More Challenges


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. :)


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
Am I missing something?

Edited: 2 Feb 2009, 9:08 p.m.


I think it gets interesting for 0 < x < 1. As you've discovered, for all x > 1, f(x) = 0



perhaps I am having a hard time reading the f'(x).. This looks almost exactly like f(x) in times new roman font.


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].



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 <? u <=? N+1
-2 < -1.732050... <= -1
That works.

Last but not least, we solve for x:

u = 1/x
-sqrt(3) = 1/x
x = -1/sqrt(3)


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.



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.
So f'(x) = -N*u^2
And if f'(x)=3 then u = -sqrt(-3/N)
So for N=-1, we get u = -sqrt(3) which is outside the interval [-1,0). But for N=-2, we get u=-sqrt(3/2) which is within the interval [-2,-1].

So with u=1/x, that means the final answer is x=-1/sqrt(3/2)

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