# Continuous and nowhere differentiable

1. Apr 2, 2005

### saltydog

The Harvey Mudd College Math dept presents the Weierstrass' function:

$$f(x)=\sum_{n=0}^{\infty} B^nCos(A^n \pi x)$$

as an example of a continuous function nowhere differentiable if 0<B<1 and $AB>1+\frac{3\pi}{2}$. Surely it converges to a continuous function if 0<B<1 regardless of the value of A. I base this on the Weierstrass M-test for this series.

However, I can differentiate the series leading to:

$$f^{'}(x)=-\pi \sum_{n=0}^{\infty}(AB)^n Sin(A^n \pi x)$$

Doesn't this series also converge uniformly (same convergence test as above) for all x if 0<AB<1? If so, why are they requiring $AB>1+\frac{3\pi}{2}$? I would think you could say it's nowhere differentialbe if AB>1 although greater than $1+\frac{3\pi}{2}$ still qualifies.

I haven't asked them, I guess I could, maybe someone here knows though.

2. Apr 3, 2005

### matt grime

I suspect you half answer your own question. I would guess that the 3pi/2 thing is to force divergence of the resulting differentiated series at some point, the sin bit may be troublesome.

Tests for converegence are like that: just because we know it converges when property P is satisfied doesn't mean it diverges if P isn't satisfied - they are not usually necessary and sufficient conditions.

3. Apr 3, 2005

### saltydog

Hello Matt,

I belive the function has a fractal dimension. Note the 3 attached plots of the sum of just the first five terms. The plots successively zoom-in to a small area (0,0.0001). It's self-similar and at the limit, it must have a fractal dimension although I don't know how to show this. I wish to make the following un-proven claim:

Any function with fractal dimension is differentiable nowhere.

My Analysis book includes a technique to prove another more simple "tent function" is no where differentiable and I think I'll try to attack this problem from that same perspective. I tend to think the $\frac{3\pi}{2}$ has some specific "limiting" relevance and perhaps my analysis will uncover it. Suppose, as a last effort, I can contact Mudd but not for now.

Anyone here know how to calculate fractal dimension and can this be applied to this function?

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4. Apr 3, 2005

### saltydog

Alright, I'll make even a stronger claim:

A continuous function is nowhere differentialbe iff it has a fractal dimension.

Might be interesting to try and prove this . . .

5. Apr 3, 2005

### Hurkyl

Staff Emeritus
The line segment y = 0, 0 < x < 1 has fractal dimension 1, and it's everywhere smooth. :tongue2:

Back to the original question, note that you failed to consider differentiability at x = 0.

Last edited: Apr 3, 2005
6. Apr 3, 2005

### saltydog

Well wait a minute Hurkyl, by definition "fractal dimension" is of "non-integer" values like 1.3 or 0.98 and so fourth so I'm not sure what you mean by a straight line having a fractal dimension of 1. I'll look into your second comment.

Alright, I guess I'll dig up Mandelbrot's book . . . where is it . . . oh yea, up there, kinda dusty. But I didn't understand it back then and nothing's changed since . . .

Last edited: Apr 3, 2005
7. Apr 3, 2005

### Hurkyl

Staff Emeritus
Well, I was specifically speaking about capacity dimension, one of the various things to which fractal dimension refers.

Now, fractal dimension is not, and should not, be limited to nonintegral values. For example, consider Cantor's dust. In the default construction, we replace each box with 4 copies at a scale of 1/3. If we instead use a scale of 1/4, the resulting fractal has dimension exactly 1.

8. Apr 3, 2005

### saltydog

Well, Cantor's dust is not continuous (and I suspect it has measure 0) and for the moment, I'd like to remain focused on continuous functions. In that regard, I have the following two questions:

1. Is Weierstrauss' curve a "fractal curve"?

2. If so, what is it's fractal dimension?

Mandelbrot, in his book which I still don't understand, states, "curves which the fractal dimension exceeds the topological dimension 1 be called "fractal curves". I belive he's referring to continuous curves.

Last edited: Apr 3, 2005
9. Apr 3, 2005

### Hurkyl

Staff Emeritus
I haven't looked all that much into this subject, so I can easily have terminology wrong.

One thing I'm pretty sure you can prove is this:

If a curve C is given by a continuous, differentiable map from [0, 1], then the length of C is defined and finite. (hurray for compact sets)

So that any such curve does have Hausdorff dimension 1.

I'm pretty sure the Wierstrauss function will have infinite length, but, a curve can have infinite 1-dimensional measure (i.e. length) and still be Hausdorff dimension 1.

