Provide an example, or prove it's impossible

[davitykale]

Homework Statement

A differentiable function g: R --> R such that f' is unbounded on any interval of length one

Homework Equations

The Attempt at a Solution

It's probably easiest to look at [0,1]...maybe something that goes to infinity as it approaches 1? But not sure if it would be differentiable then...

 

[Dick]

Give me an example of a simple function that has a unbounded derivative at some point. Then splice a lot of them together. Define the function piecewise on intervals.

 

[olivermsun]

So how can you make a bounded function's derivative really big?
(How do you make f'(x) = df/dx blow up?)

 

[davitykale]

Um...I'm not quite sure...have the function shoot up? Almost as if with a horizontal asymptote at x=1?

 

[Mark44]

Um...I'm not quite sure...have the function shoot up? Almost as if with a horizontal asymptote at x=1?

That would have to be a vertical asymptote. I think that's where olivermsun is going.

And there would be have to be a bunch of them so that f' is unbounded on every interval of length 1. Maybe a periodic function?

 

[davitykale]

Sorry, I meant vertical asymptote. So it is possible, then?

 

[Mark44]

I think so. For example, f(x) = 1/(x - 1) is differentiable on the half-open interval [0, 1)(and elsewhere, but I'm trying to make a point). This interval is of length 1, and f' is defined at all points in this interval.

Intervals between two points all have the same length, either with or without the endpoints included.

 

[davitykale]

But is f' unbounded on EVERY interval of length one?

 

[LCKurtz]

And note that f' is differentiable on R means it is continuous so no vertical asymptotes.

 

[davitykale]

So then it's not possible, right?

 

[LCKurtz]

And note that f' is differentiable on R means it is continuous so no vertical asymptotes.

So then it's not possible, right?

If you are addressing me, no, I didn't say that.

 

[Mark44]

A differentiable function g: R --> R such that f' is unbounded on any interval of length one

Should this be: A differentiable function g: R --> R such that g' is unbounded on any interval of length one ?

And note that f' is differentiable on R means it is continuous so no vertical asymptotes.

I'm not sure what we know here, with uncertainty about the problem statement.

 

[davitykale]

Sorry, completely my mistake! g' is unbounded on any interval of length one

 

[Dick]

Think about a function like g(x)=sin(1/x^3)*x^2 and define g(0)=0. Tell me some stuff about the derivative of that function. Like is it differentiable at x=0? Is is differentiable everywhere? Is the derivative bounded near x=0?

 

[davitykale]

The derivative doesn't exist at x=0, right? And I don't think it's bounded..

 

[Dick]

The derivative doesn't exist at x=0, right? And I don't think it's bounded..

Since sin(x) is between 1 and -1, -x^2<=g(x)<=x^2. Are you sure the derivative doesn't exist at 0? But you are right that it's not bounded near x=0. How do you know that? Give reasons for what you think.

 

[davitykale]

is g'(0)=0, then?

 

[Dick]

is g'(0)=0, then?

Yes, it is. Again it's a good idea to say why you think so.

 

[davitykale]

Isn't the derivative a combination of sin and cos, each with an x term...so when x=0, you just get 0. Those same x terms make g' unbounded, right?

 

[Dick]

Isn't the derivative a combination of sin and cos, each with an x term...so when x=0, you just get 0. Those same x terms make g' unbounded, right?

Are you guessing? Sure the derivative is a combination of sin and cos and if you look at the limit, they are unbounded. But that's only when x is not 0. At x=0 you had better think about a more direct proof using the difference quotient. The derivative function gets pretty badly behaved near 0.

 

[davitykale]

So, I'm probably wrong, but:

With teh difference quotient, we get lim(x-->0) x sin(1/x^3)...so I want to say that the derivative is zero, because even though 1/x^3 goes to infinity, overall sin is bounded by 1

 

[Dick]

So, I'm probably wrong, but:

With teh difference quotient, we get lim(x-->0) x sin(1/x^3)...so I want to say that the derivative is zero, because even though 1/x^3 goes to infinity, overall sin is bounded by 1

Yes, that's it. It's a squeeze theorem limit. That's good. So the function is differentiable at 0 and the derivative is 0. Can you tell me why it's unbounded near 0? And can you figure out how to make a function that has that nasty behavior at more places than just x=0?

 

[davitykale]

Sorry, why the function is unbounded near 0 or why g' is unbounded near 0?

 

[Dick]

g(x) isn't unbounded. |g(x)|<=x^2. g'(x) is. What is g'(x)? Give me a formula for g'(x) and think about why it might be unbounded. You might think about putting x^3=1/(2*pi*n) where n is a large integer. Which would make x near 0.

 

[davitykale]

g'(x) = 2x sin(1/x^3) - 3/x^2 cos(1/x^3)

So, even though sin and cos are bounded, at large values of x, it will still be really large and continue to grow...?

 

[Dick]

g'(x) = 2x sin(1/x^3) - 3/x^2 cos(1/x^3)

So, even though sin and cos are bounded, at large values of x, it will still be really large and continue to grow...?

We aren't talking about large values of x. You want to talk about small values of x near 0. Your first term approaches 0. What does the second one do?

 

[davitykale]

Sorry, my mistake. Well the second term approaches infinity, and faster than the first term approaches 0

 

[Dick]

Sorry, my mistake. Well the second term approaches infinity, and faster than the first term approaches 0

True. But doesn't really appoach infinity. It oscillates between +infinity and -infinitity, right? That proves g'(x) is unbounded near x=0. Now you got a function that has the bad behavior you want near x=0. Can you think of a way to that to make a function that does that in any interval of length 1?

 

[davitykale]

g(x) = x^2 * sin (1/(x-1)^3) ?

 

[Dick]

g(x) = x^2 * sin (1/(x-1)^3) ?

That's not a very good guess. It's your worst of the night, and you are pretty sharp. If g(x) is badly behaved at x=0 in the way you want, then g(x-1) is badly behaved at x=1. Sum them. Now you've got a function that's badly behaved at both points. How might you create an infinite number of badly behaved points? Think about geometric series.

 

[davitykale]

Sorry sometimes I cannot help but feel like an absolute idiot when it comes to this stuff.

$$\sum_{n=0}^{\infty} x^2 \sin(1/x^3 -n)$$

But even if this is close to being correct, won't that only cover x>0?

 

[Dick]

Sorry sometimes I cannot help but feel like an absolute idiot when it comes to this stuff.

$$\sum_{n=0}^{\infty} x^2 \sin(1/x^3 -n)$$

But even if this is close to being correct, won't that only cover x>0?

You aren't an absolute idiot, so don't apologize. You are actually pretty good. I feel good talking to you. But that doesn't do it. Because you haven't said why think it does. What about what about sum n=0 to infinity of g(x-n)/(2^n). That gets you half way to what you want. Do think it converges?

 

[davitykale]

It converges because g(x-n)/(2^n) < x^2/(2^n) < 1/2^n which converges...right?

 

[Dick]

It converges because g(x-n)/(2^n) < x^2/(2^n) < 1/2^n which converges...right?

Sort of. But then you have to show the derivative converges as well. This is starting to get too complicated. I have another suggestion. Why don't you try to define the full function piecewise?

 

[davitykale]

I'm not quite sure what you're asking... are we still talking about g(x-n)/2^n?