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Homework Help: Proof for Pi

  1. Aug 22, 2008 #1
    can we prove the ratio of circumference to diameter is same for all circles & is equal to pi?
     
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  3. Aug 22, 2008 #2

    CompuChip

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    Well, the circumference of a circle is [itex]L = 2\pi r[/itex] source and by definition the diameter is twice the radius (d = 2r). So [itex]L/d = 2 \pi r / (2 r) = \pi[/itex].
     
  4. Aug 22, 2008 #3

    tiny-tim

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    Hi mite!:smile:

    It depends what axioms (basic definitions) you start with.

    Euclid regarded the similarity of two circles as an axiom, so there was nothing to prove!

    And π is defined as the ratio.
     
  5. Aug 22, 2008 #4

    HallsofIvy

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    Historically, the fact that the ratio of circumference to diameter is a constant was a numerical observation. The Greeks proved it by calculating the ratio of the perimeter of a regular n-gon to its "diameter" and then seeing what happened as n got larger and larger (a limit process). As for the fact that that ratio is equal to pi- that's essentially the definition of pi.

    A modern proof would be something like this: Since sin2(t)+ cos2(t)= 1 for all t, x= Rcos(t) and y= Rsin(t) are parametric equations for a circle of radius R. The circumference, then, is given by
    [tex]\int_0^{2\pi}\sqrt{\left(\frac{dx}{dt}\right)^2+ \left(\frac{dy}{dt}\right)^2}dt[/tex]
    [tex]= \int_0^{2\pi}\sqrt{R^2sin^2(t)+ R^2cos^2(t)}dt= \int_0^{2\pi}Rdt= 2\pi R[/tex]
    Since the circumference is [itex]2\pi R[/itex] and the diameter is 2R, the ratio of circumference to diameter is [itex]2\pi R/(2R)= \pi[/itex].

    (The fact that sin(t) and cos(t) have period [itex]2\pi[/itex], which is critical to this proof, can be shown by using the fact that the second derivative of sin(t) is -sin(t) and the second derivative of cos(t) is -cos(t).)
     
  6. Aug 22, 2008 #5

    tiny-tim

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    Hi HallsofIvy! :smile:

    No, surely that's how they calculated π …

    they were already convinced that it was the same for all circles? :smile:
     
  7. Aug 22, 2008 #6

    HallsofIvy

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    I didn't say that was how they calculated it. I said that was how they proved it was the same ratio for all circles. It was Archimedes who did that. I am sure that Greeks before that just assumed it was a constant.
     
  8. Aug 22, 2008 #7

    tiny-tim

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    And they were right to do so!

    From their point of view, because it was axiomatic and/or obvious …

    from our point of view, because of the scalar symmetry of Euclidean space.

    (Of course, circumference/diameter isn't a constant in non-Euclidean space. :wink:)

    (If you'd said to them "you've proved that it's a constant", they'd have replied "no we haven't, we've only calculated the constant … we implicitly used a symmetry theorem on polygons in the course of that calculation, and that applies to circles anyway" :smile:)
     
  9. Aug 22, 2008 #8

    dynamicsolo

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    The arclength integration doesn't really constitute a proof. The choice of integration limit for the "angle parameter" of [tex]2 \pi[/tex] is based on the definition of angle as the ratio of arclength to radius for a circle: [tex]\theta = \frac{s}{R}[/tex]. So it is completely unsurprising that the result of the integration for the circumference of a circle is [tex]2\pi R[/tex].

    I haven't explored the history of the number thoroughly (although there are at least two or three histories of pi out there now), but I believe that pi is simply defined as the ratio of circumference to diameter for a circle (the specific letter was chosen somewhere around the 18th Century), it having already been understood in antiquity that the ratio is a constant. So there is no proof involved for this. (One of the continuing mysteries is why [tex]\pi[/tex] is so deeply imbedded in the structure of mathematics and turns up in other relations which has little to do with circles...)
     
  10. Aug 22, 2008 #9
    real world circles or mathematical circles? the circumference of the latter depends on your metric.
     
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