Suppose u₀ = 10 and u_n+1 = 0.8u_n + 6 for n = 0, 1, 2, … . Start with 10, multiply by 0.8, then add 6. The next values are 14, 17.2 and 19.76.

The subscript records the step; it is not multiplication. u₃ means the value after three updates from u₀. Students who start counting at u₁ must adjust their labels consistently.

Did you know? The recurrence rule alone does not specify this sequence. With the same rule but u₀ = 40, the next value is 38. The starting condition matters because it determines which journey through the rule you follow.

Imagine a container in a school demonstration. At each step, 20% of its contents are removed and then 6 units are added. If u_n measures the contents before the next update, the model gives 0.8u_n + 6.

This is an invented example, not a measurement of Punggol Waterway. A real water model would need evidence about flows, rainfall, units and changing conditions. The purpose here is to make the order of the recurrence understandable.

Reverse the actions and the formula changes: adding 6 before retaining 80% gives 0.8(u_n + 6) = 0.8u_n + 4.8. Write the verbal process before calculating. It protects you from attaching the right numbers to the wrong model.

A fixed point L satisfies L = 0.8L + 6. Rearranging gives 0.2L = 6, so L = 30. Beginning at 30 means every later value is also 30.

Finding a fixed point does not, by itself, prove that every starting value approaches it. To examine that, track the difference from 30. Subtract 30 from both sides of the recurrence to get u_n+1 − 30 = 0.8(u_n − 30).

Each update multiplies the difference by 0.8. Its magnitude therefore shrinks. This is the reason for approach to 30 in this model; “the terms look closer” is an observation, while the difference equation explains it.

Since the initial difference is 10 − 30 = −20, repeated multiplication gives u_n − 30 = −20(0.8)ⁿ. Thus u_n = 30 − 20(0.8)ⁿ.

Check n = 0: the formula gives 10. Check n = 1: it gives 14. Finally substitute into the update rule: 0.8[30 − 20(0.8)ⁿ] + 6 = 30 − 20(0.8)ⁿ⁺¹. Both the starting condition and the recurrence agree.

For n = 5 the value is 23.4464. It remains below 30. The formula approaches 30 as n increases; it does not reach 30 after finitely many steps from this starting value. Approximate equality and exact equality need different wording.

For a rule u_n+1 = au_n + b with a ≠ 1, the fixed point is L = b/(1 − a). The difference obeys u_n+1 − L = a(u_n − L).

When |a| < 1, the difference shrinks. If a is negative in that range, its sign alternates, so the sequence moves from one side of the fixed point to the other. When |a| > 1, a nonzero starting difference grows instead. Starting exactly at L remains fixed.

Keep the boundary cases separate. When a = 1 and b ≠ 0, each step adds b and there is no fixed point. When a = −1, a nonzero difference generally alternates without shrinking. Conditions turn a useful formula into a trustworthy claim.

Try v₀ = 5 and v_n+1 = 0.5v_n + 4. The fixed point is 8; the direct formula is v_n = 8 − 3(0.5)ⁿ. The first update gives 6.5, matching the formula at n = 1.

Explain why the sequence approaches 8 without saying only “because of the formula”. Track the difference from 8 and show that it halves. Then change the starting value to 11 and predict the direction of approach.

A three-student session can compare a verbal model, a table and a difference equation, with each student rebuilding all three independently afterwards. Parents can ask “What changes each step, and what stays the same?” That question opens the connection between a repeated procedure and a mathematical system.