Solve $x^2-2x+1=\log_2( \frac{x+1}{x^2+1})$

Question

Solve in $\mathbb R$ the following equation $$x^2-2x+1=\log_2 (\frac{x+1}{x^2+1})$$

First of all from the existence conditions of the logarithm, we have $x > -1$. Analyzing $x^2 - 2x - 1$ , we get that for $x \in (-1,1]$ , the function $ x \mapsto x^2 - 2x - 1$ is strictly decreasing and $ x^2 - 2x - 1 \in [-2,2)$ and for $x \in [1,\infty)$ the function $ x \mapsto x^2 - 2x - 1$ is strictly increasing and $ x^2 - 2x - 1 \in [-2, \infty)$.I don't know if it helps, but at least we know the monotony of the left member. Now it seems quite complicated what I have to do next. I don't know how many solutions there are, but I assume there are 2. We'll probably have to rely on the convexity/concavity of the functions(IDEA : as the coefficient of $x^2$ is strictly positive, and the discriminant of the equation of degree 2 is greater than 0, the function $ x \mapsto x^2 - 2x - 1$ is strictly convex , maybe we can show that the right member is strictly concave so we have at most 2 solutions), but I don't really know how.If you have any idea or a solution, I'm here to listen. Thanks!

Answer 1

Let's start with your observation: The LHS is strictly convex while the RHS is strictly concave - hence there can be at max two intersections!

Let's note that $x^2-2x+1=(x-1)^2$. Identify here that at $x=1$, these two intersect.

Note that at $x=0$, $LHS=1$ and $RHS=0$, so $LHS>RHS$.

At $x=1$ where both are equal, the derivative of LHS is zero, while the derivative of the RHS is negative. Which means, there is some point $0<p<1$ where $LHS<RHS$

Because these functions are "nice", there must be at least one intersection between 0 and $p$! And we got 2 intersections :)

Answer 2

Let $x\in\Bbb{R}$ be a real number satisfying $$x^2-2x+1=\log_2\left(\frac{x+1}{x^2+1}\right).\tag{1}$$ The left hand side equals $(x-1)^2$ and so it is nonnegative. Then the right hand side is nonnegative, and so $$\frac{x+1}{x^2+1}\geq1,$$ from which it follows that $x\geq x^2$, or equivalently, that $0\leq x\leq1$. On this interval the left hand side of $(1)$ is strictly decreasing and convex. For the right hand side, note that $$\frac{d}{dx}\log_2\left(\frac{x+1}{x^2+1}\right)=-\frac{1}{\ln{2}}\frac{x^2+2x-1}{(x+1)(x^2+1)}.$$ Clearly its only zero in the interval $[0,1]$ is at $x=-1+\sqrt{2}$, and so the right hand side of $(1)$ is strictly increasing for $x\leq-1+\sqrt{2}$, and strictly decreasing for $x\geq-1+\sqrt{2}$. Moreover $$\frac{d^2}{dx^2}\log_2\left(\frac{x+1}{x^2+1}\right)=\frac{1}{\ln{2}}\frac{x^4+4x^3-2x^2-4x-3}{(x+1)^2(x^2+1)^2},$$ which is easily verified to be strictly negative on the interval $[0,1]$, meaning that the right hand side of $(1)$ is concave. This means there are at most two solutions to $(1)$.

Next note that at the maximum of the right hand side, at $x=-1+\sqrt{2}$, we have \begin{eqnarray} x^2-2x+1&=&6-4\sqrt{2}&\approx&0.34315\ldots\\ \log_2\left(\frac{x+1}{x^2+1}\right)&=&\log_2\left(\frac{1+\sqrt{2}}{2}\right)&\approx&0.27155\ldots \end{eqnarray} and so there are no solutions with $x\leq-1+\sqrt{2}$.

Of course $x=1$ is a solution, and at $x=1$ we have $\frac{d}{dx}(x^2-2x+1)\big\vert_{x=1}=(2x-2)\big\vert_{x=1}=0$ and $$\frac{d}{dx}\log_2\left(\frac{x+1}{x^2+1}\right)\Bigg\vert_{x=1}=-\frac{1}{\ln{2}}\frac{x^2+2x-1}{(x+1)(x^2+1)}\Bigg\vert_{x=1}=-\frac{1}{2\ln{2}}<0.$$ This shows that for some $x<1$ we have $$x^2-2x+1<\log_2\left(\frac{x+1}{x^2+1}\right),$$ and so there must be a second solution with $-1+\sqrt{2}<x<1$. It does not seem to have a nice closed form.

Answer 3

These are Geogebra graphs of the respective expressions. The points of intersection are the solutions.