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Posers and Puzzles

Posers and Puzzles

  1. 08 May '04 21:01
    if pie is irratinal, and it is equal to the diamiter/cercumfrence, then either the cercumphrence or the diamiter or both must always be irratinal as well. so how dose one calculate pie?
  2. Donation Acolyte
    Now With Added BA
    08 May '04 21:54
    Originally posted by fearlessleader
    if pie is irratinal, and it is equal to the diamiter/cercumfrence, then either the cercumphrence or the diamiter or both must always be irratinal as well. so how dose one calculate pie?
    Well, here's one way to calculate pi:

    1) Calculate the perimeter of a regular n-gon of maximal diameter 1, for suitably large n. This gives an lower bound for pi.

    2) Do the same with a regular n-gon of minimal diameter 1, for suitably large n. This gives an upper bound for pi.

    Choose n large enough, and your approximation will be as good as you want it to be. I'll leave someone else to either prove this or demonstrate a better way of approximating pi (which I'm sure exists, this is something people have put a lot of research into).
  3. Subscriber Brother Edwin
    7 edits
    08 May '04 22:55
    minus pie wont work in the above
  4. Standard member royalchicken
    CHAOS GHOST!!!
    08 May '04 23:01
    I made up a bunch of serieses to do this a while ago, but only a few converge fast enough to be useful for making an approximation.

    Squish it!
  5. Standard member jimmyb270
    Top Gun
    09 May '04 07:13
    Mmmmm...pie *drools*


    Sorry
  6. 09 May '04 18:37 / 1 edit
    Originally posted by fearlessleader
    if pie is irratinal, and it is equal to the diamiter/cercumfrence, then either the cercumphrence or the diamiter or both must always be irratinal as well. so how dose one calculate pie?
    well, first of all, it's spelled pi, not as the confection; the "perimeter" of the circle is "circumference"; and the type of number is "irrational".
    that said, very few occasions demand the preceise measurement of both diamerter and circumference. if you construct a circle using a compass, usually the diameter will be rational, though of course this will be in doubt since it is impossible to make it exact. the circumference will then be the irrational part.
    the case of needing the circumfrerence only is less common.
    of course, there is nothing that says that both components can't be irrational. e. g., construct a square of side length 1, and circumscribe a circle. the diameter of said circle is thus sqrt(2) and circumference is pi*sqrt(2), both of which are irrational quantities, though of different classes (sqrt(2) is algebraic, pi (both by itself and times that) is transcendental).
    it would be a more difficult, and possibly not even feasible, problem to construct a circle in which both diameter and circumference are transcendental numbers.
    in any event, i think i'm going in circles so will end this discussion.
  7. 12 May '04 12:39
    Originally posted by Acolyte
    Well, here's one way to calculate pi:

    1) Calculate the perimeter of a regular n-gon of maximal diameter 1, for suitably large n. This gives an lower bound for pi.

    2) Do the same with a regular n-gon of minimal diameter 1, for suitably large n. This gives an upper bound for pi.

    Choose n large enough, and your approximation will be as good as you w ...[text shortened]... oximating pi (which I'm sure exists, this is something people have put a lot of research into).
    so they calculate pi by finding the limet of this prosess as n reaches infinity.
  8. Standard member Souska
    Evil Loonie
    12 May '04 19:43
    Originally posted by jimmyb270
    Mmmmm...pie *drools*


    Sorry
    I agree...drools withe jimmyb
  9. Standard member TheMaster37
    Kupikupopo!
    13 May '04 13:20
    If Pi is trancendental then so is Pi^2, so a circle with diameter/circumference both transcendental isn't hard at all...

    sin(x) = x- x^3/3! + x^5/5! - ...

    Use the graph of sine to approximate Pi as follows.

    Take a first approximation, say 3. From that point on the x-axis draw a line parallel to the line x=y. This will be the line y=x-3. Intersect that line with the graph of sine. From there draw a line perpendicular to the x-axis to find the x-coordinate of that intersectionpoint. This is your second approximation a.

    Now continue the process;
    -intersect the line y=x-a1 with sin(x)
    determine the x-coordinate of that point
    -this gives your new approximation a2

    and so on.

    the series a1, a2, a3, a4, ... converges to Pi
  10. 24 May '04 18:35
    hello
    i like pie
    and the awnser is mellon pie