More Challenges - Printable Version +- HP Forums ( https://archived.hpcalc.org/museumforum)+-- Forum: HP Museum Forums ( https://archived.hpcalc.org/museumforum/forum-1.html)+--- Forum: Old HP Forum Archives ( https://archived.hpcalc.org/museumforum/forum-2.html)+--- Thread: More Challenges ( /thread-146472.html) |

More Challenges - Chuck - 02-02-2009
Since I didn't get to play with the latest challenges a whole lot, I thought I'd make one up. Here's a little problem I gave my students today... Consider: f(x) = Floor[1/x]/x (or the greatest integer of 1/x, divided by x). We were talking about the derivative of such a beast, and found some interesting things. Can you find if or where f'(x) = 1, 1.5, 2, or 3. We'll need to bend the rules of differentiation a bit and consider one sided limits on a few of the values (graph it and you'll see why). I'd be curious if one could write/run a program on an Hp to solve f'(x)=3 to an accuracy enough to "guess" the exact value. So far, even Mathematica hasn't found the solution explicitly (bisection method yes, but no Solve), but it's more of a user error than a software limitation error.
My luck, someone here will get the answer(s) in 7 minutes or less. :)
Re: More Challenges - Allen - 02-02-2009
not sure I understand the question.. Can you give an example of the "Floor" function, since it does not behave the same way I expect.
Here's they way Excel and HP calculators interpret this problem: F(3) = Floor(1/3)/3 = Floor(.3333...)/3= 0/3= 0Am I missing something?
Re: More Challenges - Allen - 02-02-2009
perhaps I am having a hard time reading the f'(x).. This looks almost exactly like f(x) in times new roman font.
Re: More Challenges - David Hayden - 02-02-2009
I think it gets interesting for 0 < x < 1. As you've discovered, for all x > 1, f(x) = 0
Dave
Re: More Challenges - Chuck - 02-02-2009
Agree, the font is not great. Also, the "Floor" function is Mathematica's name for the greatest integer function. Here's another go at it...
let f(x)=Int[1/x]/x David is correct, it's not very interesting for x>1, but is interesting for x<1. Our interest actually lies on the interval [-1,0]. CHUCK
Re: More Challenges - David Hayden - 02-02-2009
Okay, my calculus is really rusty, but here's what I came up with. There's a good chance that I've gotten something horribly wrong, in which case, corrections will be cheerfully accepted.
To keep my head from hurting, I recast the problem: Let u(x) = 1/x Re: More Challenges - Chuck - 02-03-2009
Aha. Very close, but off by a small factor. I think there may be a slight error with the floor function on negative intervals with how you defined g. However, your method got me the answer analytically, or at least agrees with what I found earlier.
CHUCK
Re: More Challenges - David Hayden - 02-03-2009
Thanks, Chuck. I suspect I've got the definition of floor() wrong. If floor(x) is "the largest integer that's less than x" rather than "the integer formed by lopping off the fractional part" then I get: If N is a negative integer then For u = [ N , N+1 ), floor(u) is N and g(u) = N*u. |