1) Introduction: Functional Equations with Calculus 📚
Functional equations involve finding functions that satisfy given relations, often requiring calculus techniques. This topic explores equations where derivatives, integrals, or limits play a role, common in Olympiad problems and advanced mathematics. We’ll develop methods to solve these equations systematically.
We’ll cover:
- Basic Functional Equations: Equations like \( f(x+y) = f(x) + f(y) \).
- Calculus-Based Equations: Involving derivatives or integrals.
- Techniques: Differentiation, substitution, and trial solutions.
Quick Recap: Differential equations involved functions and derivatives; now we tackle functional relations.
2) Basic Functional Equations 🎓
These equations define \( f \) based on its values.
Definition 29.1: Cauchy’s Functional Equation
\( f(x + y) = f(x) + f(y) \). For continuous \( f \), \( f(x) = kx \) if \( f(1) = k \).
Example 1: Linear Solution
Solve \( f(x + y) = f(x) + f(y) \) with \( f(1) = 2 \).
- Assume \( f(x) = kx \), then \( f(1) = k = 2 \).
- Check: \( f(x+y) = 2(x+y) = 2x + 2y = f(x) + f(y) \).
Answer: \( f(x) = 2x \).
Example 2: Non-Linear Case
Solve \( f(x + y) = f(x)f(y) \) with \( f(0) = 1 \).
- Try \( f(x) = e^{kx} \), \( f(0) = e^0 = 1 \).
- \( e^{k(x+y)} = e^{kx} e^{ky} \), holds for any \( k \).
Answer: \( f(x) = e^{kx} \).
3) Calculus-Based Functional Equations 📐
These involve derivatives or integrals.
Definition 29.2: Differential Functional Equation
E.g., \( f'(x) = f(x) \) leads to \( f(x) = Ce^x \).
Example 3: Differential Equation
Solve \( f'(x) = 2f(x) \) with \( f(0) = 1 \).
- Solution: \( f(x) = Ce^{2x} \), \( f(0) = C = 1 \).
- Answer: \( f(x) = e^{2x} \).
Answer: \( f(x) = e^{2x} \).
Example 4: Integral Equation
Solve \( f(x) = x + \int_0^x f(t) \, dt \).
- Differentiate: \( f'(x) = 1 + f(x) \).
- Solve: \( f'(x) - f(x) = 1 \), use integrating factor \( e^{-x} \).
- \( (f e^{-x})' = e^{-x} \), \( f e^{-x} = \int e^{-x} \, dx = -e^{-x} + C \).
- \( f = Ce^x - 1 \), \( f(0) = 0 \), \( C = 1 \).
Answer: \( f(x) = e^x - 1 \).
4) Advanced Techniques and Olympiad Problems 🔍
Olympiad problems often combine calculus and functional equations.
Definition 29.3: Substitution Method
Substitute \( f(x) = g(x) + h(x) \) to simplify.
Example 5: Substitution
Solve \( f(x + y) + f(x - y) = 2f(x)f(y) \).
- Try \( f(x) = a x^2 + b \), expand and equate.
- Leads to \( a = 0 \), \( f(x) = b \), or \( f(x) = \frac{1}{2}x^2 + c \) (non-linear case).
Answer: \( f(x) = c \) (constant) or \( f(x) = \frac{x^2}{2} + c \).
Example 6: Olympiad Challenge
Solve \( f(x) + f'(x) = x \) with \( f(0) = 0 \).
- \( f'(x) = x - f(x) \), homogeneous solution \( f_h = Ce^x \).
- Particular: \( f_p = Ax + B \), \( A = 1 \), \( B = 0 \).
- \( f = Ce^x + x - 1 \), \( f(0) = 0 \), \( C = 1 \).
Answer: \( f(x) = e^x + x - 1 \).
5) Applications and Limits 🔍
Functional equations model growth and constraints.
Definition 29.4: Limit Constraints
Use \( \lim_{x \to 0} \) or \( \lim_{x \to \infty} \) to find constants.
Example 7: Limit Application
Solve \( f(x + 1) - f(x) = x \) with \( \lim_{x \to \infty} f(x) = 0 \).
- Sum: \( f(n+1) - f(1) = \sum_{k=1}^n k = \frac{n(n+1)}{2} \).
- \( f(n+1) = f(1) + \frac{n(n+1)}{2} \).
- Limit suggests divergence unless \( f(1) \) adjusts, try \( f(x) = \frac{x(x-1)}{2} \).
Answer: \( f(x) = \frac{x(x-1)}{2} \).
Example 8: Real-World Model
Solve \( f'(x) = f(x^2) \) with \( f(0) = 1 \).
- Try \( f(x) = e^{kx} \), leads to \( k = 0 \) or complex \( k \).
- Series: \( f(x) = 1 + x + x^2 + \cdots \), but adjust for \( x^2 \).
Answer: \( f(x) = e^{x + x^2/2 + \cdots} \) (approximate).
6) Practice Questions 🎯
Fundamental Practice Questions 🌱
Instructions: Solve the functional equations. 📚
\( f(x + y) = f(x) + f(y) \), \( f(1) = 3 \)
\( f(x + y) = f(x)f(y) \), \( f(0) = 2 \)
\( f'(x) = f(x) \), \( f(0) = 1 \)
\( f(x) = x + \int_0^x f(t) \, dt \)
\( f(x + 2) = f(x) + 2 \), \( f(0) = 0 \)
\( f'(x) = 2f(x) + 1 \), \( f(0) = 0 \)
\( f(x + y) + f(x - y) = 2f(x)f(y) \)
\( f(x) + f'(x) = x^2 \), \( f(0) = 1 \)
\( f(x + 1) - f(x) = x^2 \), \( f(0) = 0 \)
\( f'(x) = f(2x) \), \( f(0) = 1 \)
\( f(x + y) = f(x) - y^2 \), \( f(0) = 0 \)
Challenging Practice Questions 🌟
Instructions: Solve these advanced functional equations. 🧠
Solve \( f(x + y) + f(x - y) = 2f(x)f(y) - 2xy \) with \( f(0) = 0 \).
Determine \( f(x) \) where \( f'(x) = f(x^3) + x \) and \( f(0) = 1 \).
Find \( f(x) \) satisfying \( f(x) = x + \int_0^x (f(t))^2 \, dt \) with \( f(0) = 0 \).
Solve \( f'(x) + f(x) = \int_0^x f(t) \, dt \) with \( f(0) = 1 \).
Analyze \( f(x + y) = f(x)f(y) - f(x+y) \) with \( f(0) = 1 \) and discuss its limit as \( x \to \infty \).
7) Summary & Cheat Sheet 📋
7.1) Basic Functional Equations
\( f(x+y) = f(x) + f(y) \) often yields \( f(x) = kx \).
7.2) Calculus-Based
Use derivatives or integrals to solve (e.g., \( f'(x) = f(x) \)).
7.3) Techniques
Substitution, differentiation, and limit constraints.
You’ve mastered functional equations! Next, we’ll explore vector calculus. 🎉