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Topic: 2017 Nobel prize in Math        o-:^>___? goes
to Franz of Germany;; For- reverse-math

Replies: 2   Last Post: Oct 4, 2017 4:44 AM

 Messages: [ Previous | Next ]
 Me Posts: 1,716 Registered: 1/23/16
Re: 2017 Nobel prize in Math        o-:^>___?
goes to Franz of Germany;; For- reverse-math

Posted: Oct 4, 2017 3:29 AM

On Wednesday, October 4, 2017 at 8:49:27 AM UTC+2, Archimedes Plutonium wrote:

> Reverse Math is ...

Actually, "Reverse-Math" already has an established meaning:
https://en.wikipedia.org/wiki/Reverse_mathematics

> ... is there a need for such a mathematics?

Obviously. See:
https://en.wikipedia.org/wiki/Reverse_mathematics

> Franz has now eclipsed all fake sentences except one. See in Franz's
> classical report below.

Huh?! Where did you spot an error? Mind to point out at least ONE?

> > Hint:
> >
> > (1/ab)y^2 + (4/h^2)(x - h/2)^2 = 1
> >
> > is [an] equation for an ellipse.

it might help -in your case- to look up the general equation for an ellipse.
See: https://en.wikipedia.org/wiki/Ellipse#Equation_of_a_shifted_ellipse

> > If a = b = r and h = 2r

[which represents the case where the cutting plane is perpendicular to the axis of the cone/cylinder]

> > 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

Again, knowledge of the general equation for a circle might be helpful here:
https://en.wikipedia.org/wiki/Circle#Equations

> > Are you really too dumb to understand these simple things, Archie?

Obviously.

The following diagram might be helful befor looking at the actual proof.

^ x
E|
-+- <= x=h
.' | `.
/ | \
. | .
G | +c | H
. | .
\ | /
`. | .´
y <----------+ <= x=0
F

> > 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 "curve":
> > > >
> > > > 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 x = h/2

Hence Ec = cF. (See diagramm above.)

With other words, GH is an axis of symmetrie of the figure (besides EF).

So it's clearly NOT an "oval" as defined by AP.

> > > > => y(h/2) = sqrt(ab) (= Gc = cH)

Look, Archie, the fact that you are not capable of understanding a (rather simple) mathematical proof, does not invalidate that proof.

Date Subject Author
10/4/17 plutonium.archimedes@gmail.com
10/4/17 Me
10/4/17 Me