Provide an example, or prove it's impossible

[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?

 

[Dick]

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

No, I don't like that idea so much anymore. It's starting to get too complicated. I'm thinking more about cutting out a piece of g(x) around x=0 and repeating it over and over again (maybe with a little variation) along the real line to get the infinite number of bad derivative points that you need. You just have to make sure it's differentiable at points where you splice the pieces together. Do you see what I mean?

 

[davitykale]

Um...I'm really confused...but something like (g_n)? Where you have
g(0)=x^2 * sin(1/x^3)
g(1)= x^2 * sin(1/(x-1)^3)
...etc.?

 

[Dick]

Um...I'm really confused...but something like (g_n)? Where you have
g(0)=x^2 * sin(1/x^3)
g(1)= x^2 * sin(1/(x-1)^3)
...etc.?

No, that's not it. Let's change the definition of g(x) to x^2*cos(1/x^3). It has the same basic properties as the other g(x). But now it's an even function. Sketch a rough graph of the function. There are plenty of points 0<x0<1/2 such that g(x0)=g(-x0) and g'(x0)=g'(-x0)=0. Do you agree? I'm suggesting you extend g(x) on [-x0,x0] to a periodic function on the whole real line by just repeating it over and over again. The picture here is much more important than any exact formulas.

 

[davitykale]

Okay, I think I understand now. Sorry about that. I'm not exactly sure HOW to make that function, though. My first instinct is to make a series of functions and throw in an n, but I guess we're not doing that anymore...?

 

[Dick]

Okay, I think I understand now. Sorry about that. I'm not exactly sure HOW to make that function, though. My first instinct is to make a series of functions and throw in an n, but I guess we're not doing that anymore...?

You can make a series of functions and throw in a 2^n. I think it will work, but I think the technical complications of showing it and it's derivative converge make it not worth the trouble. A function that's periodic but just as nasty at a lot of points as x^2*cos(1/x^3) is at x=0 has less complications. If I were grading this course, I would be really happy if you could just describe how to make the function in words, and not give me a lot of formulas to decipher.

 

[davitykale]

I'm honestly just curious at this point...what sort of periodic function are you talking about? How would I make such a function?

Btw thank you so much for your help! You're a genius!

 

[Dick]

I'm honestly just curious at this point...what sort of periodic function are you talking about? How would I make such a function?

Btw thank you so much for your help! You're a genius!

I told you. If g(x) is a function on [-x0,x0] such that g(-x0)=g(x0) and g'(-x0)=g'(x0)=0, then if I define g(x) on [x0,3*x0] by g(x-2*x0), I'm just periodically extending the function on [-x0,x0] to [x0,3*x0], right? Is it clear to you that the resulting function is differentiable on [-x0,3*x0]? Now keep extending it to the whole real line. The genius needed here isn't coming up with the function, it's trying to figure out a way to explain it. Wish I were better at it. If you can draw a picture and get it you can jump over my poor attempts to explain it. I'm just trying to copy the part of g(x) on [-x0,x0] over and over again.

 

[davitykale]

And so...going back to the original problem...this function has a derivative that is unbounded on intervals of length one?

 

[Dick]

And so...going back to the original problem...this function has a derivative that is unbounded on intervals of length one?

g(x) on [-x0,x0] has an unbounded derivative near x=0. If the length of [-x0,x0] is less than one and the full function is a periodic extension of the original g(x), then every interval of length one will contain a problem point, yes? This is picture drawing talk.

 

[Redbelly98]

Thread locked.