Hi Tuom,

On Wed, Feb 26, 2014 at 5:26 PM,  <[email protected]> wrote:
> Hi!
>
> I'm trying to understand how the root finding of quartic polynomials works
> but I came to a dead end.
>
> The method discovered by Euler is described here:
> http://mathforum.org/dr.math/faq/faq.cubic.equations.html
>
> To quote the relevant parts:
>
> z3 + (e/2) z2 + ((e2-4 g)/16) z - f2/64 = 0 (*)
> r = -f/(8 p q)
> x = p + q + r - a/4
> x = p - q - r - a/4
> x = -p + q - r - a/4
> x = -p - q + r - a/4
>
> But "_roots_quartic_euler" function from SymPy which claims to use the
> Descartes-Euler solution contains (very) different procedure:
>
> 64*R**3 + 32*p*R**2 + (4*p**2 - 16*r)*R - q**2 = 0 (this is the same as
> above)  (*)
> but:
> p = -2*(R + A); q = -4*B*R; r = (R - A)**2 - B**2*R
> x1 = sqrt(R) - sqrt(A + B*sqrt(R))
> x2 = -sqrt(R) - sqrt(A - B*sqrt(R))
> x3 = -sqrt(R) + sqrt(A - B*sqrt(R))
> x4 = sqrt(R) + sqrt(A + B*sqrt(R))
>
> and later:
> c1 = sqrt(R)
> c2 = sqrt(A + B)
> c3 = sqrt(A - B)
>
> Please, could someone explain a few things to me?
>
> Where do all the definitions of x1, x2, x3, x4 come from? Is there somewhere
> a paper which derives them? And how was "p" derived?
>
> In particular, what I'm trying to understand is how to go from using two
> roots of (*) to using just one (c1)? I.e., the description of Euler's method
> says, that one has to pick two roots of (*) - how is it possible to pick
> just one and still be able to calculate x1...x4?

Great questions. Chris Smith is the expert here, I CCed him.

Once you understand it,
it would be a big help if you could send us PRs, documenting the code more,
possibly adding a nice documentation page about this into our Sphinx, with
equations etc. That would be a great resource about quartic equations.

Ondrej

>
> Thanks in advance!
>
> Tuom Larsen
>
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