# Talk:Weierstrass function

## Blancmarge

Is this the one known as the "blancmage"? Can't remember... Dysprosia 05:32, 27 Jan 2004 (UTC)

Perhaps we can believe the MathWorld page on Blancmange function... --Pt 22:56, 9 Oct 2004 (UTC)

Can anyone say why you can't differentiate the function, couldn't you just use the chain-rule?

Of course you can. Just the derivative doesn't (usually --Pt) converge as the function itself does. Pt 21:44, 6 Sep 2004 (UTC)

I think that more accurate plots are needed. Right now there are some obviously incorrect straight lines visible. Pt 21:46, 6 Sep 2004 (UTC)

## Other definitions

It's interesting to note that "Kaoseraamat" ("The Chaos Book") by Ülo Lepik and Jüri Engelbrecht (Tallinn 1999; ISBN 9985-50-235-3) gives on page 135 the Weierstrass function as such complex function: $W(t)=\sum _{n=-\infty }^{\infty }{1-e^{ib^{n}t} \over b^{(2-D)n}},$ where $1 "To visualize it," they take its real part, the Weierstrass-Mandelbrot function: $C(t)=\Re W(t)=\sum _{n=-\infty }^{\infty }{1-\cos b^{n}t \over b^{(2-D)n}}.$ I think these different definitions need further checking. Pt 22:13, 6 Sep 2004 (UTC)

MathWorld gives yet another definition:

$f_{a}(x)=\sum _{k=1}^{\infty }{\sin(\pi k^{a}x) \over \pi k^{a}}$ Someone should really check some reliable sources — which definition is correct?!

By the way, MathWorld also says that there are actually some points (namely, at $x={2A+1 \over 2B+1}$ for $\forall A,\,B\in \mathbb {Z}$ ), where the derivative is finite ($1 \over 2$ ). Anywhere else the function is really undifferentiable.

Thus, MathWorld is an interesting source for Wikipedia, but nevertheless needs checking.

--Pt 22:53, 9 Oct 2004 (UTC)

I think the current definition $f(x)=\sum _{n=0}^{\infty }a^{n}\cos(b^{n}\pi x),01+{\frac {3}{2}}\pi$ is correct. The original paper by Weierstrass, Abhandlungen Aus Der Functionenlehre (1886), shows this version, but with the a and b changed around and positive b less than 1. -- KittySaturn 14:02, 2005 Apr 16 (UTC)
According to this MSc thesis it seems that $f_{a}(x)$ from MathWorld is more likely the Riemann function. See also MathWorld redirect. --147.230.151.146 11:54, 5 February 2007 (UTC)

## An edit by 217.136.97.96 at 00:49 EET, 10 Oct 2004

probably the second known was also one of Weierstrass : sigma cos(n²x)/n² and the third its generalization cos(n^c x)/n^c.

I moved the sentence here because it certainly needs some discussion before going to the encyclopedia article. --Pt 23:18, 9 Oct 2004 (UTC)

## Is the definition correct?

The definition states: "Almost all continuous functions are nowhere differentiable,..."

But shouldn't it say: "Almost all continuous functions are differentiable."

This looks to be what the MathSource states as well. If you look up the word "differentiable", MathSource states: "Amazingly, there exist continuous functions which are nowhere differentiable." And then lists the Weierstrass and Blancmange functions.

I don't know, maybe i'm talking wacky talk. Sorry if i am.

Sincerely, Mark

• My initial reaction was that the original statement of "nowhere continuous" makes more sense to me. And apparently Oleg Alexandrov thinks that one is correct, too. Eric119 02:06, 9 May 2005 (UTC)
Sorry, I did miss this thing so far (have no idea why). Now, the statement:
Almost all continuous functions are nowhere differentiable
is indeed correct.
Now, until 19th century or so, people thought all continous functions are differentiable. So, that's why it was so amazing that people found examples of functions which are not differentiable anyware, and the examples of such functions are actually quite exotic.
It was only later realized that actually, even if most if not all continuous functions we know are differentiable, nevertheless, the functions which are continuous but nowhere differentiable vastly outnumber the former. I could think of a proof if you wish. Oleg Alexandrov 03:11, 9 May 2005 (UTC)

Alexandrov and All,

I beg your pardon, i was not aware of that!
Hopefully my simple additions make up for my ignorant mistake.
I did come here to learn - and learning i am!
Thanks for the correction.
Sincerely,
Mark

Mark 17:56, 2005 May 20 (UTC)

