[41088] in North American Network Operators' Group
RE: What is the limit? (was RE: multi-homing fixes)
daemon@ATHENA.MIT.EDU (Alex Bligh)
Wed Aug 29 14:18:48 2001
Date: Wed, 29 Aug 2001 19:18:17 +0100
From: Alex Bligh <alex@alex.org.uk>
Reply-To: Alex Bligh <alex@alex.org.uk>
To: Roeland Meyer <rmeyer@mhsc.com>,
"'smd@clock.org'" <smd@clock.org>, bicknell@ufp.org
Cc: cmartin@gnilink.net, nanog@merit.edu,
Alex Bligh <alex@alex.org.uk>
Message-ID: <684762168.999112697@[169.254.198.40]>
In-Reply-To: <EA9368A5B1010140ADBF534E4D32C728069E58@condor.mhsc.com>
MIME-Version: 1.0
Content-Type: text/plain; charset=us-ascii; format=flowed
Content-Transfer-Encoding: 7bit
Content-Disposition: inline
Errors-To: owner-nanog-outgoing@merit.edu
Roeland,
--On Wednesday, 29 August, 2001 10:13 AM -0700 Roeland Meyer
<rmeyer@mhsc.com> wrote:
>|> Draw two curves, the first y=x/2, the second y=x^2
>|> Move the value of x for y=1 for the first curve left by 2, 5 or 10
>|> and it will still be surpassed by the second curve.
>|> You will even see this for a second curve of y=x*2 or y=x.
>
> Prove it.
Prove that y1=A(x^2)+Bx+C always exceeds y0=Dx+E
for positive A and D, for all x>x0 for
some value x0?
Um, y1-y0 = A(x^2) + (B-D)x + (C-E) [1]
This is a positive parabola with standard
solutions. To the right of it's higher root,
it's always positive, so y1>y0.
Now, I take it you don't want proof of
the roots to quadratic equations?
--
Alex Bligh
