Increasing, decreasing and periodic recurrences
導入済み| English | 意味 |
|---|---|
| periodic sequence/ˌpɪərɪˈɒdɪk ˈsiːkwəns/ | 周期数列 |
The same update rule can increase one starting value and decrease another. What decides its behaviour?
- The same update rule can increase one starting value and decrease another. What decides its behaviour?
- This lesson studies periodic sequence 周期数列: A sequence whose terms repeat after a fixed positive number of steps.
Choose the mathematical structure
- A recurrence needs an initial term as well as a rule. Compute each new term from the previous one, preserving exact values where useful. Compare u_(n+1)−u_n and check bounds before claiming increasing or decreasing behaviour. A periodic sequence repeats after a fixed number of steps; its smallest positive repeating step is its period.
- State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.
For u next=u/2+3 starting at 1, why does the sequence increase toward 6 without passing it?
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
While u_n<6, the update adds half the remaining gap, so terms rise and stay below 6. Starting above 6 the same rule decreases toward it: the fixed point attracts from both sides.
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.
For u₁=1 and u_(n+1)=u_n/2+3, the first terms are 1,7/2,19/4,43/8. The fixed value solves L=L/2+3, giving L=6. If u_n<6, the next term remains below 6 and u_(n+1)−u_n=(6−u_n)/2>0. Thus this sequence increases toward 6. Starting instead at 11 gives 11,17/2,29/4,…, decreasing while staying above 6. The exact form is u_n=6+(u₁−6)/2^(n−1). Separately, v₁=2 and v_(n+1)=6−v_n gives 2,4,2,4,… with period 2 because applying the rule twice returns the original term. Starting that second rule at 3 gives a constant sequence with smallest period 1.
Increasing, decreasing and periodic recurrences
A recurrence needs an initial term as well as a rule
Check whether a sequence or sum argument preserves all its defining information.
For u₁=1 and u_(n+1)=u_n/2+3, find u₂.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
u₂=1/2+3=7/2=3.5.
Test a tempting shortcut
- Early numerical terms suggest behaviour but do not prove it for every n. A fixed-point equation alone does not prove convergence: the update must keep approaching it. Alternating values need not be periodic; (−2)^n alternates in sign while its magnitude grows. State whether “increasing” means strictly increasing or merely nondecreasing.
- When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.
Every recurrence with a fixed value must converge to it from every starting value. This claim is false. Explain which definition or assumption it violates.
Find u₃ for that sequence.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
u₃=(7/2)/2+3=19/4=4.75.
Every recurrence with a fixed value must converge to it from every starting value.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
Early numerical terms suggest behaviour but do not prove it for every n. A fixed-point equation alone does not prove convergence: the update must keep approaching it. Alternating values need not be periodic; (−2)^n alternates in sign while its magnitude grows. State whether “increasing” means strictly increasing or merely nondecreasing.
Interpret a new situation
- Verify the explicit form by substituting it into the recurrence and checking the initial term. The distance from 6 halves at every update, explaining convergence in this example. A different initial term changes the direction or gives a constant sequence, even though the recurrence formula stays the same.
- A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.
Find the smallest period of 2,4,2,4,… .
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
Two updates return to 2, while one gives 4; the smallest period is 2.
Match each recurrence behaviour to its test.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
Behaviour claims need the difference or bound argument; a period is the SMALLEST positive repeat step.
Use this in your course
- 7357 · A-level · D. 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 sequence whose terms repeat after a fixed positive number of steps. Choose the relationship, show the method, check its assumptions and interpret the result.