A curve with Hausdorff dimension greater than 1 obviously cannot be everywhere differentiable, but I would be surprised if you can get nowhere differentiability -- surely it's possible for a fractal to be differentiable at a single point, or maybe even a dense set of measure zero! (or even better, a dense set of d-dimensional measure zero, where d is the fractal dimension)

Oh, BTW, I'm pretty sure 3-dimensional fractal curves may have fractal dimension 2.

10. Apr 3, 2005

### saltydog

I wish to revise my unproven claim above:

A continuous function is nowhere differentiable iff its fractal dimension exceeds 1.

Now, how do I calculate the fractal dimension of a continuous curve in the plane?

11. Apr 3, 2005

### Hurkyl

Staff Emeritus
Hrm... there don't seem to be any local definitions of dimension.

For example, consider this:

Let C be your favorite fractal curve with an endpoint. Then, attach a straight line segment to that endpoint. By the definitions of dimension I could find, this leaves the dimension of C unchanged, but we have clearly added a part that "ought" to be dimension 1...

(at least the line segment will have d-dimensional measure zero)

In any case, this gives you a differentiable portion.

Anyways, this proves your claim is false. Maybe your text has a definition of "local" dimension... I had taken a stab in the dark at what that might mean (meaning I tried proving something, and defined local dimension to be the thing I wanted to assume) and can prove that if f is differentiable at P, then it has dimension 1 at P, but I don't think I had anything meaningful -- I think I could prove that any continuous function had local dimension 1.

I don't know of an easy way to calculate the dimension of anything except the self-similar type of fractal, such as the Koch Snowflake. I could only suggest hammering away at the definitions.

12. Apr 3, 2005

### saltydog

Thanks Hurkyl.

I've been looking at Peitgen's book on fractals and my initial efforts seem to indicate that Weierstrass' function has in fact a "self-similarity" dimension of 1. If that's the case then my proposal above is false. I'm not confident of this however. My initial reason for pursuing this line was the possibility that dimension had something to do with the $\frac{3 \pi}{2}$ requirement. I'll spend some more time with it and if all else fails I'll e-mail the Mudd Math dept for help.

13. Apr 4, 2005

### saltydog

I've since learned that the Weierstrass function is a fractal and does indeed have a fractal dimension between 1 and 2:

http://

According to this reference, for:

$$w(x)=\sum_{n=0}^\infty a^n Cos[2 \pi b^n x]$$

where:

$$0<a<1<b$$

$$D_w=2+\frac{ln(a)}{ln(b)}$$

Note since a is less than b, the dimension is between 1 and 2.

Thus I stand by my original claim . . .

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14. Apr 4, 2005

### Hurkyl

Staff Emeritus
Just a possibility -- maybe it will help to rewrite f as:

$$f(x) := \Re \sum_{n=0}^\infty b^n e^{i a^n \pi x}$$

15. Apr 4, 2005

### saltydog

I'm sorry Hurkyl but that's just not happening for me. I realize that notation represents taking the real part of an infinite sum of complex numbers but I don't understand the relevance. Would you kindly elaborate a bit further please?

16. Apr 5, 2005

### Hurkyl

Staff Emeritus
Sometimes, exponentials are just easier to manipulate than trig functions -- I didn't have any particular line of attack in mind, though. This was how I planned on looking at the problem next, but I hadn't really had a chance to work on it.

Last edited: Apr 5, 2005
17. Apr 5, 2005

### saltydog

Hello Hurkly,

I've been working on it . . . very interesting. What I found especially so was the claim that functions like these are the RULE and NOT the exception! You know, like there's many more irrational numbers than rationals. Apparently same diff with "nice" functions opposed to "monster" functions. We just work with nice ones because they are ammendable to analysis. Hummm . . . another words, there's a whole world of mathematical objects out there and we only use very nearly an "infinitessimal" part of them by restricting ourselves to "differentiable" ones. Not saying you don't already know this, I didn't.

Mark Nielsen of Univ. of Idaho calls this the “Law of mathematical unapproachability”. Simply stated, “most objects in the mathematical universe are too wild for humans to describe.”

Here's the reference if anyone is interested:

Reference

And so, as those guys in philosophy over there like to talk about, I wonder what the philosophical significance of this is: we use only a small part of it to very successfully describe nature. How much better would that description be if we used a larger part!

Really, I would like to just know how to prove "nowhere differentiable". Apparently it's not as simple as just taking the derivative and showing the sum converges nowhere.

18. Apr 5, 2005

### Hurkyl

Staff Emeritus
Yah, that's pretty nifty, isn't it?

