challenge for programmable calculators

[Thomas Klemm]

(12-26-2013 07:17 AM)Han Wrote:  So if you did not use any software (solver or grapher), I fail to see how you can conclude "obviousness."

Obvious in the sense that \(f\) is continuous: it will switch the sign only at roots. Maybe too obvious to note?

Quote:I'm a bit confused here. I thought the point was (in trying to go for the fastest program) to try to reduce the complexity as much as possible prior to doing any programming.

I thought the point was having some fun with your calculators on a rainy day.

Quote:I am also not seeing how luck factors in, either, given that everything was deduced mathematically.

There's nothing wrong with your deduction. It's just that it works only up to the inflection point at ~ 8.2081. But we're lucky that we only have to show it up to 7. So we're fine! This kind of luck.
I mean there's no way you could tell beforehand it will work.
In essence it's a further reduction of the search space: instead of 7 we're down at 1. Well done!

Quote:Given that \( a \) can only be 7 possible integer values, brute-force-by-hand would also lead to the conclusion that the descriminant is positive only for \( a = 1 \).

Not sure when brute-force turns into intelligent search. Really only when we're able to enumerate the solutions without further checks?

Cheers
Thomas