Talk:Singular function

Properties

This statement about it's properties doesn't seem to make any sense to me: "there exists a set N of measure 0 such that for all x outside of N the derivative f ′(x) exists and is zero." Does this just contain grammatical errors and lack of meaningful punctuations, or am I missing something? --MattWatt 23:23, 7 November 2006 (UTC)

Seems grammatical and perfectly clear to me. And phrased just the way I would expect something like that to be phrased, quite prosaically. Specifically where do you have a problem with it? Michael Hardy 23:25, 7 November 2006 (UTC)

Let's try it more long-windedly:

There is a certain subset N of the domain of f with the following properties:

• The measure of N is 0,
• If x is any number in the domain of f and x is not in N, then f ′(x) = 0.

Michael Hardy 23:29, 7 November 2006 (UTC)

I can't keep up with your quick responses! I had typed the following, but it didn't seem to stick from your changes. Hmmm "N of measure 0" is what I mostly not understanding. I thought there may be a better way of stating this property that may be clearer. I'll study it a bit more and perhaps it will begin to become clearer. Also, don't these two properties; "f(x) is nondecreasing on [a, b]" and "f(a) < f(b)" mean the same thing? Do they really need seperate bullets? This is the first time I've ever heard of this topic. I wouldn't want to overstep my bounds and make changes on something that I perceive is wrong. --MattWatt 23:38, 7 November 2006 (UTC)

It says "...a set N of measure 0". That means "a set of measure 0, to which we give the name N (so that we can refer to it later by that name)". Writing "...a set N of measure 0" exemplifies a standard way of writing mathematics. Everyone (except perhaps non-mathematicians) does that every day.

don't these two properties; "f(x) is nondecreasing on [a, b]" and "f(a) < f(b)" mean the same thing?

No. Nowhere near it. "Nondecreasing on [a,b] means that for any two points u and v in the closed interval from a to b (not just the two endpoints a and b), if u < v then f(u) ≤ f(v). Notice that it says "≤", not "<". They certainly do need separate bullets; they don't say the same thing at all. Michael Hardy 23:59, 7 November 2006 (UTC)

Actually what I meant was that the latter was a logical deduction of the former, but I see where you are right and I'm wrong: "≤", not "<". I'm not a mathematician but a graduate Engineering student who is now knee deep in more statistical mathematics than I've ever come across before. Thanks for the clarificiations though. --MattWatt 00:10, 8 November 2006 (UTC)

I'm a little confused by property two: "there exists a set N of measure 0 such that for all x outside of N the derivative f ′(x) exists and is zero."

shouldn't this be "there exists a set N of measure 0 such that for all x element of N the derivative f ′(x) exists and is zero."? Or am I missing something? 203.144.32.165 21:48, 19 March 2007 (UTC)

No, the derivative is zero almost everywhere. In other words, there exists a set M of measure b − a such that the derivative f'(x) exists and is zero for all x in M; the set M is the complement of the set N in the article. -- Jitse Niesen (talk) 23:43, 19 March 2007 (UTC)

I know I'm two years late, but it isn't just more comprehensible to say that: "The subset of the domain where the derivative is nonzero has zero measure"?

--Lucas Gallindo (talk) 21:34, 12 March 2009 (UTC)

Minkowski's question mark function

Singular function mentions it as a (strictly increasing) singular function; However, according to Minkowski's question mark function, it is differentiable only on the rationals, not almost everywhere, as is required by the definition here. Am I missing something? -- Meni Rosenfeld (talk) 00:16, 28 October 2007 (UTC)

Minkowski's question mark function says "The derivative vanishes on the rational numbers" which is similar to the position here. It also says "It does not have a well-defined derivative, in the classical sense, on the irrationals; however, there are several constructions for a measure that, when integrated, yields the question mark function" which is not the same as here, and makes Minkowski's question mark function slippery. --Rumping (talk) 17:06, 8 January 2008 (UTC)

Measure Zero!?!?

Heh. Regarding the above conversation, I believe that careful analysis shows that the question mark has non-zero derivative on a set of non-zero measure. However, I've seen literature refer to it as a "singular function" even though it clearly doesn't meet the definition given in this article. Which means that either there's a common usage that is incorrect, or that there is some broader definition of singular function that allows for a non-zero-derivative on a set of measure greater than zero. Also, FWIW, I have never seen any sort of proof that the circle-map winding number has non-zero derivative only on a set of measure zero -- in fact, its probably *not* a set of measure zero: as K is varied from 0 to 1, the flat parts of circle map winding number get wider i.e. the measure of the flat parts gets bigger, and, at K=0, the measure of the flat bits is exactly zero (since there's no phase-locking at all).

Last but not least: only the "canonical" Cantor set has a measure of zero, but there are other Cantor sets one can construct with measure that is greater than zero (e.g. instead of removing the middle 1/3 each time, one removes geometrically less each time. This results in a fat Cantor set, for which one can define a distribution function which looks like the devils staircase, but fails to meet the strict definition of this article.

Thus, we have three different examples that seem to be "devil's staircases", but don't seem to meet the "measure zero" part of the current definition given in this article. I suspect the "measure zero" part of the definition given here is wrong. Seems to me the measure-zero requirement is not germane to the definition -- rather, that the proper definition probably should be "the derivative is non-zero only on a cantor set (not necessarily the canonical cantor set), and that the measure of the set where the derivative is non-vanishing could be anything, from zero to one, thus allowing both the question mark, and erasing doubts about whether the circle-map winding number is or is not "singular". Alternately, perhaps the correct definition of a singular function is *any* function that behaves like a measure, and is a singular measure with respect to the Lebesgue measure. linas (talk) 05:17, 11 February 2010 (UTC)

Also -- the article singular distribution seems to have the same measure-zero mindset, which seems wrong to me.

Also, I don't believe that any of the examples of "devil's staircases" in physics (the Frenkel stuff, the ANNNI thing) have been proven to be singular only on a set of measure zero -- or are even provable-- these are mostly just voltage readings taken from an instrument, and usually not well-grounded mathematical functions. On the contrary, just like the phase-locked loop (circle map), they're probably singular on a set of measure greater than zero. What makes them devils-staircasey is that the sets seem disconnected (i.e. are cantor sets). linas (talk) 05:47, 11 February 2010 (UTC)

Probabilistic Interpretation

While it's ok to interpret singular functions as cdf, the arguments given for why it's neither a discrete nor a continuous distribution sound wrong to me:

"...neither a discrete random variable (since the probability density is zero everywhere it exists)..."

It's not a discrete random variable because for any single point x, its probability is zero, P(X=x)=0. This is a consequence of f being continuous (and therefore left-continuous). It's not a consequence of the probability density (presumably pdf) being zero; in fact, discrete random variables all have the property that the derivative of their cdf is zero wherever it exists in the classical sense.

"...nor an absolutely continuous random variable (since the probability is zero for each point)..."

A continuous random variable would have zero probability for each individual point, P(X=x)=0, so this property does not contradict the random variable being continuous -- it is actually a property of continuous random variables. Rather, the fact that the pdf is zero almost everywhere shows that the random variable defined by the cdf f (where f is singular), is not a continuous random variable, i.e. because the associated pdf integrates up to zero.

A rephrasing might be " neither a discrete random variable (since the probability is zero for each point) ..." "...nor an absolutely continuous random variable (since the probability density is zero almost everywhere)..."

Comments? —Preceding unsigned comment added by 128.40.24.185 (talk) 15:03, 5 October 2010 (UTC)