Increasing, decreasing and periodic recurrences
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| periodic sequence/ˌpɪərɪˈɒdɪk ˈsiːkwəns/ | dãy tuần hoàn |
What decides its behaviour?
- The same update rule can increase one starting value and decrease another.
- This lesson studies periodic sequence 周期数列: A sequence whose terms repeat after a fixed positive number of steps.
The method
- 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.
For u next=u/2+3 starting at 1, why does the sequence increase toward 6 without passing it?
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
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.
A worked case
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.
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
u₂=1/2+3=7/2=3.5.
Where it goes wrong
- 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.
Not true: Every recurrence with a fixed value must converge to it from every starting value. A fixed point can be unstable; iterations from many starting values move away from it.
Find u₃ for that sequence.
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
u₃=(7/2)/2+3=19/4=4.75.
A fixed point can be unstable.
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
True. A fixed point can be unstable; iterations from many starting values move away from it.
Applying it
- 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.
Find the smallest period of 2,4,2,4,… .
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
Two updates return to 2, while one gives 4; the smallest period is 2.
Match each recurrence behaviour to its test.
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
Behaviour claims need the difference or bound argument; a period is the SMALLEST positive repeat step.
In your course
- AQA A-level Mathematics (7357), specification section D.
- Every paper allows a calculator, but method marks still need the expression or equation written before its value.
- Periodic sequence: A sequence whose terms repeat after a fixed positive number of steps.
- A recurrence needs an initial term as well as a rule.
- Watch for: early numerical terms suggest behaviour but do not prove it for every n.