Circles generated by three-fold iterations $f(x)=\frac{1}{1-x}$

Question

I came across a weird property of the function $f(x)=\dfrac{1}{(1-x)}$

Observe the following:

$$f(x) = \frac{1}{(1-x)}, \quad\quad f^2(x) = f(f(x)) = \frac{(x-1)}{x}, \quad\quad f^3(x) = f(f(f(x))) = x$$ ultimately implying that $f^2(x)=f^{-1}(x)$.

(Mini question: Do you know of any other functions $g(x)$ where $g \circ (g \circ g(x)) = g^3(x)=x$ aside from $f(x)$ and aside from the trivial case where $g(x)=x$? I was pretty shocked when I noticed this pattern with $f(x)$.)

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Anyway, notice for every $x$, there is a set of triplets generated by repeatedly applying the function $f(x)$.

Specifically $\langle x\rangle =\{x,f(x),f^{-1}(x)\}=\{x,\frac{1}{(1-x)},\frac{(x-1)}{x}\}$

For an illustrative example let $x=2$, so then $\langle 2\rangle=\{2, -1, \frac{1}{2}\}$. See now that this can be thought of as 3 points on the graph of the function $f(x)$, where

Point $A$: $x \mapsto f(x)$

Point $B$: $f(x) \mapsto f^{2}(x)=f^{-1}(x)$

Point $C$: $f^{-1}(x) \mapsto x$

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Explicitly, still using $x=2$ as the example:

Point $A$: $(x, f(x)) = (2,-1)$

Point $B$: $(f(x), f^{-1}(x)) = (-1,\frac{1}{2})$

Point $C$: $(f^{-1}(x),x) = (\frac{1}{2},2)$

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OK so now my question!

Since 3 points uniquely define a circle, I'd like to know if we can derive a closed-form function $r(x)$ that calculates the radius of circle $R$, where circle $R$ is the circle uniquely defined by the 3 points $A$, $B$ and $C$ generated by $\langle x\rangle$.

Continuing the example where $x=2$, circle $R$ has center at Point $R=(\frac{3}{4},\frac{1}{4})$ (i.e. the circumcenter of points $A$, $B$ and $C$). The radius of circle $R$ is then simply:

$$|\overline{AR}|=\sqrt{{\left(2-\frac{3}{4}\right)}^2+{\left(-1-\frac{1}{4}\right)}^2}= \frac{5\sqrt{2}}{4}.$$

So evaluating $r(x)$ at $x=2$ gives us $r(2)=\dfrac{5\sqrt{2}}{4}\approx1.76777$.

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Another cool example to consider is $x=\phi$, where $\phi=\dfrac{1+\sqrt{5}}{2}\approx1.61803$ (the Golden Ratio). Some cool characteristics that make $\phi$ unique among all numbers are:

$$\phi-1=\frac{1}{\phi}\quad\text{and}\quad \phi+1=\phi^2$$

You can calculate this on your own, but applying $f(x)$ on $x=\phi$ repeatedly results in $\langle\phi\rangle=\{\phi,-\phi,\frac{1}{\phi^{2}}\}$.

With the help of Wolfram Alpha, I was able to calculate $r(\phi)\approx1.93649$

(Circumcenter: https://tinyurl.com/y59trfn5 | Radius: https://tinyurl.com/y6jxs9sn)

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Calculating the circumcenter seems to be the biggest issue, but maybe there's a cleaner way with the help of linear algebra? I was reading that there's a way to calculate the formula of a circle using matrices and determinants, but that seemed too complex for this. Maybe circles and triangles aren't the way to approach this at all -- I'd be happy to take suggestions and hear your thoughts!

Just some last conceptual thoughts...

1) $r(x)$ should always be positive (i.e. there is no $x$ where $r(x)$ is $0$ or negative), and therefore somewhere hit some positive minimum value for $r(x)$ (assuming/implying that $r(x)$ is smooth and differentiable on the interval $x \in (-\infty,1)\cup(1,+\infty)$).

2) $\lim\limits_{x \to 1^-}r(x)=+\infty$ and $\lim\limits_{x \to 1^+}r(x)=+\infty$

3) $\lim\limits_{x \to -\infty}r(x)=+\infty$ and $\lim\limits_{x \to +\infty}r(x)=+\infty$

4) $r(x)$ is NOT symmetric around $x=1$. Just as a quick check, $r(3)\approx2.12459$ and $r(-1)\approx1.76777$

5) $r(x)$ is actual VERY NOISY as a function since for any 1 value of $r(x)$, there are at least 3 unique variables that result in that value (i.e. all $x \in \langle x\rangle$)(e.g. $r(2)=r(-1)=r(\frac{1}{2})\approx1.76777$)

That last point makes me feel there's no true closed-form function for $r(x)$. Regardless, I'd be really curious to find out what the minimum radius is... (placing \$1 on $r(x)$ for $x \in \langle\frac{\pi^2}{4}\rangle$!)

