How to know some equation can be solved or not arithmetically?

Question

Problem

I was given the following equation to solve for x

$$ A = 2(x-1)^{-N} - B x^{-N} \sqrt{(x^{2N}-1)/(x^2 - 1)} $$

A, B, N is constant, N is positive integer, and $x>2$,

I want to solve the equation for x

But don't get hung up on it.

My solution

...was simply to use a calculator, which gave me a seemingly irrational number, which is understandable given the presence of pi.

But at the same time, I was wondering if this equation was solvable by typical algebraic methods, like substitutions yielding a quadratic equation for example.

But more generally, I started wondering how I could determine whether or not it is solvable algebraically.

Question

Do we have any "if such-and-such is not satisfied, go numerical" rules of thumb? Or do we just make a judgment call?

Answer 1

Set $n \in \mathbb{N}_0$, if the goal is to solve the equation:

$$ a = \frac{2}{(x-1)^n} - \frac{b}{x^n}\sqrt{\frac{x^{2n}-1}{x^2-1}} \tag{1} $$

• if $b = 0$, $(1)$ reduces to an elementary polynomial equation:

$$ (x-1)^n = \frac{2}{a} \quad \quad \Rightarrow \quad \quad x = 1 + \sqrt[n]{\frac{2}{a}} $$

• if $b \ne 0$, $(1)$ can still be reduced to a polynomial equation:

$$ \frac{\left(x^{2n}-1\right)\left(x-1\right)^{2n}}{x^2-1} = \frac{x^{2n}\left(2-a(x-1)^n\right)^2}{b^2} \tag{2} $$

but, unlike the previous case, we cannot avoid fixing the value of $n$. Furthermore, due to the squaring, spurious solutions may arise, so the solutions of $(2)$ necessarily will be verified by replacing them in $(1)$.

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As a first example, let us consider the case $a=1$, $b=1$, $n=1$ which leads to:

$$ \left(x^2 - 2x - 1\right)\left(x^2 - 4x + 1\right) = 0 $$

which is verified for $x = 1 \pm \sqrt{2}$ or $x = 2 \pm \sqrt{3}$, but only $x = 1 \pm \sqrt{2}$ also satisfy $(1)$.

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As a second example, let us consider the case $a=1$, $b=1$, $n=2$ which leads to:

$$ \underbrace{x^8 - 4x^7 + x^6 + 8x^5 - 6x^4 + 8x^3 - 7x^2 + 4x - 1}_{p(x)} = 0 $$

which being a non-special polynomial of degree greater than four doesn't allow to express the roots through radicals. So, once the eight first-attempt roots on the unit circle have been calculated:

$$ x_k = \cos(2\,k\,\pi/8) + \text{i}\,\sin(2\,k\,\pi/8) \quad \quad \text{with} \; k = 1,2,\dots,8 $$

it's sufficient to update them about ten times by applying the Aberth-Ehrlich method:

$$ x_k' = x_k - \left(\frac{p'(x_k)}{p(x_k)} - \sum_{j=1;\,j\ne k}^8 \frac{1}{x_k-x_j}\right)^{-1} $$

from which:

$$ \begin{aligned} & x_1 = 0.356214 + 0.443627\,\text{i}\,; \\ & x_2 = 0.356214 - 0.443627\,\text{i}\,; \\ & x_3 = -0.231134 + 0.805661\,\text{i}\,; \\ & x_4 = -1.64147 - 1.19851\cdot 10^{-94}\,\text{i}\,; \\ & x_5 = 2.79585 + 2.96462\cdot 10^{-20}\,\text{i}\,; \\ & x_6 = -0.231134 - 0.805661\,\text{i}\,; \\ & x_7 = 0.445743 - 2.06795\cdot 10^{-25}\,\text{i}\,; \\ & x_8 = 2.14971 + 3.06534\cdot 10^{-46}\,\text{i}\,; \\ \end{aligned} $$

among which only $x_3,x_6,x_7,x_8$ also satisfy $(1)$.

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On the other hand, if $a,b,n$ can only assume values such that $(1)$ is always verified for a real value close to $x = \overline{x}$, which is the only solution of interest, then all this work can be avoided by defining:

$$ f(x) := a - \frac{2}{(x-1)^n} + \frac{b}{x^n}\sqrt{\frac{x^{2n}-1}{x^2-1}} $$

and applying the Newton-Raphson method:

$$ x_0 = \overline{x}, \quad \quad x_i = x_{i-1} - \frac{f(x_{i-1})}{f'(x_{i-1})} \quad \quad \text{with} \; i=1,2,\dots $$

I hope I have given an idea of the reasoning that is usually done in these circumstances.

Answer 2

By rearranging and squaring we obtain $$ \frac{(2(x-1)^{-N}-A)^2}{B^2x^{-2N}}=\frac{x^{2N}-1}{x^2-1}=x^{2N-2}+x^{2N-4}+\cdots +x^2+1. $$ This gives a polynomial equation in $x$, which one can solve.