irregular summations in Home/CAS

[Wes Loewer]

(10-20-2024 03:27 PM)parisse Wrote:  The rule is comply with Chasles relation:
sum(..,a,b)+sum(...,b+1,c)==sum(..,a,c)

Thank you for pointing me in the right direction. It still took me a little while to wrap my head around this. I kept getting off by 1 on one of the indices, but I think I've got it now, but my "proof" seems like circular reasoning.

Proof that sum(...,a,b) = -sum(...,b+1,a-1)
0. Assume that sum(...,a,b) = -sum(...,b+1,a-1)
1. sum(...,a,b) + sum(...,b+1,a-1) = 0
2. sum(...,a,a-1) = 0
But how do we know that sum(...,a,a-1) = 0? Because if you apply the original assumption
3. sum(...,a,a-1) becomes -sum(...,(a-1)+1,(a)-1)
4. sum(...,a,a-1) = -sum(...,a,a-1)
The only way for x=-x is if x=0
So sum(...,a,a-1) is indeed = 0

Since we have arrived at a true statement, the conclusion is that the assumption was true. So the assumption is internally consistent, but a it seems like circular logic because I used the assumption in step 3.

Is there a non-circular proof, or is internal consistency the best we can hope for?

For instance, assuming that if a<b then sum(...,b,a)=0
sum(...,a,b) + sum(...,b,a) = sum(...,a,b) + 0
sum(...,a,a) = sum(...,a,b)
which doesn't make sense, so the assumption that sum(...,b,a)=0 is not consistent with Chasles relation. However, could it be argued that it is consistent with some other logical assumption?

I guess the implications of Chasles relation makes sense if you use the analogy of vectors. But if you are thinking in terms of a for-loop, then sum(...,b,a)=0 makes sense, and might be more practical.

Thank you for the enlightening mental exercise.


Edit:
Actually, my statement that
sum(...,a,b) + sum(...,b,a) = sum(...,a,b) + 0
sum(...,a,a) = sum(...,a,b)
is not quite true since sum(...,a,b) + sum(...,b,a) has a b in both sums, but the point is still true that this leads to an inconsistency with Chasles relation.