Wolfram Alpha != HP Prime CAS result: who is wrong?

[Albert Chan]

(06-04-2024 01:08 AM)robve Wrote:  On Wolfram Alpha:
\( \sum_{k=0}^n k=\frac12n(n+1) \)
but it also says that:
\( \sum_{k=0}^{-10} k=0 \)
which surprised me, because the sign of \( n \) should not matter. What am I doing wrong here?

It return 0 because there is no elements to sum

(08-26-2020 09:58 PM)Albert Chan Wrote:  Mathematica also does closed end limit, but generated a list. (conceptually)
If the list is empty, there is nothing to sum.
We lost the symmetry, but it is simple to understand.

(06-04-2024 01:08 AM)robve Wrote:  HP Prime CAS:
\( \sum_{k=0}^{-10}k=45 \)
which checks out \( \frac12(-10)(-10+1)=45 \)

Actually, limits get changed.

sum(k, k= 0 .. -10) = - sum(k, k=-9 .. -1) = sum(k, k = 1 .. 9) = 9*10/2 = 45

The reason may be due to symmetry

(08-26-2020 09:58 PM)Albert Chan Wrote:  This may explain the reason for (a,b) → (b+1,a-1), if a>b+1
Say, we have a function F(x) = sum(f(t), t=-inf .. x-1)

If a ≤ b, S1 = sum(f(t), t=a .. b) = F(b+1) - F(a)

If a > b, we could flip the limit, just like doing integrals.

S2 = - sum(f(t), t=b .. a) = -(F(a+1) - F(b)) = F(b) - F(a+1)

But, S1 != -S2. To aim for symmetry, we shift the limit a bit:

S2 = - sum(f(t), t=b+1 .. a-1) = F(a) - F(b+1) = -S1