Re: area of circle

From: shedar (nobody_at_nonesuch.com)
Date: 10/26/04 

Date: Tue, 26 Oct 2004 13:46:59 GMT

"David C. Ullrich" <ullrich@math.okstate.edu> wrote in message
news:08bsn0d53dij4r1sqmp2ve4anbmrdspqe2@4ax.com...
> On Tue, 26 Oct 2004 07:09:41 GMT, "shedar" <nobody@nonesuch.com>
> wrote:
>
> >"David C. Ullrich" <ullrich@math.okstate.edu> wrote in message
> >news:f3vmn0dhauiaqb5bbmgt0tbrb2cibaime6@4ax.com...
> >> On Sun, 24 Oct 2004 08:05:51 GMT, "shedar" <nobody@nonesuch.com>
> >> wrote:
> >>
> >> >"The World Wide Wade" <waderameyxiii@comcast.remove13.net> wrote in
> >message
> >> >news:waderameyxiii-7877BB.21582323102004@news.supernews.com...
> >> >> In article <ccsln0thnjnqkpfvv5fqbnq7imgvljl0lq@4ax.com>,
> >> >> Tim923 <juggler923@verizon.net> wrote:
> >> >>
> >> >> > Does anyone have the formal calculus based proof for area of a
> >circle?
> >> >>
> >> >> Actually there isn't one, at least judging by most calculus texts.
Take
> >> >the
> >> >> unit circle. Its area, via the calculus, is 4*int_[0,1] sqrt(1-x^2)
dx.
> >> >But
> >> >> embedded in the evaluation of this integral is the fact that the
> >> >derivative
> >> >> of sin(x) is cos(x), which is derived from lim_x->0 sin(x)/x = 1.
This
> >> >> limit is usually obtained by invoking the formula for the area of a
> >> >sector,
> >> >> to wit, the area of the sector subtended by the arc x is x/2. But
that
> >> >> means, taking x = 2Pi, that we already know, prior to any calculus
> >> >> considerations, that the area within the unit circle is Pi. So,
> >proceding
> >> >> in the manner of most calculus texts, you'll find there is really no
> >> >> calculus derivation for the area of a circle; the area formula for
the
> >> >> circle has been used in deriving the calculus answer.
> >> >
> >> >I agree with you. The formula for the area of a sector is employed to
> >prove
> >> >
> >> > lim_x->0 sin(x)/x = 1,
> >> >
> >> >which, in turn, is used to prove that
> >> >
> >> > the derivative of "sine" is "cosine",
> >> >
> >> >which, in turn, is needed in order to evaluate the following definite
> >> >integral:
> >> >
> >> > I(r) = int_[x=0, x=r] sqrt(r^2-x^2) dx
> >> >
> >> >So even though the derivation of the formula for the area of a circle
via
> >> >4* I(r) is not exactly circular reasoing, it would seem rather silly
> >*not*
> >> >to directly appeal to the formula for the area of a sector (together
with
> >> >ratios and proportions) to quickly derive the area of a circle without
> >> >directly invoking integrals.
> >> >
> >> >It seems interesting to ask whether it is possible to derive the
formula
> >for
> >> >the area of a sector via definite integration without running into
> >circular
> >> >reasoning. If I still remember it correctly, I think it IS possible to
> >> >derive the area of a sector using definite integration if one is
allowed
> >to
> >> >appeal to the arc length formula (s(theta)=r*theta). The arc length
> >formula
> >> >seems to only require the knowledge of the circumference formula and
the
> >> >knowledge of ratios and proportions.
> >> >
> >> >I have a feeling that any use of definite integration to derive the
> >> >circumference formula would be in danger of circular reasong. Any
> >thoughts
> >> >on this?
> >>
> >> Of course it's possible to prove all this without any circularity.
> >>
> >> One way: Start by defining cos(x) by its power series. It's not hard
> >> to show that cos has a smallest positive zero alpha; define
> >> pi = 2*alpha. (See for example the Prologue to Rudin "Real and
> >> Complex Analysis" for the details of what's happened so far and
> >> for at least motivation for what's below.)
> >>
> >> Similarly define sin by the power series. Then sin' = cos and
> >> cos' = -sin, hence sin^2 + cos^2 is constant, so sin(pi/2) is
> >> plus or minus 1. Also, cos > 0 on (0, pi/2) implies that sin
> >> is strictly increasing on [0, pi/2], so sin(pi/2) = 1 and
> >> sin is a bijection from [0,pi/2] onto [0,1].
> >>
> >> Hence the change of variables x = sin(t) shows that
> >>
> >> int_0^1 sqrt(1 - x^2) dx = int_0^pi/2 cos^2(t) dt.
> >>
> >> It's not hard to show that sin and cos have various
> >> symmetry properties, hence
> >>
> >> int_0^pi/2 sin^2(t) dt = 1/2 int_0^pi cos^2(t) dt
> >>
> >> = 1/4 int_0^pi (sin^2 + cos^2)
> >>
> >> = pi/4.
> >>
> >> ************************
> >>
> >> David C. Ullrich
> >
> >Yes, I should have remembered that "power series" approach of defining
pi. I
> >must have forgotten about it because it didn't sit too well with the
> >geometry cells in my stomach, :-), and still doesn't. In the context of
my
> >original question, I did not have a re-definition of pi in mind.
>
> I get the impression from other posts that you're a logician or set
> theorist or something which was the reason for my "of course" - you
> believe that essentially all math can be formalized in ZFC, but your
> comments seemed to express doubt on this one issue...

I guess years of "Platonic schooling" has taken its "toll", :-) Seriously,
perhaps my last lingering doubts of whether all of mathematics can be
formalized in ZFC are the result of my not having gone through a rigorous
examination of how to formalize the classical geometrical definition of pi
(preferably without re-defining it). How did Hilbert define it in his
"modern" axiomatization of geometry?

>
> Suppose that we're trying to define everything formally, starting
> with set theory and giving proofs that could be checked by a machine,
> in particular without relying on the way some piicture looks.
> What definition of pi would you use?

I would still prefer not to define it as twice the first positive zero of
some "power series" unless there is no other way. I would prefer to leave it
as a consequence of some theorem which proves and defines it as the constant
ratio of the circumference of any circle to its diameter as in classical
geometry [footnote 1]. Can geometry be formalized in ZFC without a
redefinition of pi?

Shedar

[1] I was enlightened some time ago by Ken Pledger in this newsgroup that in
classical Euclidean geometry, this "theorem" is actually Archimedes
"Measurement of a Circle" Proposition 1, combined with Euclid XII.2.

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