at some specific sum we have for example:
In [121]: S = Sum(1, (i, -1, 1))
In [122]: S.doit()
Out[122]: 3
In [123]: S = Sum(1, (i, 1, -1))
In [124]: S.doit()
Out[124]: -1
These contribution do not cancel out nor do the sum up as if there were
some "abs".
Interchanging limits works as follows:
In [127]: S = Sum(i, (i, a, b))
In [128]: Sr = reverse_order(S, i)
In [129]: Sr
Out[129]: Sum(-i, (i, b + 1, a - 1))
In [130]: S.doit()
Out[130]: -a**2/2 + a/2 + b**2/2 + b/2
In [131]: Sr.doit()
Out[131]: -a**2/2 + a/2 + b**2/2 + b/2
Simply exchanging a and b without shifting misses boundary terms and does
NOT conserve the value of the sum.
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