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    • CommentAuthorWill Jagy
    • CommentTimeNov 10th 2010 edited
     
    Gerald, I asked you a question in a comment to your answer at

    http://mathoverflow.net/questions/45477/closed-form-functions-with-half-exponential-growth/45479#45479

    about local solutions only to f(f(x)) = sin x around x = 0. Essentially the question is whether there is an (odd) analytic solution, i.e. nonzero radius of convergence for the formal power series solution. I screwed up, I was assuming my series would have a similar graph to one that has already been posted. That cannot happen for very long, as within the radius of convergence, should that be nonzero, the formal power series gives a genuine solution to f(f(x)) = sin x. There is a trichotomy here, given Ax, there is a functional square root given by x sqrt(A), where the case A=1 is special, so for the local analysis A >, <, = 1 matters.

    I wonder if Danica McKellar looks at MO? Her Erdos number is 4, her Erdos-Bacon number is 6.
    http://en.wikipedia.org/wiki/Danica_McKellar
    • CommentAuthorCam McLeman
    • CommentTimeNov 10th 2010 edited
     

    Off-topic: I've always wondered if summing one's Bacon number and one's Erdos number was really the most appropriate way to define the Erdos-Bacon number. Maybe the 2-norm of the [Erdos-number, Bacon-number] vector?

    Edit: Presumably we could axiomative enough properties of the number to uniquely define it. For example, you should clearly have an Erdos-Bacon number of 0 if and only if you are simultaneously Erdos and Kevin Bacon.

    • CommentAuthorpowerpuff
    • CommentTimeNov 10th 2010 edited
     
    @Cam, I disagree. In fact, I think that Erdos should be considered to have Erdos number 2 --- after all, Erdos has no joint papers with himself, although he has many joint papers with people who have joint papers with Erdos.

    EDIT: Apologies to those who missed the (possibly not very funny) joke and who seemed instead to think that I was trying to reformulate the foundations of graph theory.
    • CommentAuthorHJRW
    • CommentTimeNov 10th 2010
     

    powerpuff, I'm glad I'm not taking your course on metric spaces.

    • CommentAuthorWill Jagy
    • CommentTimeNov 10th 2010
     
    Random note, I deleted my comment to Gerald's MO answer and made it an MO question.
  1.  

    @powerpuff: Erdos has the empty joint paper with himself! By your definition, an isolated vertex wouldn't constitute a connected component of a graph.

  2.  
    Right, the correct definition is minimum path length (as Henry slyly hinted). The minimum length of paths from Erdos to himself is 0. There are various formalizations of this which can be left as an exercise to the reader.