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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 $y\in [0,\pi]$.

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    $\begingroup$ I very much doubt you'll find a closed form solution. Your link to the trinomial equation solution suggests that you'd be happy with a series or other approximation method, perhaps numerical. If so, say so; maybe someone will step up to help you. $\endgroup$ – Ethan Bolker Oct 24 '16 at 13:45
  • $\begingroup$ You are right, thanks for the suggestion. I have just edited my question. $\endgroup$ – user2983638 Oct 24 '16 at 13:49
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    $\begingroup$ I suggest you to normalize the equation to the form $x^\alpha(1-x)=a$ for easier study. $\endgroup$ – Yves Daoust Oct 24 '16 at 14:16
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    $\begingroup$ Maybe you can try to find something using a $log$ $\endgroup$ – H.C. Lefevre Oct 24 '16 at 14:26
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    $\begingroup$ Notice that for $n$ even and $a < 0$ the equation has no real solution. $\endgroup$ – Alex Silva Oct 24 '16 at 15:10
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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

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  • $\begingroup$ Wow, I was aware of that article, but I thought I had to use nested radicals, which I don't like too much. But I didn't realize it is possible to use theorem 2. So this solves the case with $a\geq0$ and $n$ even. I guess I have to understand how to get the remaining $m-1$ solutions, since eq. (13) basically gives only one solution, if I understand it correctly. Let me check the calculations before accepting your answer. By the way, the case with $n$ odd should still be an open problem. $\endgroup$ – user2983638 Oct 24 '16 at 15:33
  • $\begingroup$ I'm not sure that the final equation you get corresponds to the first one. It seems to me that the term $1$ has the wrong sign compared to the term $y^{m+1}$. But ok, maybe for the solution with the $+\sqrt a$ it works fine! $\endgroup$ – user2983638 Oct 24 '16 at 16:17
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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 $(a,n)$

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  • $\begingroup$ Hi, thanks for your comments. I know it's not straightforward to find a solution for any $n$, but I was wondering if through some series expansion like the one I suggested (see en.wikipedia.org/wiki/Bring_radical#Glasser.27s_derivation) it is possible to solve the equation, at least approximately. Also the trinomial form of the link has arbitrary $n$, but a solution does exist. $\endgroup$ – user2983638 Oct 24 '16 at 15:00
  • $\begingroup$ It is easy to prove that this equation has always at least one solution $x\in\mathbb{R}$ as whatever $n$, $x^n(x-2)^2$ takes at least all the values in $\mathbb{R}_{+}^{*}$. The thing is there are many cases in some of them you have one only solution in some other you have $n$ different solutions and in some other you have two solutions etc. The intuition is that you cannot look for a methodology other than computer based to approach or explicitely compute this solution. $\endgroup$ – H.C. Lefevre Oct 24 '16 at 15:07
  • $\begingroup$ I see your point. Also observe that for $a<0$ and $n$ even there is no solution. $\endgroup$ – user2983638 Oct 24 '16 at 15:12
  • $\begingroup$ That is true, I considered $a\geq 0$ as you stated in your first post. $\endgroup$ – H.C. Lefevre Oct 25 '16 at 8:06
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Excel can provide some intuition. Here are graphs of $x^n(2-x)^n$ on $[-1,1]$ for $n=6$ and $n=7$.

enter image description here

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.

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  • $\begingroup$ Interesting. Indeed a very special case of the equation I need to consider (for reasons I don't explain here) is $n\rightarrow\infty$. $\endgroup$ – user2983638 Oct 24 '16 at 15:47

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