Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Back to the basics: Universality of π

  1. Dec 20, 2011 #1

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    OK, a proof for this simple result.

    Let's take a circle of center O1 (point in the plane) and radius R1 (length of the radius). Call this circle C1.

    We define the number π1 as (circumference of the circle C1)/ (2R1).

    Let's take another circle of center O2 and radius R2 called C2.

    Define π2= (circumference of the circle C2)/ (2R2).

    Now prove that π12.

    Ideas ?

    My idea was to show that the equality (perimeter/side) holds for the 2 squares inscribed in the 2 circles. Then it holds for the 2 squares circumscribed to the 2 circles.

    Then take hexagons, octogons,..., generally regular-n'gons. Then grow n arbitrarily and get a proof for the 2 circles.

    Is this ok as a proof ?
     
    Last edited: Dec 20, 2011
  2. jcsd
  3. Dec 20, 2011 #2

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Hmmm, I don't like that proof. You'll have to write out the details before seeing if it holds. Mistakes are very easily made by this approach, see the famous:

    http://chzmemebase.files.wordpress.com/2010/11/9e7c48aa-1823-4d5f-aa13-40699c72d508.jpg [Broken]

    If you would define pi as something pertaining the surface area, then the proof would hold. Indeed, you could use the familiar tools of measure theory.

    I know that Billinsley in his book "probability and measure" has this as an exercise. The idea is not to define pi as something in a circle. The idea is to define pi as a series and then proving that it is the pi of the circle. The good thing is that you then can use the tools of trigonometry and analysis.

    If you want to say anything about the circumference of a circle, then you first need to define what the circumference even means. This requires you to define the tools of the Hausdorff measure. That pi is then unique for every circle can then be proven usen analytic tools. See Billingsley exercise 19.5.

    I know that Euclid also does these things with pi in his elements, but I don't know how rigorous that is.
     
    Last edited by a moderator: May 5, 2017
  4. Dec 20, 2011 #3

    AlephZero

    User Avatar
    Science Advisor
    Homework Helper

    If you draw n-gons inscribed in the circle (n = 3, 4, 5 ...) you can prove the perimeters form a monotonic bounded increasing sequence.
    N-gons circumscribed around the circle have perimieters forming a monotonic bounded decreasing sequence.
    Both sequences have the same limit, and everything is proportional to the radius of the circle.

    However, you haven't proved that the limit really IS "the circumference of the circle". Think what might happen if you replaced the circle with a fractal curve, for example.

    This argument was good enough to convince Archimedes, who used it to show that 220/71 < pi < 220/70, but the Greek geometers relied on "common sense" and "looking at pictures" for the hard part of the proof.
     
  5. Dec 20, 2011 #4

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    It is very easy to show that the limit IS the circumference of the circle. That is not the difficult part. The convergence is even uniform!!

    What the problem is, is that the limit of the lengths of the curves will not equal the length of the circle. That is: we can show that [itex]P_n\rightarrow C[/itex] (where [itex]P_n[/itex] and C are the polygons and the circle), but this does not imply that

    [tex]\text{length}(P_n)\rightarrow \text{length}(C)[/tex]

    That is the main problem. The problem that is equivalent is that if [itex]f_n\rightarrow f[/itex] as functions, then it doesn't mean for the derivatives that [itex]f_n^\prime\rightarrow f^\prime[/itex].
     
  6. Dec 20, 2011 #5

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I think circumference here is being used to refer to the length of the perimeter, not the perimeter itself
     
  7. Dec 20, 2011 #6

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Aah, that would actually make sense :tongue2: Thanks
     
  8. Dec 20, 2011 #7

    AlephZero

    User Avatar
    Science Advisor
    Homework Helper

    The dictionary defintion of "circumference" can mean either the curve, or the length of the curve. Sorry for the confusion!

    http://www.thefreedictionary.com/circumference
     
  9. Dec 21, 2011 #8

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    This is very interesting. So what I'm essentially getting from your post is that for n growing arbitrarily large, the polygonal line will approximate the curve (circle) better and better, but the (the sequence of) length(s) of the polygonal line(s) (defined as n times the length of a side (question: how do you define the length of a side of a polygon ?)) will not equal in limit the length (circumference) of the closed curve called <circle> (again, how can one define this, too ?).

    If my understanding is right, where could I find a rigorous proof ?

    Thanks!
     
  10. Dec 21, 2011 #9

    Deveno

    User Avatar
    Science Advisor

    for myself, i am content with the definition:

    [tex]\pi = \int_{-1}^1 \sqrt{1 - x^2}\ dx[/tex]

    true, it does depend on a limit, and the trigonometric functions are involved in a sneaky way, but pi is, after all, irrational, and to even have an idea of what irrational MEANS, we need some notion of convergence, so if we're going to the trouble of developing all that machinery (limits, sup's, inf's, and the rest of the real number paraphanelia), we might as well formalize archimede's "method of exhaustion" while we're at it.

    an alternate formulation, which is also satisfactory to me, is:

    [tex]\pi = \text{min}(\{x \in \mathbb{R}^+: f(x) = 0\})[/tex]

    where f is defined by:

    [tex]f + f'' = 0, f(0) = 0, f'(0) = 1[/tex]

    although a proper explanation of this definition requires some knowledge of the complex exponential (probably best defined in terms of a power series), not to mention the justification of term-by-term differentiation of a convergent power series.

    i am amazed that the mathematicians of antiquity were sure that pi was a constant, and even more amazed that they took for granted that "circumference" was well-defined.
     
    Last edited: Dec 21, 2011
  11. Dec 21, 2011 #10

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    You missed the second derivative in the ODE*. Since you mentioned power series to define real and complex exponentials, how do you define [itex] \sin \pi [/itex] ??

    *Corrected.
     
  12. Dec 21, 2011 #11

    Deveno

    User Avatar
    Science Advisor

    darn shift key.

    the good news is, the circle is a smooth curve (in both the mathematical sense, and the everyday english sense of the word), so it is rectifiable.

    polygons are piece-wise smooth, so they, too, are rectifiable. so in this particular case, we actually do have:

    limit(lengths) = length(limit).

    the "problem archimedes?" proof that pi = 4, shows why rectifiability is important. arc-length, for curves that are not piece-wise smooth, might not even make sense (the koch snowflake curve doesn't have a "perimeter length", the limit of the lengths of its approximations doesn't exist, that is, it is unbounded).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Back to the basics: Universality of π
  1. Back induction (Replies: 2)

Loading...