Iterative processes with repeated deposits · Higher
| English | 中文 | Pinyin |
|---|---|---|
| recurrence relation/rɪˈkʌrəns rɪˈleɪʃn/ | 递推关系 | dì tuī guān xì |
A saver receives 10% interest and then adds 50 yuan each year. The deposit comes after interest, so the order changes the balance.
- A saver receives 10% interest and then adds 50 yuan each year. The deposit comes after interest, so the order changes the balance.
- This lesson studies recurrence relation 递推关系: A rule obtaining a later value from the preceding value.
Choose the mathematical structure
- 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.
- State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.
Which description correctly defines recurrence relation?
A rule obtaining a later value from the preceding value.
Work through a checked case
- Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.
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.
1.1×1000+50=1150.
Test a tempting shortcut
- 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.
- When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.
Depositing before interest always gives the same result as depositing after interest. This claim is false. Explain which definition or assumption it violates.
Find B₂ using the same update.
1.1×1150+50=1315.
Depositing before interest always gives the same result as depositing after interest.
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.
Interpret a new situation
- 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.
- A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.
Find the fixed point of V next=0.8V+20.
Solve V=0.8V+20: 0.2V=20.
Match each part of a complete solution to its purpose.
An assumption justifies the model; a check tests the result; interpretation connects it to the question.
Use this in your course
- 8300 · Higher · 3.3. Match the target tier and specification before assigning extensions.
- Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.
A rule obtaining a later value from the preceding value. Choose the relationship, show the method, check its assumptions and interpret the result.