Iterative processes with repeated deposits · Higher
導入済み| English | 意味 |
|---|---|
| recurrence relation/rɪˈkʌrəns rɪˈleɪʃn/ | 再帰関係式 |
The deposit comes after interest, so the order changes the balance.
- A saver receives 10% interest and then adds 50 yuan each year.
- This lesson studies recurrence relation 递推关系: A rule obtaining a later value from the preceding value.
The method
- Write an update rule and an initial value.
- Apply the operations in their stated order, using the previous output as the next input.
- A recurrence B next=1.1B+50 differs from 1.1(B+50).
- Tables can locate a first whole-period threshold; verify both the preceding and crossing values.
- Iteration can model growth, decay or other repeated processes, not just solve equations.
Why does B next=1.1B+50 differ from B next=1.1(B+50)?
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
From B₀=1000, the first gives B₁=1150 and the second B₁=1.1×1050=1155. The five-yuan gap then grows, so operation order must be preserved exactly.
A worked case
With B₀=1000 and Bₙ₊₁=1.1Bₙ+50, B₁=1150, B₂=1315 and B₃=1496.5. The balance first exceeds 1400 after three years because 1315≤1400<1496.5. If the deposit preceded interest, B₁=1.1×1050=1155, five yuan greater. A decay-and-top-up rule V next=0.8V+20, starting at 200, gives 180 then 164. A fixed point satisfies V=0.8V+20, hence V=100; values above 100 decrease toward it. A fixed point is a value preserved by the update, not a claim that every finite step reaches it exactly.
Iterative processes with repeated deposits
Write an update rule and an initial value
Classify the worked-case statements, then explain the units or invariant that justifies each decision.
Find B₁ for B₀=1000 and B next=1.1B+50.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
1.1×1000+50=1150.
Where it goes wrong
- Preserve operation order and use the updated value each time.
- Do not round early or confuse the initial value with the first updated value.
- A continuous fractional-period estimate does not answer a whole-period question by itself.
Not true: Depositing before interest gives the same result as depositing after interest. A deposit made before interest earns interest for that period, so the balance is larger.
Find B₂ using the same update.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
1.1×1150+50=1315.
Depositing before interest gives the same result as depositing after interest.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
A deposit made before interest earns interest for that period, so the balance is larger.
Applying it
- AQA R16 Higher extends growth/decay to general iterative processes.
- State the model assumptions, initial value, recurrence and threshold interpretation.
- Use substitution to check a proposed fixed point without calculus.
Find the fixed point of V next=0.8V+20.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
Solve V=0.8V+20: 0.2V=20.
Match each recurrence concept to its meaning.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
A recurrence needs an update rule and an initial value; a fixed point satisfies V=g(V), and whole-period answers need bracketing values, not a fractional estimate.
In your course
- AQA GCSE Mathematics (8300), Higher tier, specification section 3.3.
- Paper 1 is non-calculator: keep fractions, surds and π exact and show each step. On calculator papers, still write the expression before its value.
- Recurrence relation: A rule obtaining a later value from the preceding value.
- Write an update rule and an initial value.
- Watch for: preserve operation order and use the updated value each time.