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Growth Models: Recursive Sequences and y' = ay + b

A growth model says how a quantity changes step by step: u(n+1) = a·u(n) + b. If b = 0 it is geometric, if a = 1 it is arithmetic. For 0 < a < 1 the sequence moves to the fixed point L = b/(1 − a). The continuous version is y' = ay + b with solution y = C·e^(ax) − b/a. Euler's method turns y' = f(y) back into steps, and the logistic model slows growth near a limit.

🎬 Step-by-step story

  1. Add the same number every time: u(n+1) = u(n) + 2. The bars grow by equal steps. This is an arithmetic sequence.
  2. Multiply by the same number every time: u(n+1) = 1.5 × u(n). The steps get taller and taller. This is a geometric sequence.
  3. Now do both: u(n+1) = 0.8 × u(n) + 4. The bars rise fast, then slow down and creep toward the green line, the limit 20.
  4. Do not look only at steps. Let time flow smoothly. The red curve y' = ay + b touches every bar: the discrete model sits on the continuous one.
  5. Logistic growth. Growth is fast when small, slow near the limit 100. The bars make an S-shape, because each step is a little push (Euler's method).
  6. Free play. Change a, b and the start value. Try a = 1, a bigger than 1, and a smaller than 1. Where do the bars go?

Tip: drag the 3D scene to turn it. Use two fingers to zoom.

🤔 Common doubts, cleared

Why are the first bars so equal in height steps?

Each step adds the same number, 2, so the bars climb like stairs.

Does a geometric sequence always grow?

Only if q is bigger than 1 and u(0) is positive. If q is between 0 and 1 it shrinks.

Why does the sequence stop near 20?

At 20 the rule gives 0.8 × 20 + 4 = 20 again, so nothing changes. That is the fixed point.

Why does the red curve touch every bar?

We chose the curve y' = ky + c with a = e^k, so it passes exactly through every step. Discrete and continuous agree at those points.

Why does growth slow near 100?

The bracket (1 − u/100) shrinks as u nears 100, so each push gets smaller.

What happens when a = 1 in free play?

The sequence becomes arithmetic, adding b every step, and there is no limit.

Recursive sequence models

A recursive rule gives each term from the one before it, plus a start value. Two simple ones:

Use them for equal yearly savings (arithmetic) or interest and population growth (geometric).

Arithmetico-geometric sequences

Many real models do both: u(n+1) = a·u(n) + b. If a ≠ 1, look for a fixed point L, a value that does not change: L = a·L + b, so L = b ÷ (1 − a).

Now write v(n) = u(n) − L. Then v(n+1) = a·v(n): geometric! So v(n) = an·v(0) and

u(n) = L + an·(u(0) − L)

If |a| < 1 the term an goes to 0, so u(n) tends to L. If a > 1 it grows without limit (unless u(0) = L).

The equation y' = ay + b

Now let time be continuous. The rate of change of y is y' = a·y + b. The solutions are

y(x) = C·eax − b/a (a ≠ 0), where C is found from the start value.

The constant solution y = −b/a is the equilibrium, like the fixed point. For a < 0 every solution tends to it; for a > 0 they move away.

Discrete versus continuous models

A discrete model changes at steps (once a year). A continuous model changes all the time. If u(n+1) = a·u(n) + b is matched with y' = k·y + c where a = ek, the curve passes exactly through all the points. Use discrete when you measure at fixed times (yearly census); use continuous when change is smooth (cooling, decay).

Logistic model

Plain growth can not go on forever. In a limited place the growth slows near a limit K. A simple logistic rule is

u(n+1) = u(n) + r·u(n)·(1 − u(n)/K)

When u is small, the bracket is near 1, so it grows almost geometrically. When u is near K, the bracket is near 0 and growth stops. The graph is an S-shaped curve. The continuous form is y' = r·y·(1 − y/K).

Euler's method

Most equations have no neat formula. Euler's method walks along the slope in small steps h:

y(n+1) = y(n) + h·f(y(n)) for y' = f(y).

Smaller h gives a better answer but needs more steps. The logistic bars in the 3D are exactly Euler steps with h = 1.

Try it: predict then check

In the 3D free play set a = 0.8, b = 4, start = 2. Predict the limit using b ÷ (1 − a). Check the green line. Now try a = 0.5, b = 4: predict, then check. Finally set a = 1. What kind of sequence is it?

Key formulas and definitions

Worked examples

1. u(0) = 2 and u(n+1) = u(n) + 3. Find u(5).

Arithmetic with r = 3. u(5) = 2 + 5 × 3 = 17.

2. u(0) = 5 and u(n+1) = 2·u(n). Find u(4).

Geometric with q = 2. u(4) = 5 × 2⁴ = 5 × 16 = 80.

3. u(0) = 2 and u(n+1) = 0.8·u(n) + 4. Find the fixed point and u(3).

L = 4 ÷ (1 − 0.8) = 20. u(n) = 20 − 18 × 0.8ⁿ. u(3) = 20 − 18 × 0.512 = 20 − 9.216 = 10.784. Check by steps: u1 = 5.6, u2 = 8.48, u3 = 10.784.

4. You save Rs 1000, earn 5% a year and add Rs 100 each year. Write the rule and find the amount after 2 years.

u(n+1) = 1.05·u(n) + 100. u1 = 1050 + 100 = 1150. u2 = 1.05 × 1150 + 100 = 1207.5 + 100 = 1307.5. Rs 1307.50.

5. Solve y' = 2y + 4 with y(0) = 1, and find y(1).

y = C·e^(2x) − 4/2 = C·e^(2x) − 2. y(0) = 1 gives C − 2 = 1, so C = 3. y(1) = 3e² − 2 ≈ 3 × 7.389 − 2 ≈ 20.17.

6. Use Euler's method with h = 0.5 for y' = y, y(0) = 1, to estimate y(1). Compare with the exact value e ≈ 2.718.

y1 = 1 + 0.5 × 1 = 1.5. y2 = 1.5 + 0.5 × 1.5 = 2.25. The estimate is 2.25; the exact value is 2.718, so Euler underestimates by about 0.47. A smaller h gives a closer value.

Common mistakes

Practice quiz

1. For u(n+1) = 0.5·u(n) + 3 the limit is:
2. u(n+1) = u(n) + 4 is a sequence that is:
3. The solution of y' = ay + b is:
4. In Euler's method the step size is called:
5. Logistic growth is fast at first and slows because:

Practice: answer these yourself

Type or choose your answer, then press Check. Use a hint if you are stuck; the full solution appears after you answer.

Frequently asked questions

What is an arithmetico-geometric sequence?

A sequence with u(n+1) = a·u(n) + b, which both multiplies and adds. It is geometric when b = 0 and arithmetic when a = 1.

How do I find the limit of a sequence u(n+1) = a·u(n) + b?

Solve L = a·L + b to get L = b ÷ (1 − a). The sequence tends to L only if |a| < 1.

Why use Euler's method?

Many equations cannot be solved with a neat formula. Euler's method gives numbers step by step, which a computer can do quickly.

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