A Function That Points Back at Itself

Solving a functional equation by substitution

October 19, 2026 · John Peach

A Functional Equation

I saw a problem recently that is a function defined from itself, called a functional equation: A functional equation is an equation whose unknown is a whole function rather than a number. The goal is to find every function that makes the equation true.

for all $x \neq 0$, $x \neq 1$. The problem asked for the value of $f(2).$ How do you solve something like this? We solved a similar problem in Circular Equations.

Let’s start by solving (1) for $f(x)$,

The solution can be found in a few iterations of (2). First, substitute $1 - 1/x$ for $x$ everywhere,

Simplifying, we find

Substitute (4) into (2) to get:

Apply the substitution trick again,

Substitute this into (5):

so

and

From (9) we see that $f(x)$ is undefined at $x=0$ and $x=1$. We can solve for the value of $f$ at $x = 2$,

Graphical Illustration

Plotting the solution in Desmos shows the discontinuities at $x=0$ and $x=1$, and by hovering over $x=2$ on the plot we see that the solution is indeed $f(2) = 2$. Now you know — when a function is defined in terms of itself, follow the inner map until it returns to x. The number of steps tells you how many equations you need.

Why it works

Why does this method work? Suppose $g(x) = 1 - 1/x$. Then,

and

which brings us back to the start The map $x \mapsto 1 - 1/x$ is a Möbius transformation: a function of the form $(ax+b)/(cx+d)$. Composing it with itself three times gives back $x$, so mathematicians say it has order 3. . For $x = 2,$ the orbit of 2 for $g(x)$ is $2 \rightarrow \frac{1}{2} \rightarrow -1 \rightarrow 2$, so

Let $S = f(2) + f(\tfrac12) + f(-1)$ be the sum of all three unknown values. Adding the three equations counts each value twice, so $2S = 16 + 4 - 8 = 12$, which gives $S = 6$.

To isolate $f(2)$, subtract the two values you don’t want. Their sum is exactly the second equation, $f(\tfrac12) + f(-1) = 4$:

Could some other function also satisfy (1)? No, and the three-step cycle shows why. Write $a = f(x)$, $b = f\left(1 - \frac{1}{x}\right)$ and $c = f\left(\frac{1}{1-x}\right)$ for the three unknown values. Applying (1) at $x$, at $1 - 1/x$, and at $1/(1-x)$ gives three linear equations:

Any function that satisfies (1) must satisfy all three, so its values $a$, $b$ and $c$ must solve this system In linear-algebra terms, the system (10) has a unique solution because its coefficient matrix has a nonzero determinant (here it equals 2). If (1) had a minus sign instead, $f(x) - f(1 - 1/x) = 8x$, the determinant would be 0, and the system would have either no solution or infinitely many. . Adding the three equations gives $2(a + b + c) = 8x - 8 + \frac{8}{x} + \frac{8}{1-x}$, which fixes the total $a + b + c$. Subtracting the middle equation, $b + c$, from that total leaves exactly one possible value for $a$, and that value is (9). Because this works for every allowed $x$, there is only one function that solves (1), and the substitutions didn’t just find a solution: they found the solution.

Experiments

Try these with pencil and paper or in Desmos:

1. Any right-hand side. Replace $8x$ with any function $h(x)$. The same three equations give

With $h(x) = 8x$ this reproduces (9) exactly.

2. Flip the sign. Try $f(x) - f(1 - 1/x) = 8x$. The determinant becomes 0, and a solution exists only if $h(x) + h(g(x)) + h(g(g(x))) = 0$. For $h = 8x$ that sum is $8(x^3 - 3x + 1)/(x(x-1)) \neq 0$, so there is no solution.
3. Other cycling maps. $x \mapsto 1 - x$ and $x \mapsto 1/x$ undo themselves in two steps, so $f(x) + f(1/x) = \dots$ problems need only two equations. Look for maps that cycle in four steps.

Software

• Desmos - Desmos is an advanced graphing calculator implemented as a web application and a mobile application. In addition to graphing both equations and inequalities, it also features lists, plots, regressions, interactive variables, graph restriction, simultaneous graphing, piece wise function graphing, polar function graphing, two types of graphing grids – among other computational features commonly found in a programmable calculator.

Image credits

• Hero: Emerald Green Geometric Print, Etsy.
• Function plot: Desmos.