Answer 1

The equation of a circle through $A$, $B$, $C$ is given by: $$\left|\begin{array}{cccc} x^2+y^2 & x & y & 1 \\ A_x^2+A_y^2 & A_x & A_y & 1 \\ B_x^2+B_y^2 & B_x & B_y & 1 \\ C_x^2+C_y^2 & C_x & C_y & 1 \end{array}\right| = 0 \tag{1}$$ For $$A = \left(p,\frac{1}{1-p}\right)\qquad B = \left(\frac{1}{1-p},\frac{p-1}{p}\right) \qquad C = \left(\frac{p-1}{p},p\right)$$ this becomes (with the help of a computer algebra system) $$\left(x^2+y^2\right)p(p-1) - x\left(1- 3 p + p^3\right) -y\left(1 - 2 p - p^2 + p^3\right) = -1 - p + 4 p^2 - p^3 \tag{2}$$ Completing the square yields $$\left(x -\frac{p^3-3p+1}{2p(p-1)} \right)^2+\left(y - \frac{p^3-p^2-2p+1}{2p(p-1)} \right)^2 = \frac{\left(p^2 + 1\right) \left(p^2-2p+2\right) \left(2p^2-2p+1\right)}{4 p^2(p-1)^2} \tag{$\star$}$$

Thus, the circumcenter $K$ and radius $r$ are

$$\begin{align} K &= \left(\frac{p^3-3p+1}{2p(p-1)}, \frac{p^3-p^2-2p+1}{2p(p-1)}\right)\\[6pt] r &= \frac{\sqrt{\left(p^2+1\right) \left(\,(p-1)^2+1\,\right) \left(\,p^2+(p-1)^2\,\right)\;}}{2 |p(p-1)|} \end{align}$$

Interestingly, we can rewrite $r$ as $$r = \frac12 \;\sqrt{\left(\frac{p}{1} + \frac{1}{p}\right) \left(\frac{p - 1}{1} + \frac{1}{p - 1}\right) \left(\frac{p}{p - 1} + \frac{p - 1}{p}\right)}$$

Animation courtesy of @Jyrki:

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By way of minimization ... We have $$\frac{d(r^2)}{dp} = \frac{(p-2) (p+1) (2p-1) (p^2-p+1)^2}{2 p^3(p-1)^3}$$ so that the non-extraneous critical points occur at $p=2,-1,\frac12$, which correspond exactly to OP's $\langle 2\rangle$ triangle, as can be seen in the animation.

Answer 2

Hint:

Let us derive the formula for the circumradius of three points.

WLOG, one of the points is the origin (otherwise, translate all three points). The equation of the circle must be of the form

$$x^2+y^2-2x_cx-2y_cy=0.$$

When we plug the coordinates of the two other points, we get a system of two equations in two unknowns giving the center. Finally, as the radius is the distance of the center to the origin,

$$r=\frac{\sqrt{\begin{vmatrix}x_1^2+y_1^2&2y_1\\x_2^2+y_2^2&2y_2\end{vmatrix}^2+\begin{vmatrix}2x_1&x_1^2+y_1^2\\2x_2&x_2^2+y_2^2\end{vmatrix}^2}} {|\begin{vmatrix}2x_1&2y_1\\2x_2&2y_2\end{vmatrix}|}.$$

Now you can plug the coordinates of the three points, not forgetting the translation.

