🔄 Functional Equations (Light)

Basic functional equation techniques for AMC 12. Focus on substitutions, invariants, and common patterns.

🎯 Key Ideas

Functional Equation: Equation where the unknown is a function, not a number.

Substitution: Replace variables with specific values to find function properties.

Invariants: Properties that remain constant under certain transformations.

🔄 Basic Techniques

Substitution Method

  1. Substitute specific values for variables
  2. Look for patterns in the resulting equations
  3. Use symmetry or special properties

Example: Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that $f(x+y) = f(x) + f(y)$

Solution:

  1. Substitute $x = 0$: $f(y) = f(0) + f(y)$ → $f(0) = 0$
  2. Substitute $y = -x$: $f(0) = f(x) + f(-x)$ → $f(-x) = -f(x)$ (odd function)
  3. Substitute $y = x$: $f(2x) = 2f(x)$
  4. By induction: $f(nx) = nf(x)$ for all integers $n$
  5. For rationals: $f(\frac{p}{q}x) = \frac{p}{q}f(x)$
  6. Assuming continuity: $f(x) = cx$ for some constant $c$

📊 Common Patterns

Cauchy’s Functional Equation

$$f(x+y) = f(x) + f(y)$$

Solution: $f(x) = cx$ (assuming continuity)

Multiplicative Functional Equation

$$f(xy) = f(x)f(y)$$

Solution: $f(x) = x^k$ for some constant $k$ (assuming continuity and $f \neq 0$)

Example: Find all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ such that $f(xy) = f(x)f(y)$

Solution:

  1. Substitute $x = 1$: $f(y) = f(1)f(y)$ → $f(1) = 1$ (assuming $f \neq 0$)
  2. Substitute $y = \frac{1}{x}$: $f(1) = f(x)f(\frac{1}{x})$ → $f(\frac{1}{x}) = \frac{1}{f(x)}$
  3. For positive integers: $f(n) = f(1 \cdot 1 \cdot \ldots \cdot 1) = f(1)^n = 1^n = 1$
  4. For rationals: $f(\frac{p}{q}) = f(p)^{1/q} = 1^{1/q} = 1$
  5. Assuming continuity: $f(x) = 1$ for all $x > 0$

🔄 Invariants

Definition

An invariant is a property that remains unchanged under certain operations.

Example: Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that $f(x+1) = f(x) + 1$

Solution:

  1. Define $g(x) = f(x) - x$
  2. Substitute: $g(x+1) = f(x+1) - (x+1) = f(x) + 1 - x - 1 = f(x) - x = g(x)$
  3. So $g$ is periodic with period 1: $g(x+1) = g(x)$
  4. Therefore: $f(x) = x + g(x)$ where $g$ is periodic with period 1

🎯 AMC-Style Worked Example

Problem: Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that $f(x^2) = (f(x))^2$ for all $x$.

Solution:

  1. Substitute $x = 0$: $f(0) = (f(0))^2$ → $f(0) = 0$ or $f(0) = 1$
  2. Substitute $x = 1$: $f(1) = (f(1))^2$ → $f(1) = 0$ or $f(1) = 1$
  3. Substitute $x = -1$: $f(1) = (f(-1))^2$ → $f(-1) = \pm\sqrt{f(1)}$
  4. Case 1: If $f(0) = 0$ and $f(1) = 0$:
    • For $x > 0$: $f(x^2) = (f(x))^2 = 0$ → $f(x) = 0$ for all $x > 0$
    • For $x < 0$: $f(x^2) = (f(x))^2 = 0$ → $f(x) = 0$ for all $x < 0$
    • Solution: $f(x) = 0$ for all $x$
  5. Case 2: If $f(0) = 0$ and $f(1) = 1$:
    • For $x > 0$: $f(x^2) = (f(x))^2$ → $f(x) = x^k$ for some $k$
    • Check: $(x^k)^2 = (x^2)^k$ → $x^{2k} = x^{2k}$ ✓
    • For $x < 0$: $f(x^2) = (f(x))^2$ → $f(x) = |x|^k$ or $f(x) = -|x|^k$
    • Solution: $f(x) = x^k$ for $x \geq 0$ and $f(x) = |x|^k$ or $f(x) = -|x|^k$ for $x < 0$

Answer: $f(x) = 0$ or $f(x) = x^k$ (with appropriate domain restrictions)

🔍 Common Traps & Fixes

Trap: Assuming continuity without justification

Fix: Only assume continuity if the problem states it or if it’s reasonable in context.

Trap: Forgetting to check all cases

Fix: When you get $f(a) = 0$ or $f(a) = 1$, consider both possibilities.

Trap: Not verifying solutions

Fix: Always substitute your proposed solution back into the original equation.

Trap: Ignoring domain restrictions

Fix: Pay attention to the domain of the function (e.g., $\mathbb{R}^+$ vs $\mathbb{R}$).

📋 Quick Reference

Common Functional Equations

  • $f(x+y) = f(x) + f(y)$ → $f(x) = cx$ (Cauchy’s equation)
  • $f(xy) = f(x)f(y)$ → $f(x) = x^k$ (multiplicative)
  • $f(x+1) = f(x) + 1$ → $f(x) = x + g(x)$ where $g$ is periodic

Basic Techniques

  1. Substitute specific values (0, 1, -1, etc.)
  2. Look for patterns and use induction
  3. Define auxiliary functions to simplify
  4. Use symmetry and special properties
  5. Verify solutions by substitution

Common Substitutions

  • $x = 0$: Often gives $f(0)$
  • $y = -x$: Often gives symmetry properties
  • $y = x$: Often gives doubling formulas
  • $y = 1$: Often gives $f(1)$

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