Explicit recurrence solutions, equilibria and population models

Expand a few updates to find the structure

From uₙ₊₁=ruₙ+c, successive substitution gives u₁=rs+c, u₂=r²s+c(r+1), and u₃=r³s+c(r²+r+1). The accumulated additions form a finite geometric sum. These first cases suggest a formula; they do not prove every index

For u₀=4 and uₙ₊₁=2uₙ−3, find u₃

Shift by the equilibrium when r is not one

Solve L=rL+c to get L=c/(1−r). Subtract L from both sides of the recurrence: uₙ₊₁−L=r(uₙ−L). The deviations form a geometric sequence with starting value s−L. Thus uₙ=L+(s−L)rⁿ

For uₙ₊₁=uₙ/2+3, calculate L.

If the same recurrence starts at u₀=4, what is C in uₙ=6+C(1/2)ⁿ?

Treat multiplier one separately

For r=1 the rule is arithmetic: uₙ=s+nc. If c=0, every starting value is an equilibrium. If c≠0, no equilibrium exists and the terms diverge with the sign of c. Do not use c/(1−r) at r=1.

For u₀=−5 and uₙ₊₁=uₙ+2, which explicit rule is valid?

Prove the explicit solution, ungraded

For r≠1 derive uₙ=L+(s−L)rⁿ by shifting the recurrence. Verify the initial condition and induction step, and show equivalence to the finite-sum expression. State the r=1 case separately. Independent derivation and proof are ungraded

Model: uniqueness and induction

At n=0 the expression gives L+s−L=s. If it holds at k, the recurrence gives u_(k+1)=rL+c+(s−L)r^(k+1)=L+(s−L)r^(k+1), using rL+c=L. Induction proves the formula. Given the initial value and rule, each next value is uniquely determined. Expanding L+(s−L)rⁿ gives srⁿ+c(1−rⁿ)/(1−r), the finite-sum result. For r=1, induction instead gives s+nc

Equilibrium and attraction are different

For r≠1, compare the initial value with L. If s=L the sequence is constant for any allowed r. Otherwise |r|<1 gives convergence to L; r>1 gives divergence with the sign of s−L; r=−1 alternates between two values; r<−1 alternates with unbounded magnitude. The deviation, rather than the raw term, controls the classification

For u₀=3 and uₙ₊₁=2uₙ−3, what happens?

The table shows uₙ₊₁=−uₙ/2+6 from u₀=10. What is its equilibrium?

Infer coefficients only under a stated model

Suppose three consecutive exact terms obey a fixed affine recurrence. Subtract u₁=ru₀+c from u₂=ru₁+c to get u₂−u₁=r(u₁−u₀). If the first difference is nonzero, divide to recover r, then c=u₁−ru₀. This fits the assumed model; it does not prove a real system will continue to follow it

An affine sequence begins u₀=1,u₁=4,u₂=10. Find r.

Recurrence rule

\(\displaystyle u_{n+1}=2u_n+c\)

Known terms

\(\displaystyle u_0=1,\quad u_1=4,\quad u_2=10\)

For the recurrence and known terms shown, find c.

Constant data do not identify both coefficients

If u₀=u₁=u₂=t, any nonzero r together with c=(1−r)t fits these values and continues constantly. The difference quotient is 0/0 and cannot identify a unique multiplier. If u₀=u₁ but u₂ differs, no fixed affine recurrence can fit the three values

A nonzero multiplier allows reversal

From uₙ₊₁=ruₙ+c, recover uₙ=(uₙ₊₁−c)/r. Reversing n steps recovers the initial value uniquely. For r≠1 the explicit rearrangement is s=L+(uₙ−L)/rⁿ. If r were zero, later terms would lose information about the start

For uₙ₊₁=−2uₙ+3 and u₂=25, recover u₀

Population models need event timing and units

For expected population Pₙ at each census, retaining proportion q and then adding I organisms gives Pₙ₊₁=qPₙ+I. Adding first and then retaining q gives Pₙ₊₁=q(Pₙ+I). Define the census and time-step length. Fractional expectations may be valid even when observed organism counts are integers

Eight organisms arrive before a week with 75% retention. What is c?

Formulate and assess a population model, ungraded

At census zero an expected population is 240. Each week 80% remains, then 20 organisms are introduced before the next census. Define and solve the recurrence, predict census 3, find the first census below 150 and state the limiting value. Discuss what observations could invalidate the model

Model: prediction, threshold and limitations

Let Pₙ be expected organisms immediately after census n. Then P₀=240 and Pₙ₊₁=(4/5)Pₙ+20. The equilibrium is 100 and Pₙ=100+140(4/5)ⁿ. Thus P₃=171.68. Since 0<4/5<1 and the initial deviation is positive, the sequence decreases towards 100. P₄=157.344 and P₅=145.8752, so census 5 is the first below 150. Fractional values are expectations, not literal fractional organisms. Constant retention and arrivals, no capacity effect and identical weekly conditions are assumptions; compare predictions with held-out census observations and examine systematic errors before extending the model