Talk:Minkowski's question-mark function

Just a comment, this has to be the worst example of bad choice of notation...what was he thinking?? Revolver 07:06, 21 May 2005 (UTC)Reply

Dates back to the dawn of the discovery of pathological functions, at the turn of the 20th century. Maybe it was felt to be very confusing? And maybe he had a twisted sense of humour. linas 00:36, 22 May 2005 (UTC)Reply

Why is ? any crazier than ! (asks one who shorthands sin and cos with $ and ¢)? Kwantus 2005 June 29 17:52 (UTC)

Worse yet, it conflicts with a much more useful notation, d?k for the binomial coefficient d!/(d−k)!k!, generalizing to ?⟨i₀,…,i_n⟩ for the multinomial coefficient of a multi-index. Without regard for degree we can write

${\displaystyle ?I{\mathbf {x} }^{I}}$

as the general term of a multinomial expansion, where the multi-index exponent, as usual, means

${\displaystyle x_{0}^{i_{0}}x_{1}^{i_{1}}\cdots x_{n}^{i_{n}}}$,

and

${\displaystyle |I|=i_{0}+i_{1}+\cdots +i_{n}}$

equals the degree. Ah well; there's little chance of confusion. --KSmrq^T 20:53, 2005 August 31 (UTC)

Question

Latest comment: 19 years ago3 comments2 people in discussion

The fractal and self-similar nature of the function is unclear. Exactly how does the modular group describe the self-similarity? AxelBoldt 03:52, 29 March 2006 (UTC)Reply

Yes, its a sloppy statement; it is only the "period-doubling monoid" inside SL(2,Z) (and not the modular group PSL(2,Z)) that applies to period-doubling fractals. (Different people seem to call this monoid different names). Consider the operators R and S:

${\displaystyle [S?](x)=?(x/1+x)=?(x)/2}$ 

Note that the above is a self-symmetry for x in [0,1]: the question mark on the interval [0,1/2] is a half-size of the whole thing. Let R be a reflection:

${\displaystyle [R?](x)=?(1-x)=1-?(x)}$ 

Then R and S generate the monoid: that is, any string of the form

${\displaystyle S^{m}RS^{n}RS^{p}...}$ 

for positive integers m,n,p ... is a self-symmetry of the question mark. The requirement that m,n,p be poistive is what makes it a monoid,nt a group. linas 02:31, 30 March 2006 (UTC)Reply

I fixed the article. The relationship to SL(2,Z)/PSL(2,Z) etc. is not hard but has some subtle confusions.linas 03:12, 30 March 2006 (UTC)Reply

question

Is this function really absolutely continuous?