Deriving Exponential Models from Repeated Percent Change

From Repeated Change to an Exponential Model

A quantity changes by a fixed percentage in each period when each new change is measured against the amount at the start of that period. The amount added or removed may therefore vary from one period to the next: the percentage stays fixed, but the amount it applies to changes.

This repeated multiplication is the basic reason an exponential model appears. If the starting amount is $A_0$, and the same rate $r$ applies each period, then after $t$ whole periods the model is $A(t)=A_0(1+r)^t$, subject to an important special case when the multiplier is zero.

Definition: A fixed per-period rate $r$ means that during each period the amount changes by $r$ times its amount at the beginning of that period. Here $r$ is a decimal rate: for example, $5\%$ is $0.05$, and a $5\%$ decrease is a rate of $-0.05$.

Find the Multiplier

Suppose the amount at the beginning of a period is $A_n$. A rate $r$ changes it by $rA_n$, so the amount at the end of that period is

A_{n+1}=A_n+rA_n=(1+r)A_n.

The number $1+r$ is the per-period multiplier. A positive rate gives a multiplier greater than $1$, so the amount increases when it is positive. A negative rate greater than $-1$ gives a multiplier between $0$ and $1$, so the amount decreases while remaining positive. For instance, increasing by $6\%$ means multiplying by $1.06$, while decreasing by $6\%$ means multiplying by $0.94$.

For a decrease of $q\%$, where $q$ is written as a positive decimal, the rate is $r=-q$, and the multiplier is $1-q$. One common source of errors is to use the size of the decrease itself as the multiplier. A $20\%$ decrease leaves $80\%$ of the current amount, so the multiplier is $0.80$, not $0.20$.

Deriving the Formula

Start with amount $A_0$. After one period, multiply by $1+r$; after a second period, multiply by $1+r$ again. Continuing this process gives

\begin{aligned} A_1&=A_0(1+r),\\ A_2&=A_0(1+r)(1+r)=A_0(1+r)^2,\\ A_3&=A_0(1+r)^3. \end{aligned}

Each period contributes one more factor of the same multiplier. The following induction proves the formula for every whole-number period, including the starting period when the multiplier is positive.

Base case.
At period $n=0$, the amount is $A_0$, and $A_0(1+r)^0=A_0$ whenever the power is defined.

Inductive assumption.
Suppose that after $n$ periods, $A_n=A_0(1+r)^n$.

Advance one period.
The recurrence gives $A_{n+1}=(1+r)A_n=(1+r)A_0(1+r)^n=A_0(1+r)^{n+1}$, as required.

Thus, when $1+r>0$, the repeated-change rule gives $A_n=A_0(1+r)^n$ for every nonnegative integer $n$. The proof relies on using the same rate each period. If the rates vary, the successive factors need not be equal, so this particular power formula does not follow.

Theorem: If the same rate $r$ is applied once in each whole-number period, then $A_{n+1}=(1+r)A_n$. For $1+r>0$, the amount after $n$ periods is $A_n=A_0(1+r)^n$, for every integer $n\geq 0$.

The model is sometimes written with $t$ in place of $n$: $A(t)=A_0(1+r)^t$. In the repeated-period derivation, $t$ counts whole periods. If $t$ is allowed to be any real number, a real-valued power of the multiplier is defined in the usual way when the multiplier is positive, $1+r>0$. Using real times then describes a mathematical extension of the whole-period model; it is not itself a step in the repeated-period recurrence.

Worked Examples

Worked Example: A 5% Increase Each Year

A savings balance begins at $A_0=\$800$ and grows by $5\%$ at the end of each year. Find the balance after four years, before rounding, and check it by applying the annual rule.

Convert the rate to a decimal: $r=0.05$. The multiplier is $1+r=1.05$, so the model is $A_n=800(1.05)^n$, where $n$ is the number of years.

\begin{aligned} A_1&=800(1.05)=840,\\ A_2&=840(1.05)=882,\\ A_3&=882(1.05)=926.10,\\ A_4&=926.10(1.05)=972.405. \end{aligned}

Directly from the formula, $A_4=800(1.05)^4=800(1.21550625)=972.405$. The two calculations agree. Since this is money, rounding only the final result to the nearest cent gives $\$972.41$. As a final check, $\$926.10(1.05)=\$972.405$, the unrounded fourth-year balance.

Worked Example: A 12.5% Decrease Each Period

A quantity begins at $640$ units and decreases by $12.5\%$ each period. Find its amount after three periods.

The decimal rate is $r=-0.125=-\frac18$, so the multiplier is $1+r=0.875=\frac78$. Therefore,

A_3=640\left(\frac78\right)^3 =640\left(\frac{343}{512}\right) =\frac{42875}{100} =428.75.

Check period by period: $A_1=640(0.875)=560$, $A_2=560(0.875)=490$, and $A_3=490(0.875)=428.75$. This matches the formula. The amount lost in the third period is $490-428.75=61.25$ units, which is $12.5\%$ of $490$, since $0.125(490)=61.25$. Thus the percentage is applied to the current amount, not repeatedly to the original $640$.

