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Re: 2017 Nobel prize in Math o:^>___? goes to Franz of Germany;; For reversemath
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, "ReverseMath" 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  ((ab)/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.



