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How can the third root of $\sqrt{x} - 2x + x^a$ be expressed in terms of a when $a>1.5$? (Obviously there is always one at 1 and 0)

Christian
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  • Why do you expect there is a nice expression? When $a=3.5$, we get $y(y - 1)(y^5 + y^4 + y^3 + y^2 + y - 1) = 0$ where $y=\sqrt x$. That quintic is probably not solvable by radicals. – lhf Nov 17 '20 at 10:54
  • if we allow $y = \sqrt{x}$, the equation reduces to $y^{2a}-2y^2+y=0$, or $y^{2a-1} - 2y + 1 = 0$. We notice $y = 1$ is a solution, by as Ihf points out in his comment, when we divide this expression by $y-1$, we don't always get a solvable polynomial out of it. There's not, in fact, always a nice expression for this in terms of $a$. Certain values of $a$ give you a third root other than $1$ and $0$ when $a < 0$, but there's no promising those will give closed forms. –  Nov 17 '20 at 11:05
  • the second root is $x=0$. – user2661923 Nov 17 '20 at 11:50

1 Answers1

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Beside the trivial $x=0$ and $x=1$, the third solution of the equation $$\sqrt{x} - 2x + x^a=0$$ is close to $\frac 14$ as soon as $a >3$.

So, we can have approximations using Taylor series. For example $$0=4^{-a}+\left(4^{1-a} a-1\right) \left(x-\frac{1}{4}\right)-2^{-2 a} \left(-8 a^2+8 a+4^{a}\right) \left(x-\frac{1}{4}\right)^2+O\left(\left(x-\frac{1}{4}\right)^3\right)$$ which is a quadratic in $\left(x-\frac{1}{4}\right)$.

Using only the first term (this would be equivalent to the first iterate of Newton method $$x_1=\frac{1}{4}+\frac{1}{4^a-4 a}$$ is quite decent. For $a=3$, it would give $x=\frac 7{26}=0.269231$ while the "exact" solution is $0.269143$.

But we can continue with series reversion and have $$x_2=\frac 14+\frac{1}{4^a-4 a}+\frac{8 (a-1) a-4^a}{\left(4^a-4 a\right)^3}+\cdots$$

For $a=3$, this gives $x=\frac{2365}{8788}=0.269117$

Another one which is not bad $$x_3=\frac 14+\frac{4^a-4 a}{8 a \left(a-4^a+1\right)+4^a+4^{2a}}$$

For the case where $a$ is close to $1.5^+$, the third solution is close to $1$ but, using the same method, it can be approximated as $$x=1-\frac{4(2a-3)}{4 (a-1) a-1}-\frac{8 (2 a-3)^3 \left(4 a^2-6 a-1\right)}{3 (4 (a-1) a-1)^3}-$$ $$\frac{4 (2 a-3)^3(320 a^6-1728 a^5+3440 a^4-2784 a^3+452 a^2+300 a+27 )}{9 (4 (a-1) a-1)^5}$$

For example, for $a=2$, the above formula gives $x=\frac{6639}{16807}=0.395014$ while the exact solution is $x=\frac{3-\sqrt{5}}{2} =0.381966$.

Claude Leibovici
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