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

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.

You can see why I gave up on analysis some years ago... ;-)

> >Assuming a power series definition of
> >sin x, that we can call psin x, the limit is (of course) obvious. Is
> >it very easy to show that psin x = sin x, where sin x is the usual
> >geometric definition (with regular radians)?
>
> That's also not too hard - it's not included in the place
> in Rudin where he works out the basic properties of trig
> functions starting from power series, but it's in one
> of the appendices to my complex notes, to be published
> some day when I get around to it:
>
> Start with the definition in terms of power series.

<snip>

Thanks - that seems very clear.

> Now, c(t) = (cos(t), sin(t)). So to show
> that cos and sin are the same as the
> geometric definition we just need to show
> that (for 0 < x < 2pi, say) the length
> of the arc of the unit circle from (1,0)
> to c(x) is exactly x. But this is immediate
> from the calculus formula for the length
> of a (paramtrized) curve: ||c'(t)|| = 1,
> so the length of this arc is the integral
> of ||c'(t)|| from 0 to x, which is is
> exactly x. qed.

I still feel there is a little slight-of-hand here, as the calculus
formula for the length of a parameterised curve is only really
convincing if you "believe" the definition of arc length given above...

Regards,

--
Dr Darren Wilkinson
mailto:d.j.wilkinson@xxxxxxxxxxxxxxx
http://www.staff.ncl.ac.uk/d.j.wilkinson/
.

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