From One Update to Any Number of Updates
A fixed-period growth or decay rule describes a quantity that is updated at regular intervals. At each interval, the new amount is obtained by multiplying the current amount by the same factor. This is a natural model when observations are made at regular times and the relative change from one observation to the next stays constant.
For instance, if a quantity increases by the same percentage every month, its monthly values form a sequence. The update rule lets us calculate the amount after one month; solving the recurrence lets us find it after any number of months. It also gives a precise connection to the continuous exponential models used to describe change at every time, rather than only at the observation times.
Defining the Recurrence
Let \(P_n\) denote the amount after \(n\) update periods, where \(n\) is a nonnegative integer. The initial amount is \(P_0\). The fixed-period recurrence is
where \(a\) is the multiplier applied during each period. The period might be a day, a month, or another fixed time interval. The multiplier \(a\) has no units; \(P_n\) and \(P_{n+1}\) have the same units.
The multiplier is not itself the percentage change. If \(a=1.08\), then each step multiplies the amount by \(1.08\), which corresponds to an increase of \(8\%\) per period. If \(a=0.92\), the amount retains \(92\%\) of its previous value, so it decreases by \(8\%\) per period. In general, the relative change from one step to the next is
provided \(P_n\ne 0\). Thus \(a-1\) is the change as a fraction of the current amount.
Solving the Recurrence
Applying the update rule repeatedly gives \(P_1=aP_0\), then \(P_2=aP_1=a^2P_0\), and then \(P_3=aP_2=a^3P_0\). The exponent counts how many times the multiplier has been applied.
Here is a proof by induction for \(a\ne 0\). The formula holds at \(n=0\), since \(P_0a^0=P_0\). Now suppose it holds at a particular nonnegative integer \(n\), so \(P_n=P_0a^n\). The recurrence gives
Therefore, if the formula holds at step \(n\), it also holds at step \(n+1\). Since it holds initially, induction proves it for every nonnegative integer \(n\). If \(a=0\), the recurrence gives \(P_1=0\), and every later amount is also zero, as stated.
This result replaces a list of repeated calculations with a direct formula. It applies at any step \(n\), so long as the update rule uses the same multiplier each time.
Interpreting the Multiplier
For a positive initial amount, the size of \(a\) determines how the sequence changes. The main cases for ordinary nonnegative growth and decay quantities are:
| Multiplier | Effect at each step |
|---|---|
| \(a>1\) | The amount increases by the same percentage each period. |
| \(a=1\) | The amount stays constant. |
| \(0<a<1\) | The amount decreases by the same percentage each period. |
| \(a=0\) | The amount becomes zero at the first update and stays there. |
When \(a>1\), the amount is multiplied by a factor greater than one at every step. When \(0<a<1\), each new amount is a positive fraction of the preceding one. If \(P_0>0\), these cases keep the amount positive. A negative multiplier is mathematically possible, but it reverses the sign at every step; that alternating behavior is generally not an appropriate model for a nonnegative population or material amount.
Worked Example: Repeated Percentage Growth
A quantity starts at \(250\) units and increases by \(12\%\) each month. Find the amount after four months using the recurrence.
An increase of \(12\%\) means that each month the quantity is multiplied by \(1+0.12=1.12\). Thus \(P_0=250\) and \(a=1.12\). The recurrence formula gives
To evaluate the power, calculate \((1.12)^2=1.2544\), \((1.12)^3=1.404928\), and \((1.12)^4=1.57351936\). Therefore,
Check the last update directly: \(P_3=250(1.404928)=351.232\), and \(351.232(1.12)=393.37984\). This agrees with the four-step formula. If the amount is reported to two decimal places, it is \(393.38\) units.
Worked Example: Repeated Decay
A sample contains \(640\) milligrams. At the end of each four-day period, \(25\%\) of the amount present is lost. Find the amount after twelve days.
Losing \(25\%\) means retaining \(75\%\), so the multiplier is \(a=0.75\). Twelve days contain \(12/4=3\) update periods. The recurrence formula gives
Step by step, \(P_1=640(0.75)=480\) milligrams, \(P_2=480(0.75)=360\) milligrams, and \(P_3=360(0.75)=270\) milligrams. Equivalently, \((0.75)^3=0.421875\), and \(640(0.421875)=270\). The final update check is \(360(0.75)=270\) milligrams, as required.
