Discrete Versus Continuous Growth

Two Ways to Model Change

A fixed percentage change applied once per time unit and a change accumulated continuously are different assumptions about how an amount changes. The corresponding models can produce close predictions, but they are not identical. Comparing them is useful whenever a growth or decay rate is reported without much explanation of how it is applied.

For a fair comparison, use the same initial amount, the same elapsed time, and the same numerical rate. Then any difference in the predicted amounts comes from the compounding convention.

Define the Models Carefully

Let \(A_0>0\) be the amount at time zero, let \(r\) be the rate per time unit written as a decimal, and let \(t\geq 0\) be elapsed time measured in those same units. A rate of \(5\%\), for example, is written \(r=0.05\). The models below use the same \(A_0\), \(r\), and \(t\).

The condition \(r>-1\) makes \(1+r\) positive, so the discrete formula has a real value for every real \(t\geq0\). In the usual once-per-unit interpretation, \(t\) is a whole number of periods. Using a real value of \(t\) extends that formula mathematically between periods; it does not mean that a real-world account literally compounds for a fractional period under the same convention. The continuous model naturally allows real elapsed times.

When \(-1<r<0\), both models describe decay: their amounts decrease as time passes. When \(r>0\), they describe growth. The restriction also rules out rates at or below \(-100\%\), which cannot produce a positive discrete balance through this formula for arbitrary real \(t\).

Worked Example: Comparing Growth Predictions

Why the Continuous Prediction Is Larger

To compare the models, first compare their one-unit growth factors, \(1+r\) and \(e^r\). The required inequality is \(e^r>1+r\) whenever \(r>-1\) and \(r\neq0\).

Here is a justification using the area interpretation of the natural logarithm. For \(x>0\), \(\ln x\) is the signed area under the curve \(y=1/u\) from \(u=1\) to \(u=x\). If \(x>1\), then \(1/u<1\) between \(1\) and \(x\), so that area is less than the rectangle of width \(x-1\) and height \(1\). Thus \(\ln x<x-1\). If \(0<x<1\), then \(1/u>1\) between \(x\) and \(1\). The area from \(x\) to \(1\) is greater than \(1-x\), and reversing the limits gives \(\ln x<-(1-x)=x-1\). In either case, \(\ln x<x-1\) for \(x>0\), \(x\neq1\).

Set \(x=1+r\), which is positive because \(r>-1\). If \(r\neq0\), then \(x\neq1\), and the inequality gives \(\ln(1+r)<r\). Since the exponential function is increasing and is the inverse of the natural logarithm, exponentiating gives \(1+r<e^r\).

For \(t>0\), raising both positive factors to the same power preserves their order. Multiplication by \(A_0>0\) also preserves it. This proves the comparison:

At \(t=0\), both models give \(A_0\). If \(r=0\), both stay equal to \(A_0\) for all \(t\). Those are the cases of equality.

Worked Example: Comparing Decay Predictions

A larger predicted amount during decay means the continuous model decays less quickly than the once-per-time-unit discrete model at the same numerical rate. It does not mean that the continuous model predicts growth.

Absolute and Relative Differences

The absolute difference between the predictions is \(A_{\mathrm{c}}(t)-A_{\mathrm{d}}(t)\). Its units are the same as the amount being modeled. It answers, for example, how many dollars separate two account predictions.

The ratio of the predictions gives a relative comparison:

For \(r\neq0\), the factor \(e^r/(1+r)\) is greater than \(1\), so this ratio is greater than \(1\) for \(t>0\). It also increases as \(t\) increases. Thus the continuous prediction's relative advantage over the discrete prediction becomes larger over time. This statement concerns the ratio; it does not say the absolute difference increases forever. During decay, both amounts can become small, so their absolute difference need not keep growing.

Worked Example: Converting to an Equivalent Continuous Rate

Interpret the Rate Before Comparing Models

The same numerical \(r\) does not mean the same compounding convention. In the discrete model, \(r\) is the percentage change applied once per time unit. In the continuous model, it appears as the exponent's rate parameter. For a positive \(r\), the continuous model accumulates change throughout the period; its one-unit factor \(e^r\) is larger than the discrete model's factor \(1+r\).

If the goal is to represent the same one-period change in both conventions, convert the periodic rate \(r\) to the continuous rate \(\ln(1+r)\). Conversely, if the continuous rate is \(r\), the amount after one time unit is multiplied by \(e^r\); the equivalent discrete rate is \(e^r-1\). These conversions apply when the relevant factors are positive, as ensured by \(r>-1\) for the periodic rate.

A Common Comparison Mistake

A frequent error is to say that both models have the same rate and therefore must predict the same amount. They do not: the same numerical rate is applied under different conventions. For instance, with \(r=0.05\) and \(t=1\), the discrete factor is \(1.05\), while the continuous factor is \(e^{0.05}\approx1.051271\). Replacing \(e^{0.05}\) by \(1.05\) incorrectly makes the two one-year predictions identical.

The correction is to keep each model's factor intact, use the same initial amount and elapsed time, and compare the results only after evaluating both expressions. If identical outcomes are required instead, change the rate in one model using the conversion above.