Welcome to the World of Recurrence Relations!
In your H2 Mathematics journey, you’ve encountered sequences like Arithmetic and Geometric Progressions. In Further Mathematics (9649), we take this a step further. Think of a Recurrence Relation as a mathematical "recipe": if you know what happened yesterday, the formula tells you exactly what will happen today.
Recurrence relations are incredibly useful in the real world—from predicting how a population of rabbits grows to calculating how much interest you'll earn on your bank savings. Let’s dive in!
1. What is a Recurrence Relation?
A recurrence relation is an equation that defines a term in a sequence (\(u_n\)) using its previous terms (like \(u_{n-1}\) or \(u_{n-2}\)).
Example: \(u_n = 2u_{n-1} + 3\).
This means "to get the current term, double the previous term and add three." To find the actual numbers in the sequence, you just need a starting point, like \(u_0 = 1\).
Did you know? The famous Fibonacci sequence (\(1, 1, 2, 3, 5, 8...\)) is a recurrence relation where each number is the sum of the two before it: \(u_n = u_{n-1} + u_{n-2}\).
Quick Review:
• Order: If a formula uses only one previous term (\(u_{n-1}\)), it is First Order. If it uses two (\(u_{n-1}\) and \(u_{n-2}\)), it is Second Order.
• Linear: All terms of \(u\) are to the power of 1 (no \(u_n^2\) or \(\sqrt{u_n}\)).
• Homogeneous: If there is no extra "dangling" constant or function of \(n\) added at the end.
2. First-Order Linear Recurrence Relations
The standard form we look at is:
\(u_n = au_{n-1} + b\)
A. Homogeneous Case (\(b = 0\))
If \(b = 0\), the relation is \(u_n = au_{n-1}\). This is just a Geometric Progression (GP)!
The solution is simply:
\(u_n = u_0(a)^n\)
B. Non-Homogeneous Case (\(b \neq 0\))
When there is a constant \(b\), we use a two-step "Search and Combine" strategy. The general solution is:
Total Solution = Complementary Function (CF) + Particular Solution (PS)
Step 1: Find the CF. Ignore the \(b\) and solve \(u_n = au_{n-1}\). This gives \(A(a^n)\).
Step 2: Find the PS. Since \(b\) is a constant, we guess that the particular solution is also a constant, let's call it \(k\).
Set \(k = ak + b\) and solve for \(k\).
\(k = \frac{b}{1-a}\) (This only works if \(a \neq 1\)).
Step 3: Combine.
\(u_n = A(a^n) + \frac{b}{1-a}\)
(You can find the value of A using the initial condition \(u_0\)).
Common Mistake: If \(a = 1\), the formula \(u_n = u_{n-1} + b\) is just an Arithmetic Progression. The solution is \(u_n = u_0 + nb\). Don't try to use the fraction formula for this!
Key Takeaway: For \(u_n = au_{n-1} + b\), the sequence behaves like a GP that is "shifted" by a constant value.
3. Behaviour of a Sequence
Before solving, we often want to know what happens in the long run (as \(n \to \infty\)).
• Convergent: If \(|a| < 1\), the sequence settles down to a single value (a limit).
• Divergent: If \(|a| > 1\), the sequence explodes to infinity or negative infinity.
• Oscillating: If \(a\) is negative, the terms might jump between positive and negative values.
Analogy: Imagine a bouncing ball. If each bounce is 0.8 times the height of the previous one (\(a=0.8\)), the ball eventually stops (converges to 0). If the ball magically gained energy and bounced 1.2 times higher each time (\(a=1.2\)), it would eventually hit the moon (diverge)!
4. Second-Order Linear Homogeneous Relations
These look like this:
\(au_n + bu_{n-1} + cu_{n-2} = 0\)
To solve these, we use the Characteristic Equation. Think of this as translating the sequence into a quadratic equation that we already know how to solve!
The Trick: Replace \(u_n\) with \(m^2\), \(u_{n-1}\) with \(m\), and \(u_{n-2}\) with \(1\).
Equation: \(am^2 + bm + c = 0\)
Case 1: Two Distinct Real Roots (\(m_1\) and \(m_2\))
The general solution is:
\(u_n = A(m_1)^n + B(m_2)^n\)
Case 2: One Repeated Real Root (\(m\))
The general solution is:
\(u_n = (A + Bn)m^n\)
Note: For Further Maths, you might also encounter complex roots, which lead to solutions involving sine and cosine, but the process of solving the quadratic remains the same!
Step-by-step Process:
1. Write down the characteristic quadratic equation.
2. Solve for the roots \(m\).
3. Pick the correct general solution format based on the roots.
4. Use provided values (like \(u_0\) and \(u_1\)) to solve for the constants \(A\) and \(B\).
5. Working with Substitutions
Sometimes, the examiners will give you a scary-looking relation that doesn't fit the rules above. Don't panic! They will usually suggest a substitution to turn it into something familiar.
Example: If you see \(u_n^2 = 3u_{n-1}^2\), it's not linear because of the square.
The hint might say: "Let \(v_n = u_n^2\)".
Suddenly, the equation becomes \(v_n = 3v_{n-1}\), which is just a simple GP! Solve for \(v_n\) first, then go back to \(u_n\).
Key Takeaway: If a recurrence relation looks "weird," trust the substitution. It's there to simplify your life.
6. Modelling with Recurrence Relations
The syllabus requires you to apply these to real-life scenarios. The most common is Financial Interest or Population Change.
Example (Loan Repayment):
Suppose you owe \$1000. Every month, the bank adds 1% interest, and you pay back \$50.
Let \(u_n\) be the amount owed after \(n\) months.
The relation is: \(u_n = 1.01u_{n-1} - 50\).
(The 1.01 represents 100% + 1% interest; the -50 is your payment).
Common Mistakes to Avoid:
• Start points: Check if the sequence starts at \(n=0\) or \(n=1\). This changes your final values for \(A\) and \(B\).
• Arithmetic errors: Solving the characteristic equation incorrectly is the most common way to lose marks. Double-check your quadratic factors!
• Notation: Be careful with \(u_{n-1}\) vs \(u_{n+1}\). They represent the same relationship, just shifted.
Summary Checklist
- Can I identify the order of the relation?
- Do I know the condition for a sequence to converge (\(|a| < 1\))?
- Can I solve \(u_n = au_{n-1} + b\) using the CF + PS method?
- Can I set up and solve the characteristic equation for 2nd order relations?
- Am I comfortable applying substitutions to simplify complex relations?
Don't worry if this seems tricky at first—recurrence relations are all about practice and recognizing patterns. Once you master the "recipes," you'll find them much easier to handle!