Content deleted Content added
→Naming convention?: new section |
No edit summary |
||
| (21 intermediate revisions by 10 users not shown) | |||
Line 1:
{{talkheader}}
{{WikiProject banner shell|class=Start|
{{WikiProject Mathematics|priority=Low}}
}}
== "Thomae's function" vs "popcorn function" ==
Line 40 ⟶ 41:
:where ''f'' is Thomae's function and 2<sup>''q''</sup> is any power of 2 that is larger than ''n'', so that ''n''/2<sup>''q''</sup> is between 0 and 1. Dunham uses "ruler function" as a synonym for Thomae's function in [http://books.google.co.uk/books?id=QnXSqvTiEjYC&pg=PA175&dq=%22ruler+function%22+Thomae&sig=ACfU3U3B0h1K-42ZyOPlR3f7KbY_QZzZHQ ''The Calculus Gallery''], as does Burn in [http://books.google.co.uk/books?id=1nteiLydVpQC&pg=PA158&lpg=PA158&dq=%22ruler+function%22+Thomae&source=web&ots=LEY1ak2-qV&sig=1rSU6rtbebN1Mxs4ZxecMeqLt6I&hl=en&sa=X&oi=book_result&resnum=12&ct=result#PPA158,M1 ''Numbers and Functions'']. We could obviously extend the article to include the MathWorld/OEIS definition of ruler function as an alternative definition. [[User:Gandalf61|Gandalf61]] ([[User talk:Gandalf61|talk]]) 10:38, 24 August 2008 (UTC)
::While that is clever, the contortions needed to establish a link between Thomae's function and the ruler function as defined in MathWorld (the 2-adic valuation of positive integers) are such that it is hard to consider one as a transformation of the other. Note that you need to modify ''n'' in a way that depends essentially on the value of ''n'' itself before feeding it into ''f'', and then need to tweek the value of ''f'' similarly; if one is going to allow that kind of things, one could "transform" almost any function into any other function. The fact is the function have vastly different properties (to name a few, Thomae's function is defined on an interval, takes arbitrarily small positive values, and is bounded above by 1; the ruler function is defined on positive integers only, has natural number values values and is unbounded). I don't think the graph of Thomae's function even vaguely resembles any ruler in existence (nor does it resemble popcorn, raindrops, or stars over Babylon for that matter). It is undeniable that some authors use "ruler function" as a name for Thomae's function, but other use it to designate 2-adic valuation, for instance in [[Concrete Mathematics]], p 113. In any case the current situation with a separate section in this article describing a ''different'' function is not really acceptable. [[User:Marc van Leeuwen|Marc van Leeuwen]] ([[User talk:Marc van Leeuwen|talk]]) 18:06, 9 April 2017 (UTC)
== Formal proof of continuity ==
Line 58 ⟶ 61:
The next step would be to investigate copyright issues for these proofs. Or possibly rewrite them to avoid copyright issues.</small></small>
--[[User:Nickalh50|Nickalh50]] ([[User talk:Nickalh50|talk]]) 19:41, 14 March 2011 (UTC)
:Hi guys. I rewrote the proofs and put them in collapse boxes. I believe that the proofs are complete and precise now. Copyright problems shouldn't be an issue; I used only this article as a guide. (Any constructive criticism on my math writing would be greatly appreciated.)
:Actually I started an hour or two ago by deleting the "Integrability" section, since it made the obviously false statement that the Thomae function is only discontinuous on the rationals. I'm wiser now. 04:18, 9 September 2017 (UTC) <!-- Template:Unsigned --><small class="autosigned">— Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[User:Norbornene|Norbornene]] ([[User talk:Norbornene#top|talk]] • [[Special:Contributions/Norbornene|contribs]]) </small> <!--Autosigned by SineBot-->
Hi everyone. I think there is a minor mistake in the proof of the continuity.
On the fourth line it is stated that 0 < (k_i)/i. However I think there should be a ≤ there since for small i (e.g. i=1) k_i has to be zero. Also I get Latex compilation problems while editing it. Could someone have a look? Thanks! [[User:Mike2304|Mike2304]] ([[User talk:Mike2304|talk]]) 20:46, 26 October 2023 (UTC)
== Naming convention? ==
Line 64 ⟶ 74:
I remain uncertain that either "Thomae's function" or "popcorn function" is a proper canonical name if it is not the most common name in the literature. At the very least it should be mentioned that this function is quite often called "the Dirichlet function", regardless as to its attribution (which is correctly given to Thomae). [[User:TricksterWolf|TricksterWolf]] ([[User talk:TricksterWolf|talk]]) 00:42, 26 August 2011 (UTC)
== Amendments and tweaks ==
I tweaked the lede, added a few properties, and tweaked the proofs. I tried to keep the proofs as elementary epsilon/delta proofs, and also to be constructive and rigorous. [[User:Purgy Purgatorio|Purgy]] ([[User talk:Purgy Purgatorio|talk]]) 10:15, 19 September 2017 (UTC)
==the level is not right for a general encylopedia==
the intro is a , excuse me, shitty cause it is not written for a general audience
a def of coprime should be given and wtf are Z and N ?
