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Topic: How's this? Proof (hopefully) that pi is irrational.
Replies: 5   Last Post: Sep 12, 2009 7:10 AM

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Re: How's this? Proof (hopefully) that pi is irrational.
Posted: Sep 12, 2009 7:01 AM
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>
>
> In article <tkidd.834615695@hubcap>,
> tkidd@hubcap.clemson.edu (Travis Kidd) writes:
> |> hook@cscsun3.larc.nasa.gov (Ed Hook) writes:
> |> >|>Assume pi were rational. Then there would be
> some integer k such that
> |> >|>sin kn = 0 for all integers n. Take the
> function sin x^2 / sin kx.
> |> >|>What is the limit of this function as x
> approaches k? Well, Do not judge. Fair warning.Support a cure for childhood cancer: Alex's Lemonade


©2009 MeAmI.org "Search for the Psince sin x^2
> |> >|>and sin x are both continuous, this limit could
> be expressed by substitutiong
> |> >|>k in for x, or sin k^2 / sin k^2 = 1. However,
> since both sin x^2 and sin kx
> |> > Nope -- not gonna work. Above, you chose 'k' as
> a magical integer with
> |> > the property that sin(kn) = 0 for _all_
> integers n; in particular, you
> |> > have to accept that sin(k^2) = 0, so you've got
> an indeterminate form
> |> > here.
> |> Yes. But nowhere is either the numerator *or* the
> denominator 0 until the
> |> limit is reached. (I'll repeat what I just said
> in reply to another post...
>
> Makes no difference -- you could apply the same
> me argument to the difference
> quotient for a monotone function at any point in
> in its domain: since
> f(x+h) - f(x) and (x+h) - x are never 0 *except* at
> at h, then the derivative
> can be calculated by plugging in 0 for h ?? Put
> ut this way, it's blatantly
> wrong ... but this is exactly what you're about.
>
> |> I neglected to mention that k is positive.)
> Therefore (I hope...I agree this
> |> is the weakness in the proof), since substituting
> k for x yields equivalent
> |> functions, the limit of the quotient would be 1.
> |>
> How do you define "equivalent" here ?? Near k, the
> he two functions agree
> at precisely _one_ point. namely x = k ... if you
> ou want to evaluate the
> limit, then you're going to have to resort to
> to something like L'Hopital's
> rule. But, then, your argument vanishes in a puff
> ff of smoke.
>
> |> The fact that the limit is of an indeterminate
> form *then* allows me to use
> |> L'Hopital's rule to find a *different* limit,
> which is the contradiction.
> |>
> |> Indeed, the fact that sin(k^2) = 0 is the
> (hopeful) key to the proof!
> |>
> |> >|>
>
>
>
> QED(?)
> |> > Ed Hook |
> Coppula eam, se non posit
> |> > Computer Sciences Corporation |
> acceptera jocularum.
> |> > NASA Langley Research Center | Me? Speak
> for my employer?...<*snort*>
> |> > Internet: hook@cscsun3.larc.nasa.gov |
> ... Get a _clue_ !!! ...
> |> -Travis
>
> --
> Ed Hook | Coppula
> a eam, se non posit
> Computer Sciences Corporation |
> | acceptera jocularum.
> NASA Langley Research Center | Me? Speak for
> r my employer?...<*snort*>
> Internet: hook@cscsun3.larc.nasa.gov | ...
> . Get a _clue_ !!! ...
>
>
>
>




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