🎯 Functional Equations — Substitutions & Symmetry

Essential for AMC12 problems involving unknown functions and their properties.

🎯 Key Ideas

Functional Equations — Equations involving unknown functions, often solved by substitution or exploiting symmetry.

Substitution Strategy — Replace variables with specific values or expressions to reveal function properties.

Symmetry — Look for patterns like $f(-x)$, $f(x+a)$, or cyclic relationships.

📊 Essential Techniques

Common Substitutions

Symmetry Patterns

🎯 Micro-Examples

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

Solution:

1. Substitute $x = 0$: $f(y) = f(0) + f(y)$ → $f(0) = 0$
2. Substitute $x = -y$: $f(0) = f(x) + f(-x)$ → $f(-x) = -f(x)$ (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. For reals: $f(x) = cx$ for some constant $c$ (assuming continuity)

Example 2: Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that $f(xy) = f(x) + f(y)$ for all $x,y > 0$.

Solution:

1. Substitute $x = y = 1$: $f(1) = f(1) + f(1)$ → $f(1) = 0$
2. Substitute $y = \frac{1}{x}$: $f(1) = f(x) + f(\frac{1}{x})$ → $f(\frac{1}{x}) = -f(x)$
3. Substitute $y = x$: $f(x^2) = 2f(x)$
4. For rationals: $f(x^r) = rf(x)$ for rational $r$
5. For reals: $f(x) = c\log x$ for some constant $c$ (assuming continuity)

Example 3: Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that $f(x^2) = f(x)^2$ for all $x \in \mathbb{R}$.

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. If $f(0) = 0$: $f(x) = 0$ for all $x$ (constant zero function)
5. If $f(0) = 1$: $f(x) = 1$ for all $x$ (constant one function)
6. Answer: $f(x) = 0$ or $f(x) = 1$ for all $x$

⚠️ Common Traps & Fixes

Trap: Assuming continuity without justification

• Fix: Only assume continuity if explicitly stated or if it’s reasonable
• Example: $f(x+y) = f(x) + f(y)$ has discontinuous solutions without continuity

Trap: Forgetting to check all cases

• Fix: Consider all possible values of variables
• Example: $f(x^2) = f(x)^2$ has different cases for $x = 0, 1, -1$

Trap: Incorrect substitution

• Fix: Be careful with variable replacement
• Example: In $f(x+y) = f(x) + f(y)$, substituting $x = y$ gives $f(2x) = 2f(x)$, not $f(x^2) = 2f(x)$

🎯 AMC-Style Worked Example

Problem: Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that $f(x+1) = f(x) + 1$ and $f(x^2) = f(x)^2$ for all $x \in \mathbb{R}$.

Solution:

1. From first equation: $f(x+1) = f(x) + 1$ for all $x$
2. Substitute $x = 0$: $f(1) = f(0) + 1$
3. From second equation: $f(x^2) = f(x)^2$ for all $x$
4. Substitute $x = 0$: $f(0) = f(0)^2$ → $f(0) = 0$ or $f(0) = 1$
5. If $f(0) = 0$: $f(1) = 1$
6. If $f(0) = 1$: $f(1) = 2$
7. Check $f(1) = 1$: $f(1^2) = f(1) = 1$ and $f(1)^2 = 1^2 = 1$ ✓
8. Check $f(1) = 2$: $f(1^2) = f(1) = 2$ and $f(1)^2 = 2^2 = 4$ ✗
9. So $f(0) = 0$ and $f(1) = 1$
10. By first equation: $f(n) = n$ for all integers $n$
11. By second equation: $f(x^2) = f(x)^2$ for all $x$
12. For $x = \frac{1}{2}$: $f(\frac{1}{4}) = f(\frac{1}{2})^2$
13. Answer: $f(x) = x$ for all $x$ (assuming continuity)

Key insight: Functional equations often require multiple substitutions and careful case analysis.

🔗 Related Topics

• Functions — Functional equations involve unknown functions
• Symmetry — Look for patterns and relationships
• Substitution — Replace variables with specific values
• Continuity — Often needed for complete solutions

📝 Practice Checklist

• Master basic substitution techniques
• Practice symmetry recognition
• Learn common functional equation types
• Practice case analysis
• Understand continuity requirements
• Work on speed and accuracy

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