Worked Example: A Rate That Makes a Positive Amount Negative

Suppose a positive amount starts at $A_0=50$ units and the proposed rate is $r=-1.2$, or a $120\%$ decrease per period. Compute the first two periods and assess whether this is a sensible model for a nonnegative physical quantity.

The multiplier is $1+r=1-1.2=-0.2$. Hence $A_1=50(-0.2)=-10$ units. Applying the same recurrence again gives $A_2=(-10)(-0.2)=2$ units.

These values satisfy the algebraic recurrence: $(-10)(-0.2)=2$. But the first result is negative, so the model does not keep a positive physical amount nonnegative. A rate below $-1$ produces a negative multiplier, and multiplying a positive starting amount by it gives a negative amount after one period.

Worked Example: Zero Starting Amount

Suppose $A_0=0$ and the rate is $r=-2$. The multiplier is $1+r=-1$, which is negative. Nevertheless, the recurrence gives $A_1=0(-1)=0$, then $A_2=0(-1)=0$, and in general $A_n=0$ at every whole-number period.

This checks the special case directly: a zero starting amount stays zero under multiplication by any finite multiplier. Therefore a negative multiplier does not make the amount negative when the starting amount is zero. This is different from the positive-starting-amount case.

Which Rates Keep the Amount Nonnegative?

For a positive starting amount, the amount after one period is $A_1=A_0(1+r)$. Since $A_0>0$, this first amount is nonnegative exactly when $1+r\geq 0$, or $r\geq-1$. If that condition holds, all later whole-period amounts are also nonnegative: the multiplier is nonnegative, and each new amount is the previous amount times that multiplier.

There are two boundary cases to distinguish. If $r=-1$, the multiplier is zero, so a positive starting amount becomes zero after the first period and remains zero. This represents a $100\%$ decrease. If $r<-1$, the multiplier is negative, so a positive starting amount becomes negative after one period. For a model that must remain strictly positive after each period, require $r>-1$. For a real-time power model using arbitrary real $t$, the multiplier must also be positive, so again require $r>-1$.

Rate condition: From a positive starting amount, $r\geq-1$ is necessary and sufficient for nonnegative amounts at every whole-number period. The case $r=-1$ reaches zero after one period. Strictly positive amounts at every period require $r>-1$. If $A_0=0$, the recurrence keeps the amount zero for any finite rate $r$.

For growth, $r>0$, so $1+r>1$. For a decrease that leaves some positive amount after every period, $-1<r<0$, so $0<1+r<1$. A zero rate $r=0$ gives multiplier $1$, leaving the amount unchanged. These conclusions follow from the recurrence and the sign and size of its multiplier.

A Common Mistake: Using the Percent as the Multiplier

Suppose a quantity of $200$ units decreases by $20\%$. A mistaken calculation is

$$ A_1=200(0.20)=40. $$

This calculation finds $20\%$ of the original amount, which is the amount removed, not the amount that remains. The amount removed is $0.20(200)=40$ units, so the amount remaining is $200-40=160$ units. Equivalently, the correct multiplier is $1-0.20=0.80$, and $A_1=200(0.80)=160$. Checking the result, $200-160=40$, which is exactly $20\%$ of $200$.

Likewise, with repeated change, do not apply every period's percentage to the original starting amount. For the $12.5\%$ decrease above, the second reduction was $12.5\%$ of $560$, not $12.5\%$ of $640$. Repeated percent change multiplies the current amount each time; a repeated fixed-amount change would follow a different rule.

Why the Model Is Useful

The recurrence makes the assumption behind the model explicit: each period's change is proportional to the amount at the start of that period. Replacing that repeated rule with a power gives a compact way to calculate any whole-number period without listing all the intermediate periods. It also makes the constraints visible. A positive amount can remain nonnegative under repeated change only if the multiplier is nonnegative, and interpreting fractional periods through a real power calls for a positive multiplier.

Before using the formula, identify the starting amount, the time period, and the rate per that same period. Convert a percentage to a decimal, use $1+r$ as the multiplier, and state whether time counts whole periods or is being treated as real-valued. Finally, check at least one step using $A_{n+1}=(1+r)A_n$; this catches reversed decrease multipliers and incompatible rate values.

Check Your Understanding

Use the recurrence and the multiplier to answer the following questions.

A population begins at $1{,}200$ and increases by $3\%$ each year. What are the multiplier and the model after $n$ whole years?
A quantity decreases by $8\%$ per month. What is its multiplier, and what fraction of its current amount remains after one month?
For $A_0>0$, what rate makes the amount reach zero after exactly one period? What happens in later periods?
For a positive starting amount, explain why a rate below $-1$ gives a negative amount after one period.
If $A_0=0$, does a negative multiplier necessarily produce a negative amount? Verify using the recurrence.

Deriving Exponential Models from Repeated Percent Change

What You'll Learn

From Repeated Change to an Exponential Model

Find the Multiplier

Deriving the Formula