Actually (at least in the current context), there is a viewpoint from which one can say that everything is, in fact, describable. However, that's a far cry from, for example, being able to evaluate any real function at 0.

Anyways, back to the problem at hand... your problem is that you swapped the order of limits, an operation that isn't generally allowed. Specifically, there's a limit operation implicit in differentiating, and also with summation. So, in generally, summing then differentiating won't be the same as differentiating then summing.

19. Apr 5, 2005

### Data

Well, swapping them is ok if you have uniform convergence of the differentiated series (which he does~).

ie. if

$$f(x) = \sum_{n=0}^\infty f_n(x)$$

on a closed interval $I$ and

$$\sum_{n=0}^\infty f^\prime_n(x)$$

is uniformly convergent on $I$ and each $f^\prime_n$ is continuous on $I$ then

$$f^\prime(x) = \sum_{n=0}^\infty f^\prime_n(x)$$

on $I$. Of course, that doesn't mean that the function isn't differentiable outside of the interval.

Last edited: Apr 5, 2005
20. Apr 5, 2005

### Hurkyl

Staff Emeritus
Only when AB < 1, though. The fact the series diverges almost everywhere for AB > 1 doesn't guarantee the derivative doesn't exist.

21. Apr 5, 2005

### Data

yep, that's definitely true~

22. Apr 5, 2005

### saltydog

Thanks guys. I need to review some Analysis. So, as I see it I can swap the order of summation and differentiation as long as the series converges uniformly to a function which it does if AB<1. This I can prove via the Weierstrass M-test for both the function and the sum of the derivatives of $f_n(x)$.

So, however, when AB becomes greater than 1 this no longer applies and another approach must be used. Is that correct? I found a good reference proving "a" function exists which is nowhere differentiable which is under the title of "The Weierstrass Function":

Think I'll get a hardcopy and spend some time going over this proof. I suspect it can be applied to prove nowhere differentiable of the cosine series above.

23. Apr 5, 2005

### Data

as long as the series of derivatives converges uniformly, yes

24. Apr 6, 2005

### saltydog

In an effort to approach this problem, I've chosen to fall back to an easier one first:

$$f(x)=\sum_{n=0}^\infty \frac{GetDistance[4^n x]}{4^n}$$

Where the GetDistance[x] function returns the distance from x to the integer nearest x.

This is the saw-tooth function commonly used to demonstrate continuous and nowhere differentiable. I've attached a plot superimposing $f_0$,$f_0+f_1$ and$f_0+f_1+f_2$. My Analysis book has a proof I can follow of it's nowhere differentiable behavior. Essentially, we seek to show:

$\lim_{n\rightarrow\infty}{} \frac{f(a+h)-f(a)}{h}$

does not have a limit. My interpretation of this is that it has a "corner" at all points in its domain.

Next, I'll pursue the following:

$$f(x)=\sum_{n=0}^\infty \frac{Sin[(n!)^2 t]}{n!}$$

I'm told that T.W. Koerner in his book "Fourier Analysis" gives a detailed proof that this function is nowhere differentiable. Again, this involves an analysis of the difference quotient. Apparently using the same technique, one can prove the non-differentiability of the Weierstrass function.

So, I'll work with them a bit first myself and if nothing pans out, well, it's a pleasant ride to the University library from my home.

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25. Apr 7, 2005

### saltydog

I've retrieved the reference cited above from T.W.Koerner's book, "Fourier Analysis". I wish to focus on reviewing the proof here of nowhere differentiable for the function below (reviewing/reporting here will help me better understand it and perhaps will help others who also wish to learn more about these functions). I'll report it in parts. I should add that he states after thoroughly investigating this proof, one should be able to apply it to prove nowhere differentiable of the cosine series of the original post which is my ultimate goal.

$$f(t)=\sum_{r=0}^\infty \frac{1}{r!}Sin[(r!)^2 t]$$

In general, we seek to investigate the difference quotient:

$$\mathop {\lim }\limits_{x_n \to x} \frac{f(x)-f(x_n)}{x-x_n}$$

In order to investigate this limit, Koerner breaks the sum into three parts which turn out to be the key to the proof:

$$h_n(t)=\sum_{r=0}^{n-1}\frac{1}{r!}Sin[(r!)^2 t]$$

$$k_n(t)=\frac{Sin[(n!)^2t]}{n!}$$

$$I_n(t)=\sum_{r=n+1}^{\infty}\frac{1}{r!}Sin[(r!)^2 t]$$

And thus:

$$f(t)=h_n(t)+k_n(t)+I_n(t)$$

These function components will be used to evaluate the difference quotient. However, some Lemmas are needed in order to set bounds on the various quantities. That's next.