you math people in general do a really bad job at writing stuff for general audiences - not just this article <!-- Template:Unsigned IP --><small class="autosigned">— Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/73.119.28.38|73.119.28.38]] ([[User talk:73.119.28.38#top|talk]]) 12:22, 31 May 2019 (UTC)</small> <!--Autosigned by SineBot-->
== Reciprocal of the popcorn function ==
Do you guys think we should include the fact that the reciprocal of the function (mapping the irrationals to 1 instead of 0) takes on every real value on every interval?<br>
[[User:Joel Brennan|Joel Brennan]] ([[User talk:Joel Brennan|talk]]) 20:20, 13 April 2018 (UTC)<small> Please, "sign" your comments at the end, not at the start, and allow for my edit.</small>
: Neither I am sure about your meaning of "reciprocal", nor do I understand the "every real value on every interval" (e.g., I see no negative reals anywhere.) Please, express your idea more consistently. [[User:Purgy Purgatorio|Purgy]] ([[User talk:Purgy Purgatorio|talk]]) 06:39, 14 April 2018 (UTC)
:: I mean <math>\frac{1}{\tilde{f}}</math> where <math>\tilde{f}</math> is the same as f except that in the second piece we map to 1 instead of 0. You could achieve the same effect (and I think it would be nicer) if you took the reciprocal of the first piece of f and left the second piece alone. And of course you are right; the reciprocal function only takes on the positive (or non-negative if you leave the second piece alone) real values on every (non-degenerate) real interval. [[User:Joel Brennan|Joel Brennan]] ([[User talk:Joel Brennan|talk]]) 13:10, 15 April 2018 (UTC)
::: I am still not sure about your intentions, but -as I currently interpret your text- "your reciprocal function" would have only natural numbers as values (q is natural, 1/q is rational, and 1/(1/q) is natural again, 1 and 1/1 and 1/(1/1) are also natural). Please, carefully rethink your considerations. [[User:Purgy Purgatorio|Purgy]] ([[User talk:Purgy Purgatorio|talk]]) 13:52, 15 April 2018 (UTC)
::::Yes I'm being an idiot! It is only unbounded on every interval. That makes my supposedly 'interesting property' much less noteworthy, so it probably isn't worth remarking on in the main article. [[User:Joel Brennan|Joel Brennan]] ([[User talk:Joel Brennan|talk]]) 17:29, 15 April 2018 (UTC)
== Proof "nowhere differentiable" is (too?) complicated ==
As the proof relies on [[Hurwitz's theorem (number theory)|Hurwitz's theorem]], it is complicated. It is easy to construct a suitable rational sequence with Limit inferior of the difference quotient at least 1. I will edit the proof accordingly.
The proof:
* For rational numbers, this follows from non-continuity.
* For irrational numbers:
*:For any [[sequence]] of irrational numbers <math>(a_n)_{n=1}^\infty</math> with <math>a_n \ne x_0</math> for all <math>n \in \mathbb{N}_{+}</math> that converges to the irrational point <math>x_0</math>, the sequence <math>(f(a_n))_{n=1}^\infty</math> is identically <math>0</math>, and so <math>\lim_{n \to \infty}\left|\frac{f(a_n)-f(x_0)}{a_n - x_0}\right| = 0</math>.
*: On the other hand, consider the sequence of rational numbers <math>(b_n)_{n=1}^{\infty}</math> with <math>b_n = \lfloor nx_0\rfloor/n</math>, where <math>\lfloor nx_0\rfloor</math> denotes the [[Floor and ceiling functions|floor]] of <math>nx_0</math>. Since <math>nx_0-1<\lfloor nx_0\rfloor\le nx_0</math>, the sequence <math>(b_n)_{n=1}^{\infty}</math> converges to <math>x_0</math> using the [[Squeeze theorem]]. Also, <math>|b_n-x_0| = |\lfloor nx_0\rfloor/n - x_0| = |\lfloor nx_0\rfloor - nx_0|/n \le 1/n</math> for all <math>n</math>.
*: Thus for all <math>n</math>, <math>\left| \frac{f(b_n)-f(x_0)}{b_n-x_0} \right| \ge \frac{1/n - 0}{1/n} = 1</math>. Therefore we obtain <math>\liminf_{n\to\infty} \left| \frac{f(b_n)-f(x_0)}{b_n-x_0} \right| \ge 1 \ne 0</math> and so <math>f</math> is not differentiable at any irrational number <math>x_0</math>.
Follow-up: Using [[Roth's theorem]] one can easily show that <math>f</math> is [[Hölder condition|Hölder continuous]] for every <math>\alpha<\tfrac12</math> on the set of irrational numbers. Using [[Hurwitz's theorem (number theory)|Hurwitz's theorem]] one can easily show that <math>f</math> is not [[Hölder condition|Hölder continuous]] for every <math>\alpha\ge\tfrac12</math> on the set of irrational numbers. It might be useful to add this information and the links to [[Roth's theorem]] and [[Hurwitz's theorem (number theory)|Hurwitz's theorem]]. <!-- Template:Unsigned IP --><small class="autosigned">— Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/134.2.163.201|134.2.163.201]] ([[User talk:134.2.163.201#top|talk]]) 09:12, 22 May 2025 (UTC)</small> <!--Autosigned by SineBot-->
| |||