Talk:Singular function

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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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]