Answer 3

Given a point $x\in\Bbb{R}$, with $x\notin\{0,1\}$, the corresponding three points are $$P_1(x)=\left(x,\tfrac{1}{1-x}\right), \qquad P_2(x)=\left(\tfrac{1}{1-x},\tfrac{x-1}{x}\right), \qquad P_3(x)=\left(\tfrac{x-1}{x},x\right).$$ Then we want to find the unique circle passing through these three points. Note that these three points are not collinear for any choice of $x\in\Bbb{R}$ with $x\notin\{0,1\}$, so such a circle does indeed exist. Now we want to find $a,b,r\in\Bbb{R}$ such that $$||P_i(x)-(a,b)||^2=r^2,$$ for $i\in\{1,2,3\}$. This yields the following three equations: \begin{eqnarray*} \left(x-a\right)^2+\left(\tfrac{1}{1-x}-b\right)^2&=&r^2,\\ \left(\tfrac{1}{1-x}-a\right)^2+\left(\tfrac{x-1}{x}-b\right)^2&=&r^2,\\ \left(\tfrac{x-1}{x}-a\right)^2+\left(x-b\right)^2&=&r^2. \end{eqnarray*} Clearing all denominators then yields the following system of equations: \begin{eqnarray*} (1-x)^2(x-a)^2+(1-(1-x)b)^2&=&(1-x)^2r^2,\\ x^2(1-(1-x)a)^2+(1-x)^2((x-1)-xb)^2&=&x^2(1-x)^2r^2,\\ ((x-1)-xa)^2+x^2(x-b)^2&=&x^2r^2. \end{eqnarray*} We can rearrange these to make them look more like polynomials in $a$, $b$ and $r$: \begin{eqnarray*} (1-x)^2&a^2-2x(1-x)^2&a+(1-x)^2&b^2-2(1-x)&b+x^2(1-x)^2+1&=(1-x)^2&r^2,\\ x^2(1-x)^2&a^2-2x^2(1-x)&a+x^2(1-x)^2&b^2+2x(1-x)^3&b+(1-x)^4+x^2&=x^2(1-x)^2&r^2,\\ x^2&a^2+2x(1-x)&a+x^2&b^2-2x^3&b+x^4+(1-x)^2&=x^2&r^2. \end{eqnarray*} Multiplying the first equation by $x^2$ and subtracting it from the second shows that $$-2x^2(1-x)(x^2-x+1)a+2x(1-x)(x^2-x+1)b-(x-1)^2(x^4-x^2+2x-1)=0.$$ Note that this equation is linear in both $a$ and $b$. In the same way we can multiply the third equation by $(1-x)^2$ and subtract it from the second to find that $$-2x(1-x)(x^2-x+1)a-2x(1-x)^2(x^2-x+1)b-x^2(x^4-2x^3+x^2-1)=0,$$ which is again linear in $a$ and $b$. Now we have a system of two linear equations in two variables, which we can solve by linear algebra. Dividing by $x\neq0$ and $1-x\neq0$ yields the system: \begin{eqnarray*} -2x^2(x^2-x+1)&a&+2x(x^2-x+1)&b&=(x-1)(x^4-x^2+2x-1),\\ -2(1-x)(x^2-x+1)&a&-2(1-x)^2(x^2-x+1)&b&=x(x^4-2x^3+x^2-1). \end{eqnarray*} Because also $x^2-x+1\neq0$ for all real numbers $x$, we can also write this as \begin{eqnarray*} -x&a&+&b&=\frac{x-1}{x}\frac{x^4-x^2+2x-1}{2(x^2-x+1)},\\ -&a&-(1-x)&b&=\frac{x}{1-x}\frac{x^4-2x^3+x^2-1}{2(x^2-x+1)}. \end{eqnarray*} It follows that \begin{eqnarray*} a&=&\frac{x^5-x^4-2x^3+4x^2-4x+1}{2x(x-1)(x^2-x+1)}=\frac{x^3-3x+1}{2x(x-1)},\\ b&=&\frac{x^5-2x^4+2x^2-3x-1}{2x(x-1)(x^2-x+1)}=\frac{x^3-x^2-2x+1}{2x(x-1)}, \end{eqnarray*} and plugging this back into the original equations shows that $$r^2=\frac{2x^6-6x^5+11x^4-12x^3+11x^2-6x+2}{4x^2(x-1)^2} =\frac{(x^2+1)(x^2-2x+2)(2x^2-2x+1)}{(2x(x-1))^2}.$$ It is interesting to note that the square of the radius can also be expressed as \begin{eqnarray*} r(x)^2&=&\frac14\left|f^{0}(x)+\frac{1}{f^{0}(x)}\right| \left|f^{1}(x)+\frac{1}{f^{1}(x)}\right| \left|f^{2}(x)+\frac{1}{f^{2}(x)}\right|,\\ &=&\frac{1}{4|f^{0}(x)f^{1}(x)f^{2}(x)|} \left(f^0(x)^2+1\right) \left(f^1(x)^2+1\right) \left(f^2(x)^2+1\right)\\ &=&\frac14\prod_{i=0}^2||(f^i(x),1)||^2, \end{eqnarray*} and hence we see that $$r(x)=\frac12\prod_{i=0}^2\left|\left|(f^i(x),1)\right|\right|.$$