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 ▼ Chuck Senior Member Posts: 320 Threads: 59 Joined: Dec 2006 02-02-2009, 08:40 PM 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. :) ▼ Allen Senior Member Posts: 562 Threads: 27 Joined: Oct 2006 02-02-2009, 09:08 PM 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= 0 Excel: FLOOR(1/3,1)=0 RPN: 3 [1/x] FLOOR -> 0 Wolfram:Floor(1/3)=0 ``` Am I missing something? Edited: 2 Feb 2009, 9:08 p.m. ▼ David Hayden Senior Member Posts: 528 Threads: 40 Joined: Dec 2008 02-02-2009, 09:12 PM I think it gets interesting for 0 < x < 1. As you've discovered, for all x > 1, f(x) = 0 Dave Allen Senior Member Posts: 562 Threads: 27 Joined: Oct 2006 02-02-2009, 09:11 PM perhaps I am having a hard time reading the f'(x).. This looks almost exactly like f(x) in times new roman font. ▼ Chuck Senior Member Posts: 320 Threads: 59 Joined: Dec 2006 02-02-2009, 09:38 PM 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 find where f'(x)=3 ``` 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 ▼ David Hayden Senior Member Posts: 528 Threads: 40 Joined: Dec 2008 02-02-2009, 11:09 PM 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 Let g(u) = floor(u)*u Then f(x) = floor(1/x)/x is the same as g(u(x)) If N is a negative integer then For u = ( N , N+1 ], floor(u) is the constant N+1 and g(u) = (N+1)*u. Inside one of these intervals, the derivative of f(x) is: f'(x) = g'(u) * u'(x) = (N+1) * -1/x^2 = -(N+1) / x^2 = -(N+1) * u^2 Now we can find where f'(x)=3: 3 = -(N+1) * u^2 u = sqrt(-3/(N+1)) or -sqrt(-3/(N+1)) Since we already said that N is a negative integer and u is in the interval (N, N+1], that means u is a negative number also, so we can get rid of the positive solution and we're left with: u = -sqrt(-3/(N+1)) Now lets try some solutions. N=-1 won't work because -3/(N+1) causes division by zero. How about N=-2? u = -sqrt(-3/(-1)) = -sqrt(3) = -1.732050... Checking our assumption: N

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