Quartic formula

[tuom....@gmail.com]

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?

Thanks in advance!

Tuom Larsen

[Ondřej Čertík]

Hi Tuom,

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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[Kalevi Suominen]

Hi, Tuom

I'm trying to attach a file answering your questions.

I hope it arrives.

Kalevi

[tuom....@gmail.com]

Hi Kalevi,

you are awesome! It helps a lot, thank you!

[Ondřej Čertík]

Hi Kalevi,

Awesome writeup, thanks a lot!

Ondrej

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