Date: Sep 13, 2013 4:58 AM
Author: Thomas Nordhaus
Subject: Re: Building an Equation to find (Maximum Y) ie Highest Point on<br> a curve!
Am 13.09.2013 03:15, schrieb mervynmccrabbe@gmail.com:

>> Just interchange x and y, then get Max y for the new equation.

>

> x^4 + y^4 + A(x^2) - A(y^2) + 2(x^2)(y^2) - Bxy + C = 0

>

> By substituting x for y the above equation becomes :-

> x^4 + y^4 - A(x^2) + A(y^2) + 2(x^2)(y^2) - Bxy + C = 0

>

> So presumably the new dy/dx becomes:-

> dy/dx = -(4x^3 - 2ax + 4xy^2 - by)/(4y^3 + 2ay + 4x^2 y - bx)

>

> and in turn giving

> 4x^3 - 2ax + 4x(y^2) - by = 0

4x^3 - 2Ax + 4x(y^2) - By = 0

> as the equation to be merged with the xy-altered equation:-

> x^4 + y^4 - A(x^2) + A(y^2) + 2(x^2)(y^2) - Bxy + C = 0

>

> Even if i'm right so far, I am again lost in finding the equivalent of

> quasi's solution to the original equation.

Look closer: The only difference in the set of equations is that A is

replaced by -A. Now look at the coefficients d8, d6, d4, d2, d0 from

quasi's equation for y. What happens when you replace A by -A:

d8 -> d8 (nothing changes because of only even powers in A)

d6 -> -d6 (switch sign because of only odd powers in A)

d4 -> d4

d2 -> -d2

d0 -> d0.

So your equation for x becomes:

d8*x^8 - d6*x^6 + d4*x^4 - d2*x^2 + d0 = 0.

This is the power of symmetry ;-)

--

Thomas Nordhaus