Pi in Different Bases?

Discussion in 'Recreational Math' started by Pavel314, Mar 27, 2005.

  1. Pavel314

    Pavel314 Guest

    Pi has been computed in Base 10 to millions of decimal places. Does anyone
    know if any work has been done in computing Pi in other bases? I Googled but
    didn't find anything; any URL's you could point me to would be appreciated.
    I would assume that it would have to be a non-repeating decimal in any base,
    as it is an irrational number.

    I was fooling around with an Excel spreadsheet and came up with the Pi
    equivalencies below in Bases 2 through 16 for a limited number of digits.
    For bases over 10, I use the standard notation of symbols A=10, B=11, etc.,
    for digits 10 or greater.

    I plugged e into my spreadsheet as a base for the final entry in the series.
    Could an irrational number in an irrational base number system end up as a
    repeating decimal?

    BASE PI
    2
    11.00100100001111

    3
    10.01021101222201

    4
    3.02100333122220

    5
    3.03232214303343

    6
    3.05033005141512

    7
    3.06636514320361

    8
    3.11037552421026

    9
    3.12418812407442

    10
    3.14159265358979

    11
    3.16150702865A48

    12
    3.184809493B9186

    13
    3.1AC1049052A2C7

    14
    3.1DA75CDA813752

    15
    3.21CD1DC46C2B7A

    16
    3.243F6A8885A300



    e 10.10100202000211


    Paul
     
    Pavel314, Mar 27, 2005
    #1
    1. Advertisements

  2. Pavel314

    fiziwig Guest

    <quote>
    One of the most amazing mathematical results of the last few years was
    the discovery of a surprisingly simple formula for computing digits of
    the number pi. Unlike previously known methods, this one allows you to
    calculate isolated digits-without computing and keeping track of all
    the preceding numbers.

    "No one had previously even conjectured that such a digit-extraction
    algorithm for pi was possible," says Steven Finch of MathSoft, Inc. in
    Cambridge, Mass.

    The only catch is that the formula works for hexadecimal (base 16) or
    binary digits but not for decimal digits. Thus, it's possible to
    determine that the 40 billionth binary digit of pi is 1, followed by
    00100100001110. . . . However, there's no way to convert these numbers
    into decimal form without knowing all the binary digits that come
    before the given string.

    </quote>
    from: http://www.sciencenews.org/pages/sn_arc98/2_28_98/mathland.htm

    --gary shannon
     
    fiziwig, Mar 27, 2005
    #2
    1. Advertisements

  3. Pavel314

    Stan Liou Guest

    Computation of pi in other bases has been done; in fact, one can
    purchase whole CDs of binary (or, equivalently, base-2^n), which
    can be used as a for pseudorandom number generation. As to how
    efficient computation is done in base-2^n, Mr. Shannon's reply
    goes into this.
    Certainly. Pi in base pi is exactly one. However, no
    transcendental number (as pi is) can be repeating in an
    algebraic base. However, there is no general way to determine
    whether a number is algebraic or transcedental, and just
    because the base is trascedental (as e is also) does not
    mean it will have repeating representation.
    How about base 2i [pseudo-quaternary]? Base i-1 [pseudo-binary]?
     
    Stan Liou, Mar 28, 2005
    #3
  4. Pavel314

    Walter Baeck Guest

    Certainly. Pi in base pi is exactly one.

    Make that ten.

    --
    __________
    \~ALCATEL/~~~~Walter Baeck
    ~\~BELL~/~~~~~Alcatel Belgium
    ~~\~~~~/~~~~~~DSL Microelectronics Design
    ~~~\~~/~~~~~~~E-mail :
    ~~~~\/~~~~~~~~Phone : +32 3 240 73 86
     
    Walter Baeck, Mar 28, 2005
    #4
  5. Am 27.03.05 21:48 schrieb Pavel314:
    Only if PI and E were "algebraically dependent", so that
    (let f = 1/E , and d integers>=0 )

    d0 + d1*f + d2*f^2 ... di*f^(i-1)
    + f^i *( d0 + d1*f + d2*f^2 ... di*f^(i-1))
    + f^(2i)*( d0 + d1*f + d2*f^2 ... di*f^(i-1))
    + ...
    = PI

    [ d0 + d1*f + d2*f^2 ... di*f^(i-1) ] * ( 1+ f^i + (f^i)^2 + (f^i)^3...)
    = [ d0 + d1*f + d2*f^2 ... di*f^(i-1) ] * 1/(1-f^i)

    And this means that the following equation must be satisfied

    PI = [ d0 + d1*f + d2*f^2 ... di*f^(i-1) ] * 1/(1-f^i)
    = [ d0 + d1*f + d2*f^2 ... di*f^(i-1) ] * e^i/(e^i-1)

    with finite many digits d0..di.
    THis would express a rational proportion of a finite polynomial in E
    to PI, which also means algebraical dependence.

    Gottfried Helms
     
    Gottfried Helms, Mar 28, 2005
    #5
  6. Make it 10. By no means is it ten.
     
    Patrick Hamlyn, Mar 28, 2005
    #6
    1. Advertisements

Ask a Question

Want to reply to this thread or ask your own question?

You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.