My name is Oleg by the way.
I think you should relax a bit. :) Oleg Alexandrov 18:16, 20 May 2005 (UTC)
Oleg
I am relaxed, i just type like i'm uptight. :-)
Cheers,
Mark
--Mark 17:16, 2005 May 24 (UTC)

## Odd integer requirement

I just added the requirement that b is an odd integer. This is how I learned it, and is verified by (for instance) , , . I don't know how to verify its necessity myself, but this requirement is, at least, common. LWizard @ 01:25, 27 April 2006 (UTC)

• I am not sure about the necessity, but it is an odd integer in Weierstrass' original paper, and this does at least make the proof of the properties of this function a lot easier. Weierstrass 22:04, 8 July 2007 (UTC)

## Picture?

Would it be possible to get a graphic showing the construction of this function, just to make it clearer? -- 314159 01:05, 27 August 2006 (UTC)

## Proof of continuity of the function...

It says in the "elementary" proof that

Since the terms of the infinite series which defines it are bounded by $\pm a^{n}$ and this has finite sum for 0 < a < 1, convergence of the partial sums to the function is uniform

in the article. But why does a finite sum imply uniform convergence? It seems to me that since, for |b| < 1,

$\left|\sum _{k=0}^{\infty }b^{k}-\sum _{k=0}^{n}b^{k}\right|={\frac {|b|^{n+1}}{|1-b|}}$ ,
$\sup _{|b|<1}\left|\sum _{k=0}^{\infty }b^{k}-\sum _{k=0}^{n}b^{k}\right|=\infty$ for all n, thus it is not going to converge to anything as n goes to infinity and so it doesn't look like uniform convergence to me... or did I do something wrong? -- 129.78.64.102 11:39, 18 July 2007 (UTC)

## Proof of nowhere differentiability

After the proofsketch of the continuity of the function, it says something like

"To prove that it is nowhere differentiable, we consider an arbitrary point x \in {\mathbb R} and show that the function is not differentiable at that point. To do this, we construct two sequences of points xn and x'n which both converge to x, having the property that [...]"

This makes me expect to see more of the proof. The rest is though omitted. Just wondering if something should be added or then change the wording.

## Formating the formulae

In the section "Construction of the Weierstrass function" is:

$\lim \inf$ meant to be:

$\lim _{x\to \infty }$ and is the:

$\lim \sup$ meant to be something else?
RMFan1 (talk) 15:36, 2 January 2008 (UTC)

No. "inf" and "sup" mean "infimum" and "supremum", respectively. Ralphmerridew (talk) 02:46, 13 March 2008 (UTC)

it is described terribly in the text, see Limit superior and limit inferior — Preceding unsigned comment added by 131.111.185.4 (talk) 00:35, 6 May 2013 (UTC)

## The picture on the article

Nice picture. But what choices of a and b were made in such plot? —Preceding unsigned comment added by 67.83.187.113 (talk) 00:44, 30 March 2008 (UTC)

Because f(0)=2, it seems that a=1/2, but b is a bit harder to decipher130.234.5.138 (talk) 09:48, 7 October 2010 (UTC)

## Differentiable at integers

When I try to get the derivative using the differentiation rules, I get this function:

With most values for x, the number changes completely on every iteration of the sum operator. It doesn't converge to a number, which is expectable for a function which hasn't a derivative. But, if you use an integer for x, the argument of the sine function is a multiplycation of pi. In that case the sine function returns 0, which causes the whole function to return 0.

This means the Weierstrass function is differentiable at least some places. These places are where x is an integer, the derivative equals 0 at those places. Do you agree with me?

Paul Breeuwsma (talk) 23:40, 16 October 2008 (UTC)

You cannot in general differentiate an infinite series by differentiating each of the terms. Eric119 (talk) 16:31, 19 October 2008 (UTC)
I'm guessing that you are also letting b be an integer. You shouldn't. If b is not an integer, this won't happen since an integer multiple of a non-integer is not an integer, and

b^n will not be an integer in most cases. You COULD still choose x such that xb^n is an integer (e.g. x = m*b^(-n), m an integer), but obviously xb^(n+1) = b*(xb^n) won't be. Thus: choose b a non-integer and the argument for sin will be a non-integer almost always for any x and any n (and for every x, there will be n such that it is non-integer). Also, note the comment from Eric119. In your case, the function could well have a derivative, but it is not proven just by applying a derivative formula for each term.84.238.115.158 (talk) 21:11, 16 March 2009 (UTC)