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

On Sat, 3 Sep 2005 19:01:29 +0000 (UTC), Darren J Wilkinson
<d.j.wilkinson@xxxxxxxxx> wrote:

>David C. Ullrich <ullrich@xxxxxxxxxxxxxxxx> wrote:
>> First I should say that it seems to me that the simplest
>> thing really is to define sin(x) by the power series. Then
>> the proof that sin(x)/x -> 1 is very simple (it's not clear
>> to me whether you said in a post below that this was not
>> clear to you), and it's not too hard to show that this
>> turns out to be the same as a geometric definition.
>>
>> But if you say that's not elementary enough fine. Let's
>> talk about sin defined "geometrically", and let's ignore
>> the fact that just _defining_ what the area of a region
>> _is_ takes some not-so-elementary work; let's assume,
>> as in a typical elementary context, that we know what
>> areas are.
>
>OK - I guess I should have been more explicit - by "elementary" and
>"nice", I was meaning the kind of explanation that could be given to a
>bright 15 year old. So, techniques based on power series, integrals,
>differential equations, etc., are really not what I had in mind. The
>kind of argument you gave was exactly the kind of thing I was after.

Ok.

>Now, if I understood it correctly, you showed that sin x / x tended to
>one when x was measured in "aradians".

You might note that there was a factor of 1/2 missing in my
definition of "aradian", which was cancelled by the missing
1/2 in the formula I was using for the area of a triangle.
(Details... heh.)

>Is there an easy way to show that
>aradians and radians are the same?

In the sense in which I would use the word "show",
I doubt it - I don't think that we can even give an
easy definition of area, much less arc length.

But in the sense in which I "showed" something in
my previous post then yes, I believe there is an
easy way, although I don't see exactly how to
do it (which is why I decided to give an argument
based on area this morning).

Given our definition of aradian, the definition
we gave for pi amounts to saying that pi is
the area of the unit circle, and showing that
aradians are the same as radians then amounts
to showing that the circumference is 2pi. So
we have to show that the circumference is twice
the area. (And showing that amounts to the
same thing as giving a proof that sin(x)/x
tends to 1 based on arclength and radians
instead of area...)

You might note that if we're willing to
believe a little bit then it's _very_ easy:
Say P_n is a regular n-gon inscribed in the
unit circle. It's easy to show that the ratio
length(P_n)/area(P_n) tends to 2 as n tends
to infinity, because that ratio is exactly
the length of the segment from the center to
the midpoint of one of the sides of P_n.
So if we're willing to just believe that
the length of P_n tends to the length of
the circle and the area tends to the area
we're done. That might count as "showing"
it for your purposes - it bothers me, because
of examples where this curve approaches that
curve although the length does not approach
the length.

Seems like we should be able to do better,
giving actual inequalities as this morning,
but I don't see exactly how to do it.
The point:

Say Q_n is a regular n-gon _circumscribed_ about
the unit circle. If we could show that the length
of Q_n was greater than the circumference of the
unit circle we'd be set; the fact that a straight
line is the shortest distance between two points
shows that the length of P_n is less than the
circumference and we have the inequalities
we want. But the the fact that the length of
Q_n is greater than the circumference of the
unit circle is what I didn't see how to prove
this morning on the basis of anything plausible
as plausible as "a straight line is the shortest
distance between two points), although it certainly
looks like it's true (as of course it is). If you
can prove this inequality you're set.

Or for that matter if you can prove that
the length of Q_n is larger than some c_n
times the circumference of the unit circle,
where c_n -> 1, you're set. I thought I
had a proof of that a second ago, based on
using a larger circle in which Q_n is
inscribed, but the inequality seems to go
the wrong way.

>Regards,


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

David C. Ullrich
.

Relevant Pages

  • Re: sin x / x tends to 1...
    ... >with aradians fully justifiable. ... >with respect to the radius is the circumference ... One inequality on the circumference ... >>the area of the unit circle, ...
    (sci.math)
  • Re: sin x / x tends to 1...
    ... easier if we restrict to convex polygonal regions (since ... with aradians fully justifiable. ... >the area of the unit circle, ... >to showing that the circumference is 2pi. ...
    (sci.math)
  • Re: sin x / x tends to 1...
    ... > Say we define radians in terms of arc length, ... > pi to be half the circumference of the unit circle. ... You assume the circumference of a circle is well-defined, ...
    (sci.math)