• (disco) in reply to dkf
    dkf:
    xaade:
    But if I order the numbers from smallest to largest, it diverges, because I don't know what the largest number is.
    Not universally true. Consider the series 0, 1, 1.5, 1.75, 1.875, … where the amount you're adding each time halves. It is ordered strictly from smallest to largest, yet the average is (infinitesimally below) 2. Why is the average the limit? Because we can discard any finite prefix of the series and still get the same average from the infinite suffix ('cos it's infinite and so vastly more important).

    That argument works for any converging series.

    For a series that contains an infinitely repeating finite segment (after a finite prefix, see previous argument), e.g., 0, 1, 0, 1, 0, 1, …, the average of the series is the simple average of the repeating segment (in this case, ½). You can use a functional reduction of the series to prove that.

    A fractal series… requires a better grasp of Analysis than I possess, but I'd guess that a more sophisticated functional reduction would be the way to go.

    A further analysis of @xaade's statement shows that it only holds true for the range [n,∞], where n is a known number. As with @dkf's examples, @xaades claim must be inverted for the range [-∞,n], again with n being a known number. Further, neither @xaade's original statement nor it's inverse is true for the range [-∞, ∞]. :P

  • (disco) in reply to xaade
    xaade:
    Well, before we had some indication of progress.

    Now it's just sitting there calculating totals.

    Going to troll SE Math:

    https://math.stackexchange.com/questions/1256200/infinite-averages

    So, I went back and read that question. The accepted answer is wrong because you don't know enough about the x1+x2+⋯+xn in the numerator. If this is a converging sum, then the accepted answer is correct. If it is a diverging sum, then you end up with ∞/∞, which is undefined.

    Anyway, some ammo for your sniping, since I don't have an account over there.

  • (disco) in reply to abarker
    abarker:
    As with @dkf's examples, @xaades claim must be inverted for the range [-∞,n], again with n being a known number. Further, neither @xaade's original statement nor it's inverse is true for the range [-∞, ∞]. :P

    Most series are defined as maps from the sequence of natural numbers (i.e., 0, 1, 2, …) to the values of the series (X0, X1, X2, …). They are therefore infinite in only one direction. Bidirectionally-infinite things are possible, of course, but they're not usually thought of as series. Definitions are important. ;)

  • (disco) in reply to dkf
    dkf:
    Bidirectionally-infinite things are possible, of course, but they're not usually thought of as seriesre-mapped to the natural numbers.

    IME

  • (disco) in reply to boomzilla

    Well yeah, I know about that sort of thing. It all comes down to proving that you've got a measure partition/function that yields a sequence that converges. Converging series are trivial to reason about.

  • (disco) in reply to dkf
    dkf:
    Converging series are trivial to reason about.

    It is therefore obvious that...

  • (disco)

    Yeah. Use random big number. Like 2^32. There`s no way anyone would ever need more IP addresses than this... :smile:

  • (disco) in reply to Gandor
    Gandor:
    Like 2^32. There`s no way anyone would ever need more IP addresses than this...

    The big question is: Is 2128 really big enough?

  • (disco) in reply to Mikael_Svahnberg

    Note to self: "Bring lots of toner"

  • eric bloedow (unregistered)

    this reminds me of a story where someone had badly messed up his printer settings-or his picture settings-and stubbornly insisted on "making it print with THESE settings!" seriously, he was effectively trying to print a 10-foot by 100-foot picture on a SINGLE PAGE! the tech tried to explain, but he refused to listen to reason...

  • Axel (unregistered)

    One parsec is 30.86 trillion km, not 13. The 11 billion gigaparsecs figure was correct; just the number of km in a parsec was wrong.

Leave a comment on “Universal Printout”

Log In or post as a guest

Replying to comment #:

Log Out

« Return to Article