How to Solve Cyclic Functional Equations (2 Elegant Methods)

Functional equations can often feel abstract or intimidating. Unlike standard algebraic equations where you solve for a specific variable $x$, here you are solving for the entire rule of the function $f(x)$.

In this article, we’ll break down a classic functional equation that frequently appears in advanced algebra competitions and university entrance exams:

$$\text{Find } f(x) \text{ given that: } f(x) + f\left(\frac{1}{1-x}\right) = 1 + x$$

We will cover two distinct strategies:

1. The System of Equations Method (Cyclic Substitution) — The standard, most foolproof approach.
2. Structural Decomposition (Undetermined Coefficients) — The symmetric, structural shortcut.

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The Hidden Pattern: Period-3 Cyclic Permutation

Before diving into algebraic manipulations, notice the nested transformation:

If we let $T(x) = \frac{1}{1-x}$, let’s see what happens when we compose $T(x)$ repeatedly:

1. $T_1(x) = \frac{1}{1-x}$
2. $T_2(x) = T(T(x)) = \frac{1}{1 — \frac{1}{1-x}} = \frac{x-1}{x}$
3. $T_3(x) = T(T(T(x))) = \frac{1}{1 — \frac{x-1}{x}} = x$

The input sequence returns to $x$ after 3 iterations! This cycle ($x \to \frac{1}{1-x} \to \frac{x-1}{x} \to x$) is the secret key to solving this problem.

(Note: For these expressions to remain defined, we assume $x \neq 0$ and $x \neq 1$.)

Method 1: The System of Equations Method (Cyclic Substitution)

Step 1: Write Down Equation (1)

Our baseline given relation:

$$f(x) + f\left(\frac{1}{1-x}\right) = 1 + x \quad \text{ — — (1)}$$

Step 2: First Substitution ($x \mapsto \frac{1}{1-x}$)

Replace every instance of $x$ in Equation (1) with $\frac{1}{1-x}$:

$$f\left(\frac{1}{1-x}\right) + f\left(\frac{1}{1 — \frac{1}{1-x}}\right) = 1 + \frac{1}{1-x}$$

Simplifying the inner fraction gives $\frac{x-1}{x}$, yielding Equation (2):

$$f\left(\frac{1}{1-x}\right) + f\left(\frac{x-1}{x}\right) = 1 + \frac{1}{1-x} \quad \text{ — — (2)}$$

Step 3: Second Substitution ($x \mapsto \frac{x-1}{x}$)

Now, replace every instance of $x$ in Equation (1) with $\frac{x-1}{x}$:

$$f\left(\frac{x-1}{x}\right) + f\left(\frac{1}{1 — \frac{x-1}{x}}\right) = 1 + \frac{x-1}{x}$$

Simplifying the inner fraction brings us back to $x$, yielding Equation (3):

$$f\left(\frac{x-1}{x}\right) + f(x) = 1 + \frac{x-1}{x} \quad \text{ — — (3)}$$

Step 4: Solve the $3 \times 3$ Linear System

We now have a system of three linear equations with three unknown terms: $f(x)$, $f\left(\frac{1}{1-x}\right)$, and $f\left(\frac{x-1}{x}\right)$:

$$\begin{cases} f(x) + f\left(\frac{1}{1-x}\right) = 1 + x & \text{(1)} \\ f\left(\frac{1}{1-x}\right) + f\left(\frac{x-1}{x}\right) = 1 + \frac{1}{1-x} & \text{(2)} \\ f\left(\frac{x-1}{x}\right) + f(x) = 1 + \frac{x-1}{x} & \text{(3)} \end{cases}$$

To isolate $f(x)$, compute Equation (1) + Equation (3) — Equation (2):

$$2f(x) = (1 + x) + \left(1 + \frac{x-1}{x}\right) — \left(1 + \frac{1}{1-x}\right)$$

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$$2f(x) = 1 + x + \frac{x-1}{x} — \frac{1}{1-x}$$

$$2f(x) = 1 + x + \frac{x-1}{x} + \frac{1}{x-1}$$

Step 5: Algebraic Simplification

Combine all terms over the common denominator $x(x-1)$:

$$2f(x) = \frac{(1+x) \cdot x(x-1) + (x-1)² + x}{x(x-1)}$$

$$2f(x) = \frac{x(x² — 1) + x² — 2x + 1 + x}{x(x-1)}$$

$$2f(x) = \frac{x³ — x + x² — x + 1}{x(x-1)} = \frac{x³ + x² — 2x + 1}{x(x-1)}$$

Dividing both sides by $2$:

$$f(x) = \frac{x³ + x² — 2x + 1}{2x(x-1)} \quad (x \neq 0, x \neq 1)$$

Method 2: Structural Decomposition (Undetermined Coefficients)

Step 1: Assume a Linear Combination Model

Since the transformation components cyclicly permute between $x$, $\frac{1}{1-x}$, and $\frac{x-1}{x}$, we can assume $f(x)$ takes the form:

$$f(x) = A \cdot x + B \cdot \frac{1}{1-x} + C \cdot \frac{x-1}{x} + D$$

Then, substituting $\frac{1}{1-x}$ for $x$:

$$f\left(\frac{1}{1-x}\right) = A \cdot \frac{1}{1-x} + B \cdot \frac{x-1}{x} + C \cdot x + D$$

Step 2: Set Up the Coefficient System

Substitute $f(x)$ and $f\left(\frac{1}{1-x}\right)$ into $f(x) + f\left(\frac{1}{1-x}\right) = 1 + x$:

$$(A + C)x + (A + B)\frac{1}{1-x} + (B + C)\frac{x-1}{x} + 2D = x + 1$$

Matching terms on both sides gives:

• Coefficient of $x$: $A + C = 1$
• Coefficient of $\frac{1}{1-x}$: $A + B = 0 \implies B = -A$
• Coefficient of $\frac{x-1}{x}$: $B + C = 0 \implies C = -B = A$
• Constant term: $2D = 1 \implies D = \frac{1}{2}$

Since $A + C = 1$ and $C = A$, we immediately find:

$$A = \frac{1}{2}, \quad B = -\frac{1}{2}, \quad C = \frac{1}{2}, \quad D = \frac{1}{2}$$

Step 3: Reconstruct and Simplify

Plugging these coefficients back into our model for $f(x)$:

$$f(x) = \frac{1}{2}x — \frac{1}{2}\left(\frac{1}{1-x}\right) + \frac{1}{2}\left(\frac{x-1}{x}\right) + \frac{1}{2}$$

$$f(x) = \frac{1}{2} \left[ x + 1 + \frac{1}{x-1} + \frac{x-1}{x} \right]$$

Combining into a single rational function yields the exact same closed-form solution:

$$f(x) = \frac{x³ + x² — 2x + 1}{2x(x-1)} \quad (x \neq 0, x \neq 1)$$

Key Takeaway

When facing a functional equation of the form $a f(x) + b f(g(x)) = h(x)$, always check if $g(x)$ forms a finite periodic orbit ($g^k(x) = x$). If it does, cyclic substitution allows you to set up a linear system of equations to isolate $f(x)$ cleanly.

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