Re: maybe, maybe not

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#1186

From: C. Bond
Date: Wed, 04 Aug 2004 14:52

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Jon wrote:

> The attempt is made to find the roots to polynomials.
>
> if
> N=(a[1],a[2],..,a[n])
> T=(t,t^2,t^3,..,t^n)
> N*T+a[0]=0   is the nth degree polynomial.
>
> dT=(1,2t,3t^2,..)dt
>
> -a[0]=INT N*dT = INT(a[1]+2a[2]t+3a[3]t^2+...)dt
>
> using the virtue that
> a[1]t+a[2]t^2+...=-a[0],
>
> -a[0]=INT(-a[0]/t + a[2]t+2a[3]t^2+...)dt
>
> which results in,
>
> -a[0]=-a[0]ln|t|+(1/2)a[2]t^2+(2/3)a[3]t^3+(3/4)a[4]t^4+...
>
> again, using the virtue that
> a[1]t+a[2]t^2+...=-a[0],
>
> -a[0]=-a[0]ln|t|-a[0]-a[1]t-(1/2)a[2]t^2-(1/3)a[3]t^3-..
>
> or,
>
>              n
> a[0]ln|t|=-SUM(1/p)a[p]t^p        eqn i.
>             p=1
>
> Both sides of eqn i. are expressed as two Taylor Series
>
> f(c)+f'(c)(t-c)+.. = g(c)+g'(c)(t-c)+...
>
> Consequently, after n derivatives,
>
> c=((-1)^n a[0]/a[n])^(1/n)
>
> After n+1 derivatives, the right hand side of eqn i. becomes zero, but
> the left hand side continues to produce values and is subtracted from
> the result.

If the "right hand side of eqn i. becomes zero, but the left hand
sidecontinues to produce values" then it is *not* an equation.

--
There are two things you must never attempt to prove: the unprovable --
and the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com

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