
Re: Building an Equation to find (Maximum Y) ie Highest Point on a curve!
Posted:
Sep 13, 2013 4:58 AM


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 xyaltered 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