Connecting the Recurrence to a Continuous Model
A recurrence gives values at discrete update times. A continuous exponential model describes a quantity at all times. Suppose updates are separated by a fixed time interval \(\Delta t>0\), and let the observation times be \(t_n=t_0+n\Delta t\). A continuous exponential model with initial amount \(P_0\) at time \(t_0\) has the form
where \(k\) is the continuous rate constant, measured in inverse units of time. To make this model agree with the recurrence after one update period, its multiplier over \(\Delta t\) must equal \(a\):
For \(a>0\), taking the natural logarithm gives \(k\Delta t=\ln(a)\), so
This choice makes the continuous model agree with the recurrence at every update time, not just the first one. Indeed, at \(t_n=t_0+n\Delta t\),
The recurrence and continuous model therefore give the same values at the scheduled observations when \(a>0\) and \(k=\ln(a)/\Delta t\). The recurrence alone does not specify what happens between updates. Treating the amount between those times as \(P_0e^{k(t-t_0)}\) adds the continuous exponential-model assumption.
Worked Example: Matching a Continuous Decay Model
A continuous decay model has half-life six days. Measurements are taken every two days. If the initial amount is \(90\) milligrams, find the recurrence multiplier and the amount after three update periods.
A half-life of six days means that after six days the continuous model retains one half of its initial amount. For \(P(t)=90e^{kt}\), this gives \(e^{6k}=1/2\), so \(k=\ln(1/2)/6\) per day. The update interval is \(\Delta t=2\) days, so the multiplier over one interval is
There are three two-day periods in six days. The recurrence gives
For a check, the continuous model at six days gives \(90e^{6k}=90(1/2)=45\) milligrams as well. The two models agree at the update times; here the third update time is exactly six days.
A Common Multiplier Mistake
A frequent error is to use the percentage change itself as the multiplier. For example, if a quantity of \(100\) units increases by \(20\%\), setting \(a=0.20\) would give \(P_1=0.20(100)=20\) units. That is not a \(20\%\) increase: it is a decrease of \(80\%\).
The correct multiplier is \(1+0.20=1.20\), because the new amount includes the original \(100\%\) plus the additional \(20\%\). Thus \(P_1=1.20(100)=120\) units, and the check is that \(120-100=20\), which is \(20\%\) of \(100\). For a decrease of \(20\%\), the multiplier would instead be \(1-0.20=0.80\).
When the Continuous Match Applies
The logarithm formula for \(k\) requires \(a>0\). If \(a=1\), it gives \(k=0\), which is the constant continuous model. If \(0<a<1\), then \(\ln(a)<0\), so the matching model has a negative rate constant and represents decay. If \(a>1\), then \(\ln(a)>0\), so the continuous model represents growth.
For \(a=0\), the recurrence drops to zero after one update, whereas an exponential model with finite \(k\) and positive initial amount remains positive at every finite time. For a negative \(a\), the recurrence alternates signs, while a real-valued exponential model with positive initial amount does not. These cases cannot be matched by a real continuous exponential model of the stated form.
Fixed-period recurrences are useful whenever an amount is recorded or updated at regular intervals. Their closed form makes long-term predictions direct, and the logarithmic conversion relates discrete observations to continuous growth or decay rates. Keeping the time interval explicit is essential: the same multiplier over a month and over a year describes different rates per unit time.
Check Your Understanding
Answer each question using the recurrence formula and the stated assumptions.
- If \(P_0=80\) and \(a=1.25\), write a formula for \(P_n\) and calculate \(P_3\).
- A quantity retains \(84\%\) of its amount each week. What is its weekly multiplier, and what is the fractional relative change per week?
- If a multiplier \(a=0.90\) is applied every two hours, write the matching continuous rate constant \(k\) in terms of a logarithm.
- Why does a positive multiplier \(a\) produce the same recurrence and continuous-model values at \(t_n=t_0+n\Delta t\) when \(k=\ln(a)/\Delta t\)?
- Can a real continuous exponential model with finite rate constant match a recurrence with \(a=0\) and positive initial amount at every update time? Explain.
