Solving the power equation $A^X = \frac{(1+X)}{(1-X)}$

Question

I want to solve the following power equation (get $X$ value):

$$A^X = \frac{(1+X)}{(1-X)},$$ where $X\neq 0$, $A\in {\mathbb R}$ (a real number)

$$A \geq 0 , \quad A \leq 1$$

I think $X$ should be a complex number.

Answer 1

Observations. Let us consider only when $a > 0$. We have three observations:

1. The function $f(x) = \frac{1-x}{1+x}$ is positive if and only if $|x| < 1$.
2. $f(-x) = 1/f(x)$. So the set of solution of $a^{x} = f(x)$ must be symmetric around the origin.
3. Numerical observation shows that the set of solutions $(a, x)$ of $a^{x} = f(x)$ is

Solution. If we let $b = -\log a$, then it suffices to consider the equation

$$e^{bx} = f(-x) = \frac{1+x}{1-x} \quad \text{on } (0, 1). \tag{*}$$

We divide into several cases.

• If $b \leq 0$ then $x \mapsto e^{bx}$ is decreasing while $x \mapsto (1+x)/(1-x)$ is increasing. So there is no non-trivial solution in this case.

• If $b \leq 2$, then $b^{n}/n! \leq 2 $ for $n \geq 1$ with strict inequality for $n \geq 3$. Thus we have $$ e^{bx} = \sum_{n=0}^{\infty} \frac{b^{n}}{n!} x^{n} < (1 + 2x + 2x^{2} + \cdots) = \frac{1+x}{1-x}. $$ Thus again, there is no non-trivial solution in this case.

• If $b > 2$, then $g(x) = e^{bx} - (1+x)/(1-x)$ satisfies $ g'(0) = b - 2 > 0 $ so $g(x) > 0$ near $x = 0$. But clearly $g(1^{-}) = -\infty$. This implies that there is at least one non-trivial solution to $\text{(*)}$. (A harder part is to prove that this solution is unique, and I will consider it later.)

So the conclusion is as follows:

Proposition. When $e^{-2} \leq a$, there is no non-trivial solution of $a^{x} = f(x)$. If $0 < a < e^{-2}$, then there is a unique positive real $x_{0} \in (0, 1)$ such that $\pm x_{0}$ are the only non-trivial solution of $a^{x} = f(x)$.

Answer 2

Definitely a transcendental equation, which means you're not going to be able to solve it algebraically. You're right that there are solutions in $\mathbb{C}$, but as Abel pointed there are solutions in $\mathbb R$ as well.

I'd also like to agree that there is a real solution when $A\geq 0$. Think about the limiting behavior of each side as $x$ approaches $-\infty$, $-1$ from each side, and $\infty$. In particular there must be a solution when $x\geq-1$ because of this limiting behavior and continuity.

If $A<0$ all bets are of because $A^X$ is continuous nowhere.

Answer 3

You want to solve for $X$, but it may be more useful to first solve for $A$. Note that $\dfrac{1-X}{1+X} > 0$ if $-1 < X < 1$, $< 0$ if $X < -1$ or $X > 1$. If $-1 < X < 0$ or $0 < X < 1$, you could have $$ A = \left(\dfrac{1-X}{1+X}\right)^{1/X}$$ with $A$ real. Plotting $A$ as a function of $X$, we get this:

You can check that $\lim_{X \to 0} ((1-X)/(1+X))^{1/X} = e^{-2}$ while $\lim_{X \to 1-} ((1-X)/(1+X))^{1/X} = \lim_{X \to -1+} ((1-X)/(1+X))^{1/X} = 0$.

Also note the symmetry:

$$ \left( \dfrac{1+X}{1-X}\right)^{1/(-X)} = \left(\dfrac{1-X}{1+X}\right)^{1/X}$$

For $0 < A < e^{-2}$, it appears we have two solutions $X$, one the negative of the other. I don't believe they can be expressed in closed form.

If $X < 1$ or $X > 1$, it's trickier to get real values of $X$ for which (some branch of) $\left(\dfrac{1-X}{1+X}\right)^{1/X}$ is real. Namely you need $X$ to be rational with an odd numerator. For example, $X = 5/3$ is OK, corresponding to $\left(\dfrac{1-X}{1+X}\right)^{1/X} = (-1/4)^{3/5} =-4^{-3/5}$. On the other hand, $X = 4/3$ does not work, because $(-1/7)^{3/4}$ is not real.

Answer 4

Taking into account what has been answered and limiting to the case $A \leq \frac{1}{e^2}$ and $0 \leq x \leq 1$, it seems interesting to notice that $$A= \left(\dfrac{1-X}{1+X}\right)^{1/X}$$ can be quite well approximated by $$A \simeq \frac{1-x^{8/3}}{e^2}$$ The exponent has been empirically adjusted using a least square fit.

So, using a Newton method, the initial guess could be selected as $x_0=\left(1-A e^2 \right)^{3/8}$.

For illustration purposes, let us try using $A=0.08$; this gives $x_0=0.715067$ and, from here, Newton iterates are $ 0.722035$, $0.721973$ which is the solution for six significant figures.