Roots of equation $x^n\left(2-x\right)^{2}=a$?

Question

I need to find analytically the roots of the following polynomial equation:

$$x^n\left(2-x\right)^{2}=a$$

for an arbitrary integer $n$ and an arbitrary real parameter $a$. The only trick I can think of is in the special case when $a\geq0$ and $n=2m$ (i.e. $n$ even). In this case the equation can be decomposed into two simpler and lower dimensional equations:

$$x^{m}\left(2-x\right)-\sqrt{a}=0$$ $$x^{m}\left(2-x\right)+\sqrt{a}=0$$

but then I don't know how to proceed (I was thinking that maybe through some change of coordinates it is possible to transform these new equations into trinomial equations of the form $x^n-x+t=0$, whose solution is known, see here). I'd also like to find the roots for any $n$ and $a$. Closed-form solutions are not required: I'd really appreciate also a solution written e.g. as a series expansion.

Many thanks in advance.

P.S.: I don't know if it may help, but in my case $x\in[-1,1]$ so that you may write $x=\cos y$ for $y\in [0,\pi]$.

Answer 1

The first thing you can do is to study the trivial case $a=0$ which gives : $$x^n(2-x)^2=0$$ The roots are obviously $x=0$ with order $n$ and $x=2$ with order $2$. You can try to solve other cases to sketch the general form of the solution if it exists.

You can also try some particular values of $n$ :

For $n=0$ you have :

$$x=2\pm\sqrt{a}$$

Which is an order $2$ root.

For $n=1$ I tried with Wolfram Alpha which provides only one ugly solution, which has to be order $3$

For $n=2$ you have : $$x^2(2-x)^2=a$$ $$x(2-x)=2x-x^2=\pm\sqrt{a}$$ Hence the solutions are : $$x=1\pm\sqrt{1\pm\sqrt{a}}$$ Depending on $a$ you can have up to four different solutions with order $1$ each.

It seems complicated to find a general expression as you can have different number/orders of solutions depending on the choice of $(a,n)$

Answer 2

See the following link (theorem 2) https://www.researchgate.net/publication/262973394_Solution_of_Polynomial_Equations_with_Nested_Radicals

He solves :

$$Aqx^{p}+x^{q}=1$$

Your equation is :

$$2x^m-x^{m+1}-\sqrt{a}=0$$ If you make the following substitution :

$x=y\beta$

We have :

$$2\beta^{m}y^m-\beta^{m+1}y^{m+1}-\sqrt{a}=0$$

So divide by $-\beta^{m+1}$ you have:

$$\frac{-2}{\beta}y^m+y^{m+1}+\frac{\sqrt{a}}{\beta^{m+1}}=0$$

And make the last substitutions :

$$\beta^{1}=\frac{1}{m+1}$$ and $$\sqrt{a}=\beta^{m+1}$$

Finally we obtain :

$$(-2)(m+1)y^m+y^{m+1}+1=0$$ So you can apply the theorem 2 with $A=-2$ and $q=m+1$ Another way is to use the theorem 1 with $b=0$ but we obtain a nested radical... Ps:

It's a partial solution because there is some conditions on $\sqrt{a}$... If you want other details see this Solving 5th degree or higher equations

Answer 3

Excel can provide some intuition. Here are graphs of $x^n(2-x)^n$ on $[-1,1]$ for $n=6$ and $n=7$.

You can see the range in each case, and prove it (for $n$ odd and $n$ even). For large(r) $n$ than these modest values the shape remains the same, so for values of $a$ (not too close to $0$) you'll have one or two solutions. Those solutions approach $\pm 1$ as $n$ grows. Newton's method should work just fine.

Answer 4

There is a new special function I have been working that solves your problem and gives you insigths:

$$ x= \frac{W_{2r}\bigg(-r \bigg( \frac{a}{4} \bigg)^r \bigg)}{-r}$$ where $W_r(x)$ is the so-called Lambert-Tsallis function. In the solution above $r=1/n$. Assuming $n \in N$ and $a >0$, one has:

1. If $ r \bigg( \frac{a}{4} \bigg)^r > \bigg( \frac{2r}{2r+1} \bigg)^{2r+1}$ and $n$ is even there is only one negative root in all possible $x$ solutions. If $n$ is odd, that solution becomes positive;
2. In the equality of statement 1, $x=\frac{2}{2r+1.}$ is a double root. However, depending on the choice of $n$ some other real roots may appear as solutions also;

There are also interesting properties that if community is interested I can post it later.