Posté le 15/02/2019 22:54
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Citer : Posté le 15/02/2019 23:01 | #
Hi TLM, I don't have an fx-CG 500 to try this out, and testing it on my Graph 90+E (French equivalent of the fx-CG 50) only reveals that it computes the loop.
My wild guess is that the fx-CG 500 CAS knows the formula for the sum of squares, so naturally the result would come out in constant time. Maybe you can try with different upper bounds to check this theory?
TLM Invité
Citer : Posté le 16/02/2019 01:05 | #
I believe you're correct. I tried some ridiculous number (1,000,000,000,000) and with X^2 it still did the calculation nearly instantly but, with X^10 it took a lot longer. Thanks for clearing up my confusion Lephenixnoir!
Citer : Posté le 16/02/2019 09:25 | #
Nice! We can only regret that it still needs 0.32 seconds to compute something as simple as n(n+1)(2n+1)/6, even if there is some recognition involved.
Citer : Posté le 16/02/2019 15:14 | #
Does that mean that the sum of (x^2)^5 or the sum of (x^5)^2 might be computed faster than the sum of x^10 ? (And that both of them are computed in a different time ?)
Citer : Posté le 16/02/2019 15:16 | #
The formula in question works only for the sum of the first n squares, so unless there's something more general it won't necessarily work.