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Topic: Building an Equation to find (Maximum Y) ie Highest Point on a curve!
Replies: 14   Last Post: Sep 14, 2013 3:40 PM

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 Thomas Nordhaus Posts: 433 Registered: 12/13/04
Re: Building an Equation to find (Maximum Y) ie Highest Point on
a curve!

Posted: Sep 13, 2013 4:58 AM
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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

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