greatest integer, math terms, & derivatives

Factual Questions
1

Okay, this may end up being a rather mundane question, but I wanted to make sure I wasn’t missing something obvious.

There’s a problem from my analysis text that is either a misprint, or a new use of terminology, or I’ve read it completely wrong. As it is, I don’t think the problem is correct.

All typographical symbols are as they appear in the book (oo = infinity).

Problem: Suppose that f’ exists and is increasing on (0, oo) and that f is continuous on [0, oo) with f(0) = 0. Show that g(x) = [f(x)]/x is increasing on (0,oo).
[/quote]

Now the only use I’ve seen for the operator in this book (aside from order of operations) is to indicate the greatest integer function. But that means that g(x) isn’t increasing, since for each piece of the function, it’s actually decreasing. What makes this even more suspect is that the hint says, “Show that g’(x) > 0.” So is there another use of that might work here?

Unless I’m wrong and the problem can be solved, my guess is that maybe g(x) is just f(x)/x (which I think works out, but I haven’t fully proven).

Or this is a new use of ‘increasing’ (despite an earlier example apparently explicitly contradicting this definition) to mean that there exists for every x1 in g’s domain, an x2>x1 such that g(x2)>g(x1). Prior to this ‘increasing’ seemed to only mean ‘strictly monotone increasing’; which is the accepted meaning?

In a completely unrelated problem, I’m also trying to work this one out :

Show that f(x) is differentiable, but the derivative is not continuous at x=0.
f(x) = x[sup]2[/sup]sin(1/x), x != 0
. . . .0, x= 0

2

Although MathWorld says it used to be used that way, I’ve never seen to denote the floor function. The only other use it gives is the Iverson bracket, which clearly does not apply here. Does your book actually use for floor somewhere else? I also think your speculated alternate definition of “increasing” is not what’s going on, either.

As for your second problem, f(x) is a well-behaved function everywhere but x = 0, so you can evaluate f’(x) simply using the product and chain rules everywhere but x = 0. At x = 0, use the definition of a derivative.

3

I think that your guess that g(x) is just (f(x))/x is probably correct.

4

I’ve never seen used to indicate anything but a floor or ceiling function. If you assume that they meant g(x) = f(x)/x, then you are trying to prove something true, so maybe you should take it that way.

5

I’ll agree with RM here. I wind up with a positive g’(x) also from my work. Assuming I can remember my calc.

6

Just a guess, but I think h[y,z) means that the domain(?) of h goes from y INCLUSIVE to z EXCLUSIVE

Brian

7

We got that part, N9IWP, it was the function notation where g(x) was defined that was the question.

8

Hey panamajack any chance of giving us the answer? I’d like to see how badly I did. :stuck_out_tongue:

9

Hi, Grey. Here’s the one I came up with (disclaimer: The Weak Force makes no guarantee of the accuracy of this proof and hereby disclaims its suitability for any particular purpose): (Proof by contradiction)
Suppose g is not an increasing function. Because g is differentiable and has a connected domain, it is not monotonically increasing iff at some x in the domain of g, g’(x) < 0 [well-known result; proof omitted] (NOTE: I’m assuming that in the statement of the problem, “increasing” means “monotonically increasing” and not “strictly monotonically increasing”; the alternate proof is very similar to this one). Therefore, there must be some positive x for which g’(x) < 0.
Using the “quotient rule,” g’(x) = (xf’(x) - f(x))/x[sup]2[/sup], so if g’(x) < 0, then xf’(x) - f(x) < 0 (since x[sup]2[/sup] is always positive on the domain), so f(x)/x > f’(x). Now, remember that f(0) = 0. Using this, the left side of the inequality is (f(x) - f(0))/(x - 0). By the Mean Value Theorem, there is a number c between 0 and x for which f’© = (f(x) - f(0))/(x - 0) > f’(x). Alas, we said c < x, and we have f’© > f’(x), contradicting our premise that f’ is increasing.
Since we have reached a contradiction, our initial supposition, i.e., that g’(x) is negative for some positive x, must be false. Because g’(x) >= 0 for all positive x, g(x) must be increasing, QED.

10

That looks like a pretty good proof to me.

11

The Weak Force, thanks. My napkin has fewer words mind you. :slight_smile:

12

Sorry I missed responding to that, Grey. and thanks to The Weak Force for an excellent proof.

The book does define the floor function using the brackets. It’s a little old ( 1989, Intro. to Analysis by J. Kirkwood) but I only paid $1.25 for it so I shouldn’t complain too much.

13

Incidentally, yesterday I was looking through my own analysis book: Principles of Mathematical Analysis, 3rd ed. (1976) by Walter Rudin, and this book has the same exercise (exercise 6, chapter 5), except that it has g(x) = f(x)/x, with no brackets, naturally. And no, Rudin does not include the solution to any of his exercises :wink: I like this book a lot, but I paid about $55 for my used copy :frowning:

14

Makes you feel any better, I bought new copies of that and Herstein’s Topics in Algebra the same semester.