Re: sin x / x tends to 1...

On Sun, 4 Sep 2005 16:07:44 +0000 (UTC), Darren J Wilkinson
<d.j.wilkinson@xxxxxxxxx> wrote:

>David C. Ullrich <ullrich@xxxxxxxxxxxxxxxx> wrote:
>> On Sun, 4 Sep 2005 14:26:13 +0000 (UTC), Darren J Wilkinson
>> <d.j.wilkinson@xxxxxxxxx> wrote:
>>
>> >David C. Ullrich <ullrich@xxxxxxxxxxxxxxxx> wrote:
>> >> On Sat, 3 Sep 2005 21:17:29 +0000 (UTC), Darren J Wilkinson
>> >> That's easy. The actual _definition_ of the length of a curve
>> >> is the supremum of the lengths of inscribed polygons, and using
>> >> that it's easy to show that the circumference of the unit circle
>> >> is twice the area.
>> >
>> >True.
>> >
>> >But here the perimeter of the polygon only tends to 2pi if the sin x / x
>> >limit holds. So it has always seemed to me that defining arc length this
>> >way is essentially adopting the sin x / x limit as an axiom.
>>
>> ??? This has nothing to do with that.
>>
>> Say we define radians in terms of arc length, and we define pi to be
>> half the circumference of the unit circle. If we define the length
>> as the limit of the lengths of inscribed polygons then the length
>> of that polygon tends to 2pi _by definition_.
>>
>> And it's easy to see that length/area -> 2 for the polygons,
>> hence the area of the unit circle is pi. Hence sectors
>> have the right area even when we define radians in terms
>> of arc length, and the the argument comparing the area of
>> the sector to two triangles gives those inequalities that
>> show that sin(x)/x -> 1.
>>
>> We've been going around in circles a bit, so maybe you're
>> sort of on a different page. But honest, if we start by
>> defining the circumference as the limit of the length of
>> those polygons then everything works with no circularity.
>
>I've possibly lost the plot by now, but if you inscribe a regular
>N-agon, the perimeter is 2Nsin(pi/N), and you are telling me that
>_by definition_ this tends to 2pi as N increases...

No, that's not what I said, although I suppose that it's
not unreasonable for you to think that that's what I meant.
The fact that 2Nsin(pi/N) -> 1 _is_ going to fall out
of all this very simply, but I'm certainly not asserting
that _that_ fact is true by definition.

Start over. Clear all definitions relating to arc length,
angles, and trig functions. But assume that areas and
lengths of straight line segments work as usual.

Say P_n is an inscribed regular polygon with
n sides. Say L_n is the perimeter of P_n and A_n
is the area of P_n. Let r_n be the distance from
the center of the circle to the midpoint of one
of the sides of P_n.

The formula for the area of a triangle shows
that

(*) A_n = r_n L_n / 2.

It is clear and/or "clear" that r_n -> 1
and that A_n -> the area of the circle.
Hence L_n has a limit as n -> infinity.

Definition 1: The circumference of the unit circle is
the limit of L_n as n -> infinity.

(Note: The official definition would be the _sup_ of
the lengths of all insribed polygons, regular or not.
We could do what's below with that definition just
as well, a few things would need to be rewritten.
Let's just use the definition above for simplicity.)

Definition 2: pi is one half the circumference of the
unit circle.

The two definitions show that L_n -> 2 pi.

Now define the sine of an angle geometrically,
define radian measure as usual in terms of
arc length, and define the sine of a number
by saying that sin(x) is the sine of x radians.
Then yes, L_n = 2Nsin(pi/N), and so we've
proved that Nsin(pi/N) -> pi.

As a free bonus, note that since L_n -> 2 pi
(*) shows that the area of the unit circle is pi.
This shows that aradians are the same as radians,
clearing up the nagging doubt with the original
proof that sin(x)/x -> 1 using aradians.

I've said all along that if we adopt definition 1
everything works very simply - maybe I should have
been more explicit about that.

