Talk:Thomae's function: Difference between revisions
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What is the value of f at 0? I would assume it to be 0 but I have no reference for this. --[[User:89.12.119.20|89.12.119.20]] 18:58, 12 August 2007 (UTC) |
What is the value of f at 0? I would assume it to be 0 but I have no reference for this. --[[User:89.12.119.20|89.12.119.20]] 18:58, 12 August 2007 (UTC) |
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: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. --[[User:Cheeser1|Cheeser1]] 21:16, 12 August 2007 (UTC) |
Revision as of 21:16, 12 August 2007
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"Tomae'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)
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)