Date: Oct 4, 2017 2:49 AM
Author: plutonium.archimedes@gmail.com
Subject: 2017 Nobel prize in Math        o-:^>___?   goes<br>	 to Franz of Germany;; For- reverse-math

John Gabriel:: here is an exciting day in Germany where Franz has just won the Nobel Prize in Math, and here is Alouatta, here to explain what Franz has achieved with Reverse-Math.

Alouatta:: yes, John, Reverse Math is where you compose a proof in which everything, every statement is wrong, except for one sentence in the proof that is possibly true, all the rest of the sentences are crap.

John;; is there a need for such a mathematics? I mean, isn't it difficult enough to get someone to have a full page without errors?

Alouatta:: that is the boldness of Reverse Math, which is extremely difficult to do, to compose all lies, except for one sentence. Most math professors can only get 5 true sentences, whereas Franz has now eclipsed all fake sentences except one. See in Franz's classical report below.


On Tuesday, October 3, 2017 at 11:14:58 AM UTC-5, Me (Franz of Germany) wrote:

>
> You obviously don't know the general equation for an ellipse.
>
> Hint:
>
> (1/ab)y^2 + (4/h^2)(x - h/2)^2 = 1
>
> is the equation for an ellipse.
>
> If a = b = r and h = 2r we get the equation of a circle:
>
> (1/r^2)y^2 + (1/r^2)(x - r)^2 = 1
>
> => (x - r)^2 + y^r = r^2
>
> Are you really too dumb to understand these simple things, Archie?
>
> Here's the complete proof again:
>

> > > Cone/Cylinder (side view):
> > >
> > > / | \ (with b <= a)
> > > /b | \
> > > /---+---´ <= x = h
> > > / |´ \
> > > / ´ | \
> > > / ´ | \
> > > x = 0 => ´-------+-------\
> > > / a | \
> > >
> > > (cone: b < a, cylinder: b = a = r)
> > >
> > > r(x) = a - ((a-b)/h)x
> > > d(x) = a - ((a+b)/h)x
> > >
> > > y(x)^2 = r(x)^2 - d(x)^2 = ab - ab(2x/h - 1)^2 = ab(1 - 4(x - h/2)^2/h^2
> > >
> > > => (1/ab)y(x)^2 + (4/h^2)(x - h/2)^2 = 1 ...equation of an ellipse

>
> qed.
>
> Now lets just look at some "properties" of this ellipse:
>

> > > Some considerations:
> > >
> > > => y(h/2 + x')^2 = ab - ab(2(h/2 + x')/h - 1)^2 = ab - ab(2x'/h)^2
> > >
> > > => y(h/2 + x') = sqrt(ab) * (sqrt(1 - (2x'/h)^2) ...symmetric relative to h/2 (hence Ec = cF)
> > >
> > > => y(h/2) = sqrt(ab) (= Gc = cH)