******************************************************

The alert reader will notice a slight hole here:
We gave a definition for the circumference of a
circle but no definition for the length of a
circular arc.

Fix 1: Don't worry about it, assume we know
what we mean...

Fix 2: Go Greek and only deal with rational
numbers, or rather rational ratios:

Say an angle has measure pi/N if N copies
of the angle fit together to form a straight
line. Say an angle has measure (p/q) pi if
it can be divided into p smaller angles,
each of measure pi/q. Either ignore angles
of measure x pi for irrational pi or
mumble something about limits.

Fix 3, the right one, but the most complicated:

Give up and use the _actual_ definition of
the length of a curve:

If c: [a,b] -> R^2 is continuous then the
length of c is the sup of the sums

sum ||c(x_j) - c(x_{j-1})||

over all partitions a = x_0 < ... < x_n = b.
(This sup may be finite or infinite; if it's
finite the curve is "rectifiable".) Probably
you want to prove that if c_1:[a,b] -> R^2
and c_2:[b,c]-> R^2 are curves with c_1(b)
= c_2(b) and c:[a,c] -> R^2 is the curve
you get by putting them together then the
length of c is the sum of the lengths.

We need to modify some of what's above,
as I said was possible:

If c is a parametrization of the unit
circle, 1-1 except for the endpoints,
then sum ||c(x_j) - c(x_{j-1})|| is
simply the perimeter of an inscribed
polygon P. Say the side lengths of
P are s_1, ... s_n, and say r_j is
the distance from the center to the
midpoint of the j-th side. Say L is
the perimeter of P and A is the area.

Then

(**) A = sum r_j s_j / 2.

Note that if P' is another inscribed
polygon which includes all the vertices
of P plus more vertices then the fact
that a straight line is the shortest
distance between two points shows that
L' > L. Given any P we can always add
more vertices to obtain a P' where
all the rj' are larger than 1/2. So
in calculating the sup of L over all
polygons P we can consider only polygons
P such that r_j > 1/2 for all j.

(In other words, a polygon that has
r_j < 1/2 for some j is clearly a very
bad approximation to the circle - the
previous paragraph is an official proof
that we can ignore such polygons.)

Now A is < the area of the circle.
Since r_j > 1/2, (*) shows that

L = sum s_j < four times the area of the circle.

In particular (**) shows that L is bounded
above, so that the L's have a finite sup
as P ranges over all polygons.

So the circumference of the circle, which
we've already defined to be the sup of the
L's, is finite.

Definition: pi is half the circumference
of the circle.

Now that we've defined the length of any
curve we can define angle measure in
radians using arc length with no problem.
The rest of the proof is the same as
before, except for one technicality:

Say P_n is an inscribed regular n-gon.
Define L_n, etc, as before.
We defined the cirumference as a sup
over _all_ inscribed polygons, so we
no longer know just from the definitions
that L_n -> 2 pi; could be a priori
that there is a P such that L is much
greater than _all_ the L_n. But that
doesn't happen:

Lemma: L_n -> 2 pi.

Proof: Let eps > 0. Fix a P such that
L > 2 pi - eps (recall that L < 2 pi
for every P, by definition.) Say P
has N sides.

Having fixed P, choose n so that the
sides of P_n are much smaller than the
sides of P, and so that n is much larger
than N. Now let Q be a polygon with the
same vertices as P_n, except that the
N vertices of P_n closest to the vertices
of P have been shifted to coincide with
the vertices of P.

Since we only moved N vertices of P_n
(where N is fixed and n is much larger
than N) and we only moved them a tiny
amount, the perimeter of Q is incredibly
close to being the same as the perimeter
of P_n; say it's within eps. We can make
it within eps just by taking n large
enough that the lengths of all the 2N
sides that were changed add up to less then
eps, say. But now every vertex of Q is
also a vertex of P, and hence

perimeter(Q) > perimeter(P) = L > 2pi - eps.

This shows that L_n > 2 pi - 2 eps if n
is large enough, qed.

************************

David C. Ullrich
.

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