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Simply Beautiful Art

2:07 PM

@Secret btw I can sorta factor quintics if you want an explanation on that.

Secret

hmm? How will that not violate galois theory?

Simply Beautiful Art

@Secret don't use radicals xD

well use radicals

but also more than just radicals

Secret

demonstrate what $x^5-x+1$ be factorised to?

Simply Beautiful Art

2:24 PM

@Secret I can explain it if you want, but the substitutions and what-not are absurd

Secret

sure

Simply Beautiful Art

For that specific case, it's already in Bring-Jerrard form, so I'd just Bring radical or hypergeometric function it.

Lagrange inversion theorem

In mathematical analysis, the Lagrange inversion theorem, also known as the Lagrange–Bürmann formula, gives the Taylor series expansion of the inverse function of an analytic function. == Theorem statement == Suppose z is defined as a function of w by an equation of the form z = f ( w ) {\displaystyle z=f(w)} where f is analytic at a point a and f '(a) ≠ 0. Then it is possible to invert or solve the equation for w, expressing it in the form w = g ( z ...

Leaky Nun

bring me some radicals please

Simply Beautiful Art

>.>

It's probably a lot easier to start with a general cubic polynomial.

$x^3=ax^2+bx+c$

We want to make the substitution $y=x^2+mx+n$.

@Secret compute $y^2$ if you will for me.

Secret

$y^2= x^4+mx^3+nx^2+mx^3+m^2x^2+mnx+nx^2+nmx+n^2$

$= x^4+2mx^3+(2n+m^2)x^2+2mnx+n^2$

Simply Beautiful Art

2:30 PM

@Secret $x^4=x(x^3)=x(ax^2+bx+c)$

Simplify until you reach $y^2=\dot\ell x^2+\dot mx+\dot n$

Secret

$y^2=x(ax^2+bx+c)+2m(ax^2+bx+c)+(2n+m^2)x^2+2mnx+n^2$

$= (2n+m^2)x^2+(2mn+(ax^2+bx+c))x+(2m(ax^2+bx+c)+n^2)$

o wait I am doing stupid

retry

Simply Beautiful Art

>.>

Secret

$y^2= x (ax^2+bx+c)+2m(ax^2+bx+c)+(2n+m^2)x^2+2mnx+n^2$

Simply Beautiful Art

It's easier in my opinion to only expand the highest power

Secret

$=a(ax^2+bx+c)+bx^2+cx+2m(ax^2+bx+c)+(2n+m^2)x^2+2mnx+n^2$

$= (a^2+b+2ma+2n+m^2)x^2+(ab+c+2mb+2mn)x+(ac+2mc+n^2)$

so uh, I got some kind of quadratic

Simply Beautiful Art

2:48 PM

@Secret yeah looks good I think

Now multiply that by $x^2+mx+n$

And compute $y^3$

You'll get $y^3=\ddot\ell x^2+\ddot mx+\ddot n$ after immense simplification.

Suppose you've reached this point.

You then have the following system of equations:

$$\begin{cases}y&=x^2+mx+n\\y^2&=\dot\ell x^2+\dot mx+\dot n\\y^3&=\ddot\ell x^2+\ddot mx+\ddot n\end{cases}$$

From which you can eliminate $x$ from the system, producing $y^3+uy^2+vy+w=0$.

For some constants $u,v,w$.

It only requires solving quadratics in $m$ and $n$ to get $u=v=0$.

and the rest is trivial.

To reduce the general quintic in the form of:

$x^5=ax^4+bx^3+cx^2+dx+e$

Only two steps are required (though I recommend simplifying with easier substitutions such as $x=y+a/5$ first)

First, the substitution $y=x^2+mx+n$, just like we did in the previous problem.

Solve for $m$ and $n$ so that you end up with $y^5=\dot cy^2+\dot dy+\dot e$

Then make the substitution $z=y^4+py^3+qy^2+ry+s$

Which will allow you to reach the form $z^5-\ddot dz=\ddot e$

With a simple substitution, one can convert this to the form $t^5-t=\dddot e$, which can be dealt with using Bring radicals or hypergeometric functions.

$\ddot\smile$

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