| |  | |  | Raider of the lost time
Join Date: Nov 2003 Posts: 6,036
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05-22-2006, 01:55 PM
| | conics In the 2D subspace of an Euclidean plane in where n=2, if the position of a point changes its location such that the ratio of its distance from a stationary point known as the focus to the distance from a stationary line known as the directrix is a constant known as the eccentricity then the locus of all instantaneous positions describe a curve of conics. When the eccentricity is less than unity, the loci or traces are known as ellipses. When the eccentricity is exactly unity, the traces are known as parabolas. When the eccentricity is greater than unity, the traces are known as hyperbolas. When the eccentricity approaches relative zero 1/¥ or absolute zero then and only then do the described curves approach perfect circles. These facts were known to Apollonius of Perga over 2,200 years ago. http://mathforum.org/cgraph/history/people.html#apollonius Therefore, the inevitable question is that without an axiom of choice such as a minimum length (e.g. Planck length), could an analytical process exist that possibly determine the radii of these circles whether values are unity or infinity. However, one thing that is absolutely certain is that any radius could never take absolute zero as its value although the eccentricity must be exactly zero for any of these divine circles to exist.
__________________ Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: ¶a(t)·¶r(t)=c² | | | | The Thinker
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05-22-2006, 02:40 PM
| | Re: conics Antonio,
The psot is very interesting. Preciselly today we've been intrdouced in math class to parabolas. I have a few questions:
1.What kind of equation would represent one of those perfect circles?
2. Could the eccentricity be an imaginary number? If so, what figure or conic would be traced on the graph?
3. What if the graph is such that x is the real numbers and y is the imaginary numbers? Then the 'ratio of its distance from a stationary point known as the focus to the distance from a stationary line known as the directrix' would have anysense (of so, what sense)?
Thanks for the post, it was quite inspiring to me because it has given me several ideas about solving Goldbach's conjecture on which I was stuck. I'll restart it, having the ideas of this post in mind. | | | | Raider of the lost time
Join Date: Nov 2003 Posts: 6,036
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05-22-2006, 03:14 PM
| | Re: conics Quote: |
Originally Posted by GUILLE 1.What kind of equation would represent one of those perfect circles? | 1. The favorite one that I like to use is the unit circle x+y=1.This is possible if and only if we use the axiom of choice and made the choice for the existence of radius unity. 2. I hate complex eccentricity. Therefore I shun all complex numbers. 3. Same as #2.
__________________ Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: ¶a(t)·¶r(t)=c² | | | | The Thinker
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05-22-2006, 04:43 PM
| | Re: conics Quote: |
Originally Posted by AntonioLao 1. The favorite one that I like to use is the unit circle x+y=1.This is possible if and only if we use the axiom of choice and made the choice for the existence of radius unity. 2. I hate complex eccentricity. Therefore I shun all complex numbers. 3. Same as #2. |
Is there a finite number of formula that make perfect circles? If so, how many and which are they? Also, if the power of the x and y can be up to infinity, still, how many different types of formulas are there (different from x+y=1 not in quantity but in form)?
Fine about neglecting complexity. | | | | Raider of the lost time
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05-23-2006, 01:06 PM
| | Re: conics Quote: |
Originally Posted by GUILLE Is there a finite number of formula that make perfect circles? | In polar notation, the equation of a unit circle is just r = 1. But when we start to introduce coordinate system then the equations become more complicated.
__________________ Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: ¶a(t)·¶r(t)=c² | | | | The Thinker
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05-23-2006, 05:51 PM
| | Re: conics Quote: |
Originally Posted by AntonioLao In polar notation, the equation of a unit circle is just r = 1. But when we start to introduce coordinate system then the equations become more complicated. | Would (x^3)+(y^3)=1 be a perfect circle? Does it work for any power (or changing the plus to minus)?
This actually remembers be now of Fermat's theorem. Could there be any relationship? | | | | 4th degree Black Belt
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05-24-2006, 11:35 AM
| | Re: conics Guille how can x^3+y^3=1 be a perfect circle. I have learnt that the power of x or y describes its symmetry. Like if y^2=4ax then the conic is symmetric about x axis and since for ellipse and hyperbola the powers of both x and y is 2 they are symmetric about both axis. But a perfect circle as is symmetric about both axis it must have power 2 . However I want to know how you got this idea of raising the powers to 3.
__________________ Anything for Everyone! | | | | Raider of the lost time
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05-25-2006, 06:49 PM
| | Re: conics Quote: |
Originally Posted by GUILLE Would (x^3)+(y^3)=1 be a perfect circle? Does it work for any power | Perhaps a quantized circle, if and only if x=cosqand y=sinq. That isq = 0° or q = 90°.
__________________ Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: ¶a(t)·¶r(t)=c² | | | | The Thinker
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05-26-2006, 06:24 PM
| | Re: conics Quote: |
Originally Posted by Mohan.C Guille how can x^3+y^3=1 be a perfect circle. I have learnt that the power of x or y describes its symmetry. Like if y^2=4ax then the conic is symmetric about x axis and since for ellipse and hyperbola the powers of both x and y is 2 they are symmetric about both axis. But a perfect circle as is symmetric about both axis it must have power 2 . However I want to know how you got this idea of raising the powers to 3. | I got it, as I partly mentioned, from Fermat's theorem. It states that there is no possible true solution for any equation x^n+y^n=z^n for any n>2 (when it's n=2, it's pythagoras' theorem, so just as Fermat extended Pythagoras' theorem, I've extended Antonio's developments). | | | | Raider of the lost time
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05-31-2006, 04:07 PM
| | Re: conics Quote: |
Originally Posted by Guille I've extended Antonio's developments | But it was proven that for n>2 no solution satisfies the equation. Instead it becomes an inequality expression for LHS and RHS (left hand side and right hand side).
__________________ Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: ¶a(t)·¶r(t)=c² | | | |  | |
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