Timeline for How to simplify a complicated Sum in terms of power Sums?
Current License: CC BY-SA 3.0
9 events
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| Mar 12, 2013 at 21:35 | history | edited | Michael E2 | CC BY-SA 3.0 |
Extended code to handle more cases, added examples
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| Mar 12, 2013 at 17:40 | comment | added | whuber | @belisarius You give good hints :-). I have proposed a solution based on them. | |
| Mar 12, 2013 at 14:51 | vote | accept | colinfang | ||
| Mar 12, 2013 at 19:22 | |||||
| Mar 11, 2013 at 13:20 | history | edited | Michael E2 | CC BY-SA 3.0 |
Added another solution
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| Mar 11, 2013 at 3:25 | comment | added | Dr. belisarius |
Yep, but the whole thing is about keeping n symbolic ...
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| Mar 11, 2013 at 3:23 | comment | added | Michael E2 |
@belisarius Of course, you can get the answer like this: Eliminate[{y == a, s[3] == Sum[x[i]^3, {i, n}], s[2] == Sum[x[i]^2, {i, n}], s[1] == Sum[x[i], {i, n}]} /. n -> 3, Table[x[i], {i, 3}]], for any n >= 3.
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| Mar 11, 2013 at 3:11 | comment | added | Michael E2 | It would be nice if someone found one. None of the usual functions seem to simplify indefinite sums. | |
| Mar 11, 2013 at 3:06 | comment | added | Dr. belisarius | +1 But I really don't like none of our solutions. There must be an easier way | |
| Mar 11, 2013 at 3:03 | history | answered | Michael E2 | CC BY-SA 3.0 |