A synthetic membership program retains half its members each period and gains 10 new members. If xₙ is the count at the beginning of period n, its idealized update is xₙ₊₁=xₙ/2+10. This deterministic recurrence can describe expected or aggregate counts; fractional outputs would need interpretation if individual membership is the target.
Calculate several periods from x₀=0. Predict whether the sequence grows forever or approaches a finite level.
Predict
A fixed point satisfies x=x/2+10, giving 20. Does this algebra alone prove that every starting value approaches 20? Subtract the fixed point from the recurrence to find the missing dynamical argument.
Experiment
Let eₙ=xₙ−20. Then eₙ₊₁=eₙ/2, so eₙ=2^(−n)e₀.
Period n
Stock xₙ from zero
Error from 20
0
0
−20
1
10
−10
2
15
−5
3
35/2
−5/2
Change the multiplier from 1/2 to −1/2 while keeping an appropriate fixed point. Errors now alternate in sign but shrink. Change it to 3/2 and errors grow. The magnitude controls attraction; the sign controls alternation.
Understand
For xₙ₊₁=rxₙ+b with r≠1, the fixed point is x⋆=b/(1−r). Subtraction gives xₙ=x⋆+rⁿ(x₀−x⋆). This proves global attraction on the real line when |r| is below 1. At r=1, a nonzero b gives linear drift; when b=0 every point is fixed. At r=−1, deviations alternate without decay. Mathematical stability does not automatically imply that negative states or negative retention factors are physically meaningful.
A recurrence derived from a repeated process differs from a numerical approximation to an ODE. For continuous decay x′=−kx, exact sampling every h units gives multiplier e^(−kh). Forward Euler instead gives 1−kh. The latter can become negative or have magnitude above 1 for large h, even though the exact model stays positive and decays. This is numerical failure, not evidence that the continuous system oscillates.
For a genuinely weekly retention model, changing the period requires re-deriving or estimating the corresponding retention law. One cannot keep the same coefficient while relabeling weeks as days. If retention varies with stock or time, the affine formula becomes an assumption to test, not a universal law.
Residuals against time and stock can reveal departures from constant retention. A pattern may suggest capacity limits, seasonality or omitted arrivals. Fit revisions using development observations and preserve a separate evaluation period; an improved fit on the same observations does not independently validate the new mechanism.
Apply
Derive a recurrence for a repeated resource process, solve its fixed point and transient, and state its admissible parameter range. Compare two starting states analytically. Then explain whether your recurrence is a physical update, an expectation model or a numerical approximation, since each interpretation requires different validation evidence.
Continue
Continuous compartments retain the balance logic while allowing transfers throughout time. Conservation and positivity can often be proved before any numerical trajectory is drawn.
Apply · explain · revise
Practice and reasoning
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For x next=3x/4+6 and x=8, find the next value.
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
Find the fixed point of x next=3x/4+6.
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
Solve x next=rx+b for r≠1.
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Classify fixed-point stability for the affine recurrence by r.
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Explain the difference between a weekly retention law and Euler applied to x′=−kx.
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Design a test of the constant-retention assumption.
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Reflect and continue
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