Talk:Thomae's function

"Thomae's function" vs "popcorn function"

I don't have enough analysis books to do a comprehensive count, but in every book I have, this is called Thomae's Function. I know it isn't that big a deal, but "popcorn function" is so terribly informal, whereas "Thomae's function" falls in line with so many other functions whose names are called "[name]'s function" or "[name] function" or similar. Look at List of mathematical functions and tell me how many are named after people vs how many are named after food that they (apparently) resemble. Even the closest relative to this function is named properly, as the Dirichlet function (although it is in nowhere continuous function due to a merge of some sort). That section in the list includes several functions of this exact type (canonical examples/counterexamples in elementary analysis), all of which are named after people. So anyway, sorry about rambling a bit, I propose that we move it to Thomae's function, and redirect this page there. I just feel like maybe next we'll move Error function to ski-slope function (yes, I know, I'm being sarcastic). Any opinions? --Cheeser1 05:01, 27 July 2007 (UTC)

OK, I did the move. Oleg Alexandrov (talk) 02:05, 24 August 2007 (UTC)

f(0)?

What is the value of f at 0? I would assume it to be 0 but I have no reference for this. --89.12.119.20 18:58, 12 August 2007 (UTC)

Notice that 0 is rational, and in least terms, 0=0/1. Least terms are ensured by the stipulation that gcd(p,q)=1. Thus f(0) = 1. If f(0)=0, then the function would be continuous there - this would violate the conclusion that f is discontinuous at rational points. --Cheeser1 21:16, 12 August 2007 (UTC)

Actually, by that reasoning f(0) = 1 which doesn't contradict discontinuity at all. Olaf Davis (talk) 23:46, 19 June 2009 (UTC)

Not only is f(0)=1, but f(z)=1, for any integer z.

Why? 0= 0/1 in lowest terms(intuitively), and Thomae only uses the denominator, f(0)=1/1 Similarly, z/1 -> f(z)=1/1 Nickalh50 (talk) 19:28, 14 March 2011 (UTC)

Newer image with higher resolution

I uploaded a newer version with a higher resolution (Image:Popcorn function plot bars.png, even a SVG version up to denomiator 750 (Image:Popcorn function plot bars.svg), but this is not working at the moment, see commons:Commons:Help_desk#SVG_too_big.2C_.svg.gz__or_.svgz_was_not_accepted. What do you think? --84.72.190.27 (talk) 10:01, 27 November 2007 (UTC)

Ruler function??

I suspect this is wrong, in what way is this a modification of the ruler function?

According to MathWorld the ruler function is The exponent of the largest power of 2 which divides a given number 2n.

It looks like this

-- teadrinker (talk) 08:53, 24 August 2008 (UTC)

The relation is that if p is the exponent of the largest power of 2 that divides n then

$p=\log _{2}\left(f\left({\frac {n}{2^{q}}}\right)\right)+q$

where f is Thomae's function and 2^q is any power of 2 that is larger than n, so that n/2^q is between 0 and 1. Dunham uses "ruler function" as a synonym for Thomae's function in The Calculus Gallery, as does Burn in Numbers and Functions. We could obviously extend the article to include the MathWorld/OEIS definition of ruler function as an alternative definition. Gandalf61 (talk) 10:38, 24 August 2008 (UTC)

Formal proof of continuity

Would it make much sense to include a formal proof of the evaluation of continuity for this article? The informal proof seems quite lacking. --129.138.220.140 (talk) 16:25, 7 October 2009 (UTC)

Yes, that seems reasonable. A formal proof wouldn't be too long so I'd say feel free to go ahead. Olaf Davis (talk) 17:17, 7 October 2009 (UTC)

A more formal proof of continuity on the irrational numbers, would be essential for the article to be considered quality. Possible sources: See the last few pages of Excellent proof for proofs of continuity on the irrationals and discontinuity on the rational numbers.

A second, Proof but with a Mistake is easier to follow but needs to be rearranged. It defines m, then later assumes a property about m. A better ordering starts,

Let epsilon > 0, Let m be an integer such that, 1/m < epsilon.

The second proof may have other mistakes. It could use a clearer transition from proof of continuity proof of discontinuity. Replacing variables in the 2nd half to prevent confusion with the first half, would also be useful. The next step would be to investigate copyright issues for these proofs. Or possibly rewrite them to avoid copyright issues. --Nickalh50 (talk) 19:41, 14 March 2011 (UTC)