Welcome to Further Sequences and Series!
In your standard A Level Maths, you’ve likely met sequences where you just add or multiply by a number. In Further Mathematics, we take this a step further. We explore recurrence relations—where the next term depends on the previous one—and learn how to find a "master formula" (a closed form) so we can jump straight to any term in the sequence without doing all the steps in between!
Don't worry if this seems a bit abstract at first. Think of it like a video game: the recurrence relation is the rule for "leveling up," and the closed form is the cheat code that lets you jump straight to Level 100.
1. First Order Recurrence Relations
A recurrence relation is a way of defining a sequence where each term is defined as a function of the previous terms. A first order relation means the next term \( u_{n+1} \) only depends on the current term \( u_n \).
The standard form you will see is:
\( u_{n+1} + f(n)u_n = g(n) \)
Breaking down the notation:
1. \( u_n \): The "current" term.
2. \( u_{n+1} \): The "next" term.
3. \( f(n) \): A function (often just a constant number) multiplying the current term.
4. \( g(n) \): An extra bit added on (like a constant or a function of \( n \)).
Analogy: The Savings Account
Imagine you have \( £100 \) and every month the bank gives you \( 5\% \) interest, and you add another \( £10 \).
The rule is: Next Month = (1.05 × This Month) + 10.
In math speak: \( u_{n+1} = 1.05u_n + 10 \). This is a first order recurrence relation!
Quick Review: A recurrence relation tells you how to get from one step to the next. "First order" means you only need to look back one step.
2. Finding the "Closed Form" (The Master Formula)
To solve a recurrence relation like \( u_{n+1} - au_n = g(n) \), we need to find a formula for \( u_n \) that doesn't depend on \( u_{n-1} \). We do this in three main steps. This is very similar to how you might solve differential equations!
Step A: The Complementary Function (CF)
First, we ignore the \( g(n) \) part and solve the "homogenous" version: \( u_{n+1} - au_n = 0 \).
This is based on an auxiliary equation. For a first order relation, the solution is always in the form:
\( u_n = A(a)^n \)
(Where \( A \) is a constant we find later, and \( a \) is the number multiplying \( u_n \)).
Step B: The Particular Solution (PS)
Now we look at the \( g(n) \) part we ignored. We "guess" a form for the solution based on what \( g(n) \) looks like:
1. If \( g(n) \) is a constant (e.g., \( 8 \)), try \( u_n = \lambda \).
2. If \( g(n) \) is linear (e.g., \( 3n + 2 \)), try \( u_n = \lambda n + \mu \).
3. If \( g(n) \) is polynomial (e.g., \( n^2 \)), try \( u_n = \lambda n^2 + \mu n + \nu \).
Substitute your guess into the original recurrence relation to find the values of \( \lambda, \mu, \) etc.
Step C: The General Solution (GS)
The full "Master Formula" is simply:
General Solution = Complementary Function + Particular Solution
\( u_n = A(a)^n + \text{PS} \)
Example Walkthrough: Solve \( u_{n+1} - 5u_n = 8 \) given \( u_1 = 1 \).
1. CF: Solve \( u_{n+1} - 5u_n = 0 \). The CF is \( A(5)^n \).
2. PS: Since \( g(n) = 8 \) (a constant), try \( u_n = \lambda \).
Sub into the rule: \( \lambda - 5\lambda = 8 \).
\( -4\lambda = 8 \), so \( \lambda = -2 \). The PS is \( -2 \).
3. GS: \( u_n = A(5)^n - 2 \).
4. Find A: Use \( u_1 = 1 \).
\( 1 = A(5)^1 - 2 \)
\( 3 = 5A \), so \( A = 0.6 \).
Final Formula: \( u_n = 0.6(5)^n - 2 \).
Key Takeaway: Solve the "easy" version first (CF), guess the "extra" part (PS), and add them together!
3. Real-World Modeling
Recurrence relations aren't just for exams; they are used to model real-life changes over time.
Common Applications:
- Population Growth: \( u_{n+1} = Ru_n - M \) (where \( R \) is growth rate and \( M \) is migration/death).
- Financial Interest: Borrowing money and paying it back in installments.
- Medicine: How much of a drug remains in the bloodstream if a new dose is taken every 6 hours.
Did you know? Computer scientists use these relations to figure out how fast an algorithm will run. If a task doubles in size each step, that's a recurrence relation!
4. Proof by Induction
Once you have found a closed-form formula, the exam will often ask you to prove it is correct using Mathematical Induction. Think of Induction like falling dominoes: if the first one falls, and each one knocks down the next, they all must fall!
The 4-Step Process:
1. The Basis: Show the formula works for \( n = 1 \).
2. The Assumption: Assume the formula is true for \( n = k \).
3. The Inductive Step: Use your assumption and the original recurrence relation to show it must then be true for \( n = k + 1 \).
4. The Conclusion: Write the standard "closing sentence" (e.g., "Since it is true for \( n=1 \) and \( n=k \implies n=k+1 \), it is true for all \( n \in \mathbb{Z}^+ \)").
Common Mistake to Avoid: In the Inductive step, students often forget to use the original rule given in the question. Always start with: \( u_{k+1} = \text{Rule}(u_k) \), then plug in your assumed formula for \( u_k \).
Quick Review Box:
- Basis: Test \( n=1 \).
- Assumption: Let \( n=k \).
- Step: Find \( u_{k+1} \) using the recurrence rule.
- Goal: Make it look like the closed form with \( (k+1) \) inside it!
Summary Checklist
Can you:
- Identify the components of a first order recurrence relation?
- Find the Auxiliary Equation and Complementary Function?
- Choose a correct "guess" for the Particular Solution?
- Combine them into a General Solution and solve for constants?
- Prove the result using Mathematical Induction?
Don't worry if this seems tricky at first! The more you practice the "Guess and Substitute" method for the Particular Solution, the more natural it will feel. You're building powerful tools to predict the future of a sequence!