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)