Friday, April 03, 2009
Parametric Pell's Equation and circle
Intriguingly, in rationals, my parametric solution for Pell's Equation, can be used to give the familiar parametric equation result for the circle:
With x^2 - Dy^2 = 1
I have proven:
y = 2[f_2*v - 1]/[D - (f_2*v - 1)^2]
and
x = [D + (f_2*v - 1)^2]/[D - (f_2*v - 1)^2]
where f_2 is an integer factor of D-1, so with D=-1, you have x^2 + y^2 = 1, and I'll give the circle result in its familiar form:
x = (1 - t^2)/(1 + t^2)
y = 2t/(1 + t^2)
See: http://mathworld.wolfram.com/Circle.html eqns. 16 & 17
Cool! Isn't mathematical research wonderful?
The parametric form for Pell's Equation was clearly previously unknown or a lot of textbooks are hiding a beautiful result.
Also I'll remind that I've proven that Pell's Equation itself is more easily solved using alternates to it, like the negative Pell's Equation:
j^2 - Dk^2 = -1, where x = 2Dk^2 - 1 = 2j^2 + 1
or
j^2 - Dk^2 = -2, where x = Dk^2 - 1 = j^2 + 1
or
j^2 - Dk^2 = 2, where x = Dk^2 + 1 = j^2 - 1
and, of course, x^2 - Dy^2 = 1.
One of the alternates is always true if D is a prime, where the negative Pell's Equation is the dominant one, and is true if D is a prime that is 1 mod 4 or is the product of primes that all are 1 mod 4, while the other alternates are true for cases when D is -1 mod 4, or is the product of primes where all either have 2 as a quadratic residue or have -2 as a quadratic residue, else there is yet another alternate equation which is just not as pretty, so I'm not giving it here.
Because the first x that solves Pell's Equation is roughly j^2, it is mathematically naive to solve Pell's Equation directly, rather than use the alternates which are also solvable by the same techniques available for Pell's Equation, for instance, a continued fraction solution must exist.
It is amazing that a parametric equation for Pell's Equation which gives the circle parametric equation was previously unknown, but I think it is a beautiful "pure" result which maybe just didn't want to be born into this world until the 21st century.
With x^2 - Dy^2 = 1
I have proven:
y = 2[f_2*v - 1]/[D - (f_2*v - 1)^2]
and
x = [D + (f_2*v - 1)^2]/[D - (f_2*v - 1)^2]
where f_2 is an integer factor of D-1, so with D=-1, you have x^2 + y^2 = 1, and I'll give the circle result in its familiar form:
x = (1 - t^2)/(1 + t^2)
y = 2t/(1 + t^2)
See: http://mathworld.wolfram.com/Circle.html eqns. 16 & 17
Cool! Isn't mathematical research wonderful?
The parametric form for Pell's Equation was clearly previously unknown or a lot of textbooks are hiding a beautiful result.
Also I'll remind that I've proven that Pell's Equation itself is more easily solved using alternates to it, like the negative Pell's Equation:
j^2 - Dk^2 = -1, where x = 2Dk^2 - 1 = 2j^2 + 1
or
j^2 - Dk^2 = -2, where x = Dk^2 - 1 = j^2 + 1
or
j^2 - Dk^2 = 2, where x = Dk^2 + 1 = j^2 - 1
and, of course, x^2 - Dy^2 = 1.
One of the alternates is always true if D is a prime, where the negative Pell's Equation is the dominant one, and is true if D is a prime that is 1 mod 4 or is the product of primes that all are 1 mod 4, while the other alternates are true for cases when D is -1 mod 4, or is the product of primes where all either have 2 as a quadratic residue or have -2 as a quadratic residue, else there is yet another alternate equation which is just not as pretty, so I'm not giving it here.
Because the first x that solves Pell's Equation is roughly j^2, it is mathematically naive to solve Pell's Equation directly, rather than use the alternates which are also solvable by the same techniques available for Pell's Equation, for instance, a continued fraction solution must exist.
It is amazing that a parametric equation for Pell's Equation which gives the circle parametric equation was previously unknown, but I think it is a beautiful "pure" result which maybe just didn't want to be born into this world until the 21st century.
# posted by James Harris